added
stringdate 2025-03-12 15:57:16
2025-03-21 13:32:23
| created
timestamp[us]date 2008-09-06 22:17:14
2024-12-31 23:58:17
| id
stringlengths 1
7
| metadata
dict | source
stringclasses 1
value | text
stringlengths 59
10.4M
|
|---|---|---|---|---|---|
2025-03-21T14:48:31.948252
| 2020-09-03T17:38:13 |
370773
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Pavel Safronov",
"Qiaochu Yuan",
"https://mathoverflow.net/users/123079",
"https://mathoverflow.net/users/18512",
"https://mathoverflow.net/users/290",
"user513784"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632679",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370773"
}
|
Stack Exchange
|
Gelfand–Kirillov dimension of the first Weyl algebra by using the definition
$\DeclareMathOperator\GKdim{GKdim}$Here I am trying to find the Gelfand–Kirillov dimension of the first Weyl algebra just by using the definition of the Gelfand–Kirillov dimension.
Let $A$ be an affine $\mathbb K$-algebra, and let $V\subseteq A$ be a finite dimensional $\mathbb K$ subspace (containing $1$) which generates $A$ as algebra. Let $V^n$ denote the subspace spanned by all products of $n$ elements from $V$. There is a chain of subspaces
$$\mathbb K\subseteq V\subseteq V^2\subseteq\ldots\subseteq\cup_{n\geq0}V^n=A$$
and we define the Gelfand–Kirillov dimension of $A$, denoted $\GKdim(A)$, by $$\GKdim(A)=\lim\limits_{n\to\infty}\sup\frac{\log\dim(V^n)}{\log n}.$$
The first Weyl algebra $A_1=A_1(\mathbb K)$ is the ring of polynomials in 2 variables $x$ and $y$ with coefficients in $\mathbb K$, subject to the relation $yx-xy=1$.
Let us take $V=\mathbb K+\mathbb Kx+\mathbb Ky$. Hence it is the vector space generated by $1$, $x$ and y, and hence $\dim V=3$.
Let $V^2$ denote the subspace spanned by all products of 2 elements from $V$, i.e., generated by $\{ 1, x, y, x^2, xy, y^2\}$. And hence $\dim V^2=6$.
Now let $V^3$ denote the subspace spanned by all products of 3 elements from $V$, i.e. generated by $\{ 1, x, y, x^2, xy, y^2, x^3, x^2y, yx^2, yxy, y^3\}$. And hence $\dim V^3=11$.
But I think I am doing something wrong because I can see no recursive relation!?
I have to say that at the end we should get $\GKdim(A_1)=2$.
I will appreciate if someone can help me with this.
I would suggest to PBW order the monomials. Then you will see that $\dim V^3 = 10$ (you have $x^2y$ and $yx^2$ which have the same leading term). In general, $\dim V^n = (n+1)(n+2)/2$.
Thank you very much for your comment! Yes, you are right. I was not seeing this point. Thanks!
If you're familiar with the fact that the associated graded with respect to this filtration is $k[x, y]$ with the usual grading then $\dim V^n - \dim V^{n-1} = n+1$ must be the number of monomials in $k[x, y]$ with total degree $n$ which gives Pavel's answer.
@QiaochuYuan Thank you very much for your comment!
|
2025-03-21T14:48:31.948407
| 2020-09-03T19:43:51 |
370780
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Bazin",
"fedja",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/21907"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632680",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370780"
}
|
Stack Exchange
|
On Integrals of the Airy function
Let $Ai$ be the classical Airy function and let $(a_j)_{j\ge 1}$ be the strictly decreasing sequence of its zeroes: we have $a_{j+1}<a_j<\dots <a_2<a_1<0$, $\lim_{j\rightarrow +\infty}a_j=-\infty$. I believe that for all $k\ge 0,$
$$
\int_{-\infty}^{a_{2k+1}} Ai(t) dt<0<\int_{-\infty}^{a_{2k}} Ai(t) dt,\tag{$\ast$}
$$
say with the convention $a_0=0$ (we have then $\int_{-\infty}^{a_0} Ai(t) dt=2/3$). However, it does not appear very simple to prove and not so easy to find a reference in the literature. A Mathematica drawing of the curve of the antiderivative of the Airy function vanishing at $-\infty$ leaves no doubt that $(\ast)$ holds true.
There is nothing special about the Airy function here. The situation becomes immediately clear if you recall the following standard lemma (which can be found in some form in most textbooks dealing with second order linear ODE):
If $u(0)=v(0)=0, 0<u'(0)\le v'(0)$ and $u''=-f(x)u, v''=-g(x)v$ where $f>g$ on $(0,+\infty)$, then the graph of $v$ lies above the first hump of the graph of $u$ (i.e., $v>u$ between $0$ and the first positive zero of $u$).
The standard proof is by noticing that $u>v$ slightly to the right of $0$ (Taylor with the usual trick of considering $(1+\varepsilon)v$ instead of $v$ to break the tie at $0$) and considering the Wronskian $W(u,v)=\det\begin{bmatrix}u&v\\u'&v'\end{bmatrix}$ of $u$, $v$. We have $W'=(f-g)uv>0$ and, thereby, $W>W(0)=0$ as long as both $u,v>0$ and if $v$ tries to go below $u$ as long as $u>0$, it has to hit from above, so at the hitting point $v=u>0,v'\le u'$, i.e., $W(u,v)\le 0$, which is impossible.
Now just look at two consecutive humps of the Airy function and think of them as shot from the zero separating them. The lemma shows that a couple of reflections put one hump below the other one. The rest should be clear.
Thanks for your answer. I am not completely sure of understanding your point: the antiderivative $z$ of the Airy function does not satisfy a second-order equation, but a third-order one, $z'''(t)= t z'(t)$. I am probably missing something.
@Bazin I'm talking about the Airy function itself in the lemma. The antiderivative never appears in the argument. The integral you are interested in is just viewed as the sum of the areas of positive and negative humps of $Ai(x)$ with alternating signs and the lemma allows you to compare two adjacent humps showing that the one further away from the origin is smaller. Is it clearer now? If not, feel free to ask more questions :-)
Yes, fine, thank you very much.
@Bazin You are most cordially welcome :-)
|
2025-03-21T14:48:31.948620
| 2020-09-03T20:29:06 |
370782
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"Joe Silverman",
"Mare",
"Qiaochu Yuan",
"YCor",
"https://mathoverflow.net/users/11926",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/290",
"https://mathoverflow.net/users/61949",
"https://mathoverflow.net/users/7410"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632681",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370782"
}
|
Stack Exchange
|
Sum of the coefficients of the characteristic polynomial of periodic matrices
Let $M$ be an integer matrix with determinant equal to one (or maybe also minus one, but I did not do any tests for this case) and assume that $M$ is periodic, that is $M^n$ is the identity matrix for some $n$. Let $p_M$ denote the characteristic polynomial of $M$.
Question 1: Is it true that then $p_M(1) \geq 0$ for periodic matrices $M$?
Question 2: Let $a_n$ be the largest period of a matrix in $Sl_n(\mathbb{Z})$. What is $a_n$? (or is there a good bound?)
For $n=2$ it should be $a_n=6$. Is $a_n$ attained at a matrix with entries only in $\{-1,0,1\}$?
For Q1 one could perhaps show that $p_A$ is a product of cyclotomic polynomials and then the result would follow from https://en.wikipedia.org/wiki/Cyclotomic_polynomial#Polynomial_values
@AbdelmalekAbdesselam I forgot whether this holds in general, but it holds for acyclic quiver algebra and thus your comment answers Q3!
Usually people say that $M$ has finite order, rather than calling it periodic. In any case, the fact that $M^n=I$ means that the eigenvalues are roots of unity, which in turn implies that its characteristic polynomial is a product of cyclotomic polynomials (since the characteristic polynomial is in $\mathbb Z[x]$). Hence as noted earlier, your question is whether cyclotomic polynomials are non-negative when evaluated at 1, it has nothing ot do with matirces.
@JoeSilverman That answers Q1 together with the wikipedia link I guess. If you want you can turn it into an answer.
If $\Phi(t)$ is a cyclotomic polynomial other than $t\pm 1$ then $\Phi$ has no real root, hence has a constant sign on the reals. Since it's monic, it's then positive on reals, and hence $\ge 1$ on integers.
Q1: This was already given in the comments, but: a matrix $M \in GL_k(\mathbb{Z})$ of finite order $n$ must have rational normal form a block-diagonal matrix with blocks the companion matrices of cyclotomic polynomials $\Phi_d$ for $d | n$, so the problem reduces to the case of a single such matrix, which is to say the problem reduces to asking whether we always have $\Phi_d(1) \ge 0$. This is true, and in fact:
Proposition: $\Phi_n(1)$ is equal to $p$ if $n = p^k$ is a prime power and equal to $1$ otherwise.
Proof. $\Phi_p(x) = \frac{x^p - 1}{x - 1}$ and $\Phi_{p^k}(x) = \Phi_p(x^{p^{k-1}})$ so the computation in the prime power case is clear. For a general $n$ we have that if $p \nmid m$ then
$$\Phi_{pm}(x) = \frac{\Phi_m(x^p)}{\Phi_m(x)}$$
and hence that $\Phi_n(1) = 1$ as soon as $n$ has more than one prime factor. $\Box$
Q2: As before it suffices to consider block sums of companion matrices of cyclotomic polynomials. A block sum of companion matrices of cyclotomic polynomials $\Phi_{d_i}(x)$ is an element of $GL_n(\mathbb{Z})$ where $n = \sum \varphi(d_i)$ of order $\text{lcm}(\{ d_i \})$ so the problem is to optimize this (the cyclotomic polynomials satisfy $\Phi_n(0) = 1$ for $n \ge 2$ so all these block matrices lie in $SL_n(\mathbb{Z})$ also). This looks hard in general.
The corresponding problem of finding the largest order of an element of $S_n$ is a similar optimization problem but where $n = \sum d_i$. That sequence is Landau's function (A000793) but I don't know if this one has a name or is in the OEIS.
Edit #1: If $L(n)$ denotes this largest order then we have
$L(1) = 2$ ($1 = \varphi(2)$)
$L(2) = 6$ ($2 = \varphi(6)$)
$L(3) = 6$ ($3 = \varphi(6) + \varphi(2)$)
$L(4) = 12$ ($4 = \varphi(6) + \varphi(4)$)
$L(5) = 12$ ($5 = \varphi(6) + \varphi(4) + \varphi(2)$)
$L(6) = 30$ ($6 = \varphi(10) + \varphi(6)$)
which, if I haven't messed up, already shows that this sequence is not in the OEIS. On the other hand, it's not hard to see that $L(2k+1) = L(2k)$ for $k \ge 1$ since $\varphi(d)$ is even for $d \ge 2$ and $\varphi(2d) = \varphi(d)$ if $d$ is odd so it never helps us to add a $\varphi(2) = 1$ term to the sum. (We need to rule out the possibility that $L(n)$ is a power of $2$ but this shouldn't be hard.) So perhaps the OEIS has only the even terms $L(2n)$ somewhere; I haven't ruled that out yet.
An easy upper bound is that we can compute the exponent $E(n)$, namely the lcm of all orders of elements of finite order in $GL_n(\mathbb{Z})$, so that $L(n) | E(n)$. By considering each prime separately we have that
$$\nu_p(E(n)) = \text{max} \left\{ k : \varphi(p^k) = (p - 1) p^{k-1} \le n \right\} = \left\lfloor \log_p \frac{n}{p-1} \right\rfloor + 1$$
and hence
$$E(n) = \prod_p p^{ \left\lfloor \log_p \frac{n}{p-1} \right\rfloor + 1}.$$
This sequence is much easier to compute although the bound becomes increasingly bad. It does have the virtue of also being a bound on the exponent of any finite subgroup of $GL_n(\mathbb{Z})$. We again have $E(2k+1) = E(2k)$ for $k \ge 1$, and
$E(1) = 1$
$E(2) = E(3) = 6$
$E(4) = E(5) = 2^3 \cdot 3 \cdot 5 = 120$
$E(6) = E(7) = 2^3 \cdot 3^2 \cdot 5 \cdot 7 = 2520$
which also does not appear to be in the OEIS, with or without the terms doubled. The corresponding sequence for $S_n$ is $\text{lcm}(1, 2, \dots n)$ which is A003418 and the formula is similar except the exponent is more simply just $\lfloor \log_p n \rfloor$.
Edit #2: Okay, I computed that $L(8) = 60$ which was finally enough terms for me to find it: $L(2n)$ appears to be (up to some indexing issues) A005417 on the OEIS. A comment there suggests the following argument which makes $L$ a bit easier to compute than I had been thinking: if $\gcd(n, m) = 1$ and $\varphi(n), \varphi(m) \ge 2$ (so neither $m$ nor $n$ is equal to $2$) then we can always replace a $\Phi_{mn}(x)$ block with a $\Phi_n(x)$ block and a $\Phi_m(x)$ block, since $\varphi(mn) = \varphi(m) \varphi(n) \ge \varphi(m) + \varphi(n)$. So we only ever need to consider $\Phi_d(x)$ blocks where $d$ is a prime power or twice an odd prime power. A similar argument works in $S_n$. It follows (this is the OEIS comment) that
$$L(n) = \text{max} \left\{ \prod p_i^{e_i} : \sum (p_i - 1) p_i^{e_i - 1} \le n \right\}.$$
Edit #3: The observation in the previous paragraph answers Q3: yes, the maximum is attained for a matrix with entries in $\{ -1, 0, 1 \}$, since the same is known about the cyclotomic polynomials $\Phi_d(x)$ where $d$ is a prime power or twice an odd prime power. Famously the cyclotomic polynomials are known to not always have coefficients in $\{ -1, 0, 1 \}$ and $\Phi_{105}(x)$ is the smallest counterexample, but that doesn't matter here.
Edit #4: Okay, here are some bounds. For a lower bound we clearly have $g(n) \le L(n)$. For an upper bound let $r_i = (p_i - 1) p_i^{e_i - 1}$, so that we can write the optimization problem defining $L(n)$ as
$$L(n) = \text{max}\left\{\prod \frac{p_i}{p_i - 1} r_i : \sum r_i \le n \right\}.$$
We can bound this factor $\prod \frac{p_i}{p_i - 1}$ as follows. The primes occurring in this product are at worst the primes up to $n+1$, and I believe the asymptotic behavior of $\prod_{p_i \le n+1} \frac{p_i}{p_i - 1}$ should be $\log n$ but I don't see an extremely clean proof so I'll settle for the worse bound
$$\prod_{p_i \le n+1} \frac{p_i}{p_i - 1} \le \prod_{k=2}^{n+1} \frac{k}{k-1} = n+1$$
which gives
$$L(n) \le \text{max} \left\{ (n+1) \prod r_i : \sum r_i \le n \right\}.$$
We can now relax this optimization problem so that the $r_i$ can take real values, and then a standard Lagrange multiplier argument shows that, for any number $k$ of terms (which we've left unspecified), we want to take $r_i = r$ for some fixed $r$. This gives
$$L(n) \le \text{max} \left\{ (n+1) r^k : kr \le n, k \in \mathbb{N}, r \in \mathbb{R} \right\}$$
and if we further relax $k$ to a real number then a standard calculus argument gives $r = e, k = \frac{n}{e}$, so
$$\boxed{ L(n) \le (n+1) \exp \left( \frac{n}{e} \right) }$$
exactly paralleling the analogous but slightly simpler argument for Landau's function which gives $g(n) \le \exp \left( \frac{n}{e} \right)$. I would guess that in fact like $g(n)$ we should also have $\log L(n) \sim \sqrt{n \log n}$. I think the starting point is that the relaxation we used above is very inaccurate for large primes and for $p$ such that $(p-1)p > n$ the corresponding exponent is at most $1$.
I can't resist mentioning the following fact related to these bounds: there is a fairly explicit and nice bound on the size of a finite subgroup of $GL_n(\mathbb{Z})$ (here we bound the size of a finite cyclic subgroup), and in fact $GL_n(\mathbb{Z})$ has "Sylow subgroups": maximal finite $p$-subgroups such that every finite $p$-subgroup is conjugate to a subgroup of one of these. For details and much more see this beautiful paper by Serre which I came across recently: https://citeseerx.ist.psu.edu/viewdoc/download?doi=<IP_ADDRESS>7.8897&rep=rep1&type=pdf
|
2025-03-21T14:48:31.949132
| 2020-09-03T20:29:47 |
370783
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632682",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370783"
}
|
Stack Exchange
|
Conservation law for the Camassa-Holm equation
For $k>0$, consider the Camassa-Holm equation: $$
u_t-u_{txx}+2k u_x=-3uu_x+2u_xu_{xx}+uu_{xxx}, \quad (t,x)\in\mathbb{R}^2.
$$
I've been trying to (formally) recover the second of its well-known conservation laws, that is to say, to prove that the following functional $$
F(u):=\int \big(u^3+uu_x^2+2k u^2\big)dx,
$$
is conserved. In other words, considering smooth solutions decaying sufficiently fast at $\pm\infty$, proving that the time derivative of $F(u(t))$ is equivalently zero. It is very intriguing for me that I have tried by multiplying the equation by all sort of factors, like $u^2$, $u_x^2$,... and then integrating in $\mathbb{R}$, and performing several integration by parts, but it has been imposible for me to recover the conservation law. Does anyone know how to do it? It also intrigues me that the first two terms in the conservation law are cubic, but the last one is just quadratic, that makes me thinks that this is not just multiplying the equation (not at least once) and performing some integration by parts.
The following should work.
First, denote by $U(t,x) = \int_{-\infty}^x u(t,y) ~dy$.
If you write the equation as
$$ u_t-u_{txx}+2k u_x=-3uu_x+u_xu_{xx}+(uu_{xx})_x$$
and take the primitive in $x$, you find
$$ U_t - u_{tx} + 2k u + \frac32 u^2 - \frac12 (u_x)^2 - u u_{xx} = 0. $$
Now, multiply the entire equation by $u_t$ and integrate over $x$ over the entire $\mathbb{R}$.
The first two terms vanish since they are integration of a real function against its first derivative.
The final term you integrate by part in $x$.
This should yield you
$$ \int 2 k u u_t + \frac32 u^2 u_t + \frac12 (u_x)^2 u_t + u u_{tx} u_{x} ~dx = 0 $$
which gives
$$ \frac{d}{dt} \int k u^2 + \frac12 u^3 + \frac12 u (u_x)^2 ~dx = 0 $$
|
2025-03-21T14:48:31.949398
| 2020-09-03T21:24:23 |
370786
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"GH from MO",
"Peter LeFanu Lumsdaine",
"Q_p",
"https://mathoverflow.net/users/11919",
"https://mathoverflow.net/users/2273",
"https://mathoverflow.net/users/480516"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632683",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370786"
}
|
Stack Exchange
|
On Soundararajan's explicit formula
I'm reading Soundararajan's https://arxiv.org/pdf/0705.0723.pdf, and on page 5, one has
$$\sum_{n\leq x} \frac{\Lambda(n)}{n^z} \log (x/n) = -\frac{\zeta'}{\zeta}(z)\log x - \Big(\frac{\zeta'}{\zeta}(z) \Big)' -\sum_{\rho} \frac{x^{\rho-z}}{(\rho-z)^2} + O(1/T),$$
where $\Lambda$ denotes the von Mangoldt function, $\zeta$ the Riemann zeta function, $\zeta(\rho)=0, \Re(z) \in (1/2, 2], x\geq 2$ and $|\Im(\rho)|\leq T$. Maybe i'm missing something, but doesn't the right hand side of the above formula have a pole at $z=1$, which would render the formula meaningless at $z=1$?
The formula you quote is on page 8 (not page 5). The sum is over all zeros $\rho$, not just those with $|\Im(\rho)|\leq T$. On the other hand, there is the assumption $\Im(z)\in[T,2T]$.
@user164760: As you move $z$ close to $1$, you also need to move $T$ close to $0$, since the formula is proved under the assumption $\Im(z)\in[T,2T]$. So, yes, the sum of the terms involving $z$ blows up, but so does $O(1/T)$ as well. The two blowups cancel each other.
I see...$z$ must not be a real number. Thanks.
On page 8 Soundrarajan also unconditionally has $$\sum_{n\leq x} \frac{\Lambda(n)}{n^{\sigma+it}\log n}\log(x/n) = (\log x)\log \zeta(\sigma + it) + \frac{\zeta'}{\zeta}(\sigma+it) + F (\rho)$$ for $\sigma>1/2$, where $F(\rho)=O(1)$ is a sum over the zeros of zeta. Doesn't this entail the RH ?
@user164760: Soundararajan will know when he proves the Riemann Hypothesis.
most certainly he would...but what am i missing above ?
Well, he does not say anything like "$F(\rho)=O(1)$. There is a $O(1)$ term on the LHS, but also a sum over $\rho$'s on the RHS, which he bounds by a product involving $F(\sigma+it)$. Now $F(s)$ (defined on page 7) blows up at every $\rho$, so it can easily counteract the blowup of $\zeta'/\zeta$.
great explanation. Thanks ! I guess it would be fun exploring that formula a little further...
$F(\sigma+it)$ was defined on page 7. He does not bound $F(\rho)$, because $F(\rho)$ is undefined. You mixup his notation and thoughts with your notation and thoughts. That's rather dangerous. What Sound proves is this: $$\sum_{n\leq x} \frac{\Lambda(n)}{n^{\sigma+it}\log n}\log(x/n) + O(1) = (\log x)\log \zeta(\sigma + it) + \frac{\zeta'}{\zeta}(\sigma+it) - \sum_\rho\dots,$$ where $\sum_\rho\dots$ is an explicit sum over the zeros. And, in the next line, he bounds $|\sum_\rho\dots|$ by a product involving $F(\sigma+it)$.
let $z\neq \rho$. and $\Theta$ be the sup of the real parts of the $\rho$s. We have $$-\int_{\Re(w)=c} \frac{\zeta'}{\zeta}(z+w) \frac{x^w}{w^2} \mathrm{d}w = -\frac{\zeta'}{\zeta}(z)\log x - \Big(\frac{\zeta'}{\zeta}(z)\Big) - \sum_{\rho} \frac{x^{\rho-z}}{(\rho-z)^2} + O(1/T)$$. ...(to be continued in the next message).
...(continuation)...This entails that $| \sum_{\rho} \frac{x^{\rho-z}}{(\rho-z)^2}|=\Omega(x^{\Theta-z-\varepsilon})$ for every $\varepsilon>0$ and infinitely many $x\rightarrow \infty$, else by Mellin inversion we would find that $-\frac{\zeta'}{\zeta}(z+w)$ must be holomorphic in $\Re(w)>\Theta-z-\epsilon$ for some fixed $\epsilon>0$, contradicting our definition of $\Theta$. So for fixed $z$ with $\Re(z)>1/2$, the left hand side of the expression in the previous message is $O(1)$, whereas the RHS is $\Omega(x^{\Theta -z-\varepsilon})$. Thus we deduce that $\Theta \leq 1/2$.
@user164760: Sighs.
@GHromMO, sorry, i will delete my latest comments in a moment, i just thought you or some other person might want to have a quick look...it seems the above argument works.
@user164760: I can assure you that it does not. Please don't try to prove the RH until you have basic questions like the ones above. Start with small things, and grow them big over a long time.
Okay thanks for the advice. But is your assurance based on belief or fact ? Because i'm yet to see how it fails...
@user164760: Just write it up, and submit to a journal. You will get a report.
Sure, but what's your quick opinion ? I would be honoured and extremely grateful to have it.
@user164760: I told you my quick opinion. It is wrong. And here I am finished with this topic.
I mean why are you saying it's wrong ? What/where do you think is the error..?.Surely that's not too hard to say...
@user164760: You misunderstand the purpose of this site. It is not for discussions, where I say something and then you say something, and we cycle. It is also not for checking proofs, especially of major conjectures. Discuss your proofs with your friends, professors or colleagues, or submit them to journals. That's how these things work.
Okay, thank you for your time (though if you would have pinpointed exactly where the error is, i would have stopped and deleted the RH comments right away).
Just finished polishing up my argument and submitted the paper to the Mathematics Research Letters. Note that the above argument is not in Soundrarajan's paper, though it was inspired by the explicit formula therein.
since our RH comments are probably irrelevant for the post, can we delete them ?
@user164760: Just to explain perhaps why GH wouldn’t “pinpoint the error”: Asking a more experienced mathematician to find problems in your proofs is like asking a plumber to explain how to fix your plumbing. You’re asking them to spend time doing part of their job, for free — so they’ll often be happy to chat for a bit, especially if it’s an interesting topic (which is why MO works well), but if you come back repeatedly, then they will get tired at some point and want to move on. If you keep asking questions as long as they keep answering, then the only way to move on is to stop answering.
@PeterLeFanuLumsdaine, well, i was simply asking for a little favour. Maybe i could email you the full paper and you or your number theory colleague(s) could help me by a quick review ? I don'have anyone around me who can check the argument.
@user164760: Asking for a favour is fine. But if you’re asking a stranger, then they’re under no obligation to do the favour if they don’t want to. And if you keep pushing your request after they politely decline it, then you are soon crossing the line from “simply asking for a favour” to “pestering”.
Alternatively, one can apply the fact that
$\frac{\zeta'}{\zeta}(z) = \sum_{\rho} \frac{1}{z-\rho} + O(\log T).$ Plugging this into the expression in previous comments reveals that for any given $\varepsilon > 0$ and real $t \geq 2$, the function
\begin{equation}
F(z)=\sum_{\rho} \Big( \frac{-\log x}{z-\rho} + \frac{1}{(z-\rho)^2} + \frac{x^{\rho-z}}{(\rho-z)^2}\Big)
\end{equation} must have a holomorphic continuation at $z=\frac{1}{2} + \varepsilon + it$. This implies that $\Re(\rho) \leq \frac{1}{2}$ for each zero $\rho$. But i think we should now delete these RH comments.
@PeterLeFanuLumsdaine, it's really unfair to accuse me of ''pestering'', since i have repeatedly suggested that we should delete the RH comments.
|
2025-03-21T14:48:31.949875
| 2020-09-03T21:38:08 |
370788
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D.-C. Cisinski",
"Denis Nardin",
"Dylan Wilson",
"KotelKanim",
"Maxime Ramzi",
"Phil Tosteson",
"Rune Haugseng",
"Tim Campion",
"Zhen Lin",
"https://mathoverflow.net/users/1017",
"https://mathoverflow.net/users/102343",
"https://mathoverflow.net/users/1100",
"https://mathoverflow.net/users/11640",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/43054",
"https://mathoverflow.net/users/50409",
"https://mathoverflow.net/users/52918",
"https://mathoverflow.net/users/6936"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632684",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370788"
}
|
Stack Exchange
|
Decomposing a (co)limit by decomposing the indexing diagram
Let $D: I \to \mathcal C$ be a diagram, and suppose we have a colimit decomposition $I = \varinjlim_{j \in J} I_j$ in $Cat$. Then under certain conditions, we can decompose the colimit of $D$ as $\varinjlim_{i \in I} D_i = \varinjlim_{j \in J} \varinjlim_{i \in I_j} D_i$. But I've never seen general conditions along these lines spelled out for 1-categories.
Question 1: Is there some place where conditions making the above true are given in the 1-categorical setting?
For $\infty$-categories, there is Corollary <IP_ADDRESS> of Higher Topos Theory. Unfortunately, the formulation of the result is somewhat abstruse, being expressed in terms of the bespoke simplicial set denoted $K_F$ there (defined using 4 conditions in Notation <IP_ADDRESS>).
As a result, I'm having the following problem: it seems to me that for any cocone of $\infty$-categories $(I_j \to I)_{j \in J}$, one should be able to construct a natural map $\varinjlim_{j \in J} \varinjlim_{i \in I_j} D_i \to \varinjlim_{i \in I} D_i$, and one would expect HTT <IP_ADDRESS> to imply that under the appropriate conditions, this map is an equivalence. But the formulation doesn't seem to easily lend itself to confirming this.
Question 2: Is the natural map $\varinjlim_{j \in J} \varinjlim_{i \in I_j} D_i \to \varinjlim_{i \in I} D_i$ constructed somewhere in reasonable generality? (Or else is it easy to construct from general machinery given somewhere?)
Question 3: Is there written somewhere an account of conditions (perhaps analogous to those of HTT <IP_ADDRESS>) which ensure that this map is an equivalence?
This question seems to ask about an instance of this phenomenon.
In his answer to the question you reference, Peter seems to claim this is always true for $1$-categories.
Ah, so he does, thanks! Sanity check: it works for the canonical "bad coequalizer" in Cat given by $(\bullet) \rightrightarrows (\uparrow) \to B \mathbb N$. So I suppose I'd buy it...
But he doesn't give a reference, and the claim doesn't seem that obvious to me (it seems a bit subtle, because of the possibly weird behaviour of colimits in Cat, as you point out) , so I guess it'd be good to have some clarification (should it only be to see if the proof goes through in more generality, which you seem to seek).
(Actually, it might not be that subtle in the $1$-categorical case; I think it mostly relies on the fact that Cat is cartesian closed, and on the analysis of hom-sets in a limit of categories)
There is theorem 7.3.16 in my book on higher categories (in the spirit of the result you quote from HTT but a little bit more usable). This is what explains decompositions of diagrams with Reedy-like considerations, as explained in corollary 7.4.4 proposition 7.4.5 of loc. cit. for instance.
@Denis-CharlesCisinski Thanks -- this may actually be what I need for my purposes -- I happen to be working with a directed union of indexing diagrams, it it will actually be nice not to have to check that the union is a homotopy colimit of $\infty$-categories. Lurie makes a note of this case in Rmk <IP_ADDRESS>, and I think gives the other cases you mention later in the book, but without the naturality statement as far as I can tell.
@TimCampion you might want to check out example 2.5 here: https://arxiv.org/pdf/1705.04933.pdf. Asaf Horev and I prove here what you ask for.
Though the answers below are of course very slick and comprehensive...
@KotelKanim Thanks -- I somehow just noticed your comment here. One nice thing is that your paper is a citable source!
@TimCampion might be the only thing... At the time we didn't know much oo-category theory beyond Lurie's books. As it was not (explicitly) there, we were excited to have this as a, somewhat unexpected, application of our theorem. The other answers here show of course much more direct and reasonable proofs. Nonetheless, it's always nice to get a citation!
Let $p \colon E \to J$ be the cocartesian fibration for the diagram $j \mapsto I_j$. Then the colimit over $E$ of $F \colon E \to C$ can always (assuming the appropriate colimits exist in $C$) be written as an iterated colimit:
$$ \mathrm{colim}_E \, F \simeq \mathrm{colim}_J \, p_! F \simeq \mathrm{colim}_{j \in J} \, \mathrm{colim}_{I_j} \, F|_{I_j} $$
by first doing the colimit in two steps using the left Kan extension along $p$ and then that the inclusion $E_j \to E \times_J J_{/j}$ is cofinal since $p$ is cocartesian.
Now the colimit $I$ can be described as the localization of $E$ at the cocartesian morphisms. Since any localization is cofinal, this means there is a cofinal functor $q \colon E \to I$. For a functor $D \colon I \to C$, this means we have equivalences
$$ \mathrm{colim}_I \, D \simeq \mathrm{colim}_E \, Dq \simeq \mathrm{colim}_{j \in J} \, \mathrm{colim}_{I_j} \,D|_{I_j}. $$
Nice! Thanks, Rune. I'll probably accept this one -- though I also like Dylan's presentation of essentially the same argument in his answer below. And of course, it would be nice to work out an analog of Zhen Lin's argument in a higher setting -- those types of arguments in terms of homsets tend to generalize more easily to the enriched setting.
I wonder if the proofs in HTT of the relevant statements actually depend on the more technical statement of <IP_ADDRESS>. It would be nice if they didnt'!
I may be misinterpreting your statement/making a mistake, but I don't think that $I$ is always the localization of $E$. For instance consider the case where $J$ is a one object groupoid $BG$ and $J \to Cat$ is the constant diagram with value the trivial category. Then the colimit is the trivial category, but $E$ is $BG$, and all localizations of $E$ are isomorphic to $E$.
@PhilTosteson The colimit of the constant diagram in the $\infty$-category of $\infty$-categories is $BG$, not the point. This is just another way in which colimits are better behaved in $\infty$-categories :).
@DenisNardin Indeed, that's why I chose this example-- since the homotopy quotient disagrees with the ordinary one. I thought that the question and this answer were meant to be specifically 1 categorical
@PhilTosteson You are right. The localization of E at cocartesian morphism is literally the 2-colimit (i.e. the colimit in the $\infty$-category obtained by inverting equivalences of categories in the $1$-category of small categories). This does not coincide with the colimit (in the $1$-category of small categories) in general. Rune's answer together with Zhen Lin's show that the comparison map from the $2$-colimit to the $1$-colimit, although not an equivalence, is colimit-final. I also thought that the question was about $1$-colimits though.
The way I always remember this stuff is as follows:
Given a map $J \to \mathsf{Cat}$ form the associated cocartesian fibration $E \to J$.
By assumption, $I$ is the actual colimit (as opposed to the left lax one) so we have a (weak) localization $E \to I$. Weak localizations are final (and initial, in fact) so, to compute the colimit over $I$ is the same as computing it over $E$.
To compute the colimit over $E$ we may first left Kan extend to $J$.
Since $E \to J$ is cocartesian, the map $E_x \to E_{/x}$ is final, and we may replace $E_{/x}$ with $E_x=I_x$ in the formula for left Kan extensions.
That gives the result.
ah! Rune beat me to it
I assume $\varinjlim_{j : \mathcal{J}} \mathcal{I}_j = \mathcal{I}$ is meant in the strict sense of 1-categories. Since $\textbf{Cat}$ is cartesian closed,
$$\textstyle [\mathcal{I}, \mathcal{C}] \cong \varprojlim_{j : \mathcal{J}} [\mathcal{I}_j, \mathcal{C}]$$
where the limit on the RHS is also meant in the strict sense of 1-categories. Let $\lambda_j : \mathcal{I} j \to \mathcal{I}$ be the component of the colimit cocone in $\textbf{Cat}$. Then, we also get a limit formula for the hom-sets of $[\mathcal{I}, \mathcal{C}]$, namely,
$$\textstyle [\mathcal{I}, \mathcal{C}](D, \Delta T) \cong \varprojlim_{j : \mathcal{J}} [\mathcal{I}_j, \mathcal{C}](D \lambda_j, \Delta T)$$
so if the relevant colimits exist in $\mathcal{C}$,
$$\textstyle \mathcal{C} \left( \varinjlim_\mathcal{I} D, T \right) \cong \varprojlim_{j : \mathcal{J}} \mathcal{C} \left( \varinjlim_{\mathcal{I}_j} D \lambda_j, T \right) \cong \mathcal{C} \left( \varinjlim_\mathcal{J} \varinjlim_{\mathcal{I}_j} D \lambda_j, T \right)$$
as desired.
I think the main point of contension would be with regards to the first isomorphism you wrote in the second line. Your answer would probably benefit from an additional step, where you use that hom-sets in limits of categories are the limits of hom-sets in the categories.
Right. Actually I think it might be more important to emphasise that I am working with strict limits of categories rather than pseudolimits.
Thanks Zhen Lin! This argument absolutely convinces me that 1-categorically, the result is true in full generality, answering Question 1 (and suggesting that the $\infty$-categorical version should be similarly general and formal). The main detail which still confuses me (and which I wouldn't bother worrying over if I didn't have my eye out to the $\infty$-categorical case) is the following: at which point in the argument do we produce a functor $\bar D: J \to \mathcal C$ which is defined on objects by $\bar D(j) = \varinjlim_{i \in I_j} D_i$ (along with a cocone $\bar D \to \Delta T$)?
The diagram $\mathcal{J}^\textrm{op} \to [\mathcal{C}, \textbf{Set}]$ given by $j \mapsto \mathcal{C} \left( \varinjlim_{\mathcal{I}_j} D \lambda_j, - \right)$ is a diagram of representables, so gives a diagram $\mathcal{J}^\textrm{op} \to \mathcal{C}^\textrm{op}$.
Note to self: the $\infty$-categorical result doesn't quite specialize to the strict result that Zhen Lin gives. Instead it should specialize to the analogous version with pseudocolimits.
I think this proof also goes through for $\infty$-categories, which suggests it should also work for pseudo-colimits in Cat - but then I'm confused about where the difference between colimits and pseudo-colimits in Cat shows up. Is it just that the canonical map between them is cofinal for 1-categories, so in $\mathcal{C}$ you just can't tell the difference (and both give the same isomorphisms of Hom-sets)?
I had the same doubt, which is why I edited to say my proof is for strict colimits. $[-, \mathcal{C}]$ will send pseudocolimits to pseudolimits but I don't know how (well, haven't tried) to calculate hom-sets of the pseudolimit.
|
2025-03-21T14:48:31.950609
| 2020-09-03T22:10:45 |
370790
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"RiezFrechetKolmogorov",
"https://mathoverflow.net/users/164762",
"https://mathoverflow.net/users/33741",
"leo monsaingeon"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632685",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370790"
}
|
Stack Exchange
|
Uniqueness for measure valued ode
Morning! Basically I'm working on a mean field scaling for some measure valued process (valued on $M_F(\mathbb{N})$). The limit turns up as a (deterministic) solution to a measure valued ODE. Let's say, given a bounded real valued function $f$ :
$$ d\langle\mu_t,f\rangle = \langle\Psi(\mu_t),f\rangle dt$$
where
$$ \langle\Psi(\mu_t),f\rangle = - \frac{\langle\mu_t,\chi^2-\chi\rangle}{\langle\mu_t,\chi\rangle}\frac{\langle\mu_t,\chi\Delta f\rangle}{\langle\mu_t,\chi\rangle} - \frac{\langle\mu_t,\chi f\rangle}{\langle\mu_t,\chi\rangle}$$
with $\chi(k) = k$ and $\Delta f(k) = f(k) - f(k-1)$
My problem arises when I try to show uniqueness for this ODE. Usually, I would just upper bound the total variation distance between two solutions with same initial state and use Gronwall's Lemma to show said distance is null. This time it is delicate since $\Psi(\cdot)$ contains first and second order moments of $\mu_.$. Ie for two solutions $\mu_t$ and $\nu_t$ we end up with
$$d|\langle\mu_t - \nu_t,f\rangle| \leq C||f|| ( |\langle\mu_t - \nu_t,\chi^2\rangle | + |\langle\mu_t - \nu_t,\chi\rangle | + |\langle\mu_t - \nu_t,1\rangle |)$$
Whereas the ideal would be to have a multiple of $|\langle\mu_t - \nu_t,f\rangle| $ on the right side. Now, using the EDO to bound $|\langle\mu_t - \nu_t,\chi^2\rangle| $ also fails since it involves the 3rd moment and so on.
My question is: Is there any way to bound $\langle\mu_t - \nu_t,\chi\rangle$ and $\langle\mu_t - \nu_t,\chi^2\rangle$ by some multiple of $d_{TV}(\mu_t,\nu_t)$? Do I have to try another method (suggestions are welcome)?
Additionally both $\langle\mu_t,1\rangle$ and $\langle\mu_t,\chi\rangle$ are both decreasing and bounded, and $\langle\mu_t - \nu_t,\chi^2\rangle$ is finite.
If, as I presume given that you used the "probability" tag, the measures are actually probability measures: How about trying to use the quadratic Wasserstein distance $W_2$? This seems very appropriate, since $W_2$ controls second moments by construction. In particular $W_2(\mu_t,\nu_t)$ controls $|m_2(\mu_t)-m_2(\nu_t)|$ (the difference between second moments)
Ah, my bad, I just noticed that $<\mu_t,1>$ is increasing, so you're dealing with measures with arbitrary and varying masses... Then I can suggest using "unbalanced optimal transport"
Related to "unbalanced optimal transport" and relatives: I suggest you check the following paper: https://arxiv.org/abs/1910.05105
First, thank you for your help! Actually, you can have an equivalent problem with $$<mu_t,1>$$ decreasing so I believe we can state the problem in terms of optimal transport.
I realise the problem was poorly formulated. Actually the biggest problem I encountered with the Wasserstein metric was that the edo terms contained like : $$<\mu_t,\chi \Delta f>$$ and $$<\mu_t,\chi f>$$ so that controling the second moment implies dealing with the third, etc etc
Wait, now I'm confused: who's $f$ and who's $\chi$? I had the impression that $f$ was an arbitrary test-function, but I'm not too sure anymore.
Now I see how badly written my statement is, sorry for my english. $f$ is a test function and $\chi$ is just the typical function such that $\chi(x) = x$, it also means that the first moment is expressed as $<\mu_t,\chi>$.
Basically, we have a measure valued ode which involves $$d<\mu_t - \nu_t,f> = ... + <\mu_t - \nu_t,\chi f> $$
So that for example, bounding $m_2$ will imply using $m_3$
Which domain are you working on? is $N$ bounded? in Euclidean spaces if the domain is bounded as $|x|\leq R,, x\in N$ then the $p$-th momment can be trivially controlled as $m_p(f)-m_p(g)=\int |x|^p d(f-g)\leq R^p \int |x/R|^pd(f-g)\leq R^p d_{TV}(f,g)$ (and by symmetry you control the absolute value).
Regardless of the English language (which is fine, don't worry!): unless you give us the means to grasp the sublte technical details of your problem by clarifying the notations, the setting etc. I doubt that we can help you as is. For example: perhaps it is worth wrting down explicitly the formula defining your "vector-field" $\Psi(\mu)$?
Really thank you for your patience. I was a newbie both on the site and on LateX. I've edited the problem to be more specific and I tried to add context.
I guess you don't have any additional information on the decay/support of your measures arising from some other features of the problem? If you knew for example that $(\mu_t)^i$ was rapidly decaying in $i\in\mathbb{N}$ at a uniform rate, then you would be in business right?
|
2025-03-21T14:48:31.950914
| 2020-09-03T23:09:33 |
370796
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632686",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370796"
}
|
Stack Exchange
|
Equivalent definitions of unramified characters
Let $G$ be a connected reductive group over a local field $F$. An unramified character of $G(F)$ is a continuous character $\chi: G(F)\to\mathbb{C}^\times$ that is trivial on all compact subgroups of $G(F)$.
It seems to be a well-known result that unramified characters are precisely those that are trivial on $G(F)^1=\bigcap_{X^*(G)_F}\ker|\chi|_F$, however I have been unable to find a proof of this in the literature. Is there a reference for this result?
The subgroup $G(F)^1$ contains all compact subgroups of $G(F)$. On page 8 of https://arxiv.org/pdf/1506.09112.pdf it is stated that there is an unpublished proof by Abe, Henniart, Herzig, and Vigneras that, in the case that $F$ is non-archimedean, $G(F)^1$ is generated by the compact subgroups of $G(F)$. Is there a reference for this result? Is it true if $F$ is archimedean?
|
2025-03-21T14:48:31.951001
| 2020-09-04T01:46:26 |
370809
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"6666",
"Francesco Polizzi",
"KReiser",
"abx",
"https://mathoverflow.net/users/112114",
"https://mathoverflow.net/users/40297",
"https://mathoverflow.net/users/7460",
"https://mathoverflow.net/users/88180"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632687",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370809"
}
|
Stack Exchange
|
Kummer surfaces which are not projective
This is a question from an online note. Let $A$ be a two-dimensional $\mathbb C$-torus. And there is an involution on $A$: $A\to A, x\mapsto -x$. The action has 16 fixed points. Let $Y:=A/\{\pm1\}$, then $Y$ is a complex surface with 16 ordinary double points. Let $X$ be the blow up of $Y$ at all 16 singular points. After some calculations, we can see $X$ is a $K3$ surface. Then it's claimed that if $A$ is not projective then $X$ is not projective and we get an example of a non-projective $K3$ surface. But I feel confused why "if $A$ is not projective then $X$ is not projective".
I know since $A\to X$ is finite, thus if $Y$ is projective, then we can pullback an ample line bundle to an ample line to $A$. But I can't see why $X$ is projective implies $Y$ is projective? Do we have contracting a rational curve on a complex surface preserves the projectivity?
Contracting a curve with negative square on a surface preserves projectivity. See this question of MO and its answer.
Crossposted from MSE. When crossposting, it is important to link all versions together to prevent duplicating work.
In fact, $X$ is projective if and only if $A$ is projective.
If $A$ is projective, then $Y$ is so, being the quotient of a projective variety by a finite group (this is a toy model of GIT, see this MO question). Then $X$ is projective, too, being the blow-up of the projective variety $Y$ at a finite number of points.
Conversely, assume $X$ projective. Then there is a double cover $\tilde{A} \to X$, where $\tilde{A}$ is the blow-up of $A$ at its $16$ points of order $2$. This shows that $\tilde{A}$ is projective, so the blow-down $A$ is projective as well (an alternative argument is noticing that $X$ projective implies $Y$ projective and so $A$ projective, since contracting a $(-2)$ curve on a projective surface preserves projectivity, as explained in abx's comment).
Thanks for the nice answer, but why is the blow down also projective? Can you explain or tell me a reference?
For surfaces, this is usually called "Castelnuovo contractibility theorem". Look for it in Griffiths-Harris, p. 476 (where it is called "Castelnuovo-Enriques criterion"), or in Beauville's book on surfaces (II.17)
Sorry, when we apply Castelnuovo contractibility theorem, how can I know the blow down (projective) surface is exactly $A$?
The contracting map must factor through the blow-down by the universal property, and from this it is an immediate exercise to show that it coincides with the blow-down (in fact, this is part of the statement of Castelnuovo's criterion).
|
2025-03-21T14:48:31.951192
| 2020-09-04T02:37:37 |
370812
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632688",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370812"
}
|
Stack Exchange
|
Mismatching degrees and # derivatives in polynomial interpolation error formula
It is well known that if $f : [a,b] \to \mathbb{R}$ is $n+1$ times differentiable and $p(x)$ denotes the polynomial interpolant to $f(x)$ in the $n+1$ points $\bigl(x_k \in [a,b]\bigr)_{k = 1}^{n+1}$, then there exists $\xi = \xi(x) \in [a,b]$ such that
$$
f(x) - p(x)
=
\frac{f^{(n+1)}(\xi)}{(n+1)!} \, \prod_{k = 1}^{n+1} (x - x_k)
.
$$
Does this result have a generalisation to the case that $f(x)$ has only $m < n+1$ derivatives?
Cross-posted from https://math.stackexchange.com/q/3810281 due to lack of traction there.
|
2025-03-21T14:48:31.951266
| 2020-09-04T02:37:41 |
370813
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Akhil Mathew",
"S. carmeli",
"Tim Campion",
"https://mathoverflow.net/users/115052",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/344"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632689",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370813"
}
|
Stack Exchange
|
Chromatic blueshift and Tate cohomology
Let $R$ be an $L_n$-local ring spectrum. Then one knows that the Tate construction $R^{tC_p}$ (with respect to the trivial $C_p$-action on $R$) is $L_{n-1}$-local; this "blueshift" result is due to Hovey-Sadofsky and Greenlees-Sadofsky, and has been generalized to the telescopic setting by Kuhn.
I am curious to what extent there is an analog of this result when $R$ is instead connective. For example, suppose $R$ has the property that its $K(i)$-localization vanishes for $i \geq n$ (not including $\infty$, so for instance the connective cover of an $L_n$-local ring spectrum is allowed). Does this imply that the $K(i)$-localization of $R^{tC_p}$ vanishes for $i \geq n-1$?
Super-ignorant question -- for the Hovey-Sadofsky result you mention, does it matter whether $R^{tC_p}$ is computed in the $L_n$-local category vs. being computed in the category of all spectra?
@TimCampion: It doesn't affect the statement, since the subcategory of $L_n$-local spectra is closed under all limits and colimits (the latter by the smash product theorem).
It seem to me the question for all ring spectra and connective ring spectra are equivalent. It's a matter of commuting localization and fixed points, and either composition kill coconective spectra since Ki localization kills cocobective stuff and fixed points preserve them. Do I miss something here?
@S.carmeli: Yes, I agree with you.
|
2025-03-21T14:48:31.951523
| 2020-09-04T02:56:18 |
370814
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mateusz Kwaśnicki",
"Riku",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/122620"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632690",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370814"
}
|
Stack Exchange
|
Fractional Laplacian and convolution $(-\Delta)^\alpha (u \ast \eta_\epsilon) = (-\Delta)^\alpha u \ast \eta_\epsilon$?
For $u \in L^\infty(\mathbb R)$ and $\eta_\epsilon$ mollifier, it is well-known that for the (distributional) derivative it holds that $(u \ast \eta_\epsilon)' = u'\ast \eta_\epsilon$.
Is it also true for the fractional Laplacian that
$$(-\Delta)^\alpha (u \ast \eta_\epsilon) = (-\Delta)^\alpha u \ast \eta_\epsilon$$
holds? Where can I find a proof of this?
Yes, it is true. An answer to the other question depends on your favourite definition of $(-\Delta)^\alpha u$. If $u$ is merely bounded, then $(-\Delta)^\alpha u$ can only be defined in the sense of distributions — and indeed $(-\Delta)^\alpha$ is an integrable distribution, so that $(-\Delta)^\alpha$ and $u$ are convolvable, and the desired equality follows from standard properties of convolution of distributions.
@MateuszKwaśnicki Hi! Sorry for the late reply. Thanks! Could you point out a specific reference? Or sketch the proof in an answer?
That's too long for a comment — see an answer below.
Answering the question asked in a comment recently: you can read more about distributional definition of the fractional Laplacian in the paper by Luis Silvestre [1] (or in his PhD thesis, if I remember correctly), in an excellent book by Stefan Samko [2], or in Section 5 of my survey [3]. I do not think any of these references has the statement that you are looking for written explicitly, though.
As for the proof, it is in fact a relatively simple consequence of the definition: the distribution $f = (-\Delta)^s u$ is given by
$$ \langle f, w \rangle = \langle u, (-\Delta)^s w \rangle $$
for all infinitely smooth $w$ such that all derivatives of $w$ are absolutely integrable (and $(-\Delta)^s w$ is defined by any of the usual definitions). Now simply apply this to $w(y) = \eta_\epsilon(x - y)$ for a fixed $x$ to get
$$ ((-\Delta)^s f) * \eta_\epsilon(x) = \langle f, w \rangle = \langle u, (-\Delta)^s w \rangle = u * (-\Delta)^s \eta_\epsilon(x) = \ldots $$
Now the right-hand side can be written as
$$ \ldots = \int_{\mathbb R} \int_{\mathbb R} u(x - y) (\eta_\epsilon(y + z) - \eta_\epsilon(y) - z \cdot \nabla \eta_\epsilon(y) \mathbb 1_B(z)) dz dy = \ldots $$
(with $B$ denoting the unit ball $(-1, 1)$), and it is not very difficult to see that we may apply Fubini's theorem to get
$$ \begin{aligned} \ldots & = \int_{\mathbb R} \int_{\mathbb R} u(x - y) (\eta_\epsilon(y + z) - \eta_\epsilon(y) - z \cdot \nabla \eta_\epsilon(y) \mathbb 1_B(z)) dy dz \\
& = \int_{\mathbb R} (u * \eta_\epsilon(x + z) - u * \eta_\epsilon(x) - z \cdot (u * \nabla \eta_\epsilon)(y) \mathbb 1_B(z)) dz = \ldots \end{aligned} $$
We already know that $u * \nabla \eta_\epsilon = \nabla (u * \eta_\epsilon)$, so we eventually get
$$ \ldots = (-\Delta)^s (u * \eta_\epsilon)(x) ,$$
as desired.
If you are familiar with distributional convolution, you can alternatively write $(-\Delta)^s u$ as $L * u$ for an appropriate distribution $L$, and use associativity of convolution to immediately get
$$ (-\Delta)^s u * \eta_\epsilon = (L * u) * \eta_\epsilon = L * (u * \eta_\epsilon) = (-\Delta)^s (u * \eta_\epsilon) .$$
Convolution is indeed associative, because $L$ and $\eta_\epsilon$ are integrable distributions, and $u$ is a bounded distribution.
References:
[1] L. Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator, Commun. Pure Appl. Math., 60 (2007), 67–112, DOI:10.1002/cpa.20153
[2] S. Samko, Hypersingular Integrals and Their Applications, CRC Press, London–New York, 2001, DOI:10.1201/9781482264968
[3] M. Kwaśnicki, Fractional Laplace Operator and its Properties, in: A. Kochubei, Y. Luchko, Handbook of Fractional Calculus with Applications. Volume 1: Basic Theory, De Gruyter Reference, De Gruyter, Berlin, 2019, DOI:10.1515/9783110571622-007
Thank you so much!
|
2025-03-21T14:48:31.951783
| 2020-09-04T03:52:14 |
370819
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Yemon Choi",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/114356",
"https://mathoverflow.net/users/763",
"user50394"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632691",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370819"
}
|
Stack Exchange
|
Diagonalizability of Gaussian random matrices
Let $X$ be an $n\times n$ matrix whose elements are i.i.d. sampled from a normal distribution of zero mean and unit variance. Is $X$ diagonalizable over $\mathbb{C}$ with probability 1? Is there a good reference for diagonalizability of random matrices?
Are you assuming that X is symmetric? Are you using real or complex scalars?
If you work in $M_n({\bf C})$ then the subset of matrices that are not diagonalizable over ${\bf C}$ has Lebesgue measure zero, and hence for any probability density on $M_n({\bf C})$ that is absolutely continuous with respect to Lebesgue measure, a random element of $M_n({\bf C})$ will be almost surely diagonalizable over ${\bf C}$
@YemonChoi X is not symmetric; every element is sampled independently from a normal distribution. Thanks for the result over C. I'd be more curious to know the density when X is real-valued.
But why would you expect a "typical" real matrix to be diagonalizable over R? (The particular Gaussian i.i.d. model you are asking about is the Ginibre ensemble, so we know that the spectrum is almost surely a disc (in the asymptotic sense as n tends to infinity. However, my point is that even ignoring probability theory, it is very easy for real matrices to have complex eigenvalues)
Ah I see, thanks. I am very new to random matrix theory. I think I am curious to know the density of real-valued diagonalizable matrices whose eigenvalues can be complex numbers.
@YemonChoi Do you have a reference for this? I'd like to know more details about what you just said.
see also https://mathoverflow.net/q/12657/11260
please clarify: the matrix $X$ is real, but do you wish to diagonalize over $\mathbb{R}$ or over $\mathbb{C}$? the probability for a random real matrix to have all real eigenvalues is vanishingly small, typically only $\sqrt n$ of the eigenvalues are real.
@CarloBeenakker The matrix $X$ is real, but I wish to know the density of $X$ diagonalizable over $\mathbb{C}$.
The measure of real matrices that are not diagonalizable
over $\mathbb{C}$ equals to 0, see for example On the computation of Jordan canonical form, so the probability for a random matrix with a continuous probability distribution to be non-diagonalizable vanishes. To put it differently, the set of real matrices without multiple eigenvalues is dense, and a matrix without multiple eigenvalues is definitely diagonalizable.
Thanks, that is helpful. Is it known whether the set of unitarily diagonalizable matrices, i.e., normal matrices, is dense in $R^{n\times n}$?
a small perturbation of a normal matrix spoils that property, so it cannot be a dense set, can it?
|
2025-03-21T14:48:31.952232
| 2020-09-04T06:58:46 |
370823
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"AgnostMystic",
"Dieter Kadelka",
"Iosif Pinelis",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/158175",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632692",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370823"
}
|
Stack Exchange
|
Anomaly with a pdf
The pdf of the range $\omega$ of $n$ identically r.v.'s random variables distributed with cdf $\mathbf{F}$ and pdf $\mathbf{f}$ is given by
$$ g(\omega)=n(n-1) \int_{-\infty}^{\infty} f(x)[F(x+\omega)-F(x)]^{n-2} f(x+\omega) dx$$
Now Consider the pdf
$$ f(x)=2x,\,0 \leq x \leq1 $$. and the corresponding cdf
$$F(x)=x^2,0 \leq x\leq 1,1 \text{else}$$
This means the pdf of the range of the given sequence of $n$ random variables should be given by :
$$ g(\omega)=4 n(n-1) \int_0^1 x(x+\omega)[(x+\omega)^2-x^2]^{n-2}\, dx$$
How can I verify that $g$ as a function of $\omega$ is indeed a probability density function.It appears $\omega $ will vary from 0 to 1. is When I solve the integral and take the integral of $g(\omega )$ from 0 to 1 I am not getting 1,which has to be the case if $g$ is indeed a pdf. Even for $n=2$ we obtain
$$g(\omega)= 8 \int _0^1 (x^2+x \omega) dx= \frac83+\frac{\omega}{2}$$,which is not a pdf for $\omega $ in the $(0,1)$
I would be most obliged if somebody can help me fix any errors with the help of Mathematica or some other CAS .Thanking you in advance
I have problems with the formulation of your problem. Initially you assume r.v. on $\mathbb{R}$ (terms pdf and cdf). But in the next sentence you assume that the random variables have values in the unit circle ${x \in \mathbb{R}^2 \colon x_1^2+x_2^2 \leq 1}$ (two dimensional).
@DieterKadelka there is no problem with the formulation of the problem .though the point s pie in the unit circle ,I am only interested in the distribution of their radial distances and radial distance is one dimensional variable.I have also explicitly mentioned the pdf ,which is in terms of a single variable .Hope this answers your query
O.K., it seems that the introduction is simply motivation, but has nothing to do with your problem. If I understand correctly you simply have $n$ independent r.v. $X_1,\ldots,X_n$ with pdf $f$, But what is then formally "the range of radial distances of n random points"?
I am applying the formula for range to the radial distances of $n$ independently ,uniformly and randomly points in unit circle and what i get seems not to make sense .When I do a sanity test whether the 'pdf' so obtained is really a pdf, we find that the function does not satisfy the conditions of a pdf. the axample i am taking is not meant for motivation only,it is my real question and the thing I want some help about
Again, you even calculated that $g$ is no pdf. It is completely irrelevant how you get the integral. There must be something wrong with $g$! So again what is formally "the range of radial distances of n random points"?
Your problem is that you are using incorrect expressions for the pdf $g$ of the range, and for the cdf $F$ and pdf $f$ of the radial distance. (Also, it is not good to denote a random variable (r.v.) and its values by the same symbol (in your case, $\omega$). Also, there is no reason to use the Greek letter $\omega$ if you can use (say) $t$ instead -- which is what we are going to do.)
Now, a correct expression for the pdf $g$ of the range $R:=\max_{1\le i\le n}X_i-\min_{1\le i\le n}X_i$ is given by
$$g(t)=h(t):=n(n-1)\int_{-\infty}^\infty f(x)[F(x+t)-F(x)]^{n-2} f(x+t)\, dx\;1(t>0) $$
for real $t$, where $X_1,\dots,X_n$ are independent r.v.'s with common cdf $F$ and pdf $f$; your expression is missing the factor $1(t>0)$.
The function $h$ is a pdf, because it is the pdf of the range $R$.
Another way to check that $h$ is a pdf is to note that (i) $h$ is obviously nonnegative and (ii)
$$\int_{-\infty}^\infty g(t)\,dt \\
=n(n-1)\int_{-\infty}^\infty dx\,f(x)\int_0^\infty dt\,[F(x+t)-F(x)]^{n-2} f(x+t) \\
=n(n-1)\int_{-\infty}^\infty dx\,f(x)\frac{[F(x+t)-F(x)]^{n-1}}{n-1}\Big|_{t=0}^\infty \\ =n\int_{-\infty}^\infty dx\,f(x)[1-F(x)]^{n-1}
=-[1-F(x)]^n\Big|_{x=-\infty}^\infty=1. \tag{1}$$
In your case, correct expressions for the cdf $F$ and pdf $f$ of the radial distance are given by formulas
$$F(x)=x^2\,1(0<x<1)+1(x\ge1)$$
(rather than $F(x)=x^2$) and
$$f(x)=2x\,1(0<x<1)$$
(rather than $f(x)=2x$), where $x$ is any real number.
So, your (incorrect) formula
$$ g(\omega)=4 n(n-1) \int_0^1 x(x+\omega)[(x+\omega)^2-x^2]^{n-2}\, dx$$
should be replaced by
$$g(t)=4 n(n-1)\int_0^1 x(x+t)1(x+t<1)[(x+t)^2\,1(x+t<1)+1(x+t\ge1)-x^2]^{n-2}\, dx\,1(t>0).$$
However, there is no need to repeat (1) for your particular $F$ and $f$.
@losif Pinelis I am so thankful to you for pointing out the fatal mistake in my approach.So kind o you for taking time
@losif Pnielis but i faiil to see how that will change $g(t)$ ..$F(x+t)=F(x)$ outside $(0,1)$.Could you please clarify?
@sajjadveeri : I have added the correct expression for $g(t)$ in your case. There is no need to use it, though.
|
2025-03-21T14:48:31.952571
| 2020-09-04T07:29:02 |
370825
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"GH from MO",
"Gérard Lang",
"Lagrida Yassine",
"P.-S. Park",
"Stanley Yao Xiao",
"Sylvain JULIEN",
"Yemon Choi",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/11919",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/164630",
"https://mathoverflow.net/users/30395",
"https://mathoverflow.net/users/763",
"https://mathoverflow.net/users/9147"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632693",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370825"
}
|
Stack Exchange
|
Goldbach conjecture and the representation number
Let $g(2n)$ be the number of representations of $2n=p+q$ with primes $p$ and $q$. Many people have asked whether $g(2n) \ge 2$ when $2n = p+q$ for some primes $p$ and $q$. That is, does $g(2n) \ge 1$ imply $g(2n) \ge 2$? From the famous Goldbach Comet, it looks probable although it was not yet proved.
Now, what can we say about the following weaker problem?
For any sufficiently large prime $p$, is there a prime $q$ such that $p+q$ has another representation $p' + q'$?
Conjecturally yes, but it's not been proved, as far as I know.
See https://en.m.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Tur%C3%A1n_conjecture_on_additive_bases
Please always include a high-level tag like "nt.number-theory".
@GHfromMO I forgot it. Thank you for your editing.
Strictly formally speaking, if (p,q) is a solution, (q,p) is another solution.
To your first question: we don't know. To your second question: we know much more, namely if $N$ is a large odd number, then the number of representations $N=p_1+p_2-p_3$ with each $p_j$ a prime from $[2N,3N]$, has order of magnitude $N^2/(\log N)^3$. This can be proved in essentially the same way as we prove that $N$ can be written as a sum of three primes in that many ways. See also Harald Helfgott's response here.
Here is a graph showing the number of representations of $2n$ as a sum of two primes.
It suggests that something much stronger than what you ask about is true. And there are heuristics that predict what is shown. but not proofs.
Let $n \in 2\mathbb{N}^*$ large enough.
$$n = p+q, \ (p,q)\in\mathbb{P}^2 \iff (p, n-p) \in \mathbb{P}^2$$
You search for the quantative version of Goldbach's conjecure, Hardy and Littlewood in there 1923 paper "Some problems of ‘Partitio
numerorum’; III : On the expression of a number as a sum of primes", conjecture that :
$$G(n) \sim 2 C_2 \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ 3 \leqslant p}} {\normalsize \dfrac{p-1}{p-2}} \Big)} \dfrac{n}{\log(n)^2}.$$
Where $G(n) = \#\{(p, n-p) \in \mathbb{P}^2 \, | \, p \leqslant n\}$, and : $C_2 = \displaystyle{\small \prod_{\substack{3 \leq p \\ \text{p prime}}} \left({\normalsize 1-\dfrac{1}{(p-1)^2}}\right)}$.
This conjecture agree perfectely with numeric checks, but unfortunately not proven up to now (and no hope to prove it soon).
You can see my try here : is there a link with the probabilistic model for prime numbers?
I don't see how this addresses the question that was asked
@YemonChoi, he ask if $g(2n) \ge 1 \implies g(2n) \ge 2$, and i say that $g(2n)$ is more much bigger (conjectured), please delete your comment you made a mistake.
The OP mentions that conjecture, but he or she is asking about "the following weaker problem". See the answer of GHfromMO. Please read carefully the question being asked
I understund exactly his 2 questions, the second about fixing a prime number $p$, then for every other prime number $3 \leqslant q \leqslant p$, we have $n_{p,q}=p+q$ is an even number and returning to Goldbach's conjecture on it's quantative form we have approximatly $G(n_{p, q})$ representation of $n_{p,q}$ as sum of 2 primes.
|
2025-03-21T14:48:31.952816
| 2020-09-04T07:39:32 |
370826
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABB",
"Stefaan Vaes",
"https://mathoverflow.net/users/159170",
"https://mathoverflow.net/users/84390"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632694",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370826"
}
|
Stack Exchange
|
When a normal functional is restricted to a vn Neumann sub-algebra
I have already asked this question and no comment(s) received up to now.
I am so curious to get feedback concerning the problem.
Let $M$ be a vn Neumann subalgebra in $B(H)$. Let $f$ and $g$ be normal functionals on $B(H)$ and $M$ respectively. Suppose that $f_{|_{M}}=g$ i.e, the restriction of $f$ to $M$ is just $g$.
Let us define the positive linear functional $\phi:M\to \mathbb{C}$ given by $\phi(x)=|f|(x)$, where $|f|$ is the absolute value of the normal functional $f$.
Q. Can we conclude (in general) that $|g|\leq \phi$ ?
No, such a property does not hold. For instance, you could take $H = \mathbb{C}^2$ and $M \cong \mathbb{C} \oplus \mathbb{C}$ the subalgebra of diagonal matrices in $B(H)$. Denoting by $E : M_2(\mathbb{C}) = B(H) \rightarrow \mathbb{C} \oplus \mathbb{C}$ the conditional expectation given by restricting a matrix to its diagonal, you are asking whether the operator inequality $|E(A)| \leq E(|A|)$ holds for all matrices $A \in M_2(\mathbb{C})$, where $|B| = (B^*B)^{1/2}$. This fails for instance for
$$A = \begin{pmatrix} 1 & 1 \\\ 0 & 0\end{pmatrix} \; .$$
It is great thanks a lot Stefaan.
By any proper property on $M$, could such a property satisfy?
No, the property only holds if $M = \mathbb{C}1$ or $M = B(H)$. Otherwise, pick projections $p \in M$ and $q$ in the commutant $M'$ such that both $pq$ and $(1-p)(1-q)$ are nonzero. Choose a unit vector $\xi \in H$ in the range of $pq$ and choose a unit vector $\eta$ in the range of $(1-p)(1-q)$. Define $f:B(H)\rightarrow \mathbb{C} : f(x) = \langle x\xi,\xi+\eta\rangle$. Then, $g(x) = \langle x\xi,\xi\rangle$ for all $x \in M$. So, $g$ is positive. Also, $|f|(x) = 2^{-1/2}\langle x(\xi + \eta),(\xi+\eta)\rangle$ for all $x \in B(H)$. So, $|g|(p)=1$, $\phi(p)=2^{-1/2}$, and $|g|\not\leq\phi$.
|
2025-03-21T14:48:31.952959
| 2020-09-04T08:05:30 |
370828
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632695",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370828"
}
|
Stack Exchange
|
Quiver algebras of Dynkin type
Let $kQ$ be one of the Dynin path algebras of type $A_n , D_n $ or $E_i$ for $i=6,7,8$.
Question 1: How many (up to isomorphism) quiver algebras are there that are derived equivalent to $kQ$?
Question 2: Is there an algorithm to obtain all such algebras for a given $Q$ in a quick way? Can all such algebras be obtained via QPA for a given (small) Q?
Such algebras can be obtained by taking iterated tilted algebras but there might be several different tilting modules that lead to the same algebra.
Question 3: Define a graph with point the algebras derived equivalent to $kQ$ and edges between two different points in case they are tilted algebras of eachohter. What is this graph for a given $Q$? Does it appear in some other context?
|
2025-03-21T14:48:31.953048
| 2020-09-04T09:12:27 |
370832
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asaf Karagila",
"Curiosity",
"Emil Jeřábek",
"JimN",
"Noah Schweber",
"Pace Nielsen",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/3199",
"https://mathoverflow.net/users/62043",
"https://mathoverflow.net/users/66018",
"https://mathoverflow.net/users/7206",
"https://mathoverflow.net/users/8133"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632696",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370832"
}
|
Stack Exchange
|
Peano axioms— mathematical induction and other axioms
The Peano axioms of $\Bbb N$ are:
$1 \in \Bbb N$, i.e. $\Bbb N$ is not empty and contains an element denoted by $1$.
Every natural number has a successor, i.e. $\forall n\in\Bbb N, \exists!s(n)\in\Bbb N$.
if $s(n)=s(m)$ then $n=m$.
$1\in\Bbb N$ is the only element that is not the successor of a natural number.
The axiom of mathematical induction is valid:
Let $S\subseteq\Bbb N$ such that
$1\in S$
$\forall n\in\Bbb N,n\in S\Rightarrow(s(n)\in S)$.
Then $S=\Bbb N$.
I am trying to find an example of a collection "$\Bbb N$'' with 1,2 that satisfies 5 but not 3 and also not 4. (It is easy to find examples satisfying 3 but not 4,5, and 4 but not 3,5. My question is about 5 but not 3,4.) In other words, is there a set "$\Bbb N$'' that has a $1$, successors exist, and induction holds, but $1$ is the successor of an element and also the successor function is not one-to-one? I can't seem to think of an example. I suspect that if 1,2,5 are satisfied, then either 3 or 4 must hold. Is there an elementary proof of this?
Well, the induction axiom is equivalent to $\mathbb N={1,s(1),s(s(1)),\dots}$ (that is, ${s^{(k)}(1):k\in\omega}$). Thus, the graph of $s$ consists of a single directed walk starting from $1$. Either this is an infinite path (in which case both 3 and 4 hold), or it eventually enters a cycle; if it is just a cycle, then 3 holds and 4 fails, whereas if there is a nonempty path leading to the cycle, then 3 fails and 4 holds. So, yes, 3 or 4 must hold.
@EmilJeřábek Thanks, I see now. However, since this statement is about basic set theory, is there a proof that uses just the bare-minimum of elementary logic/reasoning rather than more advanced notions of paths/cycles? It seems like there should be such a proof.
If you consider cycles and paths advanced, then I really have no idea what you mean by elementary.
I don't see why "an infinite path" must imply 3 and 4 to hold.
This question came up in a discussion (not homework) on elementary set theory for first year university students using a book like Enderton's Elements of Set Theory. So, elementary means using only basic facts like logic, sets, what a function is (covered in the first 3 chapters of Enderton's book).
You can rewrite the argument symbolically without using the graph-theoretical terminology (but I think this makes it less intuitive): if 4 fails, then (since $\mathbb N={s^{(k)}(1):k<\omega}$ by 5) there exists $k>0$ such that $s^{(k)}(1)=1$. Then $s^{(k+l)}(1)=s^{(l)}(1)$ for all $l$, that is, $s^{(k)}(n)=n$ for all $n\in\mathbb N$. (Or if you prefer: then $s^{(k)}(n)=n$ for all $n\in\mathbb N$ by induction on $n$.) Thus, if $s(n)=s(m)$, then $n=s^{(k)}(n)=s^{(k-1)}(s(n))=s^{(k-1)}(s(m))=s^{(k)}(m)=m$.
@PaceNielsen These are, in fact, more or less the axioms originally postulated by Peano (see https://archive.org/details/arithmeticespri00peangoog/page/n22/mode/2up). This is different from what later became to be known as the “Peano arithmetic”.
@EmilJeřábek Ah, I meant to write "not the axioms of Peano arithmetic." Sorry for the inaccuracy.
I think this question would be more appropriate at math.stackexchange.
If you imagine N with a successor function of: s(n) = the next power of 2 larger than n, then we satisfy (1) and (2) but not (3) or (4). Unfortunately, (5) also doesn't seem to hold in this setting. (As pointed-out in the comments)
Are you sure that induction holds? $A={2^n\mid n\in\Bbb N\cup{0}}$ seems to satisfy $1\in A$, since $2^0=1$, and if $n\in A$, then $n=2^k$, so $s(n)=2^{k+1}\in A$. But very obviously, $A$ does not contain any odd numbers except $1$.
Hmmm... you are right ... induction would be saying that a property holds for everything on the 1,2,4,8,16 .... path. I guess one actually has to find a different set for this, rather than a successfor function. Answer now edited to reflect this.
If 4 fails because there is something $w\ne 1$ that is not a successor, then the complement of $\{w\}$ shows that 5 also fails. So in order to get 5 but not 4, we need that 4 fails because $1$ is a successor.
So suppose 1,2,5 hold and $1$ is a successor. I claim 3 also holds. "Proof is left to the reader".
But, it is trivial that all we have to do is prove that 3 holds if 1,2,5 hold and 4 is false (for if 4 is true, we're done) so I'm asking "is there an elementary proof that 3 holds if 1,2,5 hold and 1 is a successor". I don't understand why you posted this as an "answer" when it's essentially a repeat of the question. I say this because making trivial modifications of a question and posting it as it it is an answer (or trivializing a question just so you can post an answer) is exactly the reason my colleague/advisor/mentor does not want anything to do with mathoverflow.
|
2025-03-21T14:48:31.953383
| 2020-09-04T11:37:22 |
370841
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632697",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370841"
}
|
Stack Exchange
|
Loop spaces of motivic Eilenberg-Mac Lane spaces
Consider the unstable $\mathbb{A}^1$-homotopy category (say over $\mathbb{C}$). By the loop space $\Omega X$ of an object $X$, we mean the homotopy fiber of $pt\to X$.
For an abelian group A and the motivic Eilenberg-Mac Lane space $K(A,p,q)$ for $p>q$, what can we say about $\Omega K(A,p,q)$? Is it homotopy equivalent to $K(A,p-1,q)$?
Update: According to page 5 of Voevodsky's note Motivic Eilenberg-Mac lane spaces, it is not known if the suspension isomorphism of either sphere (simplicial or Tate) holds. Therefore the answer to my question above was not known at the time these notes were written.
Is there any update to the status at the moment?
|
2025-03-21T14:48:31.953463
| 2020-09-04T11:42:31 |
370842
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Adrien",
"Christoph Mark",
"https://mathoverflow.net/users/13552",
"https://mathoverflow.net/users/66288"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632698",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370842"
}
|
Stack Exchange
|
When is the action of the braid group on tensor powers of Yetter-Drinfeld modules faithful?
Let $V$ be a Yetter-Drinfeld module over a Hopf algebra $H$ with invertible antipode. Recall that $V$ is a braided vector space with braiding $\Psi\colon V\otimes V\to V\otimes V, v\otimes w\mapsto v_{(-1)}w\otimes v_{(0)}$. The braid group $\mathbb{B}_r$ with $r-1$ generators $\sigma_1,\ldots,\sigma_{r-1}$ acts on $V^{\otimes r}$ by $\sigma_ix=\Psi_{i,i+1}x$ where $x\in V^{\otimes r}$ and where $\Psi_{i,i+1}$ is the braiding of the $i$th and $(i+1)$th tensor factor.
Question. Does there exist a Yetter-Drinfeld module $V$ over some Hopf algebra $H$ with invertible antipode such that the action of $\mathbb{B}_r$ on $V^{\otimes r}$ is faithful for some $r>1$?
A proof in special cases, e.g. $H=\mathbb{C}\Gamma$ for a group $\Gamma$, is welcome. I am interested in the ground field $\mathbb{C}$ and don't want positive characteristic examples.
If you know related literature, this is accepted as an answer as well.
EDIT. What I proved so far and which leads me to this question is as follows:
For each $r>1$, there exists a Yetter-Drinfeld module $V$ over
$\mathbb{S}_{r+1}$ such that $t(\sigma)x=x$ for all $x\in V^{\otimes r}$ and some $\sigma\in\mathbb{S}_r$ implies $\sigma=1$. Here,
$t\colon\mathbb{S}_r\to\mathbb{B}_r$ is the Matsumoto section.
The answer at this level of generality is clearly no (e.g. because the trivial module is tautologically a Yetter-Drinfeld module). Even excluding this trivial case I don't see why you'd expect these representations to be faithful. Representations coming from f.d. modules over quantum groups aren't faithful either. Constructing faithful representations of $B_n$ is a hard problem, see e.g. http://www.numdam.org/item/SB_1999-2000__42__389_0/. In particular I doubt you'll find an example of a group $\Gamma$ which works.
Thank you for your thoughtful comment and the literature @Adrien. I edited my question to make it fit better.
Note that the Yetter-Drinfeld module $V$ can well be infinite dimensional. I have no problem with that. If the answer is affirmative for all $r>1$ simultaneously, it would be even better, but I doubt that this is possible.
The Lawrence--Krammer representations which is faithful can apparently be obtained from the braid action on some infinite dimensional $U_q(\mathfrak{sl}_2)$-modules https://arxiv.org/abs/0912.2114. Modulo subtleties I guess those can be thought of as Yetter-Drinfeld modules over $U_q(\mathfrak{b}^+)$ where $\mathfrak b^+$ is the positive Borel sub-Lie algebra. Note that more generally representations coming from quasi-triangular Hopf algebras can all be interpreted as YD modules over an appropriate sub-Hopf algebra, so you could as well ask the question for those.
|
2025-03-21T14:48:31.953772
| 2020-09-04T13:00:16 |
370846
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jan Bohr",
"Jochen Wengenroth",
"Santi Spadaro",
"Taras Banakh",
"Will Brian",
"https://mathoverflow.net/users/11647",
"https://mathoverflow.net/users/126651",
"https://mathoverflow.net/users/21051",
"https://mathoverflow.net/users/61536",
"https://mathoverflow.net/users/70618"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632699",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370846"
}
|
Stack Exchange
|
Non-sequential spaces in the wild
TLDR: What are examples of (function-)spaces that are not sequential? When does this matter?
As a simple analyst, I am most happy if I can just work with sequences all the time. In most situations this is totally fine, as many many spaces one encounters in one's daily life are actually sequential (or even 1st countable, or even better metrisable). Now recently I was a bit shocked to find out that the seemingly familiar space of test-functions $\mathscr{D}(\mathbb{R}^d)$ (with its usual LF-topology) actually fails to be sequential. But hadn't I learned that one can verify whether a linear functional on $\mathscr{D}(\mathbb{R}^d)$ is a distribution by checking continuity with sequences? Well in that case it is true (Proposition 21.1 in Trèves TVS book), but only because we looked at linear functionals.
This got me thinking that there might actually be a bunch of spaces around, not pathological counterexamples, but real spaces one encounters in the wild, that fail to be sequential. In some cases, like above, this might not be problematic, but potentially for non-trivial reasons. In order to become more aware of these subtleties I would like to collect some more examples of this.
An answer should ideally contain the following:
A concrete example or a class of examples of non-sequential spaces, which are widely used or naturally show up in analysis. My main interested lies in topological vector spaces that appear as function spaces in some context. The example should not be some 'patholocial counterexample' (this is of course a bit vague).
An instance of where it matters that the space is non-sequential. Or a warning, of when one needs to be more careful with it and use filters or nets.
Loop holes or special situations where it suffices to focus on sequences nevertheless.
I'll make a start:
Test-functions: $\mathscr{D}(\mathbb{R}^d)$ (with standard LF-topology) is not sequential. In particular a function $f:\mathscr{D}(\mathbb{R}^d)\rightarrow \mathbb{C}$ might be sequentually continuous, but not continuous (example by PhoemueX). However, if $f$ is linear then sequential continuity implies continuity (Corollary after Proposition 13.1 in Trèves' TVS book). The same is true for other LF-spaces.
Distributions: $\mathscr{D}'(\mathbb{R}^d)$ (with the strong topology) is not sequential. A sequence of distributions converges strongly if and only if it converges weakly, but this is not true when sequences are replaced by nets/filters. The same result (for sequences) holds in strong duals of Montel spaces (Corollary 1 to Proposition 34.6 in Trèves)
An infinite dimensional Banach space, equipped with the weak topology is not sequential. However, despite this we have that compactness = sequential compactness (Eberlein-Smulian theorem).
Finally, here are some spaces that are sequential: $\mathscr{S}(\mathbb{R}^d)$ (Schwartz-space), $\mathscr{D}(M)$ (distributions on compact manifold $M$), the dual of a separable locally convex space with the weak$^*$-topology, ...
I don't know if this is the kind of example you're looking for, but the Stone-Cech compactification $\beta \mathbb N$ of the countable discrete space $\mathbb N$ contains no converging sequences at all (except those that are eventually constant). https://en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification#The_Stone%E2%80%93%C4%8Cech_compactification_of_the_natural_numbers
Thanks for that example, Will. This is not quite what I was after - I have tried to make the question more precise.
I would like to add a remark to the failure of $\mathscr D(\mathbb R^3)$ being sequential: It is of course true that a linear map is continuous if it is sequentially continuous, but this fails for linear maps defined on closed subspaces!
The space $C_p([0,1])$ of all continuous real-valued functions on the unit interval with the topology pointwise convergence is a pretty natural example of a non-sequential topological vector space which is a function space. More generally, for a compact space $X$, $C_p(X)$ is sequential if and only if $X$ is scattered (see Arhangel'skii's "Topological Function Spaces" book for more information).
The unit ball of the dual of a separable normed space is weak$^*$ sequentially compact, but this fail dramatically for non-separable spaces: The sequence of evaluations $\delta_n: \ell_\infty\to\mathbb R$ is probably the first sequence in $\ell^*_\infty$ that comes to mind and it has no weak$^*$-convergent subsequence: This would be a sequence of integers $n_1<n_2<\cdots$ such that for every bounded sequence $(x_n)_{n\in\mathbb N}$ of scalars one had a limit $\lim\limits_{k\to\infty} x_{n_k}$.
That's a great example. I'll add a (probably trivial) observation: The example proves at the same time that $(\ell_\infty,w^*)$ is not sequential and that in general one does not have compactness$ \Rightarrow $sequential compactness.
The sequentiality does not match well with an algebraic structure. For example, the following result of Banakh and Zdomskyy characterizes sequential topological groups with countable $cs^*$-character:
Theorem. A topological group $G$ with countable $cs^*$-character is sequential if and only if $G$ is either metrizable or contains an open $\mathcal M\mathcal K_\omega$-subgroup.
Let us recall that a topological space $X$ has countable $cs^*$-character if for every point $x\in X$ there exists a countable family $\mathcal F_x$ of subsets of $X$ such that for every neighborhood $O_x\subseteq X$ of $x$ and every sequence $\{x_n\}_{n\in\omega}\subseteq X$ that converges to $x$, there exist a set $F\in\mathcal F_x$ such that $F\subseteq O_x$ and $F$ contains infinitely many points of the sequence $(x_n)$.
A topological space $X$ is $\mathcal{MK}_\omega$ if there exists a countable cover $\mathcal C$ of $X$ by compact metrizable subspaces such that a subset $F\subseteq X$ is closed if and only if for every compact set $C\in\mathcal C$ the intersection $C\cap F$ is closed in $C$.
In this paper of Banakh and Repovs the above result of Banakh--Zdomskyy was extended to rectifiable spaces and topological left-loops.
In fact, the above theorem, is a corollary of the following result of Banakh:
Theorem. If a perfectly normal topological group $G$ contains a topological copy of the Frechet-Urysohn fan $S_\omega$ and a closed topological copy of the metric fan $M$, then $G$ is not sequential.
The metric fan is the subspace $$M=\{0\}\cup\{\tfrac1{n}+\tfrac{i}{nm}:n,m\in\mathbb N\}$$ of the complex plane. The Fr'echet-Urysohn fan is the set $M$ endowed with the strongest topology that coincides with the Euclidean topology on each subspace $K_m=\{0\}\cup\{\frac1{n}+\tfrac{i}{nm}:n\in\mathbb N\}$, $m\in\mathbb N$. It is easy to see that the Fr'echet-Urysohn fan is an $\mathcal{MK}_\omega$-space.
Thank you for this answer! Can you maybe add an example of when the Theorem applies or how it is used? Which common topological groups have countable $cs^*$-character or contain open $\mathcal{MK}_{\omega}$-subgroups?
@JanBohr All sequential non-metrizable topological vector spaces you mentioned are $\mathcal{MK}_\omega$.
@JanBohr Concerning topological groups with countable $cs^*$-character, then natural examples are LF-space or more generally, inductive limits of sequences of metrizable spaces.
Concerning possible applications, we can consider the inductive limit $\mathbb R^\infty=\lim \mathbb R^n$ of finite-dimensional Euclidean spaces. It is an $\mathcal MK_\omega$-group and hence a sequential topological vector space. On the other hand, for any infinite-dimensional metrizable space $X$ the product $X\times\mathbb R^\infty$ is not sequential since it is not metrizable and not $\mathcal{MK}_\omega$ but has countable $cs^*$-character.
I see, thank you. I was wondering whether your theorem can be used to check that $X=\mathscr{D}(\mathbb{R}^d)$ is non-sequential. You say that $X$ has countable $cs^*$-character. Further $X$ is complete and it is the countable union of nowhere dense sets, hence it cannot be metrisable. So in order to establish that $X$ is not sequential, according to the theorem, it would suffice to show that $X$ does not have any $\mathcal{MK}_\omega$-subgroups. But is this feasible to check?
@JanBohr Since topological vector spaces are connected, they contain an open $\mathcal{MK}\omega$-subgroup if and only if they are $\mathcal{MK}\omega$-spaces. Since $\mathcal D(\mathbb R^d)$ is not $\sigma$-compact, it is not $\mathcal{MK}_\omega$ neither.
@JanBohr Another way to prove that $\mathcal D(\mathbb R^d)$ is not sequential is to observe that this space has countable network and hence is perfectly normal and it contains closed topological copies of the metric fan and Frechet-Urysohn fan.
|
2025-03-21T14:48:31.954317
| 2020-09-04T13:13:18 |
370850
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben Wieland",
"Dylan Wilson",
"Thomas Rot",
"https://mathoverflow.net/users/12156",
"https://mathoverflow.net/users/4639",
"https://mathoverflow.net/users/6936"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632700",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370850"
}
|
Stack Exchange
|
loop space of a finite CW-complex
Let $X$ be a finite connected pointed CW-complex and $H_{\ast}(\Omega X)$ the integral homology of the loop space on $X$. Are the homology groups $H_{n}(\Omega X)$ finitely generated abelian groups for any $n$ ?
If the answer is negative, what are the sufficient conditions to impose on $\pi_{1}(X)$ such that the homology groups $H_{n}(\Omega X)$ turns out to be finitely generated ?
My goal is to collect different sufficient conditions on the fundamental group for which a positive answer holds.
I think the serre spectral sequence gives this immediately when X is simply connected
I guess if $X = S^1$ this is false, so definitely some $\pi_1$ conditions need to be made
$H_0(\Omega X)$ is the free abelian group on $\pi_1(X)$. You might ask if that is the only problem; you might ask about the higher homology of the connected component of $\Omega X$: $\Omega_0X=\Omega \tilde X$, but this doesn't help. If you take any $X$ with infinite $\pi_1$ and glue on a sphere $S^n$ at a point, bad things happen. For example, $S^1\vee S^2$ has universal cover a line with infinitely many $S^2$ glued on, thus infinitely generated $H_2$; and so its loop space has infinitely generated $H_1$.
This is true for finite $\pi_1$ and false for infinite $\pi_1$: Let $\widetilde{X}$ denote the universal cover of $X$, then $\Omega\widetilde{X}$ is the unit connected component of $\Omega X$, and $\Omega X = \coprod_{\pi_1(X)} \Omega\widetilde{X}$. So if $\pi_1$ is infinite, then certainly $H_0(\Omega X)$ is not finitely generated as others have noted in the comments, and indeed if $\Omega\widetilde{X}$ has any nontrivial homology group (which is true unless $\widetilde{X}$ is contractible), some higher homology group of $\Omega X$ will be an infinite direct sum of nontrivial abelian groups, so also not finitely generated.
If $\pi_1$ is finite, on the other hand, $\widetilde{X}$ is again a finite CW complex, so in that case it suffices to look at the simply-connected case. For a simply-connected finite CW-complex $X$, $H_*(\Omega X)$ indeed consists of finitely generated abelian groups, which goes back to Serre (and is proved easily using the spectral sequence named after him).
|
2025-03-21T14:48:31.954481
| 2020-09-04T13:20:38 |
370851
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben McKay",
"coudy",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/6129"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632701",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370851"
}
|
Stack Exchange
|
Klein geometry associated to a degenerate conic
In his study of non-euclidean geometries,
Felix Klein considers the group of projective transformations
acting on the real projective plane whose extensions
to the complex projective plane preserve a given conic.
When the conic is given in homogeneous coordinates
by the equation $x^2+y^2-z^2$,
he obtains the isometry group of what is now called
the Klein model of hyperbolic geometry.
The conic with equation $x^2+y^2+z^2$ is related
to the group of motions of spherical geometry.
He also considers the group of projective
transformations preserving the pair of points
given by the intersection
of the two degenerate conics $x^2+y^2 = 0$ and $z^2 = 0$.
This leads to affine conformal geometry.
What geometry do we obtain if we consider
the degenerate conic $x^2+y^2=0$ given by the two lines
$x = iy$ and $x = -iy$? Is it related to conformal geometry?
All I can see is that the transformations that preserve the conic fix the origin, which probably should be sent to infinity.
EDIT: Some computations give the following for the matrices associated to the transformations preserving the degenerate conic.
$$
\pmatrix{ \alpha & \beta & 0 \cr
-\beta & \alpha & 0 \cr
\gamma & \delta & 1 \cr }
$$
So it seems that this geometry is dual to affine conformal geometry.
Does it have a name?
These are the linear transformations that take $z=0$ to itself and act by complex linear transformation on it.
@McKay I don't understand your comment. We are looking for a group of real transformations acting on real the projective plane. Even if I interpret your comment as the group of similitudes of the plane, I don't think this is right. A transformation preserving globally the two lines may not preserve the line at infinity.
Ok, projectivized, these are the projective transformations of the real projective line that preserve its usual metric, as the circle of any radius you choose to fix. Just check the linear algebra to see that the linear transformations of 3 real variables are the ones I said, and then projectivize. The group action doesn't depend on the chosen radius, so you get to pick it.
@McKay I did the computations and put the results above. Still this geometry is unknown to me.
|
2025-03-21T14:48:31.954654
| 2020-09-04T13:36:58 |
370854
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ChrisLazda",
"Jérôme Poineau",
"https://mathoverflow.net/users/13647",
"https://mathoverflow.net/users/4069"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632702",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370854"
}
|
Stack Exchange
|
Topological and algebraic covering spaces in Berkovich geometry
Let $k$ be a complete, non-archimedean field, and $X$ a Berkovich space over $k$ (as nice as you like, for arguments sake let's say strictly $k$-analytic, good, and geometrically connected). As discussed in this article of de Jong, covering spaces of $X$ come in two slightly different flavours. One the one hand you can take finite etale covers $Y\rightarrow X$ as you would for schemes, on the other hand you can take a covering space $Y\rightarrow |X|$ of the underlying topological space of $X$, and, roughly speaking, use the Berkovich space structure of $X$ to put one on $Y$. Following de Jong, let us call the first of these 'algebraic' and the second 'topological'. A general covering space is then some kind of mixture of the two.
If $k$ is not separably closed, then one way of producing algebraic covering spaces is via finite separable extensions of $k$: if $L/k$ is such an extension then $X_L \rightarrow X$ is a finite etale map of Berkovich spaces, where $X_L$ denotes the base change of $X$ to $L$. My question is then the following:
Question: Is it possible that $X_L \rightarrow X$ is a topological covering space, for some non-trivial extension $L/k$?
It's not to hard to see this can't be the case if $X$ has a $k$-rational point (since the fibre of $X_L\rightarrow X$ over this point will have cardinality 1), but I'm particularly interested in the case when we might have $X(k)=\emptyset$. Concretely, I'm interested in the case when $X$ is (the analyitification of) a smooth projective conic over $k$, without a rational point, and $L/k$ is a quadratic extension over which $X$ does admit a rational point.
In your particular case, $X_L$ has a point, so it is isomorphic to $P^{1,\mathrm{an}}_L$, hence simply connected. If your covering $X_L \to X$ were a covering, it would then be a universal covering. But we know that Berkovich curves retract by deformation onto graphs, so the topological fundamental group of $X$ is a free group. In particular, the universal covering of $X$ is either $X$ itself or of infinite degree, and we get a contradiction.
Hi Jerome, I was actually about to email you this question, so I'm glad you've popped up here! I think I perhaps didn't explain it very well - the question was more about whether $X_L$ can ever be a topological covering space of $X$ - i.e. a covering space map on the underlying topological spaces. It boils down to the question of whether or not every point of $X$ (of any Type) has precisely $[L:k]$ preimages in $X_L$. It feels like this is unlikely - for conics, I feel as though it should be possible to cook up some Type II point with only one preimage, but I didn't manage to do so.
I was indeed answering something completely different. I will give it another try.
That's great thanks!
|
2025-03-21T14:48:31.954868
| 2020-09-04T13:47:08 |
370856
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"JonHales",
"Max Alekseyev",
"https://mathoverflow.net/users/164796",
"https://mathoverflow.net/users/7076"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632703",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370856"
}
|
Stack Exchange
|
Solutions to matrix equations in the non-negative integers
For an integer matrix $S$, and an integer vector $y$, I'm looking for solutions to $xS = y$ where the entries in $x$ are in the non-negative integers.
I've been doing this with Sage's mixed integer linear programming, but this has several disadvantages, like being forced to pick a component of the solution vector $x$ to maximize or minimize instead of just searching for a solution. The code I'm currently running is
p = MixedIntegerLinearProgram(maximization = True, solver = "GLPK")
x = p.new_variable(integer = True, nonnegative = True)
p.add_constraint( x*S == 0) #here the vector can be anything
p.add_constraint( x[25] <= 2020 )
p.set_objective( x[25] )
p.solve()
p.get_values(x)
Where $x[25]$ was chosen arbitrarily to make the mixed-integer linear program $p$ actually have a solution. Sometimes optimizing over $x[25]$ does not have a solution (or gives a solution after a very long time) while optimizing over another variable gives an almost instantaneous solution.
So my question. Is there a way to solve such systems without optimizing a specific component? (And if Sage Math isn't the best software to do this in please let me know and I'll give something else a try). If I must optimize over some condition, would it be possible to have my optimization be something more
uniform across all the variables, like picking a solution of minimal norm?
Is there any literature I could read on solving matrix equations in the non-negative integers? A simple search of MathSciNet didn't seem to find anything too relevant. Thanks for your thoughts!
To get just any feasible solution, one can set the objective to None -- like p.set_objective(None)
That's cool! I didn't know that. I'll definitely give that a try. Thanks!
It's not a secret, but a feature described in the documentation.
|
2025-03-21T14:48:31.955023
| 2020-09-04T14:29:42 |
370861
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Iosif Pinelis",
"Piotr Hajlasz",
"Qfwfq",
"https://mathoverflow.net/users/100355",
"https://mathoverflow.net/users/121665",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/4721",
"jens"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632704",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370861"
}
|
Stack Exchange
|
Is a smooth transformation of a plane domain onto a plane domain with everywhere nonzero Jacobian determinant necessarily a bijection?
Let $U$ and $V$ be connected open subsets of $\mathbb R^2$. Let $f$ be a smooth map from $U$ onto $V$ such that the Jacobian determinant of $f$ is nonzero everywhere. Does it then necessarily follow that $f$ is a bijection?
Counterexamples are easy to find if we allow $V$ to be contained in a bigger space, say $\mathbb R^3$; or if we relax enough the conditions on the smoothness and the Jacobian determinant of $f$.
I apologize in advance if this question is trivial. I have no background in differential topology (and am not even quite sure that this question belongs in differential topology).
As was noted in a comment by Moishe Kohan, the answer to this question follows from the previous answers. However, to make a bridge from those answers to an answer to this question, one needs to recall and use the Great Picard Theorem (which I, shamefully, forgot about). Also, now we have, not only the excellent answer by Alexandre Eremenko, but also a completely elementary answer, which I think shows in what specific manner the non-bijectivity can occur. For these reasons, I'd prefer that this question be kept.
Edit: In view of the discussion of the answer by Alexandre Eremenko, let me add here the condition for $U$ and $V$ to be simply connected. A further question would then be whether the thickness parameter $\epsilon$ in the answer by Qfwfq can be chosen so that $V$ be simply connected.
No. Let $C$ be the complex plane, $U=V=C\backslash\{0\}$. Transformation $z\mapsto z^2$ is smooth and has non-zero Jacobian $4|z|^2$ but it is not a bijection.
The question in the comments: no. Take $U=\{ z\in C\backslash\{0\}:|\arg z|<2\pi/3\}$,
and the same $f$.
Second question in the comments: again the answer is no. Take $f(z)=\int_0^z e^{\zeta^2}d\zeta$, and $U=V=C$. The map is surjective, every point has infinitely many perimages, and the Jacobian is
$|f'|^2=\exp2(\Re z^2)\neq 0$. To see that the map is surjective, notice that it is an entire function of order 2. If an entire function of order two omits a complex value $a$, then it has the form $f(z)=a+e^{P(z)},$
where $P$ is a polynomial of degree 2. This is of course not the case for our function. Similarly, if the value $a$ is taken finitely many times,
then $f(z)=a+Q(z)s^{P(z)}$. So our function takes every value infinitely many times.
Thank you. I should have thought a bit more. Will the answer change if we also assume that $U$ is simply connected?
Thank you again. What if we assume that both $U$ and $V$ are simply connected?
All right. How can I see that the latest map is surjective and every point has infinitely many preimages?
This is a truly amazing example I did not know. All my life I was sure that a surjectivve local diffeomorphism of $\mathbb{R}^n$ must be a covering map and hence a diffeomorphism. First I wrote an incorrect answer and only after I saw yours I added "proper".
@AlexandreEremenko : Thank you for your patience and the excellent answer. (I think the expression for $|f'|^2$ should also involve $\Im z$.)
@Iosif Pinelis: thanks. I corrected the formula.
Here is an elementary answer, which shows how the bijectivity of $f$ may be violated, even when $U$ and $V$ are both simply connected domains (and $f\colon U\to V$ is a smooth surjection with the Jacobian determinant nonzero everywhere).
The idea is very simple: First of all, it is easy to get a smooth non-bijective map, say $f_1$, from a simply connected open set $U_1\subseteq\mathbb R^2$ to a simply connected open set $V\subseteq\mathbb R^2$ with the Jacobian determinant of $f_1$ nonzero everywhere and one point in $V$ missing from the image $f_1(U_1)$. The idea then is just to extend $f_1$ to a map $f\colon U\to V$ in order to smear over the missing point -- keeping, for $f$, the smoothness and nonzero Jacobian determinant properties. Below, the smear is represented by the function $f_2$, which extends the function $f_1$ to $f$ (while the image set of $f_1$ misses a point in $V$).
Let $U:=(0,1)\times(0,4\pi+\pi/2)$ and $V:=D$, the open unit disk. For $(r,t)\in U$, let
$$f(r,t):=
\begin{cases}
f_1(r,t) & \text{ if}\quad 0<t\le2\pi, \\
f_2(r,t) & \text{ if}\quad 2\pi<t<4\pi+\pi/2,
\end{cases}
$$
where $$f_1(r,t):=r\,(\cos t,\sin t),$$
$$f_2(r,t):=(-2a(t),0)+(a(t) + r (1 - 4 a(t))) \,(\cos t,\sin t),$$
and $$a(t):=\frac14\, \exp\Big(-\frac\pi{t - 2\pi}\Big).$$
Shown below are the (properly shaded) "parametric plots", that is, the sets $S_1:=\{f_1(r,t)\colon 0<r<1,0<t\le2\pi\}$ (left), $S_2:=\{f_2(r,t)\colon 0<r<1,2\pi<t\le4\pi+\pi/2\}$ (right), as well as both of these two sets superimposed together, forming the set $S:=f(U)=S_1\cup S_2=\{f(r,t)\colon 0<r<1,0<t<4\pi+\pi/2\}$.
It is not hard to see that the map $f$ is smooth and the Jacobian determinant of $f$ is nonzero (actually, $>0$) everywhere.
It is also easy to see that $S\subseteq D=V$, $S_1=D\setminus\{(0,0)\}$, whereas $f(r_*,4\pi)=f_2(r_*,4\pi)=(0,0)$, where $r_*:=1/(4 (\sqrt e-1))=0.385\dots$. So, $f$ is surjective.
However, as seen from the pictures, $f$ is not bijective. Formally, for instance,
$f(r-(r+1/4)/\sqrt e,2\pi)=f(r,4\pi)$ for all $r\in(r_*,1)$.
This is a very nice example.
May I ask what in your counterexample violates the assumptions for the Hadamard-Cacciappoli Theorem as stated in https://en.wikipedia.org/wiki/Diffeomorphism? I can't figure out.
I am not familiar with the proof of that theorem. I would guess that the map in my example is not proper in the sense used in that theorem. You are welcome to check that.
I couldn't find an accessible proof. It seems clear to me that your function is proper since compact sets in the codomain have compact preimages (nothing escapes to infinity). So that shouldn't be the issue. I suspect, however, that your counterexample is not smooth at the transition between both parts of the curve, at $t=2\pi$. Could that be the issue? On the other hand, the existence of the bump function (see https://en.wikipedia.org/wiki/Smoothness) gives me some confidence that your counterexample could be made smooth. So, I am still puzzled.
As Alexandre Eremenko pointed out, in general the answer is in the negative. However, in a comment the OP asked a modified question:
What if we assume that both $U$ and $V$ are simply connected?
The following result is in a positive direction:
If, $V$ is simply connected and $U$ is connected, and $f$ is proper (i.e. preimages of compact sets are compact), then
then $f$ is a bijection.
Under the given assumptions $f$ is a covering map, see https://math.stackexchange.com/q/860351/798404
A covering space is a universal covering space if it is simply connected. Assuming that $V$ is simply connected, it is its universal cover and $\operatorname{id}:V\to V$ is a covering map. Now we use the fact:
If the mapping $p: D \to X$ is a universal cover of the space $X$ and the mapping $f : C \to X$ is any cover of the space $X$ where the covering space $C$ is connected, then there exists a covering map $g : D \to C$ such that $f ∘ g = p$.
In our situation $D=X=V$, $p=\operatorname{id}$, $C=U$ and $f$ is $f$. Therefore there is $g:D\to C$ i.e. $g:V\to U$ such that $f\circ g=p$ i.e., $f\circ g=\operatorname{id}$. That proves that $f$ is a bijection (and hence a diffeomorphism).
You can find basic statements about universal cover and covering maps in
https://en.wikipedia.org/wiki/Covering_space#Lifting_properties
If you want to learn more, take almost any book in topology. For example:
M. A. Armstrong, Basic Topology.
Thank you. Can you give a proof of this? This would appear to contradict the last part of Alexandre Eremenko's answer (with $f(z)=\int_0^z e^{w^2}dw$), which my plotting suggests is correct -- but I don't know how to prove that.
@IosifPinelis I had to modify my answer by adding the assumption that the map is proper. I also sketched a proof.
@Iosif Pinelis: the key word in this answer is "proper". Of course my example is not proper.
@PiotrHajlasz : Thank you again for your answer. In the elementary counterexample that I now have, it is easy to find a compact set with a non-compact preimage.
Another trivial counterexample.
Consider the rational parametrization of the real nodal plane cubic curve $y^2=x^2(x+1)$:
$$\gamma:t\mapsto (t^2-1,t(t^2-1))\;.$$
We have $\dot{\gamma}(t)=(2t,3t^2-1)$. Consider a normal $\nu(t)=(1-3t^2,2t)$.
(image from Wikipedia)
A "thickening"
$$F:(t,s)\mapsto\gamma(t)+s\nu(t)$$
will do, for example restricting to $U=(-3/2,3/2)\times(-\epsilon,\epsilon)$, $V=F(U)\subseteq\mathbb{R}^2$, for $\epsilon<<1$.
Thank you for this answer. I thought about cubic curves. However, this will not work for small $\epsilon$, as you seem to be suggesting -- because then $F(U)$ will not be simply connected. It will not work for large $\epsilon$ either -- because then the Jacobian determinant of $F$ will not be nonzero everywhere. I don't know if there is a working middle range of values of $\epsilon$.
Yes, this is literally an answer to your question (you mention "connected open subsets"), not to the more interesting version where $U$ and $V$ are simply connected which you asked in the comments to another answer.
|
2025-03-21T14:48:31.955612
| 2020-09-04T15:20:17 |
370864
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben",
"LSpice",
"Qiaochu Yuan",
"Terry Tao",
"Yemon Choi",
"https://mathoverflow.net/users/150898",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/290",
"https://mathoverflow.net/users/41291",
"https://mathoverflow.net/users/763",
"https://mathoverflow.net/users/766",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632705",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370864"
}
|
Stack Exchange
|
Is there a name for this "inner product" on projective space?
$\newcommand{\proj}{\mathbb{P}}\newcommand{\complex}{\mathbb{C}}\newcommand{\ip}[2]{\langle #1 , #2\rangle}\newcommand{\abs}[1]{\lvert #1 \rvert}$There is a natural bijection $\phi: \proj(\complex^n)\rightarrow S(\complex^n)/U(1)$, where $S(\complex^n$) is the unit sphere of $\complex^n$ with respect to the Hermitian inner product, and $U(1)=S(\complex)$ (i.e, $\phi$ identifies projective space with the unit sphere modulo complex phase). We can therefore define a map
$$
\abs{\ip{\cdot}{\cdot}}^2: \proj(\complex^n) \times \proj(\complex^n) \rightarrow \mathbb{R}^+
$$
by $\abs{\ip{[u]}{[v]}}^2=\abs{\ip{\tilde{u}}{\tilde{v}}}^2$, where $\tilde{u}, \tilde{v}$ are any representatives of the equivalence class of $\phi[u], \phi[v]$, and the bracket $\ip{\cdot}{\cdot}$ on the righthand side is the standard Hermitian inner product on $\complex^n$.
Does this map have a name? Does it fit into some well-known category of maps on projective space?
For $n=1$ this is a map $S^2\times S^2\to{\mathbb R}^+$. The only candidate that comes to mind is cosine square of the angle, is it this?
@მამუკაჯიბლაძე $S^1$ surely if we are looking at the unit "sphere" in ${\mathbb C}$?
@YemonChoi Sorry I meant $n=2$. What confused me is the nonstandard notation. The $\mathbb P(\mathbb C^n)$ of the OP is what one usually denotes by $\mathbb C\mathbb P^{n-1}$. For $n=1$ it is a single point (I think).
MathJax note: unlike in ordinary TeX, blank lines, or even spaces, after command definitions won't be ignored, so you have to butt up the beginning of the question directly against the $ ending the definitions. I have edited accordingly.
Isn't this sort of inner product-ish used in quantum mechanics for computing probabilities?
@LSpice Yes! This is the context in which it has come up for me.
Keeping with the QM theme: one can identify a projective state $[u]$ with a rank one density matrix $|u\rangle \langle u|$, and then your "inner product" is the square root of the Hilbert-Schmidt inner product between the two associated density matrices.
Is it related to the Fubini-Study metric? https://en.wikipedia.org/wiki/Fubini%E2%80%93Study_metric
|
2025-03-21T14:48:31.955922
| 2020-09-04T15:23:33 |
370865
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Joe Silverman",
"LSpice",
"Nulhomologous",
"Random",
"https://mathoverflow.net/users/11926",
"https://mathoverflow.net/users/12481",
"https://mathoverflow.net/users/158462",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/88679",
"joro"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632706",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370865"
}
|
Stack Exchange
|
Doubt about lemma for polynomial equivalence
Multivariate polynomials $f,g$ are equivalent if there exists
invertible linear transformation $A$ such that $f(X)=g(A\cdot X)$
From paper p.1:
Lemma 1.1. (Structure of quadratic polynomials). Let $F$ be an algebraically closed field of characteristic
different from 2.
For any homogeneous quadratic polynomial $f(X) \in F[X]$
there exists an invertible linear
transformation $A \in F^{n \times n}$ and a natural number $1 \le r \le n$ such that
$f (A \cdot X) = x_1^2 + x_2^2 + . . . + x_r^2$
Moreover, the linear transformation $A$ involved in this equivalence can be computed efficiently.
Furthermore, two
quadratic forms are equivalent if and only if they have the same number r of variables in the above canonical
representation.
Let $F$ be the algebraic closure of the rationals.
Basically there are more than $n$ non-equivalent $f_i$
and only $n$ possible choices for $r$.
Fix $n$ and take $n+1$ non-equivalent $f_i$.
By the pigeonhole principle there are at least two $r_i=r_j$,
contradicting non equivalence of $f_i$.
Q1 What is wrong with this argument against the lemma?
Q2 Is homogeneous quadratic polynomial equivalence
or isomorphism polynomial in $n$?
You assumption that "Basically there are infinitely many non-equivalent $f$" is false. Try writing out the proof for 2 variables and you'll see what's going on.
@JoeSilverman Thanks. I edited with "there are more than n non-equivalent f_i", is this true? I will accept as an answer correctness of the lemma.
why do you say that there are more than $n$ non-equivalent homogeneous quadratic polynomials in $n$ variables when the lemma asserts there are exactly $n$?
The lemma is correct, and there are indeed no more than $n$ non-equivalent $f_{i}$. Here is a sketch of a proof:
We can do a transformation such that the coefficient of (without loss of generality) $a_{1}^2$ is non zero, and by scaling, 1. Now by replacing $a_1$ with $a_1 - c_2 \cdot a_2 - \dotsb - c_n \cdot a_n$, where the coefficients $c_i$ are chosen appropriately, we can ensure that $a_{1}^2$ is the only place where $a_1$ appears, and now we can induct on the statement with $n-1$ variables.
Many thanks Random.
Is this related to efficient quadratic polynomial isomorphism: $f(X)=g(\pi(X))$ for permutation $\pi$?
TeX note: literal ... (as $a_1 - c_2\cdot a_2 - ... - c_n\cdot a_n$) doesn't behave well in TeX; prefer (between binary operators) \dotsb (as $a_1 - c_2\cdot a_2 - \dotsb - c_n\cdot a_n$). I have edited accordingly.
@joro I'm not sure I understand exactly what you meanby related, but a permutation is an example of polynomial equivalence.
@Lspice I see, thank you.
I mean: given quadratic $f(X)$ and $f(\pi(X))$ can you recognize it is isomorphism, not just equivalence?
|
2025-03-21T14:48:31.956115
| 2020-09-04T16:07:43 |
370867
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632707",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370867"
}
|
Stack Exchange
|
Kronecker's theorem on diophantine approximation for $\mathrm{SL}_2(\mathbb{Z})$
Kronecker's theorem on diophantine approximation gives a criterium whether the orbit of a tupple ${\underline\alpha}=(\alpha_1,\ldots,\alpha_n)\in \mathbb{R}^n/\mathbb{Z}^n$ comes arbitrarily close to another tuple $\underline x$ under the diagonal action of $a\in\mathbb{Z}$ given by multiplication by $a$, that is whether there exist $a\in \mathbb{Z}$ and $\underline k\in \mathbb{Z}^n$ such that for all $\epsilon>0$
$$|a\alpha_i-x_i-k_i|<\epsilon$$
for all $i=1,\ldots,n$.
Question Now, replace $\mathbb{R}$ by $\mathbb{R}^2$ and the diagonal action of $\mathbb{Z}$ by the action of $\mathrm{SL}_2(\mathbb{Z})$. What is the generalisation of Kronecker's theorem in this setting? So, given ${\underline\alpha},{\underline\beta}\in \mathbb{R}^n/\mathbb{Z}^n$ and $\underline x,\underline y\in \mathbb{R}^n/\mathbb{Z}^n$, does there exist $\gamma=\left( \begin{smallmatrix} a& b\\ c & d \end{smallmatrix}\right)\in \mathrm{SL}_2(\mathbb{Z})$ and $\underline k, \underline l\in \mathbb{Z}^n$ such that for all $\epsilon>0$ one has
$$|a\alpha_i+b\beta_i-x_i-k_i|<\epsilon \quad \text{and} \quad |c\alpha_i+d\beta_i-y_i-l_i|<\epsilon$$
for all $i=1,\ldots,n$?
References The paper Diophantine approximation exponents on homogeneous varieties in Section 4.3 gives estimates for the size of the matrix $\gamma$, assuming $\gamma$ exists. In the case $n=1$, a statement which can be found in for example Approximation to points in the plane by SL (2, Z)-orbits says that in case $n=1$ and the ratio of $\alpha_1$ and $\beta_1$ is irrationial, the orbit of $(\alpha,\beta)$ under $\mathrm{SL}_2(\mathbb{Z})$ is dense, hence giving an affirmative answer to my question.
|
2025-03-21T14:48:31.956240
| 2020-09-04T16:57:03 |
370868
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632708",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370868"
}
|
Stack Exchange
|
When is the dualizing sheaf globally generated?
Let $X$ be a projective integral Cohen-Macaulay variety (over $\mathbb{C}$, if that makes things easier). The Cohen-Macaulay condition says that the dualizing complex (see this answer) is concentrated in degree $-\dim X$, and so we define the dualizing sheaf to be this cohomology sheaf.
My question is simple: are there any situations where this dualizing
sheaf is known to be globally generated?
For smooth or Gorenstein curves of arithmetic genus at least one, this is true, but I am specifically interested when the dualizing sheaf is not invertible (such as Cohen-Macaulay curves or suitable surfaces).
Thanks in advance.
|
2025-03-21T14:48:31.956317
| 2020-09-04T17:03:09 |
370869
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Geordie Williamson",
"Oli Gregory",
"https://mathoverflow.net/users/76409",
"https://mathoverflow.net/users/919"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632709",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370869"
}
|
Stack Exchange
|
Cohomology classes coming from algebraic K-theory
Suppose $X$ is a smooth variety over $\mathbb{C}$. I believe there is a Chern class map
$\operatorname{ch}:K_*(X) \otimes \mathbb{Q} \to H_{sing}^*(X(\mathbb{C}),\mathbb{Q})$ for algebraic $K$-theory. By Grothendieck, $K_0(X) \otimes \mathbb{Q}$ is the Chow group, so the image of $\operatorname{ch}$ contains classes coming from algebraic cycles.
What is known about the image of $\operatorname{ch}$ when we include higher $K$-groups? Are there any obstructions against a cohomology class being in the image of this map?
One obstruction is that the image of $K_{n}(X)$ lies in the "Hodge" part of the mixed Hodge structure, i.e. $\bigoplus W_{2i}H^{2i-n}(X,\mathbb{Q})\cap F^{i}H^{2i-n}(X,\mathbb{C})$. For example, if $X$ is proper then the image is zero as soon as $n>0$.
In case someone stumbles over this question in the future, Gillet's "Comparing Algebraic and Topological $K$-Theory" is useful. In particular, from the introduction: "for $K_0$ the image is (conjecturally) described by the Hodge conjecture, while for higher $K$-theory, the image must be torsion."
Apologies for being a bit late, but let me try to expand on my comment. An excellent source for this material is James Lewis' "user-friendly" survey article [Lew14].
First recall that for $X$ smooth over a field $k$,
\begin{equation*}
K_{n}(X)\otimes\mathbb{Q}\cong\bigoplus_{i}H^{2i-n}_{\mathrm{Mot}}(X,\mathbb{Q}(i))\,.
\end{equation*}
Now suppose $k=\mathbb{C}$. Then we have the (Betti) cycle class map
\begin{equation*}
\mathrm{cl}_{i,n}:H^{2i-n}_{\mathrm{Mot}}(X,\mathbb{Q}(i))\rightarrow H^{2i-n}(X,\mathbb{Q}(i))
\end{equation*}
whose image lands in the "Hodge" part of cohomology, i.e. in $\mathrm{Hom}_{\mathrm{MHS}}(\mathbb{Q}(0),H^{2i-n}(X,\mathbb{Q}(i)))$. Note that, as per my comment, this latter group is trivial if $X$ is also projective and $n\geq 1$.
Originally it was conjectured that $\mathrm{cl}_{i,n}$ is always surjective (the so-called Beilinson-Hodge conjecture, see [Bei86, Conjecture 6]). First note that the case $n=0$ of the Beilinson-Hodge conjecture is equivalent to the Hodge conjecture (clear in the projective case, some work for quasi-projective $X$). But Jannsen [Jan90, Corollary 9.11] found counterexamples in the case $n=1$ (and counterexamples exist for all $n\geq 1$). Indeed, the surjectivity of $\mathrm{cl}_{i,n}$ is tied up with the injectivity of certain higher Abel-Jacobi maps (see [Jan90, Corollary 9.10] and [dJL13, Theorem 4.9] more generally), and we know from Mumford that Abel-Jacobi (even $\otimes\mathbb{Q}$) can drastically fail to be injective in codimension $\geq 2$. Note, however, that the Bloch-Beilinson conjectures say that the higher Abel-Jacobi maps $\otimes\mathbb{Q}$ should be injective when $X$ is defined over a number field, so one could imagine that the Beilinson-Hodge conjecture is true for $X$ defined over number fields, and indeed if the Bloch-Beilinson conjectures and the Hodge conjecture are true then $\mathrm{cl}_{i,n}$ is indeed surjective when $X$ is defined over a number field (see [Sai09]).
For an example of $\mathrm{cl}_{i,n}$ failing to be surjective for $n\geq 1$, consider a K3 surface $X$ over $\mathbb{C}$ with points $P,Q$ such that $[P]\neq [Q]\in\mathrm{CH}^{2}_{\mathrm{deg}=0}(X)$. Let $U:=X\backslash\{P,Q\}$. Then $\mathrm{cl}_{2,1}:H^{3}_{\mathrm{Mot}}(U,\mathbb{Q}(2))\rightarrow H^{3}(U,\mathbb{Q}(2))$ is the zero map. Notice that $U$ is not defined over a number field here by the assumption, because Beilinson-Bloch predicts that $\mathrm{deg}:\mathrm{CH}^{2}(X)\xrightarrow{\simeq}\mathbb{Z}$ for $X$ defined over a number field (and it is known that $\mathrm{dim}_{\mathbb{Q}}\mathrm{CH}^{2}(X)_{\mathbb{Q}}=\infty$ for $X/\mathbb{C}$).
Note as well that along the Milnor diagonal $n=i$ there is evidence that the Beilinson-Hodge conjecture might be true even outside of the situation that $X$ is defined over a number field. See [AK09], [AS07], [Sai04] and [Sai09] for examples and further discussion.
In fact, $\mathrm{cl}_{i,n}$ factors through the cycle class map $\mathrm{cl}^{\mathcal{H}}_{i,n}$ into absolute Hodge cohomology:
\begin{equation*}
\require{AMScd} \begin{CD} H^{2i-n}_{\mathrm{Mot}}(X,\mathbb{Q}(i)) @>\mathrm{cl}^{\mathcal{H}}_{i,n}>> H^{2i-n}_{\mathcal{H}}(X,\mathbb{Q}(i))\\ @. {_{\rlap{\ \mathrm{cl}_{i,n}}}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{4em}{0em}]{}}}} @VVV\\ @. \mathrm{Hom}_{\mathrm{MHS}}(\mathbb{Q}(0),H^{2i-n}(X,\mathbb{Q}(i))) \end{CD}
\end{equation*}
and the vertical arrow is a surjection (the kernel is the intermediate Jacobian $\mathrm{Ext}^{1}_{\mathrm{MHS}}(\mathbb{Q}(0),H^{2i-n-1}(X,\mathbb{Q}(i)))$). So you are really asking about the image of $\mathrm{cl}_{i,n}^{\mathcal{H}}$. There the Beilinson-Hodge conjecture says that the image of $\mathrm{cl}_{i,n}^{\mathcal{H}}$ is dense, i.e. $\mathrm{cl}_{i,n}$ is surjective and the induced Abel-Jacobi map
\begin{equation*}
\mathrm{AJ}_{i,n}:H^{2i-n}_{\mathrm{Mot}}(X,\mathbb{Q}(i))_{\mathrm{hom}}\rightarrow\mathrm{Ext}^{1}_{\mathrm{MHS}}(\mathbb{Q}(0),H^{2i-n-1}(X,\mathbb{Q}(i)))
\end{equation*}
has dense image. For the reasons given above, if $n\neq 0,i$ then this conjecture is for $X$ defined over a number field.
Finally, one could speculate that the new ideas of Clausen-Scholze (see Clausen's modified Hodge Conjecture) have something to say in this setting too. ¯$\backslash$(ツ)/¯
Bibliography:
[AK09] D. Arapura, M. Kumar, Beilinson-Hodge cycles on semiabelian varieties, Math. Res. Lett. 16 (4) (2009), 557-562.
[AS07] M. Asakura, S. Saito, Beilinson's Hodge conjecture with coefficients for open complete intersections, in Algebraic Cycles and Motives 2, Edited by J. Nagel and C. Peters, LNS 344 (2007), 3-37, London Math. Soc.
[Bei86] A. A. Beilinson, Notes on absolute Hodge cohomology, Contemp. Math., 55, American Mathematical Society, Providence, RI, 1986, 35–68.
[dJL13] R. de Jeu, J. D. Lewis, Beilinson's Hodge conjecture for smooth varieties, J. K-Theory 11 (2013), no. 2, 243–282.
[Jan90] U. Jannsen, Mixed motives and algebraic $K$-theory, Lecture Notes in Math., 1400, Springer-Verlag, Berlin, 1990.
[Lew14] J. D. Lewis, Hodge type conjectures and the Bloch-Kato theorem, Contemp. Math., 608, American Mathematical Society, Providence, RI, 2014, 235–258.
[Sai09] M. Saito, Hodge-type conjecture for higher Chow groups, Pure and Applied Mathematics Quarterly 5 (3) (2009), 947-976.
[Sai04] S. Saito, Beilinson's Hodge and Tate conjectures, in Transcendental Aspects of Algebraic Cycles, Edited by S. Müller-Stach and C. Peters, LNS 313 (2004), 276-290, London Math. Soc.
|
2025-03-21T14:48:31.956706
| 2020-09-04T17:23:01 |
370871
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"Stopple",
"Vincent Granville",
"https://mathoverflow.net/users/140356",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/6756"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632710",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370871"
}
|
Stack Exchange
|
Erroneous Wolfram result for $\sum_{k=1}^\infty (k^3 + a^3)^{-1}$, looking for correct formula
I was trying to get some interesting result for $\zeta(3)$, exploring the following function:
$$W(a) = \sum_{k=1}^\infty \frac{1}{k^3 + a^3}, \mbox{ with } \lim_{a\rightarrow 0} W(a) = \zeta(3).$$
Let $w_1, w_2, w_3$ be the three roots (one real, two complex) of $(w+1)^3+a^3=0$, with $w_1=-(a+1)$. Also, $a$ is a real number. Using Wolfram Alpha (see here), I get
$$W(a)=\frac{-1}{3}\cdot\sum_{j=1}^3 W_j(a), \mbox{ with } W_j(a) = \frac{\psi^{(0)}(-w_j)}{(w_j+1)^2}.$$
Here $\psi^{(0)}$ is the digamma function. The result is wrong because $W_1(a) \rightarrow \infty$ as $a\rightarrow 0^+$ while $W_2(a)$ and $W_3(a)$ remain bounded. Indeed using $a=0.0001$, Wolfram yields $W(a)\approx -2334.16$, see here. Surprising, with $a=0.01$ it yields $W(a)\approx 1.20206$ which is very close to the true result.
Surprisingly, Wolfram knows (see here) that
$$\lim_{a\rightarrow 0} W(a) = -\frac{\psi^{(2)}(1)}{2}.$$
Of course (this is a well known fact), $\zeta(3)=-\psi^{(2)}(1)/2$ and thus Wolfram is correct this time.
My question:
What is going on with this computation (or is it me?), and what is the correct formula for $W(a)$?
Update
See the two answers below proving that I was wrong, and that the Mathematica formula I though was incorrect, is indeed right. Kudos Mathematica! You were successful at solving a nice problem involving a few challenging steps, and coming with a somewhat unexpected but neat formula involving derivatives of the digamma function instead of the classic $\zeta(3)$.
Final note
It is possible to use a different, simpler approach that does not involve complex numbers. Consider
$$V(a) =\sum_{k=1}^\infty (-1)^{k+1}\frac{1}{k(k^2-a^2)}.$$
Wolfram is able to compute the limit of $V(a)$ as $a\rightarrow 0$, and returns the correct value $3\zeta(3)/4$, see here. It is easy to establish that
$$V(a)=\frac{1}{a^2} \Big[\int_0^\infty \frac{\cosh(ax)}{1+e^x} dx -\log 2\Big].$$
To compute $\lim_{a\rightarrow 0} V(a)$, we apply L'Hospital Rule twice to the above expression, the denominator in this case being $a^2$. This yields
$$\lim_{a\rightarrow 0}V(a) = \frac{1}{2}\lim_{a\rightarrow 0}\int_0^\infty \frac{x^2\cosh(ax)}{1+e^x}dx =\frac{1}{2}\int_0^\infty \frac{x^2}{1+e^x}dx=\frac{3\zeta(3)}{4}.$$
Here, we assume $a<1$.
This might be better on https://mathematica.stackexchange.com/
@Stopple: Thank you, I was not aware of that forum.
There is a very similar problem with $\sum_{k=1}^\infty (-1)^{k+1} (k^3 +a^3)^{-1}$.
I think the statement in the OP that $W_2(a)$ and $W_3(a)$ remain bounded when $a\rightarrow 0$ is mistaken, so that there is no inconsistency with the Mathematica result.
The three roots of $(w+1)^3+a^3=0$ are
$$w_1= -a-1,\;\; w_2= \tfrac{1}{2} \left(-i \sqrt{3} a+a-2\right),\;\;w_3= \tfrac{1}{2} \left(i \sqrt{3} a+a-2\right).$$
Then the denominator $(w+1)^2$ vanishes for all three roots when $a\rightarrow 0$, while the numerator remains finite (equal to $-\gamma_{\rm Euler}$).
And indeed, a numerical check suggets that the Mathematica output is actually correct, and the erroneous numerical result for small $a$ is a numerical instability in the computation of the digamma function. See these two plots that compare the digamma expression (blue) with a numerical evaluation of the sum (gold), as a function of $a$. For $a\gtrsim 0.01$ the two answers are nearly indistinguishable.
There is definitely one thing I agree with you: the derivative of the series (your gold curve) wrt $a$ must be equal to zero when $a=0$. Indeed it's $a^2 O(1)$.
I have to redo my limit computations. Maybe I got them wrong in the first place.
We have the partial fraction decomposition
$$\frac{ca^2}{k^3+a^3}=\frac{-\omega}{k-a/\omega }+\frac{\omega -1}{k+a}+\frac{1}{k-a \omega},$$
where $c:=3(\omega-1)$ and $\omega:=e^{i\pi/3}$.
Also,
$$\sum_{k=1}^n\frac1{k+b}=\ln n-\psi(1+b)+o(1)$$
(as $n\to\infty$), where $\psi$ is the digamma function.
Collecting the pieces, for $a\in(-1,\infty)\setminus\{0\}$ we get
$$s(a):=\sum_{k=1}^\infty\frac1{k^3+a^3}
=\frac1{ca^2}\,
\left((1-\omega) \psi(1+a)+\omega\psi\left(1-a/\omega\right)
-\psi(1-a \omega)\right).$$
For $a\to0$,
$$s(a)=-\frac{\psi ^{(2)}(1)}{2}-\frac{\pi ^6 a^3}{945}+O\left(a^4\right)
=\zeta(3)-\frac{\pi ^6 a^3}{945}+O\left(a^4\right).$$
Here is the graph $\{(a,s(a))\colon0<a\le1\}$, with $s(0)=\zeta(3)=1.2020\ldots$:
(I am not geting instability.)
Thanks. In your last formula, I assume you mean $\sum_{k=1}^\infty$, right?
@VincentGranville : Thank you for your comment. I have corrected the typo.
@ Iosif: You get the right value for $s(0)$, but not for $s'(0)$ (the derivative of $s$ wrt $a$). We should have $s'(0)=0$.
@VincentGranville : I had confused $\omega$ with $-\omega$. This mistake is now corrected. In fact, now I have $s(a)=\zeta(3)+O(a^3)$ for $a\to0$, as it of course should be.
Thank you. I am thinking about another series that would not involve complex numbers, namely $\sum_{k=1}^\infty (-1)^k (k^3 - a^2k)^{-1}$ with $a\rightarrow 0$.
@VincentGranville : I think this latter series can be dealt with quite similarly.
|
2025-03-21T14:48:31.957023
| 2020-09-04T18:16:39 |
370875
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chris Godsil",
"Ugly Duckling",
"https://mathoverflow.net/users/1266",
"https://mathoverflow.net/users/164662"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632711",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370875"
}
|
Stack Exchange
|
Showing two vertices have same degree under a certain condition
Let $L$ be the Laplacian matrix of a simple, connected graph, and $\mathcal{P}_j$ the projector into the vertex $v_j$, represented by the appropriate canonical basis vector $(0,...,1,...,0)^T$. Given the positive real parameters $t$ and $\lambda$, consider the functions $$P_j(n,t,\lambda)=v_j^T e^{-it(\lambda L-\mathcal{P}_j)}s, $$ where $s=\frac{1}{\sqrt{n}}(1,...,1)^T$ is the normalized uniform vector.
My objective is to show that the condition $P_i=P_j$ implies the degrees of the two vertices $v_i$ and $v_j$ are the same. Does anyone have any tips on how to accomplish this?
(1) Using $t$ for transpose and for time is not a good idea. (2) What’s coordination number?
@ChrisGodsil You are not wrong, I edited the post. I believe the more familiar English term for coordination number is degree of a vertex, so I changed that as well.
I'll take $\lambda=1$ and use $E_j$ for $\mathcal{P}_j$. The $k$-th time derivative of $e^{-it(L-E_j)}s$ at $t=0$ is
\[
(-i(L-E_j))^k s.
\]
Now $(L-E_j)s = -v_j$ (because $Ls=0$) and, noting that $E_j=v_jv_j^T$, we have
\[
(L-E_j)^2s = -(L-E_j)v_j = -Lv_j +v_j.
\]
Therefore
\[
v_j^T(L-E_j)^2s = -v_j^T Lv_j + 1 = - L_{j,j} + 1.
\]
Since $L_{j,j}$ is the degree of vertex $j$, it follows that if $P_i=P_j$, then $i$ and $j$ have the same degree.
[I am curious as to where this problem is coming from?]
Thank you very much, Chris. Your answer is pretty close to what I was building up to: the big difference is that I was trying to calculate directly $(L-E_j)^k$ (I got stuck there and even asked another question here). I missed the idea of taking the time derivative! [I am an undergraduate physicist doing some research in quantum computation theory, the problem comes from there :)]
I'm thinking of a more general version of the problem. Right now we have found a necessary condition for $P_i=P_j$, but I have reasons to conjecture that $P_i=P_j$ if and only if there exists $\sigma\in\text{Aut}(G)$ such that $\sigma(v_i)=v_j$ (which of course implies the two vertices have the same degree). I'm open to ideas for a proof... but of course not a complete answer!
That will fail. Choose your graph to be strongly regular and asymmetric.
That's bad news. I was guided by the path graph, for which $P_i=P_{n-i+1}$, and of course this is consistent with the action of the path-reversing $\sigma$. Also, $P$ is constant over the symmetric cycle graph, and for every non-central vertex of the wheel or star graphs. However I saw that it fails for Frucht graphs, which are $3-$regular but asymmetric, and yet $P$ is constant. Could it actually just be a sufficient, but not necessary condition?
I think the answer is that trivially when there is an automorphism all properties of the vertex, including $P$, are carried over to its image. Then, other properties like (strong) regularity may play a role. I was also thinking of studying the distance matrices of these graphs to see if they have something in common...
@Ugly Duckling: If you want to discuss this further, you can find my email address by googling.
|
2025-03-21T14:48:31.957250
| 2020-09-04T18:36:55 |
370876
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632712",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370876"
}
|
Stack Exchange
|
Lie correspondence for Ind groups
I have the following question (with a negative answer in general).
Consider an Ind algebraic group (e.g. a group object in category of ind varieties) as defined by Shafrevich. It was assumed for some time that if a subgroup of a connected Ind group has the same Lie algebra as the Lie algebra of the group then it coincides with the group. A counterexample was given in Furter and Kraft in $\S$17.3. There, however, non-tame automorphisms of $\mathbb{A}^3$ played an important role.
My question is: does this "Lie correspondence" hold in dimension $2?$ In particular, is it true that if a Lie algebra $\mathcal{Lie}(G)$ of a connected closed subgroup $G$ of the group $\operatorname{Aut}(\mathbb{A}^2)$ coincides with an algebra $\mathcal{Lie}(H)$ of another connected closed subgoup $H<\operatorname{Aut}(\mathbb{A}^2)$ then $G=H?$
|
2025-03-21T14:48:31.957330
| 2020-09-04T20:26:10 |
370883
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andy Sanders",
"Ilya Bogdanov",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/17581",
"https://mathoverflow.net/users/49247"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632713",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370883"
}
|
Stack Exchange
|
Generic linear subspaces of symmetric matrices
Let $\mathcal{S}_{n}(\mathbb{R})$ be the real vector space of symmetric $n\times n$ traceless matrices with real entries and let $L\subset \mathcal{S}_{n}(\mathbb{R})$ be a linear subspace. Noticing that every matrix $A\in L$ is diagonalizable,
suppose that for all non-zero $A\in L,$ the matrix $A$ has distinct eigenvalues.
Question: What is the optimal upper bound on the dimension of the real vector space $L?$
There is an obvious upper bound which follows from the following observation (I'll omit the numbers to be brief). Let $\Delta\subset \mathcal{S}_{n}(\mathbb{R})$ denote the $(n-1)$-dimensional linear subspace of diagonal matrices.
Then, we must have that $L\cap \Delta$ has dimension at most one. Moreover, there is nothing special about $\Delta,$ given any $g\in SO(n),$ it also needs to be true that $L\cap (g\cdot \Delta\cdot g^{-1})$ has dimension at most one. This latter condition leads to an equivalent reformulation of the question.
Question: Suppose for every $g\in SO(n),$ the intersection $L\cap (g\cdot \Delta\cdot g^{-1})$ has dimension at most one. What is the optimal upper bound on the dimension of $L?$
For Lie-theoretic minded people, there is an obvious generalization of this question. Let $\mathfrak{g}$ be a real (semi)-simple Lie algebra and $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{m}$ a Cartan decomposition. Let $L\subset \mathfrak{m}$ be a linear subspace, and suppose that for any non-zero $X\in L,$ the vector $X$ is regular. By regular, I mean that $X$ belongs to a unique Cartan subspace $\mathfrak{a}\subset \mathfrak{m}.$ Then the question is again, how big can the dimension of $L$ be?
It might be useful to give a name, say $b(n)$, to the integer defined by the question. The observation on intersection with $\Delta$ gives $b(n)\le n(n+1)/2-1- (n-2)$, and $b(2)=2$.
(I'm not sure why the word "generic" is used in the question: this is not about generic subspaces.)
@YCor, I'm not convinced naming the integer really adds anything to question, and given your comment, it should be clear to readers. As to your second comment, the title should really include "linear subspaces all of whose elements are generic," but I couldn't figure out how to say this without sounding long-winded.
In any case so far I only see $b(n)\ge 2$ for $n$ even and $b(n)\ge 1$ for $n$ odd. Any better lower bound? Do you know what is $b(3)\in{1,2,3,4}$?
There are much larger subspaces all whose elements have two equal eigenvalues --- say, the matrices where the first two rows vanish (you may also replace the $(1,1)$th and $(2,2)$th entries by two equal numbers). So the answer to the first question does nor exceed $2n-1$.
|
2025-03-21T14:48:31.957768
| 2020-09-04T20:34:36 |
370884
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632714",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370884"
}
|
Stack Exchange
|
Whether or not two distinct points in Teichmuller space induce absolutely continuous volume forms on the unit tangent bundle of a surface?
Let $S$ be a closed orientable surface of genus greater than two. Let $g$ and $g'$ be metrics two of constant curvature. I guess we an think of these as two points in the Teichmüller space $\mathcal{T}(S)$ of $S$. These metrics induce respective volume forms $vol(g)$ and $vol(g')$ on the unit tangent bundle $UTS$.
Pardon my ignorance, but what can we say on whether or not these two volume forms are absolutely continuous to each other?
Is there a way to answer this question using standard techniques/results from Teichmuller theory or dynamics?
|
2025-03-21T14:48:31.957979
| 2020-09-04T22:42:35 |
370888
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Matthew Kvalheim",
"Piyush Grover",
"Tom Copeland",
"https://mathoverflow.net/users/12178",
"https://mathoverflow.net/users/30684",
"https://mathoverflow.net/users/89166"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632715",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370888"
}
|
Stack Exchange
|
What are some foundational authors/papers in dynamical systems?
I have just begun my first dynamical systems class, and I would like to try out the advice in the top answer here. To summarize, the answer suggests that when studying a new field, one should look at the original papers, and supplement it with modern material.
My class is roughly following Introduction to Dynamical Systems by Brin and Stuck. We are starting in the second chapter, which is about topological dynamics (it is only 20 pages, but very dense). I was wondering if you could please recommend me some authors/papers which started the field?
Thank you very much.
Smale’s 1967 “differentiable dynamical systems” is one highly influential example. You might also want to have a look at the book by Guckenheimer and Holmes. Conley’s “isolated invariant sets and the Morse index” is one example of an influential work in topological dynamics specifically.
The geometrical approach was initiated by Poincare , in ' les méthodes nouvelles de la mécanique céleste'.
Philip Holmes has summarized on Scholarpedia the seminal early developments of the field of dynamical systems, from the mathematical point of view (which I understand is the view point of the OP). Starting from the classic works of Poincaré and Birkhoff, through Andronov, Pontryagin and the Moscow School, to the "era of chaos" (Smale, Levinson, Chirikov, Lorenz, KAM, Ruelle & Takens, ...).
Earlier work is related to Newton's method and extensions explored by Ernst Schröder (1871-2 and Theremin ~ 1850) in his work on iterated functions/composition as a method of determining zeroes of functions.
I would suggest to concentrate on papers that introduce interesting new dynamical systems.
Smale's 1967 paper "differentiable dynamical systems" has been mentioned in the comments. Actually everything written by Smale is highly recommended.
Smale's horseshoe appears in his 1963 pretty short article
"A structurally stable differentiable homeomorphism with an infinite number of periodic points."
See Smale, Steve "Finding a horseshoe on the beaches of Rio". Math. Intelligencer 20 (1998), no. 1, 39–44 for a short account of his discovery.
The 1963 paper by Edward Lorenz entitled "deterministic nonperiodic flow" is also quite readable. There he introduced what is now called the Lorenz attractor.
Cvitanovic's 1984 book "universality in chaos" is a collection of interesting articles in dynamical systems published prior to the eighties that is worth browsing. It contains reprints of seminal papers from Ruelle, Feigenbaum, Henon for example. There is another sourcebook compiled by a chinese mathematician that is also nice, unfortunately I can't find the reference right now.
Katok / Hasselblatt: Introduction to the Modern Theory of Dynamical Systems gives a good introduction.
|
2025-03-21T14:48:31.958205
| 2020-09-04T23:31:22 |
370892
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"Sam Hopkins",
"coolpapa",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/45745"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632716",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370892"
}
|
Stack Exchange
|
Action of Weyl group on regions of Shi arrangement
This is an elaboration of a question which was aked on MO several years ago, which was unanswered but deleted by the question-asker. I hope it is okay to elaborate on a deleted question like this; for reference the question is at https://mathoverflow.net/questions/209856/ (probably only visible if you have enough reputation).
Here's the set-up. Let $\Phi$ be a crystallographic root system in vector space $V$ with Weyl group $W$. The Shi arrangement of $\Phi$ is the collection of hyperplanes $\{H_{\alpha,k}\colon \alpha \in \Phi^+, k=0,1\}$, where $H_{\alpha,k} := \{v\in V\colon \langle v, \alpha \rangle = k\}$. The number of regions of the Shi arrangement is known to be the number of "$\Phi$-parking functions", i.e., the number of elements of the finite torus $Q/(h+1)Q$, where $Q := \mathbb{Z}\Phi$ is the root lattice of $\Phi$ and $h$ the Coxeter number. See for instance the original paper of Shi Sign types corresponding to an affine Weyl group for a proof of this.
The question is about upgrading this numerical equality to an equality of $W$-actions.
Each region $R$ of the Shi arrangement is made up of a number of alcoves, where alcoves are the regions of the infinite hyperplane arrangement $\{H_{\alpha,k}\colon \alpha \in \Phi^+, k\in\mathbb{Z}\}$ corresponding to the affine Weyl group. I believe that each region contains a "minimal" alcove, which could be defined as the one closest to the origin (and probably could also be defined in terms of minimal length in the affine Weyl group). Let me use $\mathrm{min}(R)$ to denote this minimal region.
Correction: Originally I proposed an "action" here which is clearly not a group action. Let me reformulate the question in a way which has a hope of being correct.
Question: Is there some way to define an action of $W$ on the regions of the Shi arrangement so that:
for a region $R$ and $w\in W$, we have $w\cdot R=R'$ if $w\cdot\mathrm{min}(R)=\mathrm{min}(R')$ for some region $R'$;
the action isomorphic to the action of $W$ on $Q/(h+1)Q$?
Note that when $w\cdot\mathrm{min}(R) \neq \mathrm{min}(R')$ for any $R'$, it is not clear how to define $w\cdot R$.
The original question was about the Type A case. In fact, the asker proposed that the isomorphism should be given by the bijection of Athanasiadis and Linusson (see A Simple Bijection for the Regions of the Shi Arrangement of Hyperplanes) between regions of the Shi arrangement and parking functions.
Note that in their paper Parking Spaces, Armstrong, Reiner, and Rhoades explain, following Shi and Cellini and Papi, a canonical and type-uniform way to label the regions of the Shi arrangement by parking functions (see Section 10).
But if we compare the figure on page 42 of the Armstrong-Reiner-Rhoades paper to Figure 4 in the Athanasiadis-Linusson paper, we see the labelings are not the same in Type A, and in particular the Armstrong-Reiner-Rhoades is not compatible with the Shi arrangement action I described above.
I like this idea because of how geometric it is.
Here's a picture of what the orbits of this action should look like in Type $A_2$ (the red dot is the origin, and the dominant chamber is to the upper-right):
Again, compare to Figure 4 of the Athanasiadis-Linusson paper.
Meanwhile, here's what I think it looks like in Type $B_2$ (ignore the labels, they are from the Armstrong-Reiner-Rhoades paper):
Since the question was deleted, obviously the asker doesn't want it here under their name, so it may be appropriate to remove the identity of the asker. Perhaps a compromise would be that it is perfectly possible to link even to deleted questions, as https://mathoverflow.net/questions/209856 (which, for a non-moderator, just says that the question has been voluntarily removed by its author). This both removes the need separately to mention the question number, and allows the original asker to claim credit for the original question if desired.
@LSpice: good call. Done.
Ok, I haven't gone through all the details, but now I think the distinction between the Armstrong-Reiner-Rhoades labeling and the Athanasiadis-Linusson-style labeling is: in the former, they use ceilings of regions (hyperplanes $H_{\alpha,1}$ you cross moving in the dominant direction); for the latter, you should do the same thing but using floors (hyperplanes $H_{\alpha,1}$ you cross moving in the anti-dominant direction). Viewing the poset of dominant regions of the Shi arrangement as the root poset order ideal lattice $J(\Phi^+)$, this changes label $\max(I)$ to $\min(\Phi^+\setminus I)$.
Hey, the original asker is me! I'm sure I deleted that after doing the same thing OP did - confidently declaring how the W-action SHOULD work on the minimal chambers and then realizing that just doesn't quite work. :D I've come back to think about this same question again a lot of years later....
I have what's maybe not an answer but is I hope a helpful comment for this question. I have a way to show that the "orbits" you describe here are in bijection with a set of cosets of a subgroup of $S_{n}$. (I am confident of it in Type A and have yet to work on the details for other types.) This is not as good as having an actual group action, but it does give us one important consequence that the hypothetical action would as well.
Here's the outline - pick a dominant alcove $A$. Let $X$ be the set of reflections $s_{\alpha}$ such that $H_{\alpha,1}$ is a floor of your alcove (or $H_{\alpha,m}$ if you want the $m$-Shi arrangement). Let $G_{X}$ be the group generated by $X$. Since this is a reflection subgroup, it has a set of minimal coset representatives. Moreover, by a little bit of magic about the Shi arrangement, this group has the usual property for cosets of a parabolic subgroup of a Weyl group. Namely, the set $G^{X} = \{ w \in W \, | \, \ell(wx) > \ell(w) \textrm{ for all } x \in X\}$ is the set of minimal length coset representatives. (This is not the case for all reflection subgroups AFAIK.) So then, for $g \in S_{n}$, the alcove $g \cdot A$ is another minimal alcove iff $g \in G_{X}$. This last bit can be proven using proposition 4.1 of Fishel-Vazirani's "A bijection between dominant Shi regions and core partitions", (http://dx.doi.org/10.1016/j.ejc.2010.05.014). They characterize floors of alcoves in terms of Weyl group actions on roots, and with a little bit of work, you can trace that around. The only way for $gA$ to not be minimal is if $g$ turns some $H_{\alpha,m}$ that was a floor of $A$ into a floor $H_{\alpha',m}$ with $\alpha'$ negative, and the condition for that to happen is something about inversions, which is exactly equivalent to the definition of $G_{X}$. I'm working on a related project which will hopefully be in some readable form soon.
I don't know if this will be helpful or if this is a thing that everyone knows already, but I don't think I've seen it before - I'd love to hear if it's in the literature somewhere already.
When you say "I believe that each region contains a "minimal" alcove, which could be defined as the one closest to the origin (and probably could also be defined in terms of minimal length in the affine Weyl group)" it is indeed the case, you can see it in the paper Sign types corresponding to an affine Weyl group of Shi that you cited. Shi gives actually two characterizations of this minimal element: Propositions 7.2 and 7.3.
For the way to define something that makes sense on the set of Shi regions you can use this. First of all notice that I take the notations of Shi. Let $w \in W_a$ and $s \in S \subset W$. We have the formula $k(sw, \alpha) = k(w, s(\alpha)) + k(s, \alpha)$. Since a Shi region is just the data of symbols $\{-,0,+\}$ over each position of $\Phi^+$, with some specific conditions (that Shi calls admissible sign types), you can try to see how the value of $k(sw, \alpha)$ behaves and then, maybe, by generalizing this formula for any $x \in W$, you might get something.
|
2025-03-21T14:48:31.958717
| 2020-09-04T23:59:20 |
370894
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"aglearner",
"https://mathoverflow.net/users/13441",
"https://mathoverflow.net/users/25510"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632717",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370894"
}
|
Stack Exchange
|
Cross-ratios of $4$ boundary points on a continuous family of disks in $\mathbb C^1$
Let $S^1=\mathbb R^1/\mathbb Z$. Consider a family $\varphi_t$ of pieceswise smooth injective maps $\varphi_t:S^1\to \mathbb C^1$ depending continuously on $t$. Then each curve $\varphi_t(S^1)$ is a simple closed curve in $\mathbb C^1$, i.e. it bounds an open complex disk. Using Riemann mapping theorem we can identify such a disk with the unit disk $|z|\le 1$. In particular, for each $t$, by identifying the disk bounded by $\varphi_t(S^1)$ with the disk $|z|\le 1$, we can define the cross-ratio of points $\varphi_t(0)$, $\varphi_t(\frac{1}{4})$, $\varphi_t(\frac{1}{2})$, $\varphi_t(\frac{3}{4})$ as points lying in $|z|\le 1$ . I am pretty sure that such a cross-ratio is a continuous function of $t$.
Question. Does this statement follow from some standard result? How to prove it?
What does the Riemann map have to do with this question? 2. You say $\phi_t$ continuously depends on $t$. Does not this mean that $\phi_t(0),\phi_t(1/2),...$ continuously depend on $t$ and since they are all distinct the cross ratio also continuously depend in $t$?
Sorry Alexander, it was a bit hard to phrase this question properly. I want to measure the cross ratio of these 4 points not inside $\mathbb C^1$ but inside the disk. This is why I mentioned Rimenann mapping theorem. Using this theorem we see these 4 points as points in the boundary of the disk $|z|\le 1$ and then measure the corss-ratio in the disk $|z|\le 1$. Is it more clear now?
Then set it up properly in the question text rather than in the comments.
I realised that the answer to this question follows from a different question on Mathoverflow almost 10 years ago:
Does Riemann map depend continuously on the domain?
These were good times... People were not voting to close questions just because they don't understand them...
|
2025-03-21T14:48:31.958916
| 2020-09-05T01:07:33 |
370898
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"A. Bordg",
"Aaron",
"Asaf Karagila",
"Balazs",
"Christopher King",
"D.S. Lipham",
"Dabed",
"Deane Yang",
"Denis Nardin",
"Geoff Robinson",
"Hercule Poirot",
"Hollis Williams",
"JoshuaZ",
"Matthew Daws",
"Monroe Eskew",
"Paul Siegel",
"Quarto Bendir",
"Sam Hopkins",
"Sam Nead",
"Timothy Chow",
"arsmath",
"darij grinberg",
"fedja",
"https://mathoverflow.net/users/11145",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/113606",
"https://mathoverflow.net/users/119114",
"https://mathoverflow.net/users/127690",
"https://mathoverflow.net/users/130882",
"https://mathoverflow.net/users/142708",
"https://mathoverflow.net/users/14450",
"https://mathoverflow.net/users/156492",
"https://mathoverflow.net/users/1650",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/3106",
"https://mathoverflow.net/users/3711",
"https://mathoverflow.net/users/406",
"https://mathoverflow.net/users/43054",
"https://mathoverflow.net/users/4362",
"https://mathoverflow.net/users/4435",
"https://mathoverflow.net/users/50073",
"https://mathoverflow.net/users/6107",
"https://mathoverflow.net/users/613",
"https://mathoverflow.net/users/65915",
"https://mathoverflow.net/users/7206",
"https://mathoverflow.net/users/95718",
"user21820"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632718",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370898"
}
|
Stack Exchange
|
Replication crisis in mathematics
Lately, I have been learning about the replication crisis, see How Fraud, Bias, Negligence, and Hype Undermine the Search for Truth (good YouTube video) — by Michael Shermer and Stuart Ritchie. According to Wikipedia, the replication crisis (also known as the replicability crisis or reproducibility crisis) is
an ongoing methodological crisis in which it has been found that many
scientific studies are difficult or impossible to replicate or
reproduce. The replication crisis affects the social sciences and
medicine most severely.
Has the replication crisis impacted (pure) mathematics, or is mathematics unaffected? How should results in mathematics be reproduced? How can complicated proofs be replicated, given that so few people are able to understand them to begin with?
Not quite duplicates, but closely related questions https://mathoverflow.net/questions/35468/widely-accepted-mathematical-results-that-were-later-shown-to-be-wrong and https://mathoverflow.net/questions/27749/what-are-some-correct-results-discovered-with-incorrect-or-no-proofs .
Here's a quote from a MO question along the lines of what you're asking - "As the Bing topologists familiar with these arguments retire, the hopes of reproducing the details of the proof are fading, and with it, the insight that such a spectacular proof affords." - https://mathoverflow.net/questions/108631/fake-versus-exotic Though the concern might be misguided - https://mathoverflow.net/questions/87674/independent-evidence-for-the-classification-of-topological-4-manifolds
Reading a proof and verifying all the details is the mathematical equivalent of replication. I would like to believe that most mathematicians do this for most of the results they cite. There are certainly results that are long and complicated and out of the reach for most people in a field to check in detail, but I would also like to believe that people avoid using them until experts in their field say they have checked the proof. As long as this replication is cheap and easy compared to physical experiments, our replication issues will be smaller.
As opposed to Aaron I think a vast number of mathematicians do not read the proofs, in particular do not verify the details of (all of) the results they cite. It is not even clear what this would mean since these results themselves rely upon other results etc. I do think they ensure they understand the meaning of the statements and investigate particular interesting cases and the implications on their result.
Just one ex...it is interesting that (depending on who you ask) Freedman's proof of the 4D Poincaré conjecture and Perelman's proof of the Thurston conjecture are incomplete/unclearly complete, as they are arguably the most widely renowned works in (resp.) geometric topology and geometric analysis. It seems that nobody has been able to reorganize the logic of the existing proofs in order to clarify their content, which seems to me to be the goal of an exposition. Also noteworthy that (depending on who you ask) the existing expositions lack clarity themselves or have their own errors, or both
Abbas Bahri's article "Five gaps in mathematics" is an interesting paper which seems directly relevant to the question. (To my understanding, all of his five objections can be resolved, but I believe this is irrelevant to the question.) https://twma.files.wordpress.com/2016/12/five-gaps.pdf
Whenever this topic arises, I always post this talk by Voevodsky. https://www.math.ias.edu/vladimir/sites/math.ias.edu.vladimir/files/2014_IAS.pdf
Perhaps Kevin Buzzard's recent talk is also of interest here. https://www.andrew.cmu.edu/user/avigad/meetings/fomm2020/slides/fomm_buzzard.pdf
See also https://mathoverflow.net/questions/338607/why-doesnt-mathematics-collapse-even-though-humans-quite-often-make-mistakes-in/338610#338610
See also Proofs shown to be wrong after formalization with proof assistant and Extent of “unscientific”, and of wrong, papers in research mathematics and Results that are widely accepted but no proof has appeared.
Also related: Which mathematician sampled published proofs and found one third of them to have errors?, Endless controversy about the correctness of significant papers, When have we lost a body of mathematics because errors were found?, What percentage of published mathematics papers are correct?, Nontrivially fillable gaps in published proofs of major theorems.
I believe the replication crisis in science generally refers to work which is empirical and experimental: basically, results from experimental studies can be impossible to replicate if you don't have that exact same experimental set-up.
I think this question does not make sense because mathematics does not have experiments.
One trend which I feel can be damaging to the subject is "announcing" big results, and then not publishing a proof for several years, or decades, or ever. This has a doubly damaging effect because preemptive announcements have the effect of stopping others working towards the results. Not offering a proof, even in preprint form, of course does not allow any checking by others.
@JoeT Still, it is disanalogous. The the epistemological structure of mathematics does not involve an empirical data set that is subject increased credence via replication. I think this question is an attempt at buzzword-harpooning via analogy. This is not to contradict anything that is said above regarding doubt in mathematics, but only to question the idea of lumping it together with the “replication crisis.”
@MonroeEskew : It sounds to me that you're getting hung up on the word "replication." Suppose we replace it with something like "validation" or "confirmation." It's certainly true that we gain increased confidence in the correctness of a proof every time someone independently checks the argument and confirms that it is correct. Erroneous results creep into the literature when people are careless and/or no additional person checks the result. This is highly analogous to errors creeping into the scientific literature because people are careless and/or no additional person checks the result.
Considering Blade Runner, just be thankful there's no Replicant crisis too.
@DanielD. : Are you saying that MIP*=RE does not imply a negative answer to the Connes embedding problem? If so, this is news to me. Do you have any more information about this?
@Timothy Chow no I didn't meant that but I reflecting on it I think I know too little so I shouldn't have commented on the first place so I deleted it, will try to study more and comment less from now on =)
Dear Ellie, you may find some elements of answer in my article A Replication Crisis in Mathematics?.
+1 for @A.Bordg's article. Thank you!
Mathematics does have its own version of the replicability problem, but for various reasons it is not as severe as in some scientific literature.
A good example is the classification of finite simple groups - this was a monumental achievement (mostly) completed in the 1980's, spanning tens of thousands of pages written by dozens of authors. But over the past 20 years there has been significant ongoing effort undertaken by Gorenstein, Lyons, Solomon, and others to consolidate the proof in one place. This is partially to simplify and iron out kinks in the proof, but also out of a very real concern that the proof will be lost as experts retire and the field attracts fewer and fewer new researchers. This is one replicability issue in mathematics: some bodies of mathematical knowledge slide into folklore or arcana unless there is a concerted effort by the next generation to organize and preserve them.
Another example is the ongoing saga of Mochizuki's proposed proof of the abc conjecture. The proof involve thousands of pages of work that remains obscure to all but a few, and there remains serious disagreement over whether the argument is correct. There are numerous other examples where important results are called into question because few experts spend the time and energy necessary to carefully work through difficult foundational theory - symplectic geometry provides another recent example.
Why do I think these issues are not as big of a problem for mathematics as analogous issues in the sciences?
Negative results: If you set out to solve an important mathematical problem but instead discover a disproof or counterexample, this is often just as highly valued as a proof. This provides a check against the perverse incentives which motivate some empirical researchers to stretch their evidence for the sake of getting a publication.
Interconnectedness: Most mathematical research is part of an ecosystem of similar results about similar objects, and in an area with enough activity it is difficult for inconsistencies to develop and persist unnoticed.
Generalization: Whenever there is a major mathematical breakthrough it is normally followed by a flurry of activity to extend it and solve other related problems. This entails not just replicating the breakthrough but clarifying it and probing its limits - a good example of this is all the work in the Langlands program which extends and clarifies Wiles' work on the modularity theorem.
Purity: social science and psychology research is hard because the results of an experiment depend on norms and empirical circumstances which can change significantly over time - for instance, many studies about media consumption before the 90's were rendered almost irrelevant by the internet. The foundations of an area of mathematics can change, but the logical correctness of a mathematical argument can't (more or less).
It seems that some of your points #1-4 do indicate that most major theorems are correct, but it seems that none of them address the problem of proofs failing to be understandable or replicable to readers
@QuartoBendir I think 2 and 3 provide some check against this. Modern algebraic geometry is hard to learn, but a modern student who has taken the time to learn it properly can come up with very simple proofs of results that would have been considered very hard a century ago. I think most areas of math gradually get easier over time as people discover new connections and generalizations that allow future readers to get more bang for their buck.
I agree, but I don't see how that's helpful for the problem of a present-day mathematician understanding present-day mathematics research, which is what I understand the replication crisis to be about.
@QuartoBendir By my understanding the issues that have reached crisis levels in the sciences are twofold: first, that some classical, textbook studies (e.g. the Stanford prison experiment) turned out to be fraudulent or wrong; and second, that very few results are ever replicated at all, even by experts in the same field. I do think that the practice of experts applying and generalizing each other's results helps mathematics avoid these problems even in the shorter term, though admittedly it doesn't do much for mathematicians in other fields.
I agree that the standard textbook material in math must be among the most reliable in the sciences. And I think that what you say applies to certain results and certain methods in modern practice, but that a majority of research does not directly become part of the ecosystem in that way. In part it's like Voevoedksy said, "A technical argument by a trusted author, which is hard to check and looks similar to arguments known to be correct, is hardly ever checked in detail." I think such an instance is not naturally checked against by #2 and 3, although of course in certain cases it is.
Another way a mathematical "result" can be negative is if it fails to solve a conjecture; i.e. an approach that neither proves nor disproves it. Information about this approach and why it failed can be useful to other mathematicians.
Why do you phrase it as if Mochizuki has a proof but only some experts disagree? Isn't it the case that most experts do not believe he has a proof, and only those close to Mochizuki believe that he does? In fact, the very fact that he can publish it in a journal that has himself as chief editor shows without doubt that mathematics will have a replication crisis (however small) if mathematicians do not maintain a list of reputable journals.
@user21820 I deliberately phrased my remarks about the "serious disagreement" about the "proposed proof" to remain neutral about mathematics that I am not qualified to evaluate. I have not counted the number of experts who hold one opinion or the other on the matter, nor do I intend to - the existence of the disagreement itself is enough for the argument that I'm trying to make here.
@QuartoBendir I always found Voevodsky's concerns in this matter overblown... Yes, we do not check things as often as we should and there's been a few high profile peer review failures, but it's rare that a new important result is not studied in detail in numerous reading seminars across the world, and it does happen that people find mistakes that way (usually small and easily corrected, but still). The vast majority of published wrong results are small papers of limited interest, which tend to be partially sheltered from the effects of 2-3 in this answer.
@PaulSiegel: I do not agree that your phrasing was neutral. To describe something as "foundational" or some people as "most dedicated experts" is certainly not at all neutral. I have edited it into a truly neutral version.
@JoeT: That's not the only thing. Fesenko is not even some arbitrary high-profile person, but actually has vested interest in Mochizuki's work because he was and still is a top promoter of said work. If he ever publicly admits that he doubts the work, he would effectively be saying that he has wasted a large portion of the last 5 years or so, as well as all those mathematicians to whom he publicized said work to... Maybe they know there is a flaw but they are hoping it can be fixed before enough experts catch on.
The fact that Mochizuki admits "it does not seem possible to get intermediate results in his proof of abc using IUT", as well as many other non-mathematical signals, suggests strongly that there is not much substance. I will personally accept a claimed proof of this level of obscurity only if it comes with a formal verification in some well-known formal system such as Mizar or Coq. Even the proof of Kepler's conjecture by Hales was actually quite understood and widely believed to be true though nobody could be 100% sure. Here the purported proof is described by experts as "impenetrable".
@user21820 You are reading intent here where none is present. The phrase "dedicated experts" in the original answer refers to both the critics and advocates of the work, and the term "foundational" in context refers to the role of an idea in an argument, not its correctness. I'm going to leave your edits as a show of good faith (and I genuinely don't care), but in exchange I'd like you to go grind your ax somewhere else - none of this discussion is relevant to the question or the answer, and it's pretty clear that only one of us is actually neutral toward the irrelevant dispute.
@PaulSiegel: I accept your claim that your intent was to be neutral, but I wish to point out that to be neutral you cannot say "remains obscure to all but the most dedicated experts" because it directly implies "only the most dedicated experts understand it", which is not neutral because there is currently no evidence that any expert understands it (rather than merely claims to understand it). As for neutrality, I didn't say my personal view was neutral, but my edited version was definitely neutral (in standard English), so thanks for leaving it as it is now.
@user21820 I agree with the assertion that "only the most dedicated experts understand [Mochizuki's work]" - as I understand it, the work departs significantly from the mainstream tools in its field. If a consensus emerges that the argument is incorrect then it will be thanks to the dedication of the experts who took the time to work though the argument and find a flaw. The underlying point in the second half of my answer is that mathematics is more robust against replicability issues because it naturally provides incentives for researchers to think critically about each others' work.
@user21820 But anyway, I think that underlying point still gets across, so if everyone is happy then we can leave the answer as it is now.
Just to make clear, I completely agree with everything in your answer now. =)
+1 for "some bodies of mathematical knowledge slide into folklore or arcana unless there is a concerted effort by the next generation to organize and preserve them" I think this is an incredibly important point, perhaps nearly orthogonal to "replication". Some areas of research will naturally wane, perhaps as "interesting" results to prove become rare. But I wonder about cause and effect here: does a field die because it ceases to be interesting; or does it die because getting to the "coal-face" of research becomes too hard for incomers?
How can we expect that increasingly complicated proofs are replicated when so few people can understand them in the first place?
My answer to that is that we do not expect them to be replicated in the usual sense of this word (repeated and included into textbooks with just minor cosmetic and stylistic changes). Rather we expect them to be gradually simplified and streamlined either through changing the proofs themselves by finding a shortcut or replacing the whole argument with a completely different one, or by building a theory that is locally trivial but proceeds in the direction of making the proof understandable and verifiable much faster than the currently existing one. The latter is exactly what Mochizuki tried to do though his goal was rather to just reduce the difficulty from "totally impossible" to "barely feasible" and the prevailing opinion is that he failed in the case of the ABC conjecture though he has succeeded in several other problems.
The first approach is more common in analysis (broadly understood), the second is more common in algebra (also broadly understood), but you can try to play either game in either field. My own perception of what is proved and what is not borders on solipsism: I accept the fact as proven if I've read and understood the whole argument or figured it out myself. So most mathematics remains "unproved" to me and, apparently, will stay unproved for the rest of my life. Of course, it doesn't mean that I'm running around questioning the validity of the corresponding theorems. What it means is that I just never allow myself to rely in my own papers on anything that I haven't fully verified to my satisfaction, try to make my papers as self-contained as possible within practical limits, and that I consider the activity of simplifying the existing proofs as meaningful as solving open questions even in the case when the proofs are reasonably well-known and can already be classified as "accessible". But not everybody works this way. Many people are completely happy to drop a nuke any time they have an opportunity to do it and there is nothing formally wrong with that: the underlying point of view is that our time is short, we have to figure out as many things as possible, and the simplifications, etc. will come later. Probably, we need a mixture of both types to proceed as efficiently as we can.
So I would say that the mathematics is reasonably immune to this crisis in the sense that mathematicians are aware of the associated risks, take them willingly, and try to gradually build the safe ground of general accessibility under everything though the process of this building is always behind the process of the mathematical discovery itself. The same applies to physics and medicine though the gap between the "front line" and the "safe ground" there may be wider. In fact, it applies to any science that deserves to be called by that name. As to the so called "social sciences", they are often done at the level of alchemy and astrology today in my humble opinion (and not only mine: read the Richard Feinman critiques, for example) but we should not forget that those were the precursors to such well-respected sciences as chemistry and astronomy/cosmology, so I view the current crisis there as a part of the normal healthy process of transitioning from the prevailing general "blahblahblah" and weathervane behavior with respect to political winds to something more substantial.
Edit: Paul Siegel has convinced me that things have indeed changed since the time I took (obligatory) courses of Marxist philosophy and the history of communist party, though this change may be not easily visible to the general public because it mainly happens outside academia and is driven primarily by company business interests, so a huge part of it occurs behind closed doors (Paul, please correct me if I misinterpreted what you said in any way). So my statement that the current social sciences are not capable of something beyond general blahblahblah is no longer valid and I retract it. However I still maintain the opinion that it is blahblahblah rather than hard data analysis or other scientific approach that drives many public political and social discussions and decisions of today (I don't know what happens here behind the closed doors, of course, and it may be that, like in advertising, what we see is just what shepherds choose to show to their sheep to drive them in the direction they want, but I prefer to think that it is not exactly the case). If somebody can convincingly challenge that, I would be quite interested.
Apologies to everybody for switching this discussion to a sideline.
I think the second half of this comment is typical of the low quality of commentary when it comes to the social sciences, which are a broad gamut of fields, all of which are different. Richard Feynman's thoughts on the topic 70 years ago are not the final word on the subject. There is a replication crisis in social psychology, which is not even all of psychology. Is it a general problem across all of social science? Who knows? But mention the word "social" and suddenly everyone's an expert.
@arsmath Agreed. I haven't explained myself well enough preferring to curtail the subject to a few words and to refer to the opinion of a way more eloquent and knowledgeable person. If you want a more expanded version of what I think, I can provide it later. General psychology, by the way, is a part of medicine in my classification. We just understand the words "social sciences" slightly differently, that's mainly it. Upvoting your comment but changing nothing in the main text :-)
I suspect your impression of what modern social science looks like is a couple decades out of date. The level of scientific rigor found in digital advertising, market research segmentation, forecasting consumer sentiment, and so forth would meet whatever standard you care to throw at it - there are billions of dollars riding on tiny fractions of a percent accuracy. The problem used to be "How accurately can we model human behavior?" The question has become "Is it ethical to model human behavior as accurately as we can?"
@PaulSiegel Then why is the replication crisis hitting the social sciences so hard? I think the latest analysis shows that nearly 40% of the research in social psychology (in the top journals, in fact) cannot be replicated.
@D.S.Lipham A lot of the best social science research isn't happening in universities anymore - this is because the underlying datasets are huge and proprietary. Think Amazon's consumer demand dataset, Google's clickstream data, Facebook's social graph, etc. Academic social science researchers generally work with different problems, smaller budgets, and less accountability, but a number of specific disciplines are thriving - computational linguistics, for instance, has made epoch-defining breakthroughs in the last year or two.
@PaulSiegel Thanks I understand your point now.
@PaulSiegel I would rather call all things you are talking about "applied mathematics, computer science and statistics". Have nothing to say against either of them. When people in these fields know what they are doing, the results are most impressive. Even the AI research has finally lifted off despite all scepticism about it in the 90s. Unfortunately, what you say also shows that the current primary consumers of the scientific method (in the sense of the same Feynman) are people who want to make big money (like in 50's it was people who wanted to make big weapons). cont-d
@PaulSiegel I wish I could see that level of scientific rigor, let's say, in education. The budgets are also quite noticeable there but all I have seen coming out of it as far as academic setting is concerned are various "calculus reforms". Would you argue that universities are not primary places of teaching anymore?
@fedja If you call these ideas applied math, CS, and statistics, then: applied to what? I tend to define a discipline more by its goals than its methods, and the goals in this context are to model human behaviors and relationships. That's social science.
@fedja And I don't claim that all branches of the social sciences have been equally improved over the past 30-40 years, any more than I believe that all disciplines of physics, chemistry, or mathematics have been equally successful over that period. To provide comparable results in education, for instance, you would need a comparable dataset, and that would involve collecting minute-by-minute time series on student attention / behavior, mapped to empirical outcomes. I think China is trying this sort of experiment. As above, the question is not "Can we?" but "Should we?"
@PaulSiegel There is one more point I wanted to bring up. You are talking about market analysis, big databases and Google, Amazon, and Facebook. Now, whom do those companies normally hire to do that market analysis on those big data: social psychology majors (understood broadly) or math majors (also understood broadly), or both? My opinion of whether it should be classified as progress in social sciences or as a mere takeover of traditional social science subjects by natural science hangs on the answer to this simple query. So what would you say (if you know the answer yourself, of course)?
@fedja It's a mixture of both. I work for a much smaller company on a team that does this sort of work, and the backgrounds of the people involved include psychology, economics, political science, and linguistics as well as the more expected math and computer science, and I can tell from my professional network that you see similar diversity at big tech companies. I can also testify that I had to work hard to acquire domain expertise in linguistics, market research, sociology, etc. to be effective; the math / statistics isn't that hard, but good experimental design is.
@fedja Of course the engineering teams at these companies are much larger - maybe a 7 to 1 ratio where I work - and the domain experts are still expected to develop their skills in programming and math / stats. The work is very technical and I doubt it looks much like 1970's social science, but the overarching goals and the epistemology are pretty similar.
@PaulSiegel OK, then it is a mixture of the progress and the takeover in my opinion :-). That's good actually. Probably the best option of the three. I'll update my "outdated opinion" accordingly ;-) I hope that the ethical problems you are talking about will be also decided by a mixed team like yours, not by the people who are telling us what to think and how to behave today. Thanks for your contribution to this conversation!
@fedja Thanks for the opportunity to contribute! A lot of the work going on at my company is not public, but Gary King (https://gking.harvard.edu/) is on our board and has been a leader in modernizing social science methodology for quite some time; you might check out some of his work if you're curious. And we do make a strong effort to assert ourselves on ethical matters, though usually it feels like shouting at a hurricane...
@fedja Your view here is the kind of non-scientific view that led to the replication crisis in the first place. We have results from a single field (social psychology). This is like the result from a single small-sample experiment. From this, you generalize to all of social science, just like a single social psychology experiment applied to 30 students is generalized to all of humanity through time and space. How do we know that's not just social psychology? How do we know that it's actually not all of science, including physics and chemistry? We don't, until we put in the work.
Social psychology is peculiar in that it's a route to become famous in a way that's available to few other academics. More people have probably heard of the Stanford Prison Experiment than have heard of the Michaelson-Morley experiment. You found an effect that is surprising and yet somehow confirms widespread biases, and you'll be in the New York Times in no time. The power pose research is an example. It was published in 2010, which led to a high profile TED talk in 2012, and being mentioned on Oprah. Most scientists have no route to being mentioned on Oprah.
@PaulSiegel Just want to check with you that I summarized our conversation in my edit to the post correctly. If you find that something essential is distorted, dropped, or added, just let me know and I'll fix it.
@arsmath I wouldn't say "a single field" here: the corresponding Wikipedia article (https://en.wikipedia.org/wiki/Replication_crisis), for example believes it is way more widespread and some other sources corroborate that. But, of course, I'm inclined to agree with you that I have a general tendency to overgeneralize though I disagree that this tendency was the (or even main) root of the replication crisis. As to SPE, I have rather heard of the Milgram shock experiment, though not through Oprah.
@fedja Yes, that is a reasonable summary of my argument, and those who want more detail can find it in the comment section. Your remark about your own experience with "social science" as a student interesting - I bet there are some good stories there.
Has this crisis impacted (pure) mathematics, or do you believe that
maths is mostly immune to it?
Immune to the replication problem, yes. But not immune to the attitudes which cause scientists to do unreplicable research in the first place. Some mathematicians will announce that a particular theorem has been proven, harvest the glory based on the fact that they have proved things in the past, and then never publish their results. Rota's Conjecture is one notorious example. Now we are in a situation where (a) nobody knows whether it is true and (b) nobody has worked on it for seven years, and probably (if it turns out that no proof actually exists) will not work on it for at least another decade.
How should results in mathematics be reproduced?
In science, it would be ideal if people dedicated research time to replicating published experimental results. This doesn't happen much because there is no glory to be gained by doing it.
The analogue in mathematics would be for people to publish new proofs of existing results, or expositions of existing proofs, which is happily much more common. I don't mean copying out well-known results in new language (Tom Leinster, The bijection between projective indecomposable and simple modules), I mean expository papers like this (Cao and Zhu, A complete proof of the Poincaré and geometrization conjectures, Asian J. Math. 10 (2006) pp. 165–492).
Even more noble are the people using proof assistant software to verify existing mathematics.
How can we expect that increasingly complicated proofs are replicated
when so few people can understand them in the first place?
I think our best hope is proof assistant software. Perhaps by the end of this century, we will he living in a world where no mathematician can replicate any reasonably cutting-edge proof, yet research is still happily chugging along.
|
2025-03-21T14:48:31.961420
| 2020-09-05T01:10:54 |
370899
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Donu Arapura",
"Nguyen Dang Son",
"Qiaochu Yuan",
"Will Chen",
"Zach Teitler",
"https://mathoverflow.net/users/103164",
"https://mathoverflow.net/users/117651",
"https://mathoverflow.net/users/15242",
"https://mathoverflow.net/users/290",
"https://mathoverflow.net/users/4144",
"https://mathoverflow.net/users/88133",
"user347489"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632719",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370899"
}
|
Stack Exchange
|
Which theorems in commutative algebra describe the closed property of curves (i.e. algebraic varieties) in algebraic geometry?
In $R^2$, we have those solutions of $x^2+y^2-1=0$ describe a closed unit disk and $y^2 = x^3 − x + 1$ describes an unclosed elliptic curve. When we consider corresponding algebras e.g. ${R^2}/ \langle {x^2} + {y^2} - 1 \rangle $ and ${R^2}/ \langle {y^2} - {x^3} + x - 1 \rangle $, which things in their algebras describe the closed property? It seems that homological algebras may involve, for example, as we know, dimension or multiplicity was defined.
I think what you mean is not "closed", but "compact". However, you're asking about the compactness of the set of real solutions under the topology inherited from $\mathbb{R}^n$. This type of compactness is not a purely algebraic question, so I don't see why you should be able to detect it at the level of commutative algebra (as you mention in your comment to Asvin's question). You could try to make it algebraic by asking about compactness in the Zariski topology, but then the both of your spaces are quasi-compact, though neither are Hausdorff so they are not compact (depending on your def)
You can see the shape of the elliptic curve in "https://en.wikipedia.org/wiki/Elliptic_curve#:~:text=In%20mathematics%2C%20an%20elliptic%20curve,no%20cusps%20or%20self%2Dintersections." and it is not closed. I do not talk about the topology of algebraic varieties here because I only care about the properties of its coordinate rings.
What do you mean by closed?
Closed like a Jordan curve.
So you mean "closed" as in closed path in the plane?
Would $y^2-x^2(x+1)$ count as a closed curve? It isn’t compact but it has a closed path.
The OP's question may not have been precisely stated but I think it's reasonable. We know that many geometric and topological properties of the complex points of a variety are detectable via algebra (for an extreme example, the Betti numbers via counting points over finite fields); the OP asks a similar question but for the real points.
Yes, "closed" as in a closed path, sorry for unclear here.
One way of asking this is: Does the hyperplane at infinity (in the projective plane) intersect the curve at a real point? That is, if we look at the homogeneous terms of the highest degree, does that polynomial have a real, non trivial solution? In the first case, $x^2+y^2$ has no real solution except $(0,0)$ while in the second case $x^3$ has $(0,1)$ has a solution.
But you still are using algebraic geometry, in the case, projective algebraic geometry. We have a duality of commutative algebra and algebraic geometry, so I want to see a corresponding between them.
Everything in here can be done at the level of rings: homogenize, set the variable you introduced to zero and look for solutions.
|
2025-03-21T14:48:31.961660
| 2020-09-05T01:40:56 |
370900
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632720",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370900"
}
|
Stack Exchange
|
Sheaves with specified singular support at infinity coming from hyperplane arrangements
Given a manifold $M$, we consider its cotangent bundle $T^*M$, and its cocircle bundle $T^\infty M$, quotienting out by the scaling action of the positive reals. Given a Legendrian submanifold $\Lambda \subset T^\infty M$, there is a category of "sheaves with singular support at infinity contained in $\Lambda$". (These are really complexes of sheaves, not individual sheaves, but I am trying to stay close to the language of my main source.)
I want to consider the case that $M$ is $\mathbb R^n$, which is equipped with a hyperplane arrangment. Each hyperplane in the arrangement is going to to appear in $\Lambda$, offset in the direction of a conormal. This means that the images of the hyperplanes in $T^\infty \mathbb R^n$ are not going to intersect. We choose the conormals consistently, so there is some tangent direction with respect to which they are all positive.
My question is the following: has anyone has figured out what the corresponding category of (complexes of) sheaves looks like in these examples coming from hyperplane arrangements?
The main paper I have been looking at where some of this is explained is arXiv:1512.08942 by Shende, Treumann, Williams, and Zaslow. They are interested in a somewhat different situation, where, in particular, $M$ is two-dimensional. They explain that, to understand the sheaves, it suffices to understand them on each of the regions of the arrangement (i.e., within each region, the stalks are canonically isomorphic). Further, in the case of a hyperplane arrangement in $\mathbb R^2$ with a single pair of crossing lines, the stalks of the four regions (cyclically, $S$, $E$, $N$, $W$) should be related by the fact that there is a commuting square:
$$ \begin{array}{ccc} S & \rightarrow & E\\
\downarrow &&\downarrow\\
W & \rightarrow & N\end{array}$$
and the total complex $S \rightarrow E \oplus W \rightarrow N$ should be acyclic.
Which region is which among the four has to do with the choices of the signs of the conormals.
|
2025-03-21T14:48:31.961847
| 2020-09-05T03:26:30 |
370905
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Lucia",
"Sylvain JULIEN",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/38624",
"https://mathoverflow.net/users/50073",
"user21820"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632721",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370905"
}
|
Stack Exchange
|
$π(x+y) - π(x) ≤ c·y/\ln(y)$ for some constant $c$?
(I posted this question on Math SE but it has had no answer for a year now so I would like to ask if anyone here can provide one.)
Thinking about the prime number theorem, I wondered whether it is known that there is some constant $c$ such that $π(x+y) ≤ π(x) + c·y/\ln(y)$ for every integers $x,y > 1$. I read that experts believe $π(x+y) ≤ π(x) + π(y)$ fails for some $y$, since it fails for $y = 3159$ if the k-tuple conjecture holds, but it is just barely false, so I am curious if it is known to be true if the inequality is relaxed by a constant factor. If so, is it also known that $π(x+y) ≤ π(x) + π(y) + c·\!\sqrt{y}·\ln(y)$ for some constant $c$? I simply do not know how to search for such conjectures, and neither Wikipedia nor Wolfram seem to state any results that would affirm or refute these two conjectures easily, so any references would be appreciated!
Montgomery and Vaughan (The large sieve, Mathematika, 1973) showed that $\pi(x+y)-\pi(x) \le 2y/\log y$.
@Lucia: Thank you! I see it is (1.12) in there. Do you want to post that as an answer? And do you know about the second question? The "$c·\sqrt{y}·\ln(y)$" comes from the idea that, if the conjecture that $π(x+y) ≤ π(x)+π(y)$ is barely false but the error is bounded by $O(y/\ln(y))$, the next tighter plausible natural bound is $O(\text{Li}(y)-π(y))$.
As mentioned in my comment, Montgomery and Vaughan (The Large Sieve, Mathematika 20 (1973) 119–132, doi:10.1112/S0025579300004708) showed an explicit version of the Brun--Titchmarsh inequality:
$$
\pi(x+y) - \pi(x) \le \frac{2y}{\log y}.
$$
The other question asked is likely false. Hensley and Richards (Primes in intervals, Acta Arithmetica 25 (1974) 375-391, EuDML) showed that the Hardy-Littlewood $k$-tuples conjecture contradicts the hypothesis that $\pi(x+y) -\pi(x) \le \pi(y)$. In fact their paper establishes that (on the $k$-tuples conjecture) one has (for large fixed $y$, and infinitely many $x$)
$$
\pi(x+y) -\pi(x) \ge \pi(y) + (\log 2 -\epsilon) \frac{y}{(\log y)^2}.
$$
Roughly the lower bound that they are obtaining is $2 \pi(y/2)$ (and some conjectural improvements over this are also discussed).
Thank you for your answer! That is amazing; it means that on one hand it's known that $π(x+y)-π(x) ≤ 2·π(y)$ and that it's believed that for infinitely many $x,y$ we have $π(x+y)-π(x) ≥ 2·π(y/2)$.
@Lucia: does this inequality also hold true when $y$ is a function of $x$ like, say, $y=O(\log^{C}x)$ for some absolute positive constant $C$?
|
2025-03-21T14:48:31.962066
| 2020-09-05T05:17:14 |
370909
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Daniel Li",
"Fedor Petrov",
"Giorgio Metafune",
"YCor",
"https://mathoverflow.net/users/116621",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/150653",
"https://mathoverflow.net/users/4312"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632722",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370909"
}
|
Stack Exchange
|
An "elementary" inequality
The following is left unproven in a monograph. The author refers to it as "elementary exercise" but I am unable to prove it. Any insight is appreciated.
$$
\int f \log f d\mu \le 2 \left[\int|f-1|^p d\mu\right]^{1/p}+\frac{2}{p-1}\int |f-1|^p d\mu,\quad p>1
$$
where $f$ is a probability density.
You really mean $f$ a probablity density? not $\mu$? Also please quote the monograph.
@YCor I rather expect that $\mu$ is probability measure. Because the inequality is not homogeneous with respect to $\mu$.
This follows from $(1+x) \log (1+x)\le 2x +2/(p-1)x^p$ for $x \ge 0$ writing $0 \le f=1+g$ and integrating only where $g \ge 0$.
$\mu$ is indeed a probability measure in the context of the monograph. f is probability density with respect to $\mu$. But seems that it is true in general by Metafune's proof? The inequality is from (13.6) in "Dynamic to random matrix" by Erdos and Yau.
I used that $\mu$ is a probability measure to have $|g|_1 \le |g|_p$ and $f \ge 0$.
|
2025-03-21T14:48:31.962174
| 2020-09-05T06:33:57 |
370911
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Richard Lärkäng",
"https://mathoverflow.net/users/124749",
"https://mathoverflow.net/users/49151",
"jack lion"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632723",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370911"
}
|
Stack Exchange
|
Lebesgue dominated convergence for currents
I am learning current theory presently. In Chap 3.3-Definition of Monge-Ampère Operators, J.-P. Demailly, Complex analytic and differential geometry, I am a little confused as follows.
Let $X$ be a n-dimensional complex manifold. Let $u$ be a locally bounded plurisubharmonic function on $X$ and let $T$ be a closed positive current of bidimension $(p, p),$ i.e. of bidegree $(n-p, n-p) .$ Demailly defined the wedge
$$
d d^{c} u \wedge T=d d^{c}(u T),
$$
where $d d^{c}(\quad)$ is taken in the sense of distribution (or current) theory. Then he proved the following proposition.
Propostion The wedge product $d d^{c} u \wedge T$ is again a closed positive current.
Proof. The result is local. In an open set $\Omega \subset \mathbb{C}^{n},$ we can use convolution with a family of regularizing kernels to find a decreasing sequence of smooth plurisubharmonic functions $u_{k}=u \star \rho_{1 / k}$ converging pointwise to $u .$ Then $u \leqslant u_{k} \leqslant u_{1}$ and Lebesgue's dominated convergence theorem shows that $u_{k} T$ converges weakly to $u T ;$ thus $d d^{c}\left(u_{k} T\right)$.........
My question: In the above argument, Lebesgue's dominated convergence theorem is used. As we all know, Lebesgue's dominated convergence theorem reveals something about the integral. However, $T$ may be not the current induce by an $L_{\operatorname{loc}}^{1}$ function. So How can we apply Lebesgue's dominated convergence theorem to $u_{k} T$? Anyone know something similar with Lebesgue's dominated convergence theorem in the general current theory?
A positive current is of order 0
@RichardL Why can we apply the Lebesgue's dominated convergence theorem to the current of order 0? Does the current of order 0 act on test forms by integration? Where can I find such material? Thank you.
Indeed, a current or distribution of order 0 acts by integration against a (signed) measure (and when the current is positive, the measure is positive). These questions don't really seem on topic for this site ("mathematical questions related to current research in mathematics"), but should be covered if you study distribution theory and measure theory properly (positive distributions, distributions of order 0, appropriate versions of the Riesz representation theorem).
|
2025-03-21T14:48:31.962380
| 2020-09-05T07:33:48 |
370914
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Peg Leg Jonathan",
"Ulrich Pennig",
"https://mathoverflow.net/users/137242",
"https://mathoverflow.net/users/3995"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632724",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370914"
}
|
Stack Exchange
|
Representation of $C^{*} (S_{\infty})$
I was wondering what is the group $C^{*}$-algebra of infinite symmetric group?
Mainly, I was trying to calculate the k-theory of $C^{*}$-algebra of infinite symmetric group and I found K-Theory of $C^{*}(X)$ .
But the Vershik's paper is kind of difficult to understand. Is there a simpler way or an alternative way?
Can anybody help me?
Anyone can help me?
I think the best starting point would be to understand the representation theory of the symmetric groups and what this has to do with symmetric functions.
Any particular reference?
"Symmetric Functions and Hall Polynomials" by Ian Grant Macdonald is a good reference for symmetric functions and has a chapter on characters of the symmetric groups.
|
2025-03-21T14:48:31.962591
| 2020-09-05T09:07:59 |
370917
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ennio Mori cone",
"Piotr Achinger",
"https://mathoverflow.net/users/122729",
"https://mathoverflow.net/users/3847"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632725",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370917"
}
|
Stack Exchange
|
In char zero $ \operatorname{Cox}(\operatorname{Bl}_{[1:1:1]}(\mathbb{P}(a:b:c))) $ is finitely generated, but not in char p. How?
Let $ X(a_{1}:a_{2}:a_{3}) $ be the blow-up of $ \mathbb{P}(a_{1}:a_{2}:a_{3}) $ at $ [1:1:1] $, the identity of the torus. In Steven Dale Cutkosky's paper Symbolic Algebras of Monomial Primes Cutkosky gives an example of what appears to be weights $ a_{i}, i \in \{1,2,3\} $ such that $ \operatorname{Cox}(X(a_{1}:a_{2}:a_{3})) $ is not finitely generated in characteristic zero, but is in positive characteristic. This does not make sense to me for the following reason.
Let $ R $ be a ring of mixed characteristic, and let us denote the affine coordinate ring of $ \mathbb{G}_{m}(R) $ by $ R[t]_{t} $. If $ \mathbb{A}^{3}_{R} $ has affine coordinate ring $ R[x_{1},x_{2},x_{3}] $, then the ring homomorphism $ \beta^{\sharp}: R[x_{1},x_{2},x_{3}] \to R[x_{1},x_{2},x_{3}][t] $ which sends $ x_{i} $ to $ x_{i} t^{a_{i}} $ is the co-action for an action $ \beta $ of $ \mathbb{G}_{m}(R) $ on $ \mathbb{A}^{3}_{R} $. The quotient $ (\mathbb{A}^{3}_{R} \setminus (0,0,0))//\mathbb{G}_{m}(R) $ is a scheme $ \mathbb{P}(a_{1}:a_{2}:a_{3})_{R} $ whose fibres are all isomorphic to $ \mathbb{P}(a_{1}:a_{2}:a_{3}) $ over the respective fields. The Cox ring of $ \mathbb{P}(a_{1}:a_{2}:a_{3})_{R} $ is $ R[x_{1},x_{2},x_{3}] $, and is therefore flat over $ R $.
For any $ d \in \mathbb{Z} $ and invertible sheaf $ \mathcal{O}_{\mathbb{P}(a_{1}:a_{2}:a_{3})}(d) $ the number of global sections is purely determined by the weights $ a_{1},a_{2} $ and $ a_{3} $. Since the Picard group of $ \mathbb{P}(a_{1}:a_{2}:a_{3})_{R} $ is isomorphic to $ \mathbb{Z} $, any invertible sheaf $ \mathcal{O}_{\mathbb{P}(a_{1}:a_{2}:a_{3})_{R}}(d) $ is flat over $ R $ for $ d \in \mathbb{Z} $. If $ \mathcal{I} $ is the ideal sheaf of the $ R $-valued point $ [1:1:1] $, then since $ R $ is flat over $ R $, and $ R[x_{1},x_{2},x_{3}] $ is flat over $ R $, the Hilbert polynomial of the ideal sheaf $ \mathcal{I} $ is the same for all points of $ \operatorname{Spec}(R) $. Therefore $ \mathcal{I} $ is flat over $ R $.
Let $ X(a_{1}:a_{2}:a_{3})_{R} $ be the blow-up of $ \mathbb{P}(a_{1}:a_{2}:a_{3}) $ at the $ R $-valued point $ [1:1:1] $. If $ \pi: X(a_{1}:a_{2}:a_{3})_{R} \to \mathbb{P}(a_{1}:a_{2}:a_{3}) $ is the natural projection map, then the effective divisors of the Picard group of $ X(a_{1}:a_{2}:a_{3})_{R} $ are contained within the semi-group generated by divisors of the form $ \pi^{\ast}(\mathcal{O}_{\mathbb{P}(a_{1}:a_{2}:a_{3})_{R}}(d)) \otimes \pi^{\ast}(\mathcal{I}^{m}) $ for $ (d,m) \in \mathbb{N}_{0}^{2} $. Since these sheaves are flat over $ R $, this would mean that the Cox ring of $ X(a_{1}:a_{2}:a_{3})_{R} $ would be flat over $ R $.
However, if the Cox ring of $ X(a_{1}:a_{2}:a_{3})_{R} $ is flat over $ R $, then it is finitely generated over its generic fibre if and only if it is finitely generated over its special fibre. This seems to contradict Cutkosky's result in section 3 of the aforementioned paper. What is wrong with this argument, or is there something I missed regarding Cutkosky's argument?
I don't know if this resolves the issue, but consider the divided power algebra over $\mathbb{Z}$, i.e. the subring $R$ of $\mathbb{Q}[x]$ generated by $x^n / n!$ for all $n\geq 1$. Then $R \otimes \mathbb{Q} = \mathbb{Q}[x]$ is finitely generated, but the $R\otimes \mathbb{F}_p$ are not. However, $R$ is a free $\mathbb{Z}$-module, in fact it is graded with each grading isomorphic to $\mathbb{Z}$. So the "if and only if" claim in your last paragraph is incorrect.
...The point being that the multiplication maps $R_{n_1}\otimes \ldots \otimes R_{n_k} \to R_{n_1 + \cdots n_k}$ do not have constant rank. Anyway you really want the other implication.
Just because the sheaves are flat over $R$ does not mean that the Cox ring of the generic fiber is finitely generated if this is true over the special fiber. I guess the cohomology can jump over the special fiber, and the new sections make the ring finitely generated.
|
2025-03-21T14:48:31.962858
| 2020-09-05T09:48:38 |
370918
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632726",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370918"
}
|
Stack Exchange
|
Lower semi-continuity of length-dependent functional
Let $f:\mathbb{R}\rightarrow [0,\infty]$ be a lower semi-continuous function and define the functional
$$
\begin{aligned}
F_f:&\ell^1 \rightarrow [0,\infty]\\
(x_n)_{n=0}^{\infty} &\to \sum_{n=0}^{N((x_n)_{n=0}^{\infty})} f(x_n),
\end{aligned}
$$
where $N\left((x_n)_{n=0}^{\infty}\right)=\inf\left\{
N_0\in \mathbb{N}:\, \forall n \geq N_0,\, x_{n} =0
\right\}$. Then is $F_f$ lower semi-continuous?
Let $A_n := \{x \in \ell^1 \colon x_1 \not= 0, \ldots, x_{n-1} \not= 0\}$, $g_n \colon \ell^1 \to [0,\infty]$ be defined by $g_n(x) := f(x_n)$ if $x \in A_n$ and $g_n(x) := 0$ if $x \not\in A_n$. Then $F_f(x) = \sum_{n=1}^\infty g_n(x)$. Since the sum of two l.s.c. functions and the supremum of a sequence of (nonnegative) l.s.c. functions is l.s.c. it is sufficient to show that each $g_n$ is l.s.c. But this follows from 1. $A_n$ being open in $\ell^1$, thus $g_n$ is l.s.c. at each $x \in A_n$, since $f$ is l.s.c., and 2. that $g_n \geq 0$ and $g_n(x) = 0$ if $x \not\in A_n$. Thus $F_f$ is l.s.c.
|
2025-03-21T14:48:31.962949
| 2020-09-05T10:58:05 |
370922
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"Michael Renardy",
"https://mathoverflow.net/users/12120",
"https://mathoverflow.net/users/36886"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632727",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370922"
}
|
Stack Exchange
|
Various definitions of coercivity
In this post one says that a functional $F:H\rightarrow [0,\infty]$ on an infinite-dimensional Hilbert space $H$ is (strongly) coercive if there exists a constant $k>0$ such that
$$
F(x)\geq k\|x\|_H^2.
$$
However, the definition I'm familiar with (set on an arbitrary topological space $X$) requires that $F:X\rightarrow [0,\infty]$ satisfy the following: For every $t>0$ the set $\{x \in X:\, F(x)\leq r\}$ has compact-closure.
So when applying this definition to the case where $X=H$, shouldn't $F$ only be coercive in the weak topology?
On an arbitrary topological space, you do not have a notion of a "bounded" set. I think that is why sets of compact closure were used instead.
That makes sense but are the two notions at-least equivalent when $X=H$?
No, of course not.
|
2025-03-21T14:48:31.963043
| 2020-09-05T11:50:56 |
370927
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Joseph O'Rourke",
"https://mathoverflow.net/users/115841",
"https://mathoverflow.net/users/6094",
"user929304"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632728",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370927"
}
|
Stack Exchange
|
Can prolates overlap more easily than oblates?
Context:
When modeling anisotropic particles, the two common types of shapes of interest are cylindrical and disk-like particles. For simplicity let us say we model these as prolates and oblates respectively (see below).
With increasing aspect ratio, the oblate becomes more and more disk-like, while the prolate becomes needle-like (either way, as anisotropic spheroids, both have a positional and orientational degree of freedom).
When one is interested in the connectivity properties of configurations of such particles, where the connectivity is justified on physical grounds, given the core of particles is hard (they cannot overlap or go through one another), connectivity criteria are defined as follows:
For any given particle geometry, the particle is coated by a contact shell of the same shape that covers the hard core. This shell may be very thin, but it allows us to establish a notion of connectivity, as in, when two adjacent particles have overlapping contact shells, we assume them to be connected.
Questions:
I am trying to find out, whether purely on geometric grounds, that is by solely considering the difference in shape of an oblate to a prolate, it is possible to tell which type provides a more efficient connectivity. In other words, is it easier (more probable) for two prolates to become connected or for two oblates? Instead of "probability", one can imagine it in terms of "the number of ways two such given particles can become connected".
If the question is vague, let us take an extreme scenario: suppose we have very high aspect ratios, that is, near slender needles (for prolates), and near disks (for oblates), in a box we throw two prolates, and once two oblates, can we tell which case is more likely to result in a configuration having overlaps?
I guess the question is partly also related to the packing problem of anisotropic particles, but I don't know if knowledge of how particles pack, directly tells us something about how they can connect. I understand if there's no clear-cut statement or answer that can be made here, but if you have an intuitive take on the matter, based purely on the geometry of the two particles at hand, it would be most definitely welcome as answer here. Any reference to the literature where such/similar comparison has been made would also be helpful.
Somewhat related, an earlier MO post: Packing of anisotropic objects.
@MattF. Good question: we can safely either assume them having the same volume, or the same aspect ratio! The latter is more common, but any argument requiring the former will also be definitely useful.
This is not an answer; more an excuse to post the intriguing image below.
This Phys Rev article is quite recent, and perhaps its 46 references will help
focus your question.
Jin, Weiwei, Ho-Kei Chan, and Zheng Zhong. "Shape-Anisotropy-Induced Ordered Packings in Cylindrical Confinement." Physical Review Letters 124, no. 24 (2020): 248002.
Journal link.
"Packing prolate and oblate spheroids into cylinders creates new types of packings not seen for perfect spheres."
|
2025-03-21T14:48:31.963301
| 2020-09-05T13:14:36 |
370932
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Piotr Achinger",
"https://mathoverflow.net/users/3847"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632729",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370932"
}
|
Stack Exchange
|
A silly doubt on Log structures
Let $X=\operatorname{Spec} A$ be an affine variety. Consider the log structure given by $\mathbb N\rightarrow A$ which sends $1\mapsto 0$. Also consider the log structure $\mathbb N^r \rightarrow A$ given by $e_i\rightarrow 0$ for all $1\leq i\leq r$ and $e_i$ denotes the basis of $\mathbb N^r$. Are these two log structures non-isomorphic?
My doubt is that these two log structure are basically not different.
Consider $A = k$ a field. The log structure associated to the map $P\to k$ for a sharp monoid $P$ sending $0\mapsto 1$ and everything else to $0$ is $\mathcal{M} = P\times k^* \to k$ mapping $(0, a) \mapsto a$ and everything else to $0$. It follows that $\mathcal{M}/\mathcal{O}^\times \simeq P$. In particular, different $P$'s give rise to non-isomorphic log schemes.
|
2025-03-21T14:48:31.963385
| 2020-09-05T13:31:16 |
370933
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Archie",
"Nate Eldredge",
"Stanley Yao Xiao",
"Will Sawin",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/50912"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632730",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370933"
}
|
Stack Exchange
|
Are there topics of contemporary mathematics that simply require too much knowledge to be given as 3-year PhD problems?
Just out of curiosity, is there a topic where even an excellent graduate student wouldn't be able to master all the necessary background within 3 years (I say 3 because in many European countries a PhD is meant to last three years beyond the 5-year masters, with perhaps a fourth year to finish writing-up) ?
For example, I have heard that mastering EGA or Lurie's books takes several years, but does that mean some topics are simply out of reach in those European countries, and require an american-style PhD that lasts 5 years or more ?
In every research area there are problems that don't require mastering the full theory to make progress on, and problems that do. This concern exists for the second but not the first.
The responses in this question may help: https://mathoverflow.net/questions/335347/when-is-one-ready-to-make-original-contributions-to-mathematics
@Will Sawin : thank you. But are there areas where the fraction of problems that don't require mastering the full theory is very small ?
I don't know about every area. Very few problems in algebraic geometry require mastering EGA. I don't know about higher category theory.
@Stanley Yao Xiao : thank you, yes those responses are helpful.
@Will Sawin : thanks for comment on EGA in algebraic geometry, exactly the sort of expert's opinion I was looking for.
An American-style PhD doesn't really help with this issue, because students in such programs typically come in with only a bachelor's degree and spend the first two years on basic coursework, similar to what a European student would do during their masters. So the amount of time spent on research is about the same for both.
@NateEldredge: thanks, I didn't know that. Does this mean european students that pursue a PhD in the US are expected to finish in 3 years ? Or could they game the system ?
@Archie: There is an awful lot of info about this over at http://academia.stackexchange.com, which would also be a better place to ask further questions about it. The short answer is that American PhD programs usually have special provisions for students who come in already having a masters - they may waive some or all of the coursework requirements, or let them take an exam instead, or something like that. So such students may well be able to finish in 3-4 years, but are not necessarily required to. Every institution makes its own rules.
@NateEldredge: I see, thank you!
|
2025-03-21T14:48:31.963591
| 2020-09-05T13:52:52 |
370936
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mark Wildon",
"Richard Stanley",
"https://mathoverflow.net/users/112113",
"https://mathoverflow.net/users/164847",
"https://mathoverflow.net/users/2807",
"https://mathoverflow.net/users/7709",
"lambda",
"noone "
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632731",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370936"
}
|
Stack Exchange
|
Parameter of Brauer algebra
Let $O(V)$= the set of orthogonal transformation from the vector space $V$ to $V$ where $\dim V=n$. We know that the centralizer algebra of $O(V)$ on the tensor space $V^{\otimes{f}}$ is Brauer algebra $B_{f}(\delta) $ where $\delta$ is a parameter. Could you please explain the connection between $\delta $ and $\dim V=n $
They are equal.
Wales conjecture says for integer values of $\delta $ $B_f (\delta ) $ is not semi-simple. Then Centralizer of $O(V)$ is not semi-simple ?
In Wale's conjecture that $B_f(\delta)$ is semisimple unless $\delta\in\mathbb{Z}$, the dimension $n$ is understood and $\delta$ is any real number. Wale's conjecture was proved by Wenzl at https://sites.math.washington.edu/~billey/classes/Hopf.algebra/bulletins/wenzl.1988.pdf.
But is not $\delta =n$ ?
I’m voting to close this question because it's answered in the first paragraph of the Wikipedia page on Brauer algebras.
To clarify further: normally, given $f$ one takes $n$ sufficiently large so that we get the full Brauer algebra and not a proper quotient of it. The notation $B_f(\delta)$ assumes that $n$ is sufficiently large, viz., $n\geq 2f$. Then $B_f(\delta)$ is independent of $n$ and depends on the parameter $\delta$. The discriminant $\Delta_f(\delta)$ of $B_f(\delta)$ is a polynomial in $\delta$. $B_f(\delta)$ is semisimple if and only if $\Delta_f(\delta)\neq 0$. If $B_f(\delta)$ and $B_f(\delta')$ are both semisimple, then they are isomorphic.
|
2025-03-21T14:48:31.964076
| 2020-09-05T14:19:33 |
370938
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Pita",
"https://mathoverflow.net/users/163656"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632732",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370938"
}
|
Stack Exchange
|
Representing relative homology classes orientable surfaces with boundary
Let $S$ be compact oriented surface without boundary. Then it is a classical result that a primitive class $\gamma \in H_1(S; \mathbb{Z})$ is always represented by a simple closed curve. It implies that any class $\beta \in H_1(S; \mathbb{Z})$ is represented by a disjoint union of simple closed curves (take $\beta = k \gamma$ with $\gamma$ primitive and consider $k$ parallel simple closed curves representing $\gamma$).
Let now $\Sigma$ be a compact oriented surface with non-empty boundary:
Is it true that I can always represent any element $\gamma \in H_1(\Sigma, \partial \Sigma; \mathbb{Z})$ by a disjoint union of simple closed curves and properly embedded arcs?
Yes, this can be done. You can do this directly for surfaces but it's as much a "codimension one" as a "dimension one" phenomenon and so useful to see the general argument.
For any compact oriented n-manifold with boundary, duality says that $H_{n-1}(M,\partial M) \cong H^1(M)$. From simple obstruction theory, $H^1(M)$ is identified with homotopy classes of maps to a circle. Assuming your manifold was smooth, take a smooth map $f$ representing your cohomology class and a regular value $p\in S^1$. (I guess I mean regular value for both $f$ and the restriction of $f$ to the boundary.) Then $f^{-1}(p)$ is a codimension one submanifold in the given homology class.
For surfaces, the submanifold is a union of properly embedded arcs and simple closed curves. If you want to do so, you can assume that every component is an arc.
Very nice! Thank you.
|
2025-03-21T14:48:31.964215
| 2020-09-05T14:37:22 |
370940
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"KConrad",
"LSpice",
"YCor",
"abx",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/3272",
"https://mathoverflow.net/users/40297"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632733",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370940"
}
|
Stack Exchange
|
The stabiliser group of an isotropic quadratic form over $\mathbb{Q}_p$ is non-compact?
Let $\mathbb{Q}_p$ denote the $p$-adic integers. Let $V$ be a $\mathbb{Q}_p$-vector space and $Q : V \rightarrow \mathbb{Q}_p$ be a non-degenerate integral quadratic form. We say that the pair $(Q,V)$ is $\textbf{isotropic}$ if there exists $v \in V \setminus \{0\}$ such that $Q(v) = 0$.
Let $$SO_Q(V) := \{ \sigma \in GL(V) : Q(\sigma x) = x \ \text{and} \ \det(\sigma) =1\}$$
be the special stabiliser group of the quadratic form $Q$. I would like to prove the following Lemma.
$\textbf{Lemma}$: Let $(Q,V)$ be isotropic. Then, $SO_Q(V)$ is not compact.
I know how to prove this for an isotropic quadratic form on a $\textbf{real}$ vector space. Anyone has a hint on how to prove it for $p$-adic vector spaces?
Choose $w\in V$ such that $Q(v,w)=1$, and put $u=w-\frac{1}{2}Q(w) v$. Then $u$ and $v$ span a hyperbolic plane; its automorphism group contains $\mathbb{Q}_p^*$.
It works for an arbitrary nondiscrete normed field: if $q$ is isotropic then $SO(V)$ is unbounded.
@YCor, as you know, though it is still true, one must take some care (such as not dividing by $2$, and, more to the point, thinking of quadratic forms instead of the (now-symplectic!) bilinear forms) if the field has characteristic $2$.
Since people reading this page may have a similar question about the anisotropic case, in which case $O_Q(V)$ is compact, let me give a link to an MO page discussing that: https://mathoverflow.net/questions/90117/orthogonal-group-over-local-field.
At @KConrad's request, a link to the opposite case Orthogonal group over local field.
This can be proved the same way in the $p$-adic case as in the real case!
Assume that $Q \colon V \to \mathbb Q_p$ is isotropic. It is a basic fact that $V$ contains a hyperbolic plane $H$. (For a proof we can use the comment above by abx. Let $Q(v) = 0$ where $v \not= 0$. Writing $B$ for the bilinear form associated to $Q$, there's $w$ such that $B(v,w) \ne 0$ by nondegeneracy, and we can scale $w$ to make $B(v,w) = 1$. Then $u := w - \frac{1}{2}Q(w)v$ satisfies $Q(u) = 0$ and $B(u,v) = 1$, so $Q(xu + yv) = 2xyB(u,v) = 2xy$, so the $\mathbb Q_p$-span of $u$ and $v$ is a hyperbolic plane in $V$.)
Since $Q$ is nondegenerate on hyperbolic planes in $V$, $V = H \oplus W$ where $W = H^\perp$ is the orthogonal complement to $H$. For each $c \in \mathbb Q_p^\times$, the mapping $g_c \colon H \to H$ where $g_c(x,y) = (cx,(1/c)y)$ in the basis $\{u,v\}$ is in the special orthogonal group of $Q\rvert_H$. Therefore $g_c \oplus \operatorname{id}_W \in {\rm SO}(Q)$. Since the coordinates of $g_c$ as $c$ varies over $\mathbb Q_p^\times$ are unbounded, $\operatorname{SO}(Q)$ is not compact.
It works over an arbitrary nondiscrete normed field $K$ and without assuming non-degeneracy (excluding the case when the space is 1-dimensional and the form zero).
If $q$ is an isotropic quadratic form on a finite-dimensional vector space over $K$, then $\mathrm{SO}(q)$ is unbounded (and hence noncompact), except if the dimension is $1$ and the form $0$.
The easy argument was already given in characteristic $\neq 2$:
a) if the form is degenerate, choose $x$ in the kernel. Then since one excludes the 1-dimensional case, there is a linear map $f$ whose image is $Kx$ and vanishing on $Kx$. Then $\mathrm{Id}+\lambda f$ is in the special orthogonal group for every $\lambda\in K$, which is therefore not compact.
b) if the form is nondegenerate, choose $x\neq 0$ with $B(x,x)=0$. Choose $y_0$ with $B(x,y_0)\neq 0$. Then find $y$ of the form $y_0+\lambda x$ with $B(y,y)=0$. So $B(x,x)=B(y,y)=0$ and $B(x,y)\neq 0$. Since $(x,y)$ generates a nondegenerate plane its orthogonal is a complement. Then for every $t\in K^*$, the diagonal map $x\mapsto tx$, $y\mapsto t^{-1}y$, identity on the orthogonal, is in the special orthogonal group, which is therefore unbounded.
As mentioned by @LSpice one needs to reformulate the proof to allow characteristic 2.
Here we have a quadratic form $q$ (i.e. $q(\lambda x)=\lambda^2q(x)$ for all $x$ and $b_q(x,y)=q(x+y)-q(x)-q(y)$ is bilinear), whose kernel is by definition $\{x:\forall y:q(y+x)=q(y)\}$. Isotropic means that $q^{-1}(\{0\})$ is not reduced to zero.
(a) adapts in a straightforward way. For (b) one chooses $x\neq 0$ with $q(x)= 0$. Then one chooses $y_0$ such that $q(y_0+x)\neq q(y_0)$. Then $q(y_0+tx)$ is affine in $t$ and nonconstant, hence vanishes hence one can find $y$ such that $q(y)=0$ and such that $q(x+y)\neq 0$ (observe that on the plane $Kx+Ky$, $q$ vanishes exactly on $Kx\cup Ky$).
The orthogonal $H_x$ of $x$ (resp. $H_y$ of $y$) for $b_q$ is a hyperplane meeting $Kx+Ky$ in $Kx$ (resp. $Ky$), hence their intersection $H= H_x\cap H_y$ has codimension 2 and is a complement subspace to $Kx+Ky$. Then for $t\in K^*$, the endomorphism $x\mapsto tx$, $y\mapsto t^{-1}y$, identity on $H$, is in the special orthogonal group, which is therefore unbounded.
Note that the argument consists in finding a copy of the additive group in case (a) and of the multiplicative group in case (b).
It seems that in general:
there's no copy of the additive group exactly when either the form is anisotropic, or the form is nondegenerate in dimension $\le 2$, or the form is zero in dimension 1.
there's no copy of the multiplicative group exactly when the kernel has dimension $\le 1$ and the form is anisotropic modulo the kernel.
Doesn't your argument for the characteristic-$2$ case actually handle the general case?
@LSpice yes of course, but it's less easy to read (for most readers), so I started with the $2\neq 0$ case.
|
2025-03-21T14:48:31.964555
| 2020-09-05T15:06:20 |
370941
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Lucia",
"Stanley Yao Xiao",
"Will Sawin",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/38624"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632734",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370941"
}
|
Stack Exchange
|
Equidistribution of roots of quadratic congruences
Let $f(x) \in \mathbb{Z}[x]$ be an irreducible quadratic polynomial, and let $0 \leq \alpha \leq \beta \leq 1$ be real numbers. Put
$$\displaystyle S_f(\alpha, \beta; d) = |\{v \in \{1, \cdots, d\} : f(v) \equiv 0 \pmod{d}, \alpha d \leq v \leq \beta d \}|.$$
Then it is known that
(1)
$$\displaystyle \sum_{d \leq D} S_f(\alpha, \beta; d) \sim (\beta - \alpha) \sum_{d \leq D} S_f(0,1; d).$$
The proof of this follows from Weyl's criterion, and from the estimate
(2)
$$\displaystyle \sum_{d \leq D} \sum_{f(v) \equiv 0 \pmod{d}} \exp \left(\frac{2\pi i hv}{d}\right) \ll_h D^{2/3} (\log D)^2$$
for all $h \ne 0$.
However, in all of the papers I've seen this I have not seen an explicit estimate for the error term in (1). Can one extract a power-saving error term for (1) from (2)? If so, what is the dependence on the polynomial $f$?
If you look at Hooley's paper he gives a power saving bound in (2) with the dependence on $h$ being explicit. From this and Erdos-Turan you immediately get a power saving in (1). This doesn't keep track of the dependence on the polynomial (you seem to be asking two separate questions), but I don't think that's too hard either.
You are right, my answer was about a different question and irrelevant. It seems Lucia has you covered...
@Lucia which result of Erdos-Turan are you referring to? I don't think I've looked at any of their papers in the context of this problem...
Erdos--Turan bounds the discrepancy of a sequence in terms of the Weyl sums. (Pretty standard to do, but useful that it's written down. See e.g. the first chapter of Montgomery's 10 lectures... )
|
2025-03-21T14:48:31.964689
| 2020-09-05T16:35:25 |
370943
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Robert Bryant",
"abx",
"clvolkov",
"https://mathoverflow.net/users/13972",
"https://mathoverflow.net/users/164854",
"https://mathoverflow.net/users/40297"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632735",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370943"
}
|
Stack Exchange
|
Does there exists a compact Ricci-flat K3 surface whose metric tensor is expressed in explicit formula?
Yau's theorem showed the existence. But I had difficulty finding examples other than complex tori. Any information will be appreciated.
You might be interested in A new proof of the existence of Kähler-Einstein metrics on K3 by P. Topiwala, Invent. Math. 89 (1987), no. 2, 425-454. It is somewhat more "explicit" than Yau's proof, though not as much as one could wish.
As far as I know, there is no explicit formula known for a Ricci-flat metric on a K3 surface. There have been some numerical studies done, and there are 'asymptotic' formulae for metrics near degenerations as well as some information for such metrics on K3 surfaces that have an elliptic fibration. See the papers of Gross and Wilson and later follow-up work on analogs of the SYZ conjecture in complex dimension 2.
Thanks for the helpful comments. How about the curvature tensor? Does there, by any chance, exist examples where the curvature tensor is explicit?
|
2025-03-21T14:48:31.964796
| 2020-09-05T17:17:08 |
370946
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Bill Johnson",
"Dirk Werner",
"Mikael de la Salle",
"Yemon Choi",
"https://mathoverflow.net/users/10265",
"https://mathoverflow.net/users/127871",
"https://mathoverflow.net/users/2554",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632736",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370946"
}
|
Stack Exchange
|
History of the Lewis-Stegall theorem on factorization of representable operators
The following questions are about the history of a particular result in functional analysis, hence not "mathematical questions" per se; but I think they are relevant to the business of writing research papers in mathematics and to curating the research literature, and I think that they concern something sufficiently specialized and "recent" that MO is a more suitable place for the question than HSM.SE.
A bounded operator $T$ from $L_1[0,1]$ to a Banach space $X$ is said to be representable if it can be written as
$$
T(f) = \int_0^1 f(t) {\bf h}(t)\,dt
\quad\quad\hbox(f\in L_1)
$$
for some bounded and strongly measurable function ${\bf h}: [0,1]\to X$. The following result is known as the Lewis—Stegall theorem:
Theorem A. $T:L_1[0,1]\to X$ is representable if and only if there exist bounded operators $S:L_1[0,1]\to\ell_1$ and $R:\ell_1\to X$ such that $T=RS$.
The proof can be found in various books, with varying degrees of self-contained-ness. (Three such sources known to me are the Diestel–Uhl book on Vector Measures, the appendices of the Defant–Floret book on Tensor Norms and Operator Ideals, and Ryan's Introduction to Tensor Products of Operator Spaces) However, I am having difficulty tracking down the first place in the literature where this result was explicitly stated and attributed to Lewis and Stegall.
Diestel and Uhl give as a reference the 1973 paper of Lewis and Stegall (J. Functional Analysis) on Banach spaces whose duals are isomorphic to $l_1(\Gamma)$ but the paper itself never actually mentions representable operators. It does seem that one can extract a proof of Theorem A from results in their paper, provided one knows an alternative characterization of representable operators as those whose composition with the canonical map $L_\infty[0,1]\to L_1[0,1]$ is nuclear, but the earliest explicit reference I have found for that characterization is in a 1976 paper of Linde (Math. Nachricten) although it seems to have been known to various authors at the time.
The best I have managed so far is the appendix of Rosenthal's 1975 Bull. AMS article on The Banach spaces $C(K)$ and $L^p(\mu)$ where "Theorem A" occurs, albeit with slightly different terminology, as Theorem A3. In the proof Rosenthal does not explicitly credit the result to Lewis and Stegall, but he mentions that for a representable $T$
We are indebted to D. Lewis for the observation that $T$ may be realized as an "$l^1$-sum" of compact operators, each of which may be factored through $l^1$ by the "lifting property"
of $L^1(\lambda)$-spaces.
Indeed this decomposition as a block-diagonal "sum" of compact operators does occur in the original paper of Lewis and Stegall, but as mentioned above they were working with the "$L_\infty[0,1]\to L_1[0,1]\to X$ is nuclear" formulation rather than representability by an element of $L_\infty([0,1];X)$.
Question 1. What is the earliest record in the literature where Theorem A is stated as above and attributed to Lewis and Stegall?
(I am not trying to deny them credit since there seems to be consensus that they were fully aware that methods from their paper yielded this result, but I am a little frustrated when I have to hunt down papers in the literature only to find that they do not say what the secondary literature claims they say.)
Question 2. Given that the MO readership includes some long-standing members of the community of Banach space theory, does anyone have anecdotal/verbal information of how Theorem A — rather than the headline result of the JFA 1973 paper — came to be known as the Lewis–Stegall theorem?
Well, Lewis, Stegall, and Rosenthal are still around, so maybe one of them will clarify. I was at the conference in Baton Rouge organized by Ron Retherford at which Stegall spoke on the Lewis-Stegall result. IIRC, Rosenthal was finding Stegall's somewhat abstract presentation hard to follow. He was sitting by Lewis and started questioning him, and Lewis showed him the observation he mentioned in your quote from his 1975 Bulletin article. $$$$ I have no doubt that Lewis and Stegall knew everything attributed to them.
Thanks @BillJohnson - I was tempted to try and resurrect the ask-johnson tag just for this question :)
This appears to be one instance of what Pietsch refers to by saying ``Mathematicians invent their own history.''
Another book containing the proof of the Lewis-Stegall theorem is Pisier's Martingales in Banach spaces. I am afraid that, on that specific result, the Notes do not contain more information on the history than other sources. Page 73, Pisier writes "The Lewis-Stegall theorem comes from [the 1973 paper by Lewis-Stegall]", and refers to Diestel and Uhl "for a more detailed history of the RNP and more precise references".
|
2025-03-21T14:48:31.965257
| 2020-09-05T19:04:57 |
370950
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632737",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370950"
}
|
Stack Exchange
|
Non-pencil infinite projective plane with edges of different cardinalities
A projective plane is a hypergraph $H=(V,E)$ such that
if $e_1\neq e_2 \in E$ then $|e_1\cap e_2| = 1$, and
for $v,w\in V$ there is $e\in E$ such that $\{v,w\}\subseteq e$.
Is there a projective plane $H=(V,E)$ such that
$|e|>2$ for all $e\in E$, and there are $e_1, e_2 \in E$ with $|e_1|\neq |e_2|$?
Suppose $e_1, e_2$ are two distinct hyperedges, and $v$ is a vertex that is not a member of either of them. Define the map $f_v: e_1\to e_2$ by sending a vertex $w\in e_1$ to the unique vertex $w'\in e_2$ which lies in the same hyperedge as $\{v,w\}$. The defining properties of projective planes imply that $f_v$ is a bijection, therefore $|e_1|=|e_2|$.
|
2025-03-21T14:48:31.965336
| 2020-09-05T19:36:47 |
370953
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632738",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370953"
}
|
Stack Exchange
|
multiplicative structure on the monodromy weight filtration spectral sequence
Let $f: X \to \Delta \subset \mathbb{C}$ be a projective morphism of a complex manifold to a small disc, smooth away from 0, and such that $Y=f^{-1}(0)$ is a strictly normal crossing divisor, and let $\tilde{\Delta^*}$ be the universal cover of $\Delta^*=\Delta \setminus \{0\}$. J. H. M. Steenbrink has described a certain double complex $A^{*,*}$ such that its total complex is a cohomological mixed Hodge complex that computes the limit mixed Hodge structure of W. Schmid on the nearby fibre cohomology $H^*(X \times_\Delta \tilde{\Delta^*}, \mathbb Q) \cong H^*(X_t, \mathbb Q)$ (the isomorphism is not canonical). It has the form
$$
A^{p,q} = \mathbb{C}u_{p+1} \otimes \Omega^{p+q+1}_X(\mathrm{log}\,Y) / W_p\\ \ \ d'(u_p \otimes \omega) = u_{p+1} \otimes dt/t \wedge \omega\qquad d''(u_p \otimes \omega) = u_p \otimes d\omega
$$
where $u_{p+1}$ is a formal variable, $t$ is a coordinate function on $\Delta$, and $W$ is the filtration by the number of poles. Steenbrink shows that there exists a natural quasi-isomorphism (given by wedging with $dt/t$ with some sign)
$$
\Omega^*_{X/\Delta}(\mathrm{log}\, Y) \otimes \mathcal{O}_Y \to \mathrm{Tot}(A^{*,*})
$$
where $\Omega^*_{X/\Delta}(\mathrm{log}\, Y)= \Omega^*_X(\mathrm{log} Y)/\Omega^1_\Delta(\mathrm{log}\,0) \wedge \Omega^{*-1}_X(\mathrm{log}\,Y)$ carries a natural multiplication.
It follows from the definition of the monodromy weight filtration that cup product on $H^*(X_t,\mathbb{Q})$ is compatible with it.
I would like to understand the multiplicative structure on the spectral sequence associated to the weight filtration on $\mathrm{Tot}(A^{*,*})$, so the first step is to write down a multiplication on $\mathrm{Tot}(A^{*,*})$ that would be compatible with the multiplication on $\Omega^*_{X/\Delta}(\mathrm{log}\, Y) \otimes \mathcal{O}_Y$ with respect to the aforementioned quasi-isomorphism.
My question is whether such multiplication exists and if yes, how one can find an explicit expression for it.
Note the strange shift in indices in the definition of the complex $A^{*,*}$: a product of a $p$-form and a $q$-form should be a $p+q-1$-form.
Perhaps there exists some complex $B^*$ with multiplicative structure, a quasi-isomorphism $\mathrm{Tot}(A^{*,*}) \to B^*$ and a section $B^* \to \mathrm{Tot}(A^{*,*})$ (up to homotopy)?
|
2025-03-21T14:48:31.965488
| 2020-09-05T20:30:48 |
370955
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"YangMills",
"https://mathoverflow.net/users/13168"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632739",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370955"
}
|
Stack Exchange
|
About an explicit formula of the curvature tensor by holomorphic sectional curvatures
Let $(M, g)$ be a Riemannian manifold. Define the curvature tensor convention as follows.
$$ R(X, Y) Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z$$
$$ R(X,Y,Z,W) = g(R(X,Y)Z, W)$$
It is well-known that the curvature tensor $R$ is explicitly expressed by the sectional curvatures. This can be found in Jost's
Riemannian Geometry and Geometric Analysis.
Lemma. With $K(X,Y) = R(X, Y, Y, X)$,
\begin{align} R(X,Y,Z,W) = & + K(X+W, Y+Z) - K(X+W, Y) - K(X+W, Z) \\ &- K(X, Y+Z) - K(W,Y+Z) + K(X,Z) + K(W,Y) \\ & - K(Y+W, X+Z) + K(Y+W, X) + K(Y+W,Z) \\ & + K(Y,X+Z) + K(W,X+Z) - K(Y,Z) - K(W,X).\end{align}
My question is whether there is a similar explicit formula for the Riemannian curvature tensor and the holomorphic sectional curvature when we assume that the manifold is Kähler.
I think this MO question is related to my question. From this polarization formula, at least, we know that the curvature tensor is determined algebraically by holomorphic sectional curvatures. However, I am curious about whether there is a (relatively) simpler formula describing the algebraic relation between the curvature tensor and holomorphic sectional curvatures.
Thanks!
I don't understand what you want. You get an explicit formula for $R(X,Y,Z,W)$ purely in terms of the holomorphic sectional curvature $Q(X)$ by combining the lemma that you state and my answer to the linked question. What else do you want?
There is no reason to expect a simpler formula than that.
Following the suggestion by @YangMills, I used Mathematica to combine the two formulas to get the full expression. As expected, the resulting formula is quite complicated. However, there is one neat property, which is that the formula for $R(X,Y,Z,W)$ does not contain any terms of the form $H(X), H(Y), H(Z)$ or $H(W)$.
This is intuitive in hindsight, because $R(X,Y,Z,W)$ must be anti-symmetric with respect to the $(X,Y)$-pair as well as the $(Z,W)$-pair. A non-zero term of the form $H(X)$ would break this anti-symmetry. However, I didn't realize until after computing the full expression, so thought it was worth providing an answer.
The full formula for $R(X,Y,Z,W)$ is the following. Here, $X,Y,Z,W$ are all real vectors and $H(X)=R(X,JX,JX,X)$ is the holomorphic sectional curvature.
$$R(X,Y,Z,W) = \frac{1}{32}\left( H[W-X]+5 H[W+X]-3 H[W-J X]-3 H[W+J X]-H[W-Y]+H[W+X-Y]-5 H[W+Y]-H[W-X+Y]+3 H[W-J X+Y]+3 H[W+J X+Y]+3 H[W-J Y]-3 H[W+X-J Y]+3 H[W+J Y]-3 H[W+X+J Y]-H[W-X-Z]-H[X-Z]+H[W+X-Z]+H[W-Y-Z]+H[X-Y-Z]-H[W+X-Y-Z]+H[Y-Z]-H[W+Y-Z]-H[-X+Y-Z]+H[W-X+Y-Z]-5 H[X+Z]+5 H[Y+Z]+3 \left( H[X-J Z]-H[W+X-J Z]-H[Y-J Z]+H[W+Y-J Z]+H[X+J Z]-H[W+X+J Z]-H[Y+J Z]+H[W+Y+J Z]+H[W-J (X+Z)]+H[Y-J (X+Z)]-H[W+Y-J (X+Z)]+H[W+J (X+Z)]+H[Y+J (X+Z)]-H[W+Y+J (X+Z)]-H[W-J (Y+Z)]-H[X-J (Y+Z)]+H[W+X-J (Y+Z)]-H[W+J (Y+Z)]-H[X+J (Y+Z)]+H[W+X+J (Y+Z)]) \right) \right) $$
|
2025-03-21T14:48:31.965656
| 2020-09-05T21:20:04 |
370957
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"fedja",
"https://mathoverflow.net/users/1131"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632740",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370957"
}
|
Stack Exchange
|
My hypothesis about convergence of series of independent random variable I cannot prove/disprove
Let $Y_i$, $X_i$ be sequences of independent random variables. Assume both limits exist: $$\lim_{n \to \infty} \frac{\sum_{i=1}^{n} \operatorname{Var}X_i}{\sum_{i=1}^{n} \operatorname{Var}Y_i},\quad \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \mathbb{E}X_i}{\sum_{i=1}^{n} \mathbb{E} Y_i}$$.
Does it imply this limit exists : $$\lim_{n \to \infty} \frac{\sum_{i=1}^{n} X_i}{\sum_{i=1}^{n} Y_i}$$
Any ideas whether it's true or not? The limit converging almost surely that is. Maybe Kolmogorov's two series theorem could help?
Might be worth to note, the series in the numerator/denominator alone may not converge.
"Almost surely" - definitely not: Just take $X_i$ and $Y_i$ independent Gaussians with mean zero (shift the expectations a tiny bit to have the second ratio well-defined if $0/0$ bothers you) and the same variance. Then the ratios for very different $n$ are almost independent but the distribution is all the way the same (that of the ratio of 2 standard Gaussians) and not that of a constant.
|
2025-03-21T14:48:31.965767
| 2020-09-05T22:18:33 |
370959
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"Mellon",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/58651"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632741",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370959"
}
|
Stack Exchange
|
Determining a toric GIT quotient
Consider the following $G = (\mathbb{C}^*)^{\times 3}$-action on $\mathbb{A}^6$:
$(\lambda_0,\lambda_1,\lambda_2) \cdot (x_0,x_1,x_2,y_0,y_1,y_2) = (\lambda_0x_0,\lambda_1x_1,\lambda_2x_2,\frac{\lambda_0^2}{ \lambda_1\lambda_2}y_0,\frac{\lambda_1^2}{ \lambda_0\lambda_2}y_1,\frac{\lambda_2^2}{\lambda_0\lambda_1}y_2)$.
I need some help determining the GIT quotient of $\mathbb{A}^6$ by $G$.
The union of zero-sets $Z = V(x_1,y_1) \cup V(x_2,y_2) \cup V(x_3,y_3) \cup V(x_1,x_2,x_3)$ in $\mathbb{A}^6$ appears to be the locus of non-stable points.
How does the resulting geometric quotient of $\mathbb{A}^6 \backslash Z$ by $G$ look like as an algebraic variety? How does one generally go about in determining such quotients after identifying the stable locus?
What sort of answer would be acceptable for "what does […] look like as an algebraic variety"?
I get that my question may be somewhat ambiguous, but I am looking for a description of a variety which does not have (explicit or implicit) reference to the particular action. Preferably a description would be an explicit $G$-invariant map to affine or projective space whose image realizes the quotient. Or perhaps an identification with something more familiar, such as a toric variety (with its fan), as a blow-up of $\mathbb{P}^3$, a succinct equation, etc..
I should add: the $G$-invariant map should have a target of minimal dimension.
|
2025-03-21T14:48:31.965879
| 2020-09-05T22:52:27 |
370962
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632742",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370962"
}
|
Stack Exchange
|
Deformation quantization of infinite dimensional Poisson manifolds
In 1999, Karaali wrote a review of formal deformation quantization for a class she took with Weinstein.
She ends the paper with the following remark:
Another question that remains involves the infinite dimensional case: Kontsevich‘s results settle the problem in the finite dimensional case, but there are places in physics where we have to deal with
infinite dimensional Poisson manifolds, i.e infinite dimensional manifolds with a Poisson structure on them. This case involves new problems and perhaps may shed light on a better mathematical understanding of quantum field theory.
Since then, it seems that techniques constructed by Schlichenmaier settled this problem for infinite-dimensional symplectic manifolds, as claimed in nLab.
More recently, in a paper by Hawkins and Rejzner, a deformation quantization for infinite-dimensional affine manifolds has been constructed.
They also review the problem stated above, and mention that it is still open.
What is the current status of formal deformation quantization of infinite-dimensional Poisson manifolds?
|
2025-03-21T14:48:31.965988
| 2020-09-05T23:11:48 |
370963
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABC",
"Qiaochu Yuan",
"https://mathoverflow.net/users/129583",
"https://mathoverflow.net/users/290"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632743",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370963"
}
|
Stack Exchange
|
Any group is a quotient of an acyclic group?
As far as I know, for any group $G$ there exists an acyclic group $H$ such that $G$ is a subgroup of $H$.
I am wondering about the dual situation. Is any group $A$ a quotient of an acyclic group $B$ or more simply, given a group $A$ does it exist an acyclic group $B$ and a surjective homomorphism $B\rightarrow A$ ?
Acyclic groups must in particular have trivial abelianization, so all of their quotients must be perfect.
This is the only obstruction; A.J. Berrick shows in The acyclic group dichotomy (which I just found by googling!) that every perfect group is a quotient of an acyclic group of cohomological dimension $2$ (Proposition 2.3).
Could you indicate precisely which theorem? Thanks!
It's Proposition 2.3, although the proof is given before the statement.
|
2025-03-21T14:48:31.966071
| 2020-09-06T00:30:17 |
370966
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aryeh Kontorovich",
"bolzano",
"https://mathoverflow.net/users/12518",
"https://mathoverflow.net/users/158233"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632744",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370966"
}
|
Stack Exchange
|
Chernoff-type Bounds for Continuous-space Markov Chains
Let $X_1, X_2, \dots, X_n$ be $n$ samples from a discrete-time continuous-space Markov Chain.
Are there any good references who have provided a Chernoff-type bound regarding the behaviour of the ergodic mean $\bar X = \frac 1 n \sum_{i = 1}^n X_i$ for this type of chain?
There is a good line of work about finite-state chains (e.g. http://www.columbia.edu/~khl2114/files/CLLM.pdf, https://arxiv.org/abs/1907.04467) but I cannot find any work related to continuous-space (i.e. infinite-space) Markov Chains.
The contraction-based results continue to hold for continuous state spaces (though the Doeblin contraction condition is rather strong):
https://arxiv.org/abs/math/0610427
Thank you very much for your answer! My hunch was similar but I wanted to see it formalized somewhere.
|
2025-03-21T14:48:31.966189
| 2020-09-06T00:42:02 |
370967
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alain Valette",
"Ali Taghavi",
"Jochen Glueck",
"Nik Weaver",
"Ruy",
"YCor",
"Yemon Choi",
"https://mathoverflow.net/users/102946",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/14497",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/36688",
"https://mathoverflow.net/users/763",
"https://mathoverflow.net/users/97532"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632745",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370967"
}
|
Stack Exchange
|
Fredholm $C^*$-algebras
Let $H$ be a Hilbert space. A vector subspace $W\subset B(H)$ is called a Fredholm subspace if there is an upper bound for the absolute value of Fredholm index of all Fredholm operators $T$ in $W$.
Is there a classification of all $C^*$-algebras $A$ which admit an irreducible representation $\phi:A \to B(H)$ in some Hilbert space $H$ such that $\phi(A)$ is a Fredholm subspace of $B(H)$?
Is there a classification of all $C^*$-algebras $A$ which admit a faithful representation $\phi:A \to B(H)$ in some Hilbert space $H$ such that $\phi(A)$ is a Fredholm subspace of $B(H)$?
One can consider the terminology "Fredholm algebra" for any such $C^*$-algebras.
Edit: We add an example according to comment by Yemon Choi.
Put $H=\ell^2$ let $S$ be the shift operator on $\ell^2$ and $n$ be a fixed integer. Then this is a finite dimensional Fredholm subspace of $B(\ell^2)$:
$$\{P(S)\mid \text{P is a polynomial of degree at most n}\}.$$
Do you have an example of a Fredholm subspace?
@YemonChoi Trivial examples: Finite dimensions. Or in the case of infinite dimension, the scalar 1 dimensional space. or any subspace which does not contain any fredholm operator. As another example the space of ${P(s)\mid s\text{ is the shift operator, whose index is -1 and P is an arbitrary polynomial of degree at most n}$
${P(s)\mid s\text{ is the shift operator, whose index is -1}}$
and P is an arbitrary polynomial of degree at most n
How do you exclude the trivial 1-dim rep for the first question?
@YCor Good point! one may add "non trivial representation"
Oh, but actually this trivial 1-dimensional rep doesn't always exist... maybe my comment was stupid. I see no reason in particular to single out the trivial 1-dimensional rep among all finite-dimensional irreducible reps.
@YCor I do not know. by definition, if we consider the zero representation on $H=\mathbb{C}$ as an irreducible representation. any way why should be worry about it? what do you mean by "but actually this trivial 1 dimensional representation does not exist". I think the question is clear, do you agree?
Actually I especially amended my comment. So, for the first question, this class contains all C$*$-algebras that admit a nonzero finite-dimensional representation, i.e., admit a nonzero finite-dimensional quotient.
No, irreducible is clearly defined (=nonzero and no proper nonzero closed invariant subspace). "Trivial" is slightly more ambiguous since it can be the trivial irrep (1-dimensional), or any trivial representation (which can be in dimension 0, 1, or more).
@YCor Yes I see. Any way as you said every algebra with finite dimensional quotien is a fredholm algebra.
But your questions give two definitions of Fredholm C$*$-algebra. Which one of the two questions is supposed to characterize them?
@YCor apart from these cases. it would be interesting to find some non examples.
@YCor The first one. Please see the answer by Nik Weaver.
@YCor Thanks for your recent edit of my question. But there is a question in my mind since many years ago. Some times a user edit my post. When I look at his edit I realize that some words is edited without any change. for example in the second paragraph you changed "algebras"to again "algebras". What is the reason for this situation?
@YCor However I am not sure I am convinced by answer of Nik Weaver. I should rea his answer again.
@YCor Any way one can consider both properties with two different name. But it seems that the irreducible one is more interesting(and less obvious).
Any unital $C^*$-algebra $A$ has an irreducible representation $\phi$ such that every Fredholm operator in $\phi(A)$ has index 0.
To see it, let me first repeat something from Nik Weaver's previous answer: if $\pi$ is a representation of $A$ such that $\pi(A)$ intersects the compact operators trivially, then any Fredholm operator in $\pi(A)$ is actually invertible.
Now to prove my claim, observe that we may assume that $A$ is simple (up to replacing $A$ by $A/I$ where $I$ is a maximal, proper, closed two-sided ideal). Now let $\phi$ be any irreducible representation of $A$. If $\phi$ is finite-dimensional, then the result is clear and was already mentioned. If $\phi$ is infinite-dimensional then $\phi(A)$ intersects the compacts trivially because of simplicity, so Nik's observation above applies.
Where is the assertion located in Nik Weaver's answer (or thread of comments)? (anyway I think it's true)
The proof I can think of: (1) if $A\in B(H)$ is Fredholm and has nonzero index, then either $A^A$ or $A^A$ is Fredholm, self-adjoint $\ge 0$ and non-invertible. (2) in a $C^$-algebra, if $M$ is self-adjoint invertible $\ge 0$ then $M^{-1}\in C^(M)$. Indeed, approximate $x\mapsto x^{-1}$ by polynomials on the spectrum of $M$. (3) If $N$ is self-adjoint $\ge 0$, Fredholm and non-invertible then $N$ is invertible in restriction to the orthogonal of Ker$(N)$, say with inverse $N'$, extended to $0$ on Ker$(N)$, by (2) $N'\in C^(N)$. So $I-NN'\in C^(N)$ is nonzero of finite rank.
@YCor: denote by $q$ the quotient map to the Calkin algebra. If $x\in\pi(A)$ is Fredholm, then $q(x)$ is invertible in the Calkin algebra, hence also in $q(\pi(A))$. As $q$ is injective on $\pi(A)$, the element $x$ is also invertible in $\pi(A)$.
Ah OK; actually I wasn't aware of this fact that if an element $M$ is invertible in the big $C^*$-algebra then it's invertible in the smaller one, and could only check it ((2) above) under special assumptions.
@YCor My favorite reference is Proposition 1.3.10 in Dixmier's $C^*$-book.
Ah, this proposition boils down from general case to (non-negative) self-adjoint case, for which (B2 in Appendix B) it send to the book Rickart, General theory of Banach algebras.
@AlainValette Thank you very much for your answer. Does every $C^*$ algebra admits an irreducible representation $\pi:A \to B(H)$ such that $A/\ker \pi$ is simple? I think we need to this statement to apply your proof.
Moreover I think that you use the fact that for every liminal $C^*$ algebra every irreducible representation is necessarilly a finit dimensional representation. yes?
@AlainValette I just realize that the last part of my previous two comment was not necessary since your argument is giving a complete answer to my question. So your argument also gives a complete answer to the whole question of my previous two comment. Thank you again for your attention to my question and your answer.
There's a trivial answer to the second question: every C${}^*$-algebra has such a representation. Wlog assume $A \subseteq B(H_0)$ for some Hilbert space $H_0$, then represent $A$ on $H_0 \otimes l^2$ by tensoring everything with the identity on $l^2$. All the Fredholm operators in this representation have index $0$ (in fact they would have to be invertible).
could you please explain the reason that why a fredholm operator $T\otimes Id$ must be invertible?
Sure, if $T$ has nonzero kernel then $T\otimes I$ has infinite dimensional kernel and hence is not Fredholm. Same for cokernel, so if $T\otimes I$ is Fredholm it must have no kernel or cokernel (and have closed range), which means it must be invertible.
A fancier reason is that anything nonzero in $A\otimes I$ is noncompact, so factoring out the compacts embeds $A$ in the Calkin algebra $Q(H_0\otimes l^2)$. And Fredholm in $B(H_0\otimes l^2)$ is equivalent to being invertible in $Q(H_0\otimes l^2)$.
If $\phi$ is a representation such that $\phi(A)$ is a Fredholm space then every Fredholm operator there has index zero, because $ind(T^n) = n\ ind(T)$.
Is not possible that $T$ has a dense range but $T\otimes Id$ would be surjective? I think this situation is not considered in your answer, am I right?
I don't understand this question. By definition, the range of any Fredholm operator is closed. If $T$ is not surjective then neither is $T\otimes I$.
I think the closed range is a redundant condition. But why the surjectivity of $T\otimes I$ imply that $T$ is surjective?
For surjectivity are you using the fact that $H\otimes K$ is simillar to $hom(H,K^*)$? Then $T\otimes I$ is seen as "compositi9n with $T$? But in infinite dimension, are not some non algebraic (topological) obstractions?
If an operator has fonit dimensional kernel and finite codimensional range then its range is automatically closed. One can find this fact in many references. The remaining part of my previous comment was a suggestion for proof of your statement that surjectivity(or even fredholmness) of $T\otimes I$ implies that $T$ is surjective(fredholm). I guess that your argument is based on Identifocation of tensor product space with Hom space. Right? So my comment said are you considering purely algebraic structures or you consider topological one?
Any way i think that $T\otimes I$ or $I\otimes T$ (or both) are highly non Irreducoble( according to the first part of my question).
Consider the operator of multiplication by the sequence $(1/n)$, acting on $l^2$. What is its kernel? What is the codimension of the range? Is the range closed?
... so I'd be interested in these "many references" that say this cannot happen.
The sequence of sequences $1/n(log(n)^k$ perhaps shows your map has infinite codimension. But i convonce you close range is redundant in case of banach space
So multiplication by 1/n is not fredholm
@Ruy Very interesting point, thank you!
@NikWeaver: Since this post has just popped up on the front page again: It is indeed true that if the range of a bounded linear operator between two Banach spaces has finite co-dimension, then it is closed. A reference is, for instance, Murphy's "$C^*$-Algebras and Operator Theory", namely the discussion on the bottom of page 23, along with Theorem 1.4.7 on page 24. (More generally, if a Banach space $Y$ is the (algebraically) direct sum of two operator ranges, then both operator ranges are closed; this is a consequence of the open mapping theorem.)
(But of course, this is not really relevant for your answer; I just thought it might be a good idea to point this out anyway.)
@JochenGlueck oh, you are right. I'll delete that comment.
@JochenGlueck I would like to thank you and NiK Weaver for your attention to my comments.
@JochenGlueck yes This situation is interesting: despite of existence of non closed subspace of finite codimension, the range of every Fredholm operator is necessarilly a closed subspace.
|
2025-03-21T14:48:31.966832
| 2020-09-06T01:37:39 |
370969
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Joe Silverman",
"Lev Soukhanov",
"Wilberd van der Kallen",
"https://mathoverflow.net/users/11926",
"https://mathoverflow.net/users/33286",
"https://mathoverflow.net/users/4794"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632746",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370969"
}
|
Stack Exchange
|
System of linear equations in positive characteristic
Let $K$ be a field of positive characteristic $p$. Consider the system of $\mathbb F_p$-linear equations
$$\left\{\begin{array}{ccl}
a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n&=&b_1\\
a_{11}x^p_1+a_{12}x^p_2+\cdots+a_{1n}x^p_n&=&b_2\\
\vdots&\vdots&\vdots\\
a_{n1}x^{p^{n-1}}_1+a_{n2}x^{p^{n-1}}_2+\cdots+a_{nn}x^{p^{n-1}}_n&=&b_n\end{array}\right.$$
where tha $a_{i,j}$'s and $b_i$'s are in $K$ and the unknowns are $x_1,\cdots,x_n$. Is there a tool (as the determinant in the usual case) that characterizes the fact that this system has a unique solution $(x_1,\cdots,x_n)$ in an algebraic closure of $K$. I know Moore determinant but I do not see how to apply it to my problem. I know Grobner bases, but in the theorical case like this one, it looks like it is pretty useless.
In an algebraic closure $L$ of $K$, can't you just write $a_{ij}=c_{ij}^{p^i}$ and $b_i=d_i^{p^i}$ for (unique) $c_{ij}$ and $d_i$, and then you're $i$th equation becomes $$(c_{i1}x_1+\cdots+c_{in}x_n)^{p^i}=d_i^{p^i},$$ which in turn is equivalent to $$c_{i1}x_1+\cdots+c_{in}x^n=d_i.$$ So the existence of a unique solution is $\det(c_{ij})\ne0$. (Not sure if this helps, though.)
Please correct the subscripts in the second equation.
@JoeSilverman we could also consider $p^{n-k}$-th power of the $k$-th equation, it would give a linear system on $x_1^p, ... x_n^p$. Then, det of this system would guarantee existence, and we would need to find $p^{n-1}$-th root to solve the original system.
It also gives the criterion when the solution is in the base field I think?
|
2025-03-21T14:48:31.967097
| 2020-09-06T02:17:09 |
370971
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Joe Silverman",
"M. Winter",
"Mirco A. Mannucci",
"Richard Stanley",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/11926",
"https://mathoverflow.net/users/15293",
"https://mathoverflow.net/users/2807"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632747",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370971"
}
|
Stack Exchange
|
Factorization of the characteristic polynomial of the adjacency matrix of a graph
Let $G$ be a regular graph of valence $d$ with finitely many vertices, let $A_G$ be its adjacency matrix, and let $$P_G(X)=\det(X-A_G)\in\mathbb{Z}[X]$$ be the adjacency polynomial of $G$, i.e., the characteristic polynomial of $A_G$. In some graphs that came up in my work, the adjacency polynomials $P_G(X)$ have a lot of factors in $\mathbb Z[X]$, many of them repeated factors. So my questions are:
Is it common for the adjacency polynomial of a regular graph to be highly factorizable in $\mathbb Z[X]$, and to have many repeated factors?
If not, what are the graph-theoretic consequences of having many small-degree factors?
If not, what are the graph-theoretic consequences of having factors appearing to power greater than 1?
To give an idea of the numbers involved, one example was a connected 3-regular graph with 64 vertices, and
$$
P_G(X) =
(x - 3)x^{3}(x + 1)^{3}(x^2 - 3 x + 1)^{3}(x^2 - x - 3)^{3}(x^2 - x - 1)^{6}
(x^2 + x - 3)^{3}(x^3 - 3 x^2 - x + 4)^{2}(x^3 - 4 x + 1)
(x^6 - x^5 - 11 x^4 + 9 x^3 + 31 x^2 - 19 x - 8)^{3}
$$
I've looked at a couple of references and tried a Google search, but didn't find anything relevant.
Do your graphs have a lot of automorphisms? This will induce a factorization of $P_G(X)$.
Large eigenspaces of the adjacency matrix often point to either large symmetry, or more generally, to large regularity of the graph (whatever that means precisely). You could check whether your graph is distance-regular or at least $k$-walk-regular for some small $k\ge1$ as this often comes with large eigenspaces.
Since the graphs seems to have many eigenspaces of dimension three, I would probably try to get some intiution for its structure by plotting the 3-dimensional spectral graph realizations to these eigenvalues. Maybe some of these are of a quite recognizable shape (a Platonic solid etc). See here for what I mean by "spectral realization".
Let me add that there seem to be quite a few examples of graphs for which $P_G(X)$ factors more than is explained by symmetry. An example is mathoverflow.net/questions/369515. For a different kind of example of this phenomenon of "extra factorization," see Exercise 10.9(d) of my book Algebraic Combinatorics, second ed.
@RichardStanley Thanks. I think that my graph does have a large automophism group. I'm not sure I can compute it exactly, but pulling out a large piece should give something. It's also helpful to know that although symmetry causes factorization, the converse isn't necessarily true; I'll keep that in mind.
Expanding on Richard's comment: let me rename your graph to $S$ and consider the adjacency matrix $A$ abstractly as a linear operator acting on the free vector space $\mathbb{C}[S]$ on (the vertices of) $S$, and let $G$ be its automorphism group (this is why I wanted a new name). Then $\mathbb{C}[S]$ is a completely reducible representation of $G$ and $A$ is an endomorphism of this representation. Hence if we write
$$\mathbb{C}[S] \cong \bigoplus_i n_i V_i$$
where $V_i$ are the irreducibles, then $A$ is an element of the endomorphism algebra
$$\text{End}_G(\mathbb{C}[S]) \cong \prod_i M_{n_i}(\mathbb{C}).$$
This means more explicitly that $A$ is conjugate over $\mathbb{C}$ to a block diagonal matrix with a block for each isotypic component (hence its characteristic polynomial factors accordingly). In the nicest possible case the decomposition above is multiplicity-free in which case the endomorphism algebra is a product of copies of $\mathbb{C}$ and we just have that $A$ must act by a scalar $\lambda_i$ on each $V_i$ that occurs in the decomposition, which contributes a multiplicity of $\dim V_i$ to $\lambda_i$ as a root of the characteristic polynomial and hence, over $\mathbb{Q}$, contributes a multiplicity of $\dim V_i$ to the minimal polynomial of $\lambda_i$ as a factor of the characteristic polynomial.
(I think the result of this analysis comes out the same if you work over $\mathbb{Q}$ from the beginning but it's more annoying to describe.)
I work through a few examples of this in my old blog post The Schrodinger equation on a finite graph, where I was trying to understand via a toy model the quantum mechanical phenomenon of group symmetries introducing "degeneracies," which is physics speak for eigenvalues (of the Hamiltonian in this case) of multiplicity greater than $1$.
The "most degenerate" case is the complete graph $S = K_n$, where $G = S_n$ and the corresponding representation is a copy of the trivial representation and an irreducible representation of degree $n-1$. This means the adjacency matrix $A$ must have at most two eigenvalues, one with multiplicity $1$ and one with multiplicity $n-1$, which turn out to be $n-1$ and $-1$ respectively (this is easily computed by computing $\text{tr}(A)$ and $\text{tr}(A^2)$, or just finding all the eigenvectors of $A + I$), inducing a factorization
$$\det (tI - A) = (t - n + 1)(t + 1)^{n-1}$$
with a factor of multiplicity $n-1$.
One of the "least degenerate" cases where the automorphism group still acts transitively on vertices is $S = C_n$ the cycle graph, where $G = D_n$ is the dihedral group and the corresponding representation splits up into mostly $2$-dimensional irreps. This reflects the fairly mild degeneracies of the eigenvalues of the adjacency matrix, which are $2 \cos \frac{2\pi k}{n}, k = 0, \dots n-1$, and/but which also organize themselves into nontrivial Galois orbits coming from the action of the Galois group of $\mathbb{Q}(\zeta_n)$.
Marvelous. Particularly your blog post: I wonder if there is some literature on taking your "toy model" seriously. By this I mean: can one approximate at any level of accuracy any quantum system as a graph quantum model when the finite graph grows? I bet it can. On a related note: Feynman ' s approach via paths actually becomes quantum random walks on your graph, and no infinities involved...
Thanks. This is all very helpful, especially the use of the symmetry group of the graph, and the representation that you describe, to deduce factorization. For my graphs, I expect that the symmetry group is fairly large, but not necessarily easy to compute, so I may be able to play the factorization off against potential pieces of the symmetry group. In any case, you've provided me much food for thought.
In a comment I said that large eigenspaces of the adjacency matrix may point to large symmetry or regularity in the graph.
For example, let me explain why highly symmetric graphs have large eigenspaces (that is, large factors in their characteristic polynomial).
A symmetry $\sigma\in\def\Aut{\mathrm{Aut}}\Aut(G)$ is a permutation of the vertex set $V=\{1,...,n\}$. Let $P_\sigma\in\def\RR{\Bbb R}\RR^{n\times n}$ be the associated permutation matrix. The symmetries of a graph can be characterized using its adjacency matrix as follows: a permutation $\sigma\in\mathrm{Sym}(V)$ satisfies
$$\sigma\in\Aut(G)\quad\Longleftrightarrow\quad AP_\sigma=P_\sigma A.$$
But this means, for any eigenvalue $\theta\in\mathrm{Spec}(A)$, if $u\in\RR^n$ is a $\theta$-eigenvector, then so is $P_\sigma u\in\RR^n$ for all the $\sigma\in\Aut(G)$:
$$A(P_\sigma u)=(AP_\sigma) u = (P_\sigma A)u = P_\sigma (Au)=P_\sigma(\theta u)=\theta(P_\sigma u)$$
So the $\theta$-eigenspace contains $\mathrm{span}\{P_\sigma u\mid \sigma\in\Aut(G)\}$ which can be quite large, depending on $\Aut(G)$.
Another way to say this: the eigenspaces of $A$ are invariant subspaces w.r.t. $\Aut(G)$, and if $\Aut(G)$ has no small (irreducible) invariant subspaces then $A$ cannot have small eigenspaces.
On the other hand, generic regular graphs do probably not have integer eigenvalues. So there is no reason for any such adjacency polynomial to factor over $\Bbb Z$ into many small factors (ignoring the trivial factor).
So I would say this points to some structure in your graph, but it is hard to say what it is whitout knowing more about the graph.
Thanks. I actually have a large infinite group that acts on my graph, although it is not clear (to me, at least) what the quotient of that action is, nor whether the image will be the entire automorphism group (although I suspect that it generally will be). It's also good to know, as Richard Stanley also pointed out in the comments, that although large automorphism group (often) leads to factorization of the adjacency polynomial, the converse need not hold.
Consider the category $Gph$ whose objects are directed graphs. That is the topos defined on the small category $C$ which has two objects $0,1$ and two morphisms $s,t:0\rightarrow 1$. A directed graph is thus defined by by a contravariant functor $D:C\rightarrow Set$ where we call $D(0)$ the set of nodes and $D(1)$ the set of arrows, $D(s)$ is the source map and $D(t)$ is the target map. An example of graphs is the $n$-cycle directed graph.
I have constructed with Bisson a Quillen model on the category $Gph$ such that a morphism $f:X\rightarrow Y$ is a weak equivalene if and only if for every $n>0$ the induced map $Hom(C_n,X)\rightarrow Hom(C_n,Y)$ is bijective.
Let $X$ is and $Y$ be finite graphs, we say that they are almost isospectral if the characteristic polynomial of their adjecency matrices have the same non zero eigenvalues.
The following conditions are equivalent:
Two finite directed graphs for the Quillen model above are weakly equivalent.
Two finite directed graphs have the same Zeta serie.
Two finite directed graphs are almost isospectral.
The third condition implies that two finite graphs are weakly equivalent for the Quillen model if the factors of their characteristic polynomial distinct of $X^p$ are equal.
We have also computed the homotopy category of this closed model and showed that it is equivalent to the category of periodic $\mathbb{Z}$-sets.
This shows that a finite graph is equivalent to finite set of cycles and enables to relate the degree of the factors of the characteristic polynomial of the finite graph $X$ to its image in the homotopy category which is a set of cycles.
For example, for this model, acyclic object are trees, and the characteristic polynomial of a tree is of the form $X^n$. In fact in this model if $X^p$ is a non trivial factor of the characteristic polynomial of the graph $G$, there is a weak cofibration $c:G'\rightarrow G$ (a whiskering) which is defined by attaching trees to $G'$.
https://arxiv.org/pdf/0802.3859.pdf
https://arxiv.org/pdf/0906.4087.pdf
Corollary 2.7 of Norman Biggs's book Algebraic Graph Theory says that if $d$ is the diameter of the graph, then it has at least $d+1$ distinct eigenvalues. (This bound is tight, and is achieved for example by distance-regular graphs.) So in particular, having a lot of repeated eigenvalues does not necessarily indicate a lot of automorphisms; it might just mean that the diameter is small. For example, a strongly regular graph has only three distinct eigenvalues but often has trivial automorphism group.
Thanks. I'll keep that in mind. Although I think that my graphs do, in fact, have a large automorphism group. But this will prevent me from drawing too much of a conclusion from the factorization!
An observation about graph isomorphism (GI)
since you ask about powers greater than 1.
GI is polynomial time solvable for graphs with
bounded eigenvalues multiplicity (the largest
exponent in the factorization of your polynomial).
I suspect random graphs to have bounded multiplicities
and random non-isomorphic graphs to have different polynomials.
A very extreme examples are Paley graphs:
sage: g1=graphs.PaleyGraph(37);factor(g1.characteristic_polynomial())
(x - 18) * (x^2 + x - 9)^18
If the characteristic polynomial factors (over $\mathbb{Q}$) into polynomials of degrees $d_1,d_2,\ldots,d_r$, where $d_1$ is the degree of the polynomial that the largest eigenvalue is a root of, then the controllability matrix $\begin{pmatrix}\mathbf{b} & \mathbf{Ab} & \mathbf{A}^2\mathbf{b} & \ldots & \mathbf{A}^{n-1}\mathbf{b}\end{pmatrix}$ with respect to any nonzero $0$-$1$ indicator vector $\mathbf{b}$ has rank $d_1+\sum_{i=2}^r c_id_i$, where $c_i=0$ or $1$ for all $2\leq i\leq r$. (Here, $\mathbf{A}$ is the adjacency matrix.)
In particular, if the characteristic polynomial is irreducible over $\mathbb{Q}$, then the controllability matrix is invertible for all nonzero $\mathbf{b}$.
Also note that for regular graphs, one of the factors will be $(x-\rho)$ where $\rho$ is the common degree of the graph, and this $\rho$ will be the largest eigenvalue of $\mathbf{A}$. Hence the above result would state that the rank of any controllability matrix as defined previously is $1+\sum_{i=2}^r c_id_i$ where $c_i=0$ or $1$ for all $2\leq i\leq r$. In particular for the case when $\mathbf{b}$ is the vector of all ones, it is well-known that $c_2=c_3=\cdots=c_r=0$ and the rank of the controllability matrix will be $1$.
|
2025-03-21T14:48:31.967849
| 2020-09-06T02:22:24 |
370972
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632748",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370972"
}
|
Stack Exchange
|
When is a twisted form coming from a torsor trivial?
Consider a sheaf of groups $G$, equipped with a left torsor $P$ and another left action $G$ on some $X$. Form the contracted product $P \times^G X := (P \times X)/\sim$ where $\sim$ is the antidiagonal quotient: $(g.p, x)\sim (p, g.x)$.
Q1: When is $P\times^G X$ trivial? I.e., when do we have an isomorphism $P \times^G X \simeq X$?
Partial answer: $P \times^G X \simeq X$ over $[X/G]$ iff $P \times [X/G]$ is a trivial torsor over the stack quotient $[X/G]$.
Proof: We can rewrite $P \times^G X$ as a contracted product of two torsors $(P \times [X/G])\times^G_{[X/G]} X$. Then we contract with ``$X^{-1}$'' -- the inverse to contracting with $X$ as a torsor over $[X/G]$ and we win. (as in B. Poonen's Rational Points on Varieties, section <IP_ADDRESS>)
Am I allowed to do this? This argument probably shouldn't have to appeal to algebraic stacks and may be somewhat dubious.
Q2: If I have one isomorphism $P \times^G X \simeq X$, can I choose another one that lies over $[X/G]$? Or at least is $G$-equivariant?
Q3: Is there a natural way to write the triviality of such a twisted form?
I first thought $P \times^G X \simeq X$ iff $P$ was trivial, which is clearly false for trivial actions on $X$. Then I was excited to have the pullback $* \to BG$ represent triviality of the twisted form $P \times^G X$ as well as the torsor $P$. Is there a natural representative of the sheaf of isomorphisms between $P \times^G X$ and $X$?
These can all be sheaves, although I'm primarily interested in $G = GL_n, PGL_n, SL_n$, etc. acting on $X = \mathbb{A}^n, \mathbb{P}^n$ as appropriate. More ambitious is $G = \text{Aut}(X)$ for even simple $X$. I'd be happy with answers in any level of generality.
Due Diligence Statement: I'm a novice in the area of "twisted forms" of varieties, so I apologize if the above is evident or obtuse. I checked all the "similar questions" listed here and couldn't find an answer.
This lemma might hold in a more general topos-theoretic context, but for "safety" I'm going to formulate it in a more restricted setting.
Lemma:Let $G$ be a group scheme with an action $\rho: G \times X \to X$ corresponding to a morphism $\varphi: G \to \mathrm{Aut}(X)$. Assume $G, \mathrm{Aut}(X)$ are smooth over a field $k$, let $S$ be a $k$-scheme and let $P$ be a $G$-torsor on $S$. Then $P \times^G X \simeq X \times S$ if and only if there is a morphism $\sigma: S \to \mathrm{Aut}(X)/G$ such that $P$ is the torsor of maps $\tilde{\sigma}: S \to \mathrm{Aut}(X)$ with $\tilde{\sigma} = \sigma \,\mathrm{mod} \, G$.
I'm requiring the groups to be smooth only so that the classifying stacks are algebraic. Rather than trying to precisely formulate and prove such a lemma here, I'll list 2 ways to think about it:
Morphism of classifying stacks version: The map $\varphi: G\to \mathrm{Aut}(X)$ induces a morphism $B\varphi: BG \to B\mathrm{Aut}(X)$, and one can check that the fiber of $B\varphi$ over the map $\gamma: S \to B\mathrm{Aut}(X)$ classifying $X$ is $S \times \mathrm{Aut}(X)/G$.
Exact sequence of cohomology groups version: At least when $G$ and $\mathrm{Aut}(X)$ are abelian, there will be an exact sequence
$$\cdots \to H^0(S,\mathrm{Aut}(X)/G) \to H^1(S, G) \to H^1(S, \mathrm{Aut}(X)) \to \cdots $$
In general there might be a similar exact sequence ("of sets"), but I'm not confident enought with non-abelian cohomology groups to assert that.
|
2025-03-21T14:48:31.968139
| 2020-09-06T03:53:16 |
370979
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632749",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370979"
}
|
Stack Exchange
|
Kodaira–Spencer tensor of an isoperiodic deformation
Let $\alpha$ be a holomorphic 1-form on a curve $X$ of genus $g$, which we view as a map of sheaves $\alpha \colon T \to O$. The cokernel of this map is the structure sheaf $O_Z$ of the zero locus $Z \subset X$ of $\alpha$, which is a sum of $2g-2$ skyscraper sheaves (let zeroes $z_i$ of $\alpha$ be simple). It gives rise to the exact sequence $0 \to H^0(O) \to H^0(O_Z) \to H^1(T) \to H^1(O) \to 0$, where the range of the connecting homomorphism is the tangent space of the universal isoperiodic deformation (i. e. the deformations of complex structures on $X$ s. t. the cohomology class $[\alpha] \in H^1(X,\mathbb{C})$ still lies in the corresponding $H^{1,0}$ subspace). On the other hand, the space $H^0(O_Z)/\mathrm{const}$ has natural choice of coordinates, which integrate to the coordinate system on the base of the universal isoperiodic deformation, given by $\left\{\int_{z_0}^{z_i}\alpha\right\}_{i=1}^{2g-3}$ (called in the physics literature 'the relative periods').
My question is the following: how to describe the Kodaira–Spencer tensor associated to a vector from $H^0(O_Z)$? I tried to write down the connecting homomorphism for the Čech complexes, but did not succeed, and did not bother myself again since it must be well-known anyway. However, I did not find a reference.
I suppose you are asking for how to write down pairing of basis elements $v_i\in H^0(O_Z)$ with $\omega\in H^0(\Lambda^1\otimes \Lambda^1)$. Here the simplest answer: locally around $z_i$ let our $\alpha=\alpha(z)dz$ and $\omega=\omega(z)dz^2$ then
$\langle \alpha, \omega\rangle=\oint\limits_{z_i} \frac{\omega dz^2}{\alpha dz}=\oint\limits_{z_i} \frac{\omega}{\alpha}dz$
To see this let's recall that Serre duality is in fact local statement -- it is comming from Yoneda product in $\mathcal{Ext}$'s.
We want to write local pairing in Cech terms for covering of the curve with small disks $U_i$ around $z_i$ and complement $U$ to all $z_i$:
$$\mathcal{Ext}^0(O,\Lambda^1\otimes \Lambda^1)\otimes \mathcal{Ext}^0(O,O_Z)\xrightarrow{1\otimes\delta}$$
$$\mathcal{Ext}^0(O,\Lambda^1\otimes \Lambda^1)\otimes \mathcal{Ext}^1(\Lambda^1,O)\xrightarrow{\cup} \mathcal{Ext}^1(O,\Lambda^1)$$
The first is given by connecting homomorphism. By inspecting Cech double-complex for $T\to O\to O_Z$ and down-to-earth definition of $\delta$ we obtain that $\delta v_i$ is given by $\frac{1}{\alpha}\partial_z\in T(U_i\cap U)$ as Cech cocyle.
The second map is given by cup-product of Cech cocycles, so it is $\frac{\omega(z)}{\alpha}dz$ on $U_i\cap U$.
Finally we have explicit Cech cocycle as element of $\Lambda^1(U_i\cap U)$ and we need to evaluate $\check{H}^1(\Lambda)\xrightarrow{\int\limits_{C}}\mathbb{C}$. Well, it's given by residue and could be found in general treatment somewhere in Griffits&Harris book. For our curve $C$ it's simple -- take double-complex Cech-Dolbeau resolution of the constant sheaf $\mathbb{C}$, introduce a partition of unity $\rho,\rho_i$ subordinate to $U,U_i$. Then, starting from $a=\Lambda^1(U_i\cap U),\bar{\partial}a=0$ construct by gluing cohomologous element in $\Lambda^{1,1}(C)$ as $W=\bar{\partial}(\rho a)-\bar{\partial}(\rho_i a)\in \Lambda^{1,1}(C)$, by simple complex analysys we obtain that $\int\limits_{C}W=
\lim\limits_{\varepsilon\to 0} \int\limits_{C-D_{i,\varepsilon}}W=
\lim\limits_{\varepsilon\to 0}
\int\limits_{C-D_{i,\varepsilon}}-\bar{\partial}(\rho_i a)=
\lim\limits_{\varepsilon\to 0}
\oint\limits_{\partial D_{i,\varepsilon}}\rho_i a=
2\pi i\ res_{z_i}(a)$
Here I denote by $D_{i,\varepsilon}$ small disk around $z_i$
|
2025-03-21T14:48:31.968423
| 2020-09-06T04:21:10 |
370982
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"https://mathoverflow.net/users/157762",
"https://mathoverflow.net/users/36721",
"wuhanichina"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632750",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370982"
}
|
Stack Exchange
|
Convolution of two Gaussian mixture model
Suppose I have two independent random variables $X$, $Y$, each modeled by the Gaussian mixture model (GMM). That is,
$$
f(x)=\sum _{k=1}^K \pi _k \mathcal{N}\left(x|\mu _k,\sigma _k\right)
$$
$$
g(y)=\sum _{i=1}^N \lambda _i \mathcal{N}\left(y|\mu _i,\sigma _i\right)
$$
Now, I would like to know the PDF of another variable $Z$, where $Z=X+Y$.
Is there anyone who can write the explicit PDF of $Z$?
The pdf of the sum $X+Y$ of independent random variables $X$ and $Y$ is the convolution of the pdf's of $X$ and $Y$. The convolution operation is bilinear. The convolution of Gaussian pdf's is Gaussian. So, letting $f_j(t):=\mathcal{N}(t\big|\mu_j,\sigma_j)$ and noting that $(f_k*f_i)(z):=\mathcal{N}\left(z\big|\mu_k+\mu_i,
\sqrt{\sigma_k^2+\sigma_i^2}\right)$, we find that the pdf $h$ of your $Z=X+Y$ is
$$
h=\Big(\sum _{k=1}^K \pi_k\, f_k\Big)
*\Big(\sum _{i=1}^N \lambda_i\, f_i\Big)
=\sum _{k=1}^K\sum _{i=1}^N \pi_k\lambda_i\, f_k*f_i
$$
and hence the pdf $h$ of your $Z=X+Y$ is also a Gaussian mixture given by
$$
h(z)=\sum _{k=1}^K\sum _{i=1}^N \pi_k\lambda_i\, \mathcal{N}\left(z\big|\mu_k+\mu_i,
\sqrt{\sigma_k^2+\sigma_i^2}\right).
$$
Thank you losif. Although I agree with the theory you posted, I don't think the pdf of Z is as easy as h(z).
@wuhanichina : So, you agree with the proof but don't think that what is proved is true? How can that be? Anyway, I have added some details, which may be helpful to you.
|
2025-03-21T14:48:31.968564
| 2020-09-06T07:42:00 |
370991
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632751",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370991"
}
|
Stack Exchange
|
A notion of (weak) dependency
Let $\nu$ be a probability measure over some product metric space $\mathcal{X}^n$
such that there exist some $\alpha \geq 0$ for which
for every $1\leq i \leq n$ it holds that
$\sup_{x\in \mathcal{X}^n} ||\nu_i - \nu_i (\cdot | x^{-i})||_{TV} \leq \alpha$.
The definition can be regarded as some kind of dependency bound (parameterized by $\alpha$).
Is it known? does it have a name? If not, is there any closely related notion of dependency?
|
2025-03-21T14:48:31.968633
| 2020-09-06T07:48:15 |
370992
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Bill Johnson",
"Yemon Choi",
"fedja",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/2554",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632752",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370992"
}
|
Stack Exchange
|
When does an operator from $\ell_1$ to itself factor through $\ell_p$?
I would like to know whether a given operator from $\ell_1$ to itself, given by a matrix $A$, factors through $\ell_p$, for $p>1$.Does anyone know any references/results on this topic? I am specially interested on the case where the factor from $\ell_p$ to $\ell_1$ is a diagonal operator
Thanks
Are you interested in $p\le 2$, $p\ge 2$, or both?
There is a necessary condition: for $p>1$, any operator from $\ell_p$ to $\ell_1$ is compact (Pitt's theorem) and so any operator on $\ell_1$ that factors through $\ell_p$ must also be compact.
Everything is laid out in chapter 9 of the book of Diestel, Jarchow, and Tonge. They discuss factoring through some $L_p$ space, but this is the same as factoring through $\ell_p$ for compact operators.
|
2025-03-21T14:48:31.968725
| 2020-09-06T08:51:50 |
370997
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"JimN",
"Rodrigo de Azevedo",
"https://mathoverflow.net/users/62043",
"https://mathoverflow.net/users/91764"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632753",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370997"
}
|
Stack Exchange
|
Optimal paths in set-weighted graphs
Let $G = (V,E)$ be an $n$-vertex graph, let $R$ be a finite set (to be specific, let us assume that $R = [n]$), and let $W : V \rightarrow 2^R$. Let us call the pair $(G, W)$ a set-weighted graph.
Now let us consider the following problem, which I'll call the Smallest Set Path problem. The problem is given a set-weighted graph $(G,W)$ and two vertices $s$ and $t$ of $G$, find a path $P$ between $s$ and $t$ such that the union of the sets associated with the vertices of the path has the minimum possible cardinality, i.e. $\bigcup_{v \in P} W(v)$ is minimized.
My specific questions:
Was the Smallest Set Path problem studied in the literature?
Was the model of set-weighted graphs as described above considered in the literature in other contexts?
Why not post at http://cstheory.stackexchange.com?
I have not seen this problem studied before. Sets on vertices have been used in describing graph dimension or list-colouring or preference lists in stable matching problems, for example. In general, intersection graphs are graphs where each vertex is a set and an edge represents a non-empty intersection of the two vertices. Those sets are sometimes geometrical points, or intervals of real lines, or just discrete sets.
without any structure in the graph paths or in the sets weights, essentially all paths woul have to be checked ... unless in a shortest path problem, where a path from s-t going through an intermediate u, it doesn't matter how s gets to u, just the value at u is important. But the unio nof all your sets along the way would be different for every s-to-u path, so essentially all paths would have to be considered.
|
2025-03-21T14:48:31.968880
| 2020-09-06T09:03:43 |
370998
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ron P",
"Shivin Srivastava",
"https://mathoverflow.net/users/120939",
"https://mathoverflow.net/users/85550"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632754",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370998"
}
|
Stack Exchange
|
Prove that the following running average is monotonically decreasing
Let $S_n$ be defined as $\frac{1}{n}\sum_{t=1}^{t=n} [x^2+(p-q)x]$ where $x = 1-(1-p-q)^t$. We want to find the conditions on $p$ and $q$ such that $S_n$ is monotonically decreasing for all $n$. $0 < p,q < 1$ and $0 < 1-p-q < 1$.
Note:
Till now I have tried to get a closed bound expression for $S_n$ and differentiate it w.r.t. $n$ to get the conditions for a negative slope but it is getting really complex.
Another approach was to reduce this expression to the sum $S_n = C (>0) + \frac{1}{n}\left[q\sum_{t=1}^{n} \lambda^{2t} + (p-q) \sum_{t=1}^{n} \lambda^t \right]$ where $\lambda=1-p-q$. We know the upper bounds of the two sums, and since the denominator grows more rapidly than the numerator, it is sufficient to show that the numerator is positive to get a monotonically decreasing sequence.
$p\geq q$ is a sufficient condition.
If $p\leq q$, you can rewrite the questions with $a=p+q$, and $b=q-p$, and the condition $1\geq a\geq b\geq 0$.
$S_n$ is always increasing in $n$. Note that $x$ is increasing in $t$; therefore, if $p\geq q$, then $x^2+(p-q)x$ is increasing in $t$.
If $q > p$, then you can set $a=1-p-q$, and $b=1-p+q$. The condition is $0<a\leq b<1$, and $x^2+(p-q)x=x(x-1+b)=(1-a^t)(b-a^t)$, which is increasing in $t$.
Hi Ron,
Thanks for the insights. I can't see how you have taken the $\frac{1}{n}$ factor into account. $x^2 + (p-q)x$ is increasing in $t$, therefore its summation is also increasing in $t$ but won't the $n$ factor in the denominator attenuate it?
when you have a (finite) set of numbers and you add to it another number which is greater than all the others, then the average increases.
|
2025-03-21T14:48:31.969158
| 2020-09-06T09:18:43 |
370999
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Pritam Bemis",
"Timothy Budd",
"https://mathoverflow.net/users/150564",
"https://mathoverflow.net/users/47484"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632755",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370999"
}
|
Stack Exchange
|
Average over spheres finite
Let $X_1,...,X_N$ be random variables that are iid with the uniform distribution over $\mathbb S^n.$
I am curious how to see that $f(X_1,..,X_N):=\left \lvert \sum_{i=1}^N X_i \right\rvert^{-1}$ has finite expectation, i.e.
$$\int_{(\mathbb S^n)^N} f(X_1,..,X_N) \ dS(X_1)\cdots dS(X_N)<\infty$$
is finite, where $dS$ is the surface measure.
I know it is true by this answer here, which shows it for $\mathbb S^1$ and gives even the asymptotic of the integral. So I believe that if it is true on $\mathbb S^1$ it also has to be true on all other spheres, but I am looking for a more direct argument than in the above answer to see that this is indeed the case.
I suspect you want $f$ to be the inverse distance of $X_1+\cdots+X_N$ to the origin? Note that in the case $n=1$ and $N=2$ the expectation is infinite.
Assuming you mean $f(X_1,\ldots,X_n) = |\sum_{i=1}^N X_i|^{-1}$, the case $n\geq 2$ is actually easier than $n=1$. You could use that there exists a constant $C>0$ such that for any $x\in\mathbb{R}^n$ the expectation $\mathbb{E}[ 1 / |X_1 + x| ] < C$. Then automatically also $\mathbb{E}[ 1 / |X_1 + \ldots + X_n| ] < C$ by taking $x$ to have the distribution of $X_2 + \ldots + X_N$.
Note that this argument does not work for $n=1$, because if $|x|=1$ then $\mathbb{E}[ 1 / |X_1 + x| ] = \infty$ due to a logarithmic divergence.
can the integral $\mathbb E(1/\vert X_1+x \vert)$ for $n \ge 2$ actually be worked out?
It is not necessary for the question and it is probably not possible in elementary functions, but in terms of hypergeometric functions you should be able to work out a general expression.
|
2025-03-21T14:48:31.969340
| 2020-09-06T10:37:52 |
371002
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asaf Karagila",
"Carlo Beenakker",
"Claus",
"Ilmari Karonen",
"Kevin Walker",
"Mary Sp.",
"Mike Shulman",
"Mirco A. Mannucci",
"Nate Eldredge",
"Stef",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/15293",
"https://mathoverflow.net/users/156936",
"https://mathoverflow.net/users/161374",
"https://mathoverflow.net/users/164942",
"https://mathoverflow.net/users/284",
"https://mathoverflow.net/users/29697",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/49",
"https://mathoverflow.net/users/7206",
"https://mathoverflow.net/users/7742",
"usul"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632756",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371002"
}
|
Stack Exchange
|
An intelligent ant living on a torus or sphere – Does it have a universal way to find out?
I wanted to ask a question about topological invariants and whether they are connected in a fundamental or universal way. I am not an expert in topology, so please let me ask this question by way of a simple example.
Imagine an intelligent ant living on a torus or sphere, and it wants to find out. Let’s further assume the ant does not have the capabilities to do geometrical measurements, i.e. it cannot measure length, angle, curvature, whether a line is straight, and so on. The only capabilities it has are topological, combinatorial, logical.
Now there are several ways that it can distinguish a sphere from the torus, like for example
Work with loops and determine the fundamental group.
“Comb” the surface (apply the Hairy Ball Theorem).
Triangulate the surface, count vertices, edges, and faces, and determine its Euler characteristic.
Draw the complete graph with five vertices $K_{5}$. If it can be drawn without any edges crossing, then it must be the torus.
Triangulate the surface and color the vertices. Minimize the number of colors, but make sure adjacent vertices have different colors. If more than four colors are needed, it must be the torus.
I am not an expert in this field, but I think No. 1 and 2 are fundamentally equivalent (applying the same fundamental topological concepts). I imagine that No. 3 and 4 are also fundamentally equivalent. I am not sure about No. 5, I think its relation to 3 and 4 is through Hadwiger’s conjecture.
My question, can it be shown that all these methods are, in some way, fundamentally resting on the same, deeper concept? Asking differently, is there an abstract, universal method from which all the other examples follow or can be derived?
I would be interested to learn whether category theory or homotopy type theory provide such a foundational, universal view on this classification problem. My dream answer would be if someone said something like “all your methods are examples of the universal property of …”, but maybe that’s expecting too much.
I would be grateful for any hint or reference. Thank you in advance!
EDIT: Just adding references to make the post more self-contained
Fundamental Group https://en.wikipedia.org/wiki/Fundamental_group
Hairy Ball or Poincaré Brouwer Theorem https://en.wikipedia.org/wiki/Hairy_ball_theorem
Euler Characteristic https://en.wikipedia.org/wiki/Euler_characteristic
Graphs on Torus https://en.wikipedia.org/wiki/Toroidal_graph
Four Color Theorem and generalization to torus https://en.wikipedia.org/wiki/Four_color_theorem
Hadwiger Conjecture https://en.wikipedia.org/wiki/Hadwiger_conjecture_(graph_theory)
I would say that 2 is more equivalent to 3, since the hairy ball theorem reduces to the Euler characteristic (being the sum of the indices of a vector field). The results of 4 and 5 sort of involve Euler characteristic too, but are more subtle, so I would not call them equivalent.
As Mike Shulman points out, 2 and 3 are closely related. Other than that relationship, I would say that all of the methods are distinct.
To see the difference between 1 and 2, you could step up one dimension and note that $S^3$ is simply connected (i.e. has trivial fundamental group, every loop can be shrunk to a point) but does admit nowhere vanishing tangent vector fields (so its hair can be combed).
Develop space travel, obviously.
Interestingly, the answers highlight that a finite ant cannot prove she lives on a sphere, she can only disprove it (by finding a hole). (She can also prove there are no holes of a certain size/scale, given enough time.) I guess this is reflected in the $\Pi_1$ versus $\Sigma_1$ nature of your proposed tests (for example #4).
There's a related question on Math.SE from 2014.
"Draw the complete graph with five vertices K5. If it can be drawn without any edges crossing, then it must be the torus." Deciding whether "It can be drawn" is harder than "Try to draw it for a bit. Still haven't succeeded? Must be a sphere" especially if you don't know the size of the planet.
Carlo's answer is definitely pointing in the right direction: simplicial complexes or more generally, simplicial sets, are conjured up by most points points mentioned by the PO (certainly 1 3, 4. 5 perhaps, with a twist, and as for 2, no idea) .
Unfortunately, as indicated by Carlo's comments, it falls short on one requirement, that ants do not know anything about metric spaces (nor what you can build on them, such as differential geometry).
Poor ants live in a world whose departments of math contain only three courses (*):
finite combinatorics
topology (presumably also finite)
basic logic
Topological Data Analysis starts off with a cloud set of points immersed in a metric space (mostly euclidean $R^n$, but not necessarily).
Its main tool is persistent homology, which creates a filtration of simplicial complexes, thereby providing different views of $X$ at different resolution scales.
Where do these simplicial complexes come from ? They are Vietoris-Rips Complexes (see here; essentially you use the distance between groups of points to fill your simplexes).
So, no metric no Rips complex.
But (there is always a but in life): perhaps not all is lost.
What about creating a filtration of complexes by-passing entirely the metric?
Yes, sounds good, you may say, but how? Well, in ants world they have basic topology. So, for instance, suppose an ant goes from A to B, it can tell if during her trip she met point C (ie she can tell whether C is in some 'edge" between A and B). Similarly, given a set of distinguished points $A_0, \ldots A_n$ , she can tell whether they are indipendent, ie none of them lies in some slice of ant-world which is span by some subset. The independent subsets will become higher simplices (this approach is basically the one folks in matroid theory take)
Assuming this bare bone capability, Carlo's answer can indeed be vindicated: the ants build their filtration of complexes by selecting larger and larger finite subsets of their world.
Of course, unless their world is also finite, there is no guarantee that they will ever find out its final topology.
(*) on the funny side (apologies to serious MO fellows): trying to think of Ant's World I found out that is very much to my liking, especially the Departments of Math. A non Cantorian, non Dedekind paradise . Perhaps I should move there for a change:)
Mirco last time I asked about ants and you gave a great answer. This time I am asking about spiders ;-) https://mathoverflow.net/q/371606/156936
Claus, thanks on 2 counts: for the compliment, and for the new challenge. Unfortunately, I am currently busy defending my last question from Asaf, and also I have to spend the weekend doing my own research ( I am a Data Scientist by trade, now). But, here is my promise: by next weekend, either someone has answered to your satisfaction, or I will done the spiderman suit and try my hand! :) viele Grüße, Mirco
that's a great comment, thanks a lot!! In case you have time to look into it next weekend, please also tell me about this statement in your profile: "I do not believe in natural numbers, yet I reject no math". Very interested what it means.
I presume the OP has in mind the topological distinction between a sphere and a torus, so the method should apply to deformed surfaces. A meaningful/universally valid method for this purpose must include the notion of "scale". Otherwise we would conclude that the earth is a torus, or even a surface of higher genus. Such a method is offered by the framework of persistent homology.
Space is represented by a simplicial complex with a distance function. Loops such as provided by the arch shown here can then be identified and excluded depending on the scale on which they occur. Efficient algorithms exist to identify the scale-dependent homology group, and thus obtain the topological invariants.
Aloba arch, Wikimedia Commons
I am very interested in your answer because you bring up persistent homology (and by the way your picture makes a great point!). Please let me ask two questions about your answer (1) if a distance function is required, but the ant in the OP cannot measure length - will distance function still work? (2) Does persistent homology also address the bullet points 3-5 in the OP, i.e. in the sense that it unifies all methods?
(1) if the ant cannot measure length, so if it has no notion of scale, then it has no way to exclude small perturbations of the surface (such as the arch) that increase its genus. (2) the approach gives the Betti numbers of the simplicial complex, so it fully characterizes the connectivity of the space; the coloring approach in point 5 of the OP identifies the chromatic number, which presumably is an independent characterization.
If an ant has an ability to draw different colors of line, and detect when it has crossed a line, it could start by drawing a red/green line pair, with green on the right, and wander around until it hits the line. If it encounters the green side of the line and follows the line to the left, it would reach the start point and thus be able to close the loop. If it encounters the red side, following the line to the right would let it return to the start and close the loop. In either case, it could select two new colors, arbitrarily select the red or green side of the loop, and resume wandering. If it encounters a line of the new colors, it should close that loop using the same strategy for red/green. If it encounters a previously drawn loop, it may cross it, but should keep track of which side of each loop it is on.
No amount of wandering would allow the ant to prove it was on a sphere. If, however, it were to encounter a loop from one side having last left it via the other, that would prove that the loop which was thus encountered went around a hole, and thus that the structure had at least one.
Without any metric: an ant living flat on a torus may exhibit a complete planar graph of five vertices ($K_5$). The it knows it does not live in a sphere. In fact the vertex chromatic number of the torus is 7. It works for n hole torus.
REMARK: It is not clear that living flat on the sphere you can show that $K_5$ is not planar. Meaning in general that differentiating a surface of type $A$ from a surface $B$ may not be symetrical.
|
2025-03-21T14:48:31.970391
| 2020-09-06T13:03:30 |
371009
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jiří Rosický",
"Tim Campion",
"https://mathoverflow.net/users/152679",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/73388",
"varkor"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632757",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371009"
}
|
Stack Exchange
|
Characterisation of presentations for varietal large equational theories
Let $T : \mathbf{Set}^\mathrm{op} \to \mathscr T$ be a large equational theory (i.e. a bijective-on-objects product-preserving functor). Following Linton in Some Aspects of Equational Categories, we call $T$ varietal if $\mathscr T$ is locally small. A large equational theory may be presented by a class $\Omega$ of set-indexed (i.e. possibly infinitary) operations and a class $E$ of equations.
May the presentations of varietal large equational theories be neatly characterised? If not, is there a fundamental obstacle to doing so?
In the case that there is not a precise characterisation, I would also be interested in necessary, and sufficient, conditions for a large presentations to be varietal. One sufficient condition is that $\Omega$ be bounded in that there exists some cardinal $\kappa$ such that every operation has arity less than $\kappa$, but I would like to know if there are stronger conditions: for instance, one that implies the theory of sup-lattices, or of compact Hausdorff spaces, is varietal.
I think you're asking the following: We're given a presentation $(\Omega, E)$, where $\Omega$ is a specification of a class of function symbols for each arity $n \in CARD$ and $E$ is a class of equations in the language $\Omega$. We want to determine whether the category of free algebras $\mathcal T$ for this presentation is varietal, i.e. locally small. Right? One thing to note is that it's equivalent to ask whether the underlying set of each free algebra $T \in \mathcal T$ is small. Another theory one might like to treat would be compact Hausdorff spaces (in terms of ultrafilter convergence)
@TimCampion: yes, that's right. Compact Hausdorff spaces is another example I would be interested in: I'll update the question, thanks.
The following are equivalent:
$\mathcal T$ is varietal
Each free algebra $T \in \mathcal T$ is small
For each arity, there are a small number of $E$-equivalence classes of words in the language $\Omega$.
$(1) \Rightarrow (2)$ holds because the underlying set of an algebra can be identified with morphisms from the free algebra on 1 generator. $(2) \Rightarrow (1)$ holds because there are a small set of functions between any two small sets. The equivalence between $(2)$ and $(3)$ is hopefully clear.
I think there may not be much more to say at this level of generality: In cases where condition (3) is hard to check directly, then I suspect one might need to use more information about how $(\Omega, E)$ is presented in order to get a more usable criterion.
For instance, if $(\Omega, E)$ gives rise to a varietal theory, then it must be equivalently axiomatizable in the form $(\Omega', E')$ where $\Omega'$ has a small number of operations of each arity. But this condition is not sufficient.
"I think there may not be much more to say at this level of generality" — it's also not clear to me. However, it doesn't seem unreasonable that there might be at least sufficient syntactic conditions that cover examples like the powerset monad and CHS monad. It is perhaps helpful to rephrase the condition that $T$ be varietal in these ways, but these reformulations are very direct, and it's not clear that there aren't stronger results at this level of generality.
It might be interesting o look at the old paper of J. Reiterman https://dml.cz/bitstream/handle/10338.dmlcz/106455/CommentatMathUnivCarol_027-1986-2_12.pdf
|
2025-03-21T14:48:31.970651
| 2020-09-06T13:07:19 |
371010
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex M.",
"Denis Serre",
"Federico Poloni",
"https://mathoverflow.net/users/1898",
"https://mathoverflow.net/users/54780",
"https://mathoverflow.net/users/8799"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632758",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371010"
}
|
Stack Exchange
|
How to calculate inverse of sum of two Kronecker products with specific form efficiently?
I have a matrix with specific form of $A\otimes I + B\otimes J$ where $A$ and $B$ are general dense matrices, $n\times n$. $I$ is an $m\times m$ identity matrix. $J$ is a $m \times m$ dense matrix with 1 everywhere.
Is there any efficient way to calculate $(A\otimes I + B \otimes J)^{-1}$ efficiently?
If it is impossible, what if we assume $B$ is diagonal?
In more generality, we can write down an "efficient" inverse for both
$$A_1 \otimes A_2 + B_1 \otimes B_2 \quad \text{and} \quad A_1 \otimes B_2 + B_1 \otimes A_2$$
whenever $B_1$ and $B_2$ are nonsingular. The matrix in the question is of the fist form with $A_1 = B$, $A_2=J$, $B_1 = A$, $B_2 = I$ (since, as noted by Denis Serre in his answer, the matrix can only be nonsingular when $A$ is). The idea is to simultaneously diagonalize each pair $A_i, B_i$ independently, using the generalized eigendecompositions,
$$A_i V_i = B_i V_i \Lambda_i$$
where $V_i$ is the nonsingular matrix of eigenvectors and $\Lambda_i$ the diagonal matrix of eigenvalues. It's then immediate that the "generalized" Kronecker sums have generalized eigendecompositions
$$(A_1 \otimes A_2 + B_1 \otimes B_2)(V_1 \otimes V_2) = (B_1 \otimes B_2)(V_1 \otimes V_2)(\Lambda_1 \otimes \Lambda_2 + I \otimes I),$$
$$(A_1 \otimes B_2 + B_1 \otimes A_2)(V_1 \otimes V_2) = (B_1 \otimes B_2)(V_1 \otimes V_2)(\Lambda_1 \otimes I + I \otimes \Lambda_2),$$
whence
$$(A_1 \otimes A_2 + B_1 \otimes B_2)^{-1} = (V_1 \otimes V_2)(\Lambda_1 \otimes \Lambda_2 + I \otimes I)^{-1} (V_1^{-1}B_1^{-1} \otimes V_2^{-1}B_2^{-1}),$$
$$(A_1 \otimes B_2 + B_1 \otimes A_2)^{-1} = (V_1 \otimes V_2)(\Lambda_1 \otimes I + I \otimes \Lambda_2)^{-1} (V_1^{-1}B_1^{-1} \otimes V_2^{-1}B_2^{-1}),$$
provided none of the eigenvalues are zero. This simplifies further if one or both of the matrix pencils has more structure. E.g., if $A_2$ is symmetric and $B_2$ symmetric positive definite, then we can take $V_2$ such that $V_2^T B_2 V_2 = I$ so that $V_2^{-1} B_2^{-1} = V_2^T$. This is the case for the matrix in the question with $A_2 = J$, $B_2 = I$.
In the context of numerical computation, if the matrices are not symmetric and either matrix of eigenvectors $V_i$ is ill-conditioned, it may be better to work with the QZ decomposition (generalizing the Schur decomposition) instead, which has a similar tensor product structure. If $A_i, B_i$ have QZ decompositions
$$ A_i = Q_i S_i Z_i^*, \quad B_i = Q_i T_i Z_i^*, $$
where $Q_i, Z_i$ are unitary and $S_i, T_i$ triangular, then we also have the QZ decomposition
$$ (A_1 \otimes A_2 + B_1 \otimes B_2) = (Q_1 \otimes Q_2) (S_1 \otimes S_1 + T_1 \otimes T_2) (Z_1 \otimes Z_2)^*$$
$$ B_1 \otimes B_2 = (Q_1 \otimes Q_2) (T_1 \otimes T_2) (Z_1 \otimes Z_2)^*$$
since the Kronecker product of triangular (unitary) matrices is triangular (unitary). Thus
$$ (A_1 \otimes A_2 + B_1 \otimes B_2)^{-1} = (Z_1 \otimes Z_2) (S_1 \otimes S_2 + T_1 \otimes T_2)^{-1} (Q_1 \otimes Q_2)^*,$$
requiring the inverse of a large triangular instead of diagonal matrix. This approach works even when one or both $B_i$ are singular, requiring only that the pencils $A_i - \lambda B_i$ are regular (and again that no eigenvalues, which appear along the diagonal of $S_1 \otimes S_2 + T_1 \otimes T_2$, are zero).
If you just need to compute the action of the inverse, i.e., its product with a vector, then in practice your problem is better formulated as solving a generalized Sylvester equation. There are $O(\max(m,n)^3)$ algorithms based on this QZ decomposition approach in the linked paper.
Following Nickelnine37's idea, but in the right way, let us write your matrix
$$M=(A\otimes I)(I_{nm}+C\otimes J),\qquad C:=A^{-1}B.$$
Thus
$$M^{-1}=(I_{nm}+C\otimes J)^{-1}(A^{-1}\otimes I).$$
Finally
$$(I_{nm}+C\otimes J)^{-1}=I_{nm}+K\otimes J$$
where
$$K=-(I+mC)^{-1}C=(A+mB)^{-1}B,$$
because of $J^2=mJ$.
Is this what you want ?
Remark. I implicitely assumed that $A$ is invertible. If it is not, and $m\ge2$, then $M$ is not invertible, because $\ker M$ contains $(\ker A)\otimes(\ker J)$, where each factor is non-trivial.
Why is $A$ invertible?
@AlexM. Of course, it might not be. But if $A$ is not invertible, then $M$ is not, and the question just fade away.
Assuming $B$ and $J$ are invertible and that all matrices are square:
\begin{align}
(A \otimes I + B \otimes J)^{-1} &= \big( (AB^{-1} \otimes J^{-1} + I \otimes I)(B \otimes J) \big)^{-1} \\
&= (B^{-1} \otimes J^{-1})\big(AB^{-1} \otimes J^{-1} + I \otimes I\big)^{-1}
\end{align}
Now eigen-decompose $AB^{-1}$ and $J$.
$$
AB^{-1} = U \Lambda_U U^{-1}, \quad J = V \Lambda_V V^{-1}
$$
Substitute these definitions in.
\begin{align}
(A \otimes I + B \otimes J)^{-1} &= (B^{-1} \otimes J^{-1})\big(U \Lambda_U U^{-1} \otimes V \Lambda_V^{-1} V^{-1} + I \otimes I \big)^{-1} \\
&= (B^{-1} \otimes J^{-1})\big((U \otimes V)(\Lambda_U \otimes \Lambda_V^{-1} + I \otimes I)( U^{-1} \otimes V^{-1})\big)^{-1} \\
&= (B^{-1}U \otimes J^{-1}V)(\Lambda_U \otimes \Lambda_V^{-1} + I \otimes I)^{-1}( U^{-1} \otimes V^{-1}) \\
\end{align}
And there you have it.
$J$ is rank one, thus $J^{-1}$ does not make sense.
Indeed, it makes no sense to "assume" that $J$ is invertible, given that it isn't unless $m=1$.
|
2025-03-21T14:48:31.970932
| 2020-09-06T13:18:26 |
371011
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632759",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371011"
}
|
Stack Exchange
|
Kubilius model in higher sieve dimension?
The Kubilius model, based on the fundamental lemma of sieve theory, let us approximate the probability of events depending on the variables $X_p$, $p\leq y$, where $X_p=1$ if $p|n$ ($n$ a random integer $\leq x$) and $X_p=0$ if $p\nmid n$, by the probabilities of the corresponding events depending on independent Bernoulli variables $Y_p$, $p\leq y$, where $Y_p=1$ with probability $1/p$ and $Y_p=0$ with probability $1-1/p$.
Kubilius showed that the total variation distance is $O(e^{-c u})$, where $u = \log x/\log y$; Barban and Vinogradov (1964) improved the bound to $O(u^{-u/8} + x^{-1/15})$, and Tenenbaum (1999) improved it to $O_\epsilon(u^{-u} + x^{-1+\epsilon})$.
Question: what if we are dealing with a higher-dimensional situation (in the sense of sieves)? More precisely, say that what we actually want to model is events depending on $X_{p,i}'$, $p\leq y$, $1\leq i\leq \kappa$, where $X_{p,i}'=1$ if $n\equiv a_i \mod p$, and $a_1,\dotsc,a_\kappa$ are fixed integers?
It is clear that one can still define a Kubilius-type model then. The question is whether one can bound the total variation distance by $O(e^{-c u_\kappa})$, or by $O(u_\kappa^{-c u_\kappa} + x^{-c})$, or $O(e^{-c u} + x^{-c})$, for $u_\kappa = \log x/(\kappa \log y)$. Can one?
(A very naïve approach gives a possibly spurious factor of $(\log y)^\kappa$, and thus a bound of $(\log y)^{\kappa} O(e^{9 \kappa - u})$ on the total variation distance, if we use a Brun-based fundamental lemma as in Thm 6.9 of Opera de Cribro.)
|
2025-03-21T14:48:31.971055
| 2020-09-06T13:50:32 |
371012
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Loeffler",
"Martin Brandenburg",
"R. van Dobben de Bruyn",
"https://mathoverflow.net/users/2481",
"https://mathoverflow.net/users/2841",
"https://mathoverflow.net/users/82179"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632760",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371012"
}
|
Stack Exchange
|
Etale cohomology and Kummer theory
If $K$ is a field and $n \geq 1$ is such that $n \in K^{\times}$, then $H^1_{et}(\mathrm{Spec}(K),\mu_n)=K^{\times} / (K^{\times})^n$. This is easy to prove, see for instance Tamme, Etale Cohomology, Corollary 4.4.3. I assume that this is connected, perhaps even implies the main theorem from Kummer theory stating that, if $K$ has a primitive $n$th root of unity, then there is a correspondence between abelian extensions of exponent $n$ and subgroups of $K^{\times}/(K^{\times})^n$. I assume this is because $H^1_{et}(\mathrm{Spec}(K),\mu_n) = H^1_{et}(\mathrm{Spec}(K),\mathbb{Z}/n\mathbb{Z})$ is the Galois group of the maximal abelian extension of exponent $n$ of $K$? If so, can you give me a reference for this result? Sorry if this question is too basic, I didn't touch etale cohomology for years.
$H^1(K, \mu_n)$ is not equal to $H^1(K, \mathbf{Z} / n)$, but to its Pontryagin dual -- this is an instance of Tate duality.
Serre's Galois cohomology states this relation briefly in a remark (p. 73 in §II.1.2 of the second English edition), and refers to Bourbaki's Algèbre (Chap. V) for a statement of Kummer theory (but Bourbaki doesn't use Galois cohomology). In short: yes, this does imply Kummer theory with very little work.
@DavidLoeffler I assumed that $K$ has a primitive $n$th root of unity. That should be sufficient for $\mu_n = \mathbb{Z}/n\mathbb{Z}$?
@R.vanDobbendeBruyn Cool! This seems to be an answer.
The question was answered by R. van Dobben de Bruyn in the comments.
"Serre's Galois cohomology states this relation briefly in a remark
(p. 73 in §II.1.2 of the second English edition), and refers to
Bourbaki's Algèbre (Chap. V) for a statement of Kummer theory (but
Bourbaki doesn't use Galois cohomology). In short: yes, this does
imply Kummer theory with very little work."
|
2025-03-21T14:48:31.971188
| 2020-09-06T14:06:55 |
371013
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Qiaochu Yuan",
"Thurmond",
"https://mathoverflow.net/users/290",
"https://mathoverflow.net/users/78291"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632761",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371013"
}
|
Stack Exchange
|
Nonvanishing criterion for a polynomial of polynomials
Let $P \in \mathbb{F}_q[x_1, \dots, x_n][y]$ be some fixed polynomial. Substituting polynomials for the variables in $P$ gives a map $\mathbb{F}_q[x]^{n+1} \to \mathbb{F}_q[x]$ defined by
$$(Q_1(x), \dots, Q_n(x), Y(x)) \mapsto P(Q_1(x), \dots, Q_n(x))(Y(x)).$$
Suppose that for all $Y \in \mathbb{F}_q[x]$, the resulting polynomial $P(Q_i)(Y)$ is not the zero polynomial in $\mathbb{F}_q[x]$. Is this condition equivalent to some finite set of algebraic criteria on the $Q_i$?
The finite set of criteria should of course depend on $P$; it is also fine if the set depends on the characteristic $q$, so feel free to pick your favorite one and work there. For example, writing $P(Q_i)(Y) = c_d(Q_i)Y^d + \cdots + c_0(Q_i)$, it is necessary that $c_0(Q_i) \neq 0$ (take $Y$ to be the zero polynomial) and similarly that $\sum_{j=0}^d c_i(Q_i) \neq 0$ (take $Y = 1$). But these clearly are not sufficient.
I have some reason to suspect that there is no such finite set, but the argument sketch is very roundabout and there are some finicky details. I'd like to know if there is a more direct way to see this. It seems at first glance that a simple linear-algebra-style argument should work, but Lucas's theorem on binomial coefficients mod $q$ makes me suspicious that the details here are quite a bit trickier than one might expect.
I would also be interested in versions of this question where $\mathbb{F}_q$ is replaced by other fields or rings (preferably finite or countably infinite), especially if there is a positive answer to my question.
(This question is crossposted from Math.SE, where it was originally asked over a week ago but got no responses.)
What is an "algebraic criterion" to you? For example, consider $n = 1, P(x, y) = y^2 - x$, so that $P(Q(x), y) = y^2 - Q(x)$. Then your condition is that $Q(x)$ doesn't admit a square root; does this count?
@QiaochuYuan The interpretation I had in mind (which I'll denote (I1)) allows only vanishing or non-vanishing criteria on polynomials in the $Q_i$, so your example would not count. This interpretation makes the question seem like more of a pure commutative-algebra question. There is a more permissive alternative (I2), which is to consider any first-order sentence in $\mathrm{Th}(\mathbb{F}_q[X], +, \times, 0, 1, X)$ in the variables $Q_i$. I think this is too permissive for my tastes, but perhaps a slight enlargement of (I1) would be allowable. I'll have to think.
Isn't that a counterexample then?
@QiaochuYuan Sorry, I posted in the middle of editing. I need to think a little bit about enlarging the interpretation (I1); there is a rough analogy with the divisibility predicates necessary to do quantifier elimination in Presburger arithmetic that I am looking to preserve. Your comment was very helpful in getting me to think about this!
|
2025-03-21T14:48:31.971504
| 2020-09-06T14:10:24 |
371014
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Qiaochu Yuan",
"YCor",
"Yemon Choi",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/290",
"https://mathoverflow.net/users/3759",
"https://mathoverflow.net/users/763",
"the L"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632762",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371014"
}
|
Stack Exchange
|
Why is Hochschild homology interesting if its cohomology groups are infinite-dimensional?
I am trying to understand Hochschild homology, in particular the Hochschild–Kostant–Rosenberg theorem. As far as I understand this result gives an isomorphism between the algebraic (Kähler) differential forms of a smooth commutative $k$-algebra and the Hochschild homology. This clearly implies that the homology of the complex is infinite-dimensional, which seems strange to me. How can Hochschild homology be a good invariant of an algebra if the dimensions of its homology groups are all infinite-dimensional?
Edit: Based on the answers below, perhaps the interesting invariant of HH is the length of the complex, not the dims of the cohomology groups?
The K"ahler differentials of a smooth K-algebra are finitely generated.
Kahler forms are a module over the algebra and their “size” should be understood in terms of rank as a module. Have you worked through the example of affine space?
If $A$ has finite dimension over a field $K$, the (co)homology is that of a complex with finite-dimensional spaces, hence $HH_n(A)$ has finite dimension (bounded above by $\dim(A)^n$).
BTW you write in the title "is interesting" and in the text "is a good invariant". Both terms are subjective, but I'm not sure they have the same meaning. That HH is interesting does not a priori mean that it's a good invariant (assuming "good invariant" means "efficient in distinguishing non-isomorphic algebras"); there are other good reasons it can be "interesting".
This is a tiny bit like asking why simplicial homology groups are interesting when so often they are countably infinite sets. I think the point is that when you look at integral H1 of the two-torus you do not count the number of elements in this abelian group; as Qiaochu has alluded to above, what matters is the rank (in a suitable sense)
Regarding your thoughts about length: for the polynomial ring in d variables (or affine space if you are ore geometrical) the-result-which-serves-as-the-prototype-for the HKR theorem is an identification of HH_n(R,R) as an R-module with the n'th exterior power of $\Omega_{R/k}$ as an R-module, and if $M$ is a free R-module of rank $d$ then $\bigwedge_R^n M =0$ whenever $n\geq d+1$
|
2025-03-21T14:48:31.971693
| 2020-09-06T15:27:57 |
371022
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mohan",
"abx",
"https://mathoverflow.net/users/40297",
"https://mathoverflow.net/users/9502"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632763",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371022"
}
|
Stack Exchange
|
On definition of stable vector/Higgs bundle
Recall that the slope of a holomorphic vector bundle $\mathcal{E}$ over a smooth projective variety (or rather a compact Kähler manifold) $X$ is defined as
$\mu(\mathcal{E}) :=\frac{\operatorname{deg}(\mathcal{E})}{\operatorname{rk} \mathcal{E}}$
where $\operatorname{deg}(\mathcal{E})$ is defined as $c_1(\mathcal{E}) \cdot \omega^{n-2}$ for your favourite ample (or even Kähler) class $\omega$.
The bundle is called stable if for every subsheaf $\mathcal{F} \subset \mathcal{E}$, one has $\mu(\mathcal{F}) <\mu( \mathcal{E})$.
These are my questions:
Why do we require that $\mathcal{E}$ has no subsheaves with greater slope, rather then no subbundles? Is the difference sufficient and what is the motivation for this choice? More precisely,
1a) is there an explicit example of a non-stable vector bundle, which has no subbundles with bigger slope?
1b) is there a reasonable moduli space for holomorphic vector bundles having no destabilising subbundles? If yes, how far is it from the moduli space of stable vector bundles?
2)The definition of slope a priori depends on the choice of Kähler form. How much does the moduli space of stable vector bundles depends on this choice?
I am also interested in the analogues of the questions 1a) and 1b) for the Higgs bundles. In this case we require that a Higgs bundle $(\mathcal{E}, \theta)$ has no Higgs subsheaves(?) $(\mathcal{F}, \theta|_{\mathcal{F}})$ with bigger slope.
UPD:
-cross-list with https://math.stackexchange.com/questions/3816328/slope-stability-subsheaves-vs-subbundles
-In 1a) one can ask the same for semi-stability;
-On a curve every destabilizing subsheaf is contained in a subbundle, hence there is no difference indeed;
-For Gieseker-stability the answer for 2) is the theory of wall-crossings, but a have never seen a version of it for slope-stability (and for Higgs bundles as well). Is it exist?
The motivation is GIT. And yes, there is a big difference. On a variety of dimension $\geq 2$, a vector bundle has many more subsheaves than subbundles.
You should cross list from MSE, though the titles are different, but the main questions are the same.
|
2025-03-21T14:48:31.971859
| 2020-09-06T18:23:43 |
371028
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632764",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371028"
}
|
Stack Exchange
|
On regular holonomic $D$-modules on the disk
Denote by $\Delta\subset \mathbb{C}$ a small neighborhood of $0$. Let $M$ be a regular holonomic $D_\Delta$ module. In Bjork, "Analytic D-modules and applications", section 5.2, it is proven that
$$M=\oplus_{\alpha\in F} M_\alpha$$
where $F$ is a finite set of complex numbers with real part between $0$ and $1$, and $V_0(M_\alpha)$ is generated by a subspace annihilated by a power of $t\partial_t-\alpha$.
For $\alpha\not=0$, the modules $M_\alpha$ are direct sums of modules of the form $D_\Delta/(t\partial_t-\alpha)^k$. If $\alpha=0$ there is a list of various types of indecomposable objects that make up the direct summands of $M_0$.
I find the section in Bjork not so easy to understand, so I wonder if there is an alternative source for this material? Bjork refers to this book, but I do not have access to that.
Also, what I am most interested in is understand how to compute this decomposition when given a $D_\Delta$ module, for example as a quotient of a free $D_\Delta$ module.
p.s. I am aware of a perverse-sheaves perspective on this question that is written in Dimca, but I would like an exposition purely in the language of $D$-modules.
|
2025-03-21T14:48:31.971970
| 2020-09-06T19:30:58 |
371030
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Emil Jeřábek",
"Noah Schweber",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/8133"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632765",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371030"
}
|
Stack Exchange
|
Does $\mathsf{Q}$ have any interesting provably recursive functions?
This question was asked and bountied at MSE without success.
For an appropriate theory $T$, say that an $n$-ary $T$-provably recursive function is a $\Sigma_1$ formula $\varphi$ with $n+1$ free variables which $T$ proves defines a total function. Terms and "near-terms" (e.g. "$\Delta_1$-piecewise combinations" of terms) are of course always $T$-provably recursive, but in general these account for only a miniscule fragment of the $T$-provably recursive functions (consider exponentiation in $\mathsf{PA}$).
I'm interested in whether $\mathsf{Q}$ has any such "surprising" provably recursive functions:
Is there any $n$-ary $\mathsf{Q}$-provably recursive function $\varphi(\overline{x},y)$ such that for every $\{+,\times\}$-term $t(\overline{x},\overline{z})$ we have $$\mathsf{Q}\vdash\forall \overline{z} \exists a\forall \overline{x}[\bigwedge_{1\le i\le n}x_i>a\implies \neg\varphi(\overline{x},t(\overline{x}, \overline{z}))]?$$
Intuitively, I'm asking whether there is a $\mathsf{Q}$-provably recursive function which is $\mathsf{Q}$-provably eventually different from every term. Note that "eventually different" here is used in the sense of "different on all inputs with each coordinate sufficiently large." In particular, a function like $$\varphi(x_1,x_2,y)\equiv (y=0\wedge x_1=x_2)\vee(y=1\wedge x_1\not=x_2)$$ is not an example; while there is no specific term to which it is eventually equal, both $0$ and $1$ are terms to which it is not eventually not equal. (I'm like $80\%$ sure I got that right.)
@MattF. Yes, exactly - but it's not clear to me how to do that.
The predecessor function is provably total in $Q$, and it is eventually different from each term in the standard model, but under its most natural definition, $Q$ does not prove the formula you want. I’m not sure if it can be reformulated so that $Q$ does prove the formula. In general, it’s extremely difficult for $Q$ to prove any two functions to be eventually different, because of black-hole models of $Q$ such as $\mathbb N\cup{\infty}$ here.
@EmilJeřábek That's a good point - it would probably be better to add monus to the language and go a bit beyond $\mathsf{Q}$.
Actually, the $\mathbb N\cup{\infty}$ model outright shows that the answer is NO, and you don’t even need $+$ and $\times$ (just constant terms): let $z$ be such that $\mathbb N\cup{\infty}\models\varphi(\vec\infty,z)$; then $\mathbb N\cup{\infty}\models\forall a,\exists\vec x,\bigl(\bigwedge_ix_i>a\land\varphi(\vec x,z)\bigr)$, namely $\vec x=\vec\infty$.
|
2025-03-21T14:48:31.972154
| 2020-09-06T20:32:32 |
371037
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asaf Karagila",
"Noah Schweber",
"Todd Eisworth",
"Yair Hayut",
"https://mathoverflow.net/users/18128",
"https://mathoverflow.net/users/41953",
"https://mathoverflow.net/users/7206",
"https://mathoverflow.net/users/8133"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632766",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371037"
}
|
Stack Exchange
|
Good forcings with bad squares
Consistently with $\mathsf{ZFC}$ there is a forcing which preserves cardinals but whose square does not always preserve cardinals - that is, some $\mathbb{P}$ such that for every $\mathbb{P}$-generic $G$ we have $\mathrm{Card}^{V}=\mathrm{Card}^{V[G]}$ but for some $\mathbb{P}^2$-generic $H=\langle H_0,H_1\rangle$ we have $\mathrm{Card}^{V}\not=\mathrm{Card}^{V[H]}$.
However, the only way I know how to get this is via a bit of a cheat: find two different forcings which are individually "good" but have "bad" product, and then look at their lottery sum. This construction has the drawback that the square of the resulting forcing doesn't always collapse cardinals - we're only guaranteed "bad" behavior in the extension if the two coordinates of our generic lie on different "sides" of the original lottery sum.
I recall$^*$ seeing a stronger example of this phenomenon, but I can't track it down or reconstruct it on my own:
Is it consistent with $\mathsf{ZFC}$ that there is a forcing $\mathbb{P}$ such that $\mathbb{P}$ preserves cardinals but $\Vdash_{\mathbb{P}^2}\mathrm{Card}^V\not=\mathrm{Card}^{V[\langle H_0,H_1\rangle]}$?
$^*$Actually my original memory was that even the weaker phenomenon can't happen, but after it was pointed out to me that it can, I now remember differently. I'm sure eventually my memory will have been right. :P
Consistently? Sure. Suslin trees are ccc.
@AsafKaragila Does forcing with the square of a Suslin tree (always) collapse cardinals? I wasn't aware of that.
I believe so, because $T\times T$ will collapse $\omega_1$. But now I'm not sure. But I'm pretty sure that it is at least consistently the case.
@AsafKaragila Yeah, I don't see that $T\times T$ need collapse $\omega_1$. I wouldn't be surprised if it were consistently the case, but I don't see it.
I mean, $T\times T$ definitely has a condition that collapses $\omega_1$. If $T$ is homogeneous, that should be enough.
@AsafKaragila Sorry, I'm not seeing it - why does there need to be such a condition?
I may be misremembering. Let's see what someone more knowledgeable says.
Will a self-specializing Souslin Tree do the trick here?
@ToddEisworth Yup, that looks like it does it! (I didn't know such things existed, they're neat - thanks for bringing them to my attention.) If you add that as an answer I'll accept it.
I only know about them through Mathoverflow!
It seems obligatory to give the example of Todocevic square: if $\mathbb{S}$ adds a generic $\square(\kappa)$ for $\kappa$ regular, and $\mathbb{T}$ is the threading forcing, then $\mathbb{S} \ast \mathbb{T}$ is equivalent to the $\kappa$-Cohen forcing while $\mathbb{S} \ast (\mathbb{T}^2)$ collapses $\kappa$ to be countable.
A self-specializing Souslin tree gives you a ccc notion of forcing whose square collapses $\omega_1$ (See, e.g., the answer to
Ultrafilters preserved by $\mathbb{P}$ but not by products?). Such trees exist under $\diamondsuit$, and so are consistent with ZFC.
One can indeed prove more: starting with $V=L$, there exists a tame and cardinal preserving class forcing notion $\mathbb{P}$ such that forcing with $\mathbb{P} \times \mathbb{P}$ collapses
all uncountable cardinals. See
Adam Figura, Collapsing algebras and Suslin trees, Fundamenta Mathematicae 114 (1981) 141-147, doi:10.4064/fm-114-2-141-147.
Very cool! (I've made a minor clarifying edit.)
|
2025-03-21T14:48:31.972440
| 2020-09-06T21:00:13 |
371038
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Akshat Mahajan",
"Alexander Woo",
"Anthony Quas",
"Balazs",
"C.F.G",
"Charles",
"John Coleman",
"LSpice",
"MetaSean",
"Peter LeFanu Lumsdaine",
"Piyush Grover",
"Ryan Budney",
"Stanley Yao Xiao",
"Timothy Chow",
"Yemon Choi",
"https://mathoverflow.net/users/100575",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/11054",
"https://mathoverflow.net/users/1465",
"https://mathoverflow.net/users/152049",
"https://mathoverflow.net/users/164954",
"https://mathoverflow.net/users/165131",
"https://mathoverflow.net/users/2273",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/30684",
"https://mathoverflow.net/users/3077",
"https://mathoverflow.net/users/3106",
"https://mathoverflow.net/users/6043",
"https://mathoverflow.net/users/6107",
"https://mathoverflow.net/users/763",
"https://mathoverflow.net/users/89084",
"https://mathoverflow.net/users/90655",
"user164954",
"user676464327"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632767",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371038"
}
|
Stack Exchange
|
Learning mathematics in an "independent and idiosyncratic" way
This is a question about learning mathematics outside of the standard undergraduate/graduate education.
The following is a quote from Thurston's On Proof and Progress in Mathematics:
My mathematical education was rather independent and idiosyncratic, where for a
number of years I learned things on my own, developing personal mental models
for how to think about mathematics. This has often been a big advantage for me in
thinking about mathematics, because it’s easy to pick up later the standard mental
models shared by groups of mathematicians. This means that some concepts that
I use freely and naturally in my personal thinking are foreign to most mathematicians I talk to. My personal mental models and structures are similar in character
to the kinds of models groups of mathematicians share—but they are often different models
The above quote suggests that Thurston's non-traditional mathematics education was key to his unique insight and contribution to mathematics. I know other examples of very successful mathematicians who benefited from learning outside the standard mathematical canon.
I am trying to incorporate this into my own learning, but I am struggling to get started. For one thing, the standard canons of mathematics are often written by masters of the respective fields who are also masterful writers (e.g. Thurston, Milnor, Serre, Stein). On the other hand, there are many lesser known texts which are poorly written and offer much less insight compared to the canons (even if they are idiosyncratic). Thus, one obstruction to pursuing the above "independent and idiosyncratic" approach to mathematics seems to be in judging what books are truly useful.
My Question. How specifically does one learn mathematics in an "independent and idiosyncratic" way?
Subquestions that may help answer the above:
1.) What should one look for when choosing a textbook? Again, I am looking for nonstandard ways of learning mathematics.
2.) Are there thing to keep in mind when one thinks about problems, theorems, etc.?
3.) What are the sociological obstructions to achieving the above goal? (e.g. as an undergraduate, there is an obvious push to learn mathematics in a certain way. Can this be an obstruction to achieving independence and idiosyncracy?)
I think the answer to my question depends on the audience, so for this question, let's restrict our attention to upper level undergraduates or beginning graduate students. Also, I am asking this question on MathOverflow (as opposed to stackexchange) because my goal is to become a successful research mathematician (as opposed to, say getting better grades in a math class).
Easy answer: read physics texts. As a math undergrad, take courses in classical/quantum mechanics, fluid mechanics, relativity etc. Physicists have very different mental models for many familiar math objects, and it is a 'superpower' if you can have both.
I would say your goal is a priori misguided. Mathematics are a certain way most likely because it works. Doing something different for no reason other than to be different is akin to reinventing the wheel. Only do things differently if you think it's better, and not simply because you desire to be distinct.
There is an implicit assumption of causality in your question. Thurston learned in an independent and idiosyncratic way; and Thurston was a great mathematician. I suspect it's more likely the causality went in the opposite direction: Thurston was a great mathematician; and therefore had the ability to read independently and idiosyncratically and build his own helpful but different mental models.
You may find Miles Reid's "Obituary" https://homepages.warwick.ac.uk/staff/Miles.Reid/Personal/obituary (a sort of autobiography) worth reading, both to see a non-standard path and its pitfalls. Caveat emptor: the causality issue mentioned by Anthony may well apply here also.
I was going to leave a comment along the lines of the one by @AnthonyQuas but he has put it more clearly than I would
Re mech engineering
Everything piyush said. Plus the modern field is a bit of a funny chimera of applied math methods in lots of fields; sometimes it's almost "abstract engineering". They use a lot of interesting stuff in novel ways. In particular,a lot of dynamics that's both non-trivial and very different in flavor from pure math or theoretical physics. Eg the mechanics of running and throwing-- a former colleague studies what fraction of energy when we run goes into stability versus propelling you forward, and (continued)
does a prori calculations about throwing without modelling the muscles directly, how to understand how it works and what motions maximize power eg.
Another mechE associate of mine showed me some papers that studied defects in solids or crystals or something, and they were modelling by the torsion of some connections. I forget where exactly the connection lived and all other details, pardon. But my point is that it's a sophisticated use of highly non-trivial math in a way different than I at least have seen in a lifetime of differential geometry.
Having your own questions you would like to see answered, questions that aren't currently being asked in mathematics. That would be a way to start down an independent and idiosyncratic road. Of course, these questions would have to come from you, not some textbook.
I strongly considered majoring in computer science, so I actually first seriously learned about writing proofs in the context of proving algorithms correct rather than the context of algebra or analysis. I do think compared to most mathematicians I have a more computational style of thinking - sometimes even applied to ideas that one can't actually compute with because some step requires processing an infinite amount of data - and an approach to coming up with proofs that more closely mimics the process of coming up with an algorithm. But my background isn't so uncommon these days.
Thurston (1946 –2012)
If you really want to learn mathematics in an independent and idiosyncratic way, then asking other people for advice and following that advice is precisely the wrong way to go about it.
Thurston thought about mathematical problems and structures largely on his own, and in his own way. He didn't post on MathOverflow asking people how to pick a textbook. If you want to be independent and idiosyncratic, then you should think about things your own way as much as possible instead of reading and following what others are saying.
Now, you also said that you want to become a successful research mathematician. You should be aware that this goal is somewhat in tension with your goal of being independent and idiosyncratic. There's a saying in chess, that to be original is to lose many games. Math isn't chess, but it's still true that following a conventional route is going to give you better odds of success. So you have to decide what's more important to you—worldly success, or an independent mind?
I disagree with this. I attempted to do this when younger, simply picking books off the shelves that I found interesting. I fell into the trap of the expert beginner: I learned what I was comfortable with, not what I struggled with. Education requires structure and guidance. OP should get a mentor, not just be told to go wild in a library.
Note carefully that I said "if you want to learn mathematics in an independent and idiosyncratic way." I didn't say that one should learn mathematics in an independent and idiosyncratic way. That is for the OP to decide.
I would somewhat disagree with this answer. There’s a huge middle ground between “learn in the most conventional way” and “work primarily alone and avoid listening to what others say”. The latter is very hard to be successful with — I’ve met many more cranks who’ve tried it than good mathematicians. By contrast, some of the most individual and idiosyncratic mathematicians I know are voracious readers and good listeners — they just analyse and re-interpret things according to their own worldview and tastes.
@PeterLeFanuLumsdaine : You're right, of course, but I felt it was important to point out the irony in the OP's request, which I'm not sure the OP was aware of. Independence of mind is about learning how and when to trust one's own instincts, and not about following Ten Easy Steps to Becoming an Independent Thinker. (I'm reminded of magazines for how to live a simple life. My first advice for living a simple life would be to cancel all magazine subscriptions.)
Rather than aspiring to think outside the box, I would recommend working to have a bigger box. I would say that many of the greatest mathematicians, e.g. J.-P. Serre, learned and took in mathematics very broadly, and thus had a great set of tools at their disposal when trying to crack a question, and thus were ready at the right time to have `revolutionary' ideas, and develop new fields. A great thing about trying to take in lots of mathematics is that this can continue throughout your life!
There is a general sense among educators and inspirational speakers that the best learning happens by making mistakes. (I would love to find some science that backs this up!) At the very least that means that doing exercises is as important as reading texts. I do think (having had the privilege of spending some time with them) that the most idiosyncratic and independent mathematicians are the ones most willing to think for themselves and produce their own mistakes.
For example, this can mean that you stop at every point where a statement is unclear and try to fill it in yourself. This is not always easy, but can be really rewarding. It seems harder to learn an entire subject this way; I rather think of it of straying in the woods, and occasionally having to find your way back to the main path. I certainly don't have a recipe for this (nor in fact do I claim that I master this myself).
Mathematics has this strange habit of not documenting mistakes and dead ends as well as scientific principles tell us (how often have you seen a paper whose main topic is "we tried a thing and we got stuck for the following reason"?). It is unique among the sciences (if you can call it that) in not systematically documenting the methodology; rather only the final result is presented. If you really want to understand how great mathematicians work, it would be great if they had documented the mistakes in their thinking.
Some of this is actually available in print! For example, the Serre–Tate correspondence documents a lot of the developments of group cohomology and many other topics as they were evolving, and it is a fascinating read for exactly that reason. It gives that rare insight into the minds of the mathematicians rather than merely the final product that comes out of it.
Let me end with what Goro Shimura said about Yutaka Taniyama (in the Horizon documentary on FLT):
He was not a very careful person as a mathematician. He made a lot of mistakes. But he made mistakes in a good direction. I tried to imitate him. But I've realized that it's very difficult to make good mistakes.
The type of learning that you're alluding to falls under the umbrellas of "constructivism" (as opposed to something like "behaviorism") and "situated cognition". If you want to go down the research rabbit hole related to this, check out https://en.wikipedia.org/wiki/Constructivism_(philosophy_of_education) and https://en.wikipedia.org/wiki/Situated_cognition
I vividly remember, in the PBS documentary about Wiles's proof of FLT, Shimura saying of Taniyama, "he made mistakes, but they were mistakes in the right direction." My (non-mathematician) mother was watching with me, and I think that was as close as she ever came to understanding the nature of the practice of mathematics.
(I'm sorry that I got so excited to relay that quote that I missed that it was the last paragraph of your answer. I'll leave it there because I still enjoyed the way it let me communicate the experience of my job with my mother.)
If you intend to learn mathematics in an "independent and idiosyncratic" way, do so with a certain amount of intellectual humility. For every Thurston who can successfully pull something like this off, there are several mathematical cranks who have convinced themselves that e.g. they have discovered an elementary proof of Fermat's Last Theorem. I've dealt with my fair share of them over the years. Thinking outside the box can be helpful, as long as the box in question isn't the box of careful reasoning.
A mathematical crank is not necessarily a person who is incapable of good mathematical reasoning. The most interesting crank I have personally dealt with is a person who had independently hit upon the idea of a factorial number system and worked out many of its properties. For example, he realized that in this system it wasn't obvious that a number can be represented in only one way, and had successfully worked out a good argument for the uniqueness of representation. All this came from an idea that occurred to him while listening to his high school math teacher talking about numerical bases other than 10. I was genuinely impressed with this and told him so. The transition to crankery came when he insisted that his scheme could be used to compress all data, with the idea being that computers used standard number bases and that his system allowed for the expression of all numbers using less "digits". He seemed impervious to simple pigeon-hole arguments about the impossibility of a universal compression scheme. There is no question that he had learned a nontrivial amount of mathematics in an "independent and idiosyncratic" way. Unfortunately, despite some genuine talent on his part, his idiosyncratic way was prone to error.
While otherwise agreeing, I think that "for every Thurston, there are several cranks" is off by several orders of magnitude.
@LSpice While otherwise agreeing with your comment, the subset of independent and idiosyncratic people who are interested in learning mathematics rather than independently doing research using whatever they happen to already know is probably less prone to crankery than a randomly chosen independent and idiosyncratic person.
There are many many many masters recent and old. They've written not just their most famous books, but other monographs and research papers. Nobody is reading all of them. Part of the point is not that what you learn has to be quirky, but that when confronted by problems in some domain, you have some tricks up your sleeve that others looking at the same problem may not.
So for starters, you could add to your list some masters who wrote wonderful books on varied subjects: in no particular order, Hilbert(-Courant), Weyl, Gelfand, Arnold, Kolmogorov, Hardy, Gromov, Bott, Chern, Siegel, Poincare, Gauss, Euler, Sternberg, Whitney, E. Artin, E. Cartan, Knuth, Novikov, Lefschetz, Wiener, Pontryagin, Alexandrov, Weil, Shafarevich, Manin, Hirzebruch, Atiyah, Donaldson, Ahlfors, Thom, Mumford, JH Conway, Coxeter, Berger, F. Klein.
...to say nothing of many others whose research papers may be of great use, including some that are very accessible. Someone in a different field could write a largely disjoint set of masters who wrote great books.
On top of many excellent mathematicians who wrote lovely books, both standard and idiosyncratic (hot tip: William Burke).
And many many many others from before the 20th century; I mentioned only a couple whose work I've read enough to make a firsthand claim that it's accessible and still of use. There are many nice volumes with original sources, that are chosen for accessibility and often have nice commentary.)
Also: besides individual "masters" there are many "traditions"--Russian representation theory, Polish topology and functional analysis, Hungarian combinatorics, etc. etc.
And to add to a comment: not just physics, but computer science, mechanical engineering, signals and systems, and other fields where people are using mathematics at a high level with different models than pure mathematicians.
And in any case, it's all what you make of it. Any of these sources or the authors you mentioned wrote works of incredible depth. You could take one volume of Milnor or Serre or Weyl and go off to the forest for a year, and you won't exhaust it. I went to a lecture by Steven Chu once in which he mentioned that in college, he took the minimal courseload he could and just read and reread the Feynman Lectures. You could do worse than that.
Or find a nice topic you enjoy and go deep, and you'll find it connects to many many things. Follow the connections in all directions, and presto! You understand something important from a dozen perspectives. Nobody else will have precisely the same toolbox.
I know that computer science subfields (e.g. algorithms, machine learning, and theoretical CS such as combinatorics and graph theory) and signals & systems (e.g. compressed sensing) use a lot of cool mathematics. What are mechanical engineering subfields that can be useful for mathematicians?
@user676464327 The history of mechanics is intertwined with math going back to Newton. Research in continuum mechanics, vibration, fluid mechanics, robotics all use sophisticated math ideas ..take your pick.
To effectively learn "independently" (and perhaps "idiosyncratically"), you need to first learn to critically question the material being presented. Every instructor or author will teach/write in their own "idiosyncratic" method. Learn to differentiate the standard material with the idiosyncratic, then reorganize what you learned into your own "idiosyncratic" style. This isn't (nor shouldn't) be about taking shortcuts. It is actually harder. But if you are not satisfied with the "standard" presentation, you need to ask "why?" you are not satisfied.
In other words, don't tell someone to turn off their flashlight unless you know your flashlight is better.
If you are serious about alternative methods I would suggest you try Lewis Caroll's "Alice in Wonderland" -- it is full of mathematical ideas, and very entertaining.
Dodgson (Caroll) was a third-rate mathematician; not sure how I feel about this advice.
|
2025-03-21T14:48:31.973822
| 2020-09-06T21:17:41 |
371039
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"dohmatob",
"https://mathoverflow.net/users/78539"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632768",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371039"
}
|
Stack Exchange
|
If $X \sim N(0,I_m)$, what is a necessary and sufficient condition on $u_m > 0$ such that $\lim\sup_{m\to \infty} P(\|X\|^2 \ge u_m|X_1|) = 1$
Let $m$ be a large positive integer and $X=(X_1,\ldots,X_m) \sim N(0,I_m)$. I wish to show that the squared norm of $X$ is much much bigger than the absolute value of any of the $X_j$'s. For example, one can show that$P(\|X\|^2 \ge \mathcal O(\sqrt{m})|X_1|) = 1-o(1)$. Recall that $o(1)$ is notation for any quantity which goes to $0$ in the sup-limit $m \to \infty$.
My goal is to improve the $\sqrt{m}$ to something much larger. To this end, let $(u_m)_m$ be a sequence of positive numbers.
Question. What is a sufficient (and ideally, also necessary) condition on the growth of $u_m$ such that $P(\|X\|^2 \ge u_m |X_1|) = 1-o(1)$.
For real $u_m>0$, the probability in question is
$$p_m:=P(\|X\|^2\ge u_m|X_1|)=P\Big(\frac{|X_1|}{\|X\|^2/m}\le\frac m{u_m}\Big).$$
Passing to a subsequence, without loss of generality
$$\frac{u_m}m\to c\in[0,\infty]$$
(as $m\to\infty$). By the law of large numbers, $\|X\|^2/m\to1$ in probability. So, by Slutsky's theorem ,
$$p_m\to p_\infty:=P(|X_1|\le1/c),$$
assuming $1/0:=\infty$ and $1/\infty:=0$. Thus, $p_m\to1$ iff $c=0$. That is, $P(\|X\|^2\ge u_m|X_1|)=1-o(1)$ iff $u_m=o(m)$.
Great. Thanks for the nice answer!
|
2025-03-21T14:48:31.973953
| 2020-09-06T21:44:23 |
371041
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Damien Robert",
"David Roberts",
"Marc Hoyois",
"https://mathoverflow.net/users/20233",
"https://mathoverflow.net/users/26737",
"https://mathoverflow.net/users/4177"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632769",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371041"
}
|
Stack Exchange
|
Geometric stacks, groupoids and étendues
If $(C, \tau)$ is a site with pullbacks and $\tau$ subcanonical, it is well known that these things are essentially equivalent:
Groupoids $s,t: U_1 \to U_0$ where $s,t$ are covering for the $\tau$-topology
Geometric stacks, that is stacks (in groupoids) over $C$ with a representable $\tau$-covering from an object of $C$.
Etendues, that is a topos over $\mathrm{Sh}_{\tau}(C)$ which is locally isomorphic to the small topos $\mathrm{Sh}(U_\tau)$ of coverings of $U \in C$.
Essentially to $U: U_1 \to U_0 \times U_0$ we associate the stackification of the groupoid functor associated to $U$. If $X$ is a geometric stack, and $U_0 \to X$ a presentation, we associate the groupoid $U_1=U_0 \times_X U_0 \to U_0 \times U_0$. If $X$ is an étendue, and $F \to 1$ is a well supported object such that $X/F=\mathrm{Sh}(U_\tau)$ with $U \in C$, then via $F \times F \to F$, we get (I think) that $X/(F \times F)$ also represents $\mathrm{Sh}(U'_\tau)$. And $U' \to U$ is the corresponding groupoid.
Main question: is this completely spelled out in the litterature? The closest I have been able to find is
Dorette A. Pronk, Etendues and stacks as bicategory of fractions, but she only treat some particular cases like topological stacks, differentiable stacks and algebraic stacks. What I am looking for is some kind of axiomatic treatment of geometric stacks that outlines the three approaches (and what are the exact conditions for them to be equivalent).
Second question: Pronk assumes in her statement on $DM$-stacks as étendues that
$X/(F \times F)$ comes from a $U'$. Does this not follow from the hypotheses (see also this Stack project blog post) ?
The situation is a bit subtle if we consider theses objects as part of a $2$-category. For example for groupoids we really want to invert weak isomorphisms. (For that I like the theory of anafunctors as in
David Roberts, Internal categories, anafunctors and localisations.) For étendues it is important to consider the topos over the big topos; algebraically this is the same as looking at the topos with its structure sheaf (ie see it as a locally ringed topos in the case of algebraic stacks).
The papers by David Roberts and David Carchedi seem interesting, but I have not been able to find the full link between étendues and geometric stacks completely explained. The link between stacks and groupoids is a lot more standard. For instance the post Link between internal groupoids and stacks on a topos ? by Carchedi explains well the link between (standard) stacks over a base site and internal groupoids (or categories) in the topos.
I have a (third and final) related question about geometric stacks: is the correct definition the one I gave, or should we require the stronger condition that the diagonal of $X$ has to be representable as well? For algebraic stacks this is equivalent, since a surjective representable cover implies that the diagonal is representable by effectivity of fppf descent for separated locally quasi-finite morphisms. What happens for geometric stacks when this descent condition is not satisfied? Does the iterated construction even converge?
Edit: for the third question, in HAG2, Töen and Vezzosi show that the link between geometric stacks and groupoids is still valid in derived algebraic geometry. This is their Proposition <IP_ADDRESS>. In their Assumption <IP_ADDRESS>, they do assume that a $\tau$-covering satisfy the descent condition.
I should introduce some motivations for this question: I'd like to see a treatment of algebraic (DM) stacks from the internal point of view of their étale topoi. Standard references for schemes are Hakim's thesis, and more recently
Ingo Blechschmidt thesis, Using the internal language of toposes in algebraic geometry.
It seems the étendue point of view would be the best suited from an internal logic perspective. Of course the ultimate goal would probably be an internal description of (derived) algebraic geometry expressed in the logic of their $\infty$-topos via HoTT, but meanwhile I would also be happy for just an internal description of standard algebraic stacks.
I don't think geometric stacks and étendues are equivalent. For example if $(C,\tau)$ is affine schemes with the étale topology, then étendues are arbitrary DM stacks, whereas geometric stacks are algebraic stacks with affine diagonal; neither is a special case of the other.
Yes, this is precisely the type of things that make me wish for an exhaustive treatment of when the equivalence holds. Your comment show two things: 1) we want to have some kind of saturation condition on $C$, eg a geometric stack that is a sheaf should be in $C$. 2) In my assumptions I use the same type of morphisms for the topology and for presentations. So for the étale topology, we require a geometric stack to have an étale presentation, so to be DM. A more exhaustive treatment would distinguish between the topology and the presentations (like in HAG2).
So in this case the link with an étendue is less clear. Still for general algebraic stack, we can see them as stacks for the fppf topology with an fppf presentation, or as stacks for the smooth topology (hence the étale topology) with smooth presentation. So I guess the equivalence with étendues still holds (but the only proof I know is from Pronk and she only treat the DM case).
Also the saturation condition reminds me of a comment I forgot to make on requiring $\tau$-covering to satisfy the descent condition: it looks like the key condition to bound the number of time we iterate the construction before it saturates; something like iterating n+2 time to get n-geometric stacks. We can even troncate along the way, so iterate twice to get geometric sheafs, then geometric 1-stack, 2-stacks and so on (so for algebraic stack this would go affine -> alg space with affine diagonal -> alg space -> alg stack rather than the more standard affine->schemes->alg space...).
@Marc I believe the OP is using the term "geometric stack" in a generic sense, to cover Artin, DM, differentiable, topological, etc as the site/presentation varies, rather than a technical definition. Fwiw, Damien, I prefer to call these "presentable" stacks, because of this terminology clash.
@Marc: wait, if $(C,τ)$ is affine schemes with the étale topology, then geometric stacks are stacks with an affine étale (representable) presentation, and they do correspond to étale affine groupoids and étendues (since $F$ corresponds to $U \to X$ which is étale, and representable since pullbacks are above $F$ so should be affine by definition of an étendue).
@Roberts: sorry, I'll edit!
No worries! You'd be surprised at the number of people who accidentally call me Robert, as if it's my first name... :-D
"should we require the stronger condition that the diagonal of has to be representable as well" yes, I think so.
@David: Indeed, imposing the diagonal to be representable indeed feels more "geometric". But in this case the equivalence with the groupoid approach fails no? (Or rather we then need to impose the diagonal condition everywhere.) For the terminology, I used the one from the nlab: https://ncatlab.org/nlab/show/geometric+stack. But I have also seen the presentable stack terminology (maybe from your articles?).
There's also the following idea: you don't just impose on your groupoids that their source and target come from a pullback-stable class, but also the diagonal map $(s,t)$. For instance, we can ask that $s$ and $t$ are étale, and that $(s,t)$ is proper. This is reflected in the condition placed on the diagonal of the stack.
|
2025-03-21T14:48:31.974512
| 2020-09-06T22:00:49 |
371043
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben",
"KReiser",
"https://mathoverflow.net/users/112114",
"https://mathoverflow.net/users/150898"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632770",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371043"
}
|
Stack Exchange
|
Generalization of: The dimension of a projective $\mathbb{F}$-variety equals the smallest codimension of a disjoint linear subspace
Let $\mathbb{F}$ be an algebraically closed field. Consider the following definition of the dimension of a (quasi)projective $\mathbb{F}$-variety, given in Harris Algebraic Geometry: A First Course:
It seems nonstandard to take this as the definition of a variety, so I will think of this as a theorem that should be proven from a more conventional definition of dimension, e.g. the Krull dimension.
My question is this: Does Definition 11.2 generalize to structures other than a variety over an algebraically closed field? For example, what if $\mathbb{F}=\mathbb{R}$? Or $X$ is a manifold embedded in projective space? In these cases, does the dimension of $X$ still agree with the smallest codimension of a disjoint linear subspace?
Note that Definition 11.2 holds when the ``irreducible" assumption is dropped, so this small generalization does hold.
When you say "variety", what do you mean? The answer is probably "yes, it generalizes" if you set it up the right way and "no, not quite" if you don't.
@KReiser I mean a variety in the classical sense: the common zero-locus of a set of polynomials.
@KReiser I guess what I'm looking for then is a reference request. Is there a reference to such a result when $X$ is, for example, a manifold embedded in projective space?
Such a statement is easily seen to be false: draw a small circle in $\Bbb RP^2$ that misses some line, for instance. Depending on your definitions for varieties over $\Bbb R$, this could also be a counterexample for that case too.
@KReiser Thanks, that's a nice example. In my particular case, I'm trying to find the maximum dimension of a linear subspace of $n \times n$ matrices disjoint from the real $n^2$ dimensional Lie group $U(n,\mathbb{C})$. Based on your example, this seems more difficult than I expected.
XY problems are particularly poorly suited for MO/MSE. Please consider providing context about your underlying problem to avoid this sort of issue.
Let us continue this discussion in chat.
|
2025-03-21T14:48:31.974702
| 2020-09-07T00:01:11 |
371047
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anixx",
"Circle B",
"David Roberts",
"Gerald Edgar",
"https://mathoverflow.net/users/10059",
"https://mathoverflow.net/users/164902",
"https://mathoverflow.net/users/4177",
"https://mathoverflow.net/users/454"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632771",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371047"
}
|
Stack Exchange
|
How to plot this fractal
I'm a graphic designer and my client has asked me to use this fractal in a design that I'm working on. As you can see, it's not a very good copy, so I'm trying to see if I can generate a high-resolution version for the project. I've found all sorts of fractal generators that work great and many of them will allow me to enter a specific equation. The trouble is that I'm a designer and not a mathematician! So the question is this: is it plausible to think that this could be reverse engineered, or is that an insane idea? If it's plausible, how would I achieve it?
EDIT:
I found the original from the book (below). The values @Carlo Beenakker posted got me very close to the right area. I was able to render and warp the output and achieve what we were looking for! It's not exact, but very close.
At first I was doubting this question was on-topic, but I think there's a very interesting problem here, and it seems like an applied maths kind of thing: how to model observed behaviour, especially the kind of multi-scale self-similarity we have here?
Are we sure this is not some high-order point in the Mandelbrot set?
So, I've found out a little more info on the fractal, my client found it in a book titled War in the Age of Intelligent Machines... It seems like a fellow named Robert Devaney is who generated it.
The source info (War in the Age of Intelligent Machines) identifies the fractal as a Julia set, iterates of $z\mapsto z^2+z_0$. It has evidently been distorted (warped) to give it a 3D appearance. I played around with a spiral-shaped Julia set I found at pixels.com and could create an image such as this.
The Mathematica command
Manipulate[JuliaSetPlot[a- b*I,MaxIterations->n], {a,0,1},{b,0,1},{n,100,1000}]
provides a way to find suitable values for $z_0=a-bi$ (and the iteration count $n$). I find a similar spiral for $z_0=0.202-0.55\,i$, $n=400$.
If the requirements are:
(a) fractal (image is repeated at small scales); and
(b) clockwise spiral
then the following may be of use:
This image is generated by the complex exponential function $\exp(z)+1+2.9i$. The resolution can be improved to whatever level you desire (given enough computing power).
Alternatively, if you had a high resolution image of one of the "islands" in your image, then you could make a lattice out of many copies of that, and then apply some sort of a spiral filter which I have seen in some graphics programs like GIMP and maybe Photoshop.
I think the TS asking about the exact reconstruction of the given fractal. It is obviously, machine-generated, so there should be a formula... But finding it may be enormous mathematical challenge.
It looks like a tiling of some type of julia fractals, then post-distorted with a 'spiral' map. Hence, it seems to fall in the "flame fractal" category.
The downside is that there are a lot of parameters to play around with. Also, a reverse image search gave me nothing, so this is not a picture grabbed from the internet.
There is a huge fractal art community at DeviantArt.com - if your client really want something like this, perhaps one can ask one of the artists for a commission?
Side note: I have developed some software to make this type of fractals, so I am rather familiar with the process. Also, in all fractals I generate, I hide the underlying equations as metadata in the image file, so that I can always recreate the fractal (this can be of interest for other mathematicians out there).
|
2025-03-21T14:48:31.974988
| 2020-09-07T00:58:54 |
371050
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Yemon Choi",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632772",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371050"
}
|
Stack Exchange
|
What is the completion of $L^\infty$ in the dual of BV?
Every $f \in L^\infty([0,1])$ induces a continuous linear functional on BV via $g \mapsto \int f g \mathrm{d}x$. I believe $L^\infty([0,1])$ is also separable in BV$^\ast$, while BV$^\ast$ is not separable, so it cannot be the case the BV$^\ast$ completion of $L^\infty([0,1])$ is all of BV$^\ast$.
Is there a nice characterization of the BV$^\ast$ completion of $L^\infty([0,1])$?
I have not thought through the details, but have you tried integrating by parts, since IIRC g is in BV if and only if it is an indefinite integral of a (Borel) measure on [0,1]?
|
2025-03-21T14:48:31.975073
| 2020-09-07T01:01:23 |
371051
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Robert Bryant",
"https://mathoverflow.net/users/13972"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632773",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371051"
}
|
Stack Exchange
|
Isomorphisms of Pin groups
My goal is to identify the $Pin$ group
$$
1 \to Spin(n) \to Pin^{\pm}(n) \to \mathbb{Z}_2 \to 1
$$
such that $Pin^{\pm}(n)$ are isomorphisms to other more familiar groups.
My trick is that to look at the center $Z$ of $Pin$ group, say $Z(Pin(n))$.
We see that, for $k \in \mathbb{Z}^+$,
when $n=4k+1$,
$$
Z(Pin^+(4k+1))=\mathbb{Z}_2 \times \mathbb{Z}_2,
$$
$$
Z(Pin^-(4k+1))=\mathbb{Z}_4.
$$
when $n=4k+3$,
$$
Z(Pin^+(4k+3))=\mathbb{Z}_4,
$$
$$
Z(Pin^-(4k+3))=\mathbb{Z}_2 \times \mathbb{Z}_2.
$$
when $n=4k$ or when $n=4k+2$,
$$
Z(Pin^+(n))=Z(Pin^-(n))=\mathbb{Z}_2.
$$
So my naive attempt is that we have the following $Pin$ group isomorphisms:
$$Pin^+(1)=\mathbb{Z}_2 \times \mathbb{Z}_2, \quad Pin^-(1)=\mathbb{Z}_4.$$
$$Pin^+(3)=SO(3) \times \mathbb{Z}_ 4 \text{ or } \frac{Spin(3) \times \mathbb{Z}_4}{\mathbb{Z}_2},
\quad Pin^-(3)=Spin(3) \times \mathbb{Z}_2.$$
$$Pin^+(5)=Spin(5) \times \mathbb{Z}_2,\quad Pin^-(5)=SO(5) \times \mathbb{Z}_4 \text{ or } \frac{Spin(5) \times \mathbb{Z}_4}{\mathbb{Z}_2}.$$
Question 1: Am I correct about the above statements? More generally, do we have*
$$Pin^+(4k+1)\cong Spin(4k+1) \times \mathbb{Z}_2,\quad Pin^-(4k+1)\cong SO(4k+1) \times \mathbb{Z}_4.$$
$$Pin^+(4k+3)\cong SO(4k+3) \times \mathbb{Z}_ 4, \quad Pin^-(4k+3)\cong Spin(4k+3) \times \mathbb{Z}_2.$$
I am not so sure,
$Pin^+(3)=SO(3) \times \mathbb{Z}_ 4$, could it be that instead
$Pin^+(3)=\frac{Spin(3) \times \mathbb{Z}_4}{\mathbb{Z}_2}?$
(It looks that mine agrees with Wikipedia, but I doubt Wikipedia may be wrong about $Pin^+(3)=SO(3) \times \mathbb{Z}_ 4$...) possibly only for the Lie algebra, not for the Lie group.
Could it be that:
$$Pin^+(4k+1)\cong Spin(4k+1) \times \mathbb{Z}_2,\quad Pin^-(4k+1)\cong \frac{Spin(4k+1) \times \mathbb{Z}_4}{\mathbb{Z}_2}.$$
$$Pin^+(4k+3)\cong \frac{Spin(4k+3) \times \mathbb{Z}_ 4}{\mathbb{Z}_ 2}, \quad Pin^-(4k+3)\cong Spin(4k+3) \times \mathbb{Z}_2.$$
Question 2: Are there other $Pin$ group isomorphisms for $n=4k$, or $4k+2$, say $$Pin^{+/-}(4k)\cong ?$$ and
$$Pin^{+/-}(4k+2)\cong?$$
Thanks for your comments in advance.
Have you checked F. Reese Harvey's book "Spinors and Calibrations"? That book contains a lot of information about the Pin groups.
The second version given in the question is correct:
\begin{align}
\mathrm{Pin}^+(4k+1) &\cong \mathrm{Spin}(4k+1) \times \mathbb{Z}_2 \\
\mathrm{Pin}^-(4k+1) &\cong (\mathrm{Spin}(4k+1) \times \mathbb{Z}_4)/\mathbb{Z}_2 \\
\mathrm{Pin}^+(4k+3) &\cong (\mathrm{Spin}(4k+3) \times \mathbb{Z}_ 4)/\mathbb{Z}_2 \\
\mathrm{Pin}^-(4k+3) &\cong \mathrm{Spin}(4k+3) \times \mathbb{Z}_2.
\end{align}
When $n$ is odd, the element $e_1 e_2 \cdots e_n$ of the Clifford algebra commutes with all elements. Every element of the pin group is either an element of the spin group, or is $e_1 \cdots e_n$ times an element of the spin group. If $(e_1 \cdots e_n)^2 = 1$, then $\mathrm{Pin}$ is the direct product $\mathrm{Spin} \times \langle e_1 \cdots e_n \rangle \cong \mathrm{Spin} \times \mathbb{Z}_2$. If $(e_1 \cdots e_n)^2 = -1$, then instead of a direct product, we could say $\mathrm{Pin} \cong (\mathrm{Spin} \times \mathbb{Z}_4)/\mathbb{Z}_2$. This can potentially be described neatly in other ways. For $\mathrm{Pin}^+(3)$, Harvey states that
$$\mathrm{Pin}^+(3) \cong \{ A \in U(2) : \det A = \pm 1 \}$$
which makes sense along the same lines as $\mathrm{Spin}(3) \cong SU(2)$.
|
2025-03-21T14:48:31.975258
| 2020-09-07T01:11:32 |
371052
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632774",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371052"
}
|
Stack Exchange
|
Find conditions for the following running average to be monotonically decreasing
Let $S_n$ be defined as $\frac{1}{n}\sum_{t=1}^{t=n} [px_t^2 - (p+q)x_t]$ where $x_t = 1-(1-p-q)^t$. We want to find the conditions on $p$ and $q$ such that $S_n$ is monotonically decreasing for all $n$. $0 < p,q < 1$ and $-1 < 1-p-q < 1$.
Note:
Till now I have tried to get a closed bound expression for $S_n$ and differentiate it w.r.t. $n$ to get the conditions for a negative slope but it is getting really complex.
|
2025-03-21T14:48:31.975318
| 2020-09-07T03:17:45 |
371057
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632775",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371057"
}
|
Stack Exchange
|
Can this function be minimized?
Let $X$ be a locally convex TVS, and let $A$ and $B$ be convex and compact subsets of $X$ with $A \subset B$.
Let $f: A \times B \to [0,\infty]$ have the following properties:
(1) For all $b \in B$, $f(\cdot, b)$ is lower semicontinuous.
(2) For all $b \in B$, $f(\cdot, b)$ is convex.
(3) For all $a \in A$, $f(a, \cdot)$ is continuous.
(4) For all $a \in A$, the function $f(a, \cdot)$ is minimized uniquely at $a$, i.e. $f(a,a) < f(a,b)$ for all $b \neq a$.
Does it follow that for all $b \in B$, the function $g_b: A \to [0,\infty]$ defined by $g_b(a) = f(a,b) - f(a,a)$ attains a minimum?
Note that by (4), $g_b$ is bounded below by $0$, so $\inf g_b(A)$ is finite for all $b \in B$. If $b \in A$, then the result is trivial, for then $g_b$ is minimized (uniquely) at $b$, again by (4).
In general, it would be sufficient to show that $g_b$ is lower semicontinuous, but I have not been able to see why that should be the case.
This question is cross-posted at MSE.
|
2025-03-21T14:48:31.975411
| 2020-09-07T04:02:41 |
371060
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/116950",
"user116950"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632776",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371060"
}
|
Stack Exchange
|
Permutation matrix with respect to F_p automorphisms
Let $p$ be an odd prime at least $5$. Let $a_1, \dots, a_{p-1}$ be positive integers such that
$a_i=i+p\cdot b_i$ for some integer $b_i$ for each $i$.
Let $\sigma_k$ be an automorphism on $\{1, 2, \dots, p-1 \}$ sending $i$ to $ik \pmod p$ and $\delta_k$ is the inverse of $\sigma_k$.
(So, $\sigma_1$ is the identity.)
Let $M$ be a square matrix of size $p-1$ whose $ij$-th entry is $a_{\delta_i(j)}$. (So, its first row is $(a_1, a_2, \dots, a_{p-1})$.) I think this matrix is a type of permutation matrix given by an automorphism on the finite field of $p$ elements.
Since $b_i+b_{p-i}=b_j+b_{p-j}$ for all $i\neq j$, then surely this matrix is not invertible.
However, since the permutation matrix is invertible, I wonder the same is true for $M$ under the assumption that $b_i+b_{p-i}=b_j+b_{p-j}$ for some $i \neq j$. Here is my question.
Is M invertible if $b_i+b_{p-i}=b_j+b_{p-j}$ for some $i\neq j$?
For instance, take $p=5$ and $a_1=1, a_2=2, a_3=3$ and $a_4=9$. Then, we have
$$
M=\left(\begin{array}{cccc}
1 & 2 & 3 & 9 \\
3 & 1 & 9 & 2 \\
2 & 9 & 1 & 3 \\
9 & 3 & 2 & 1 \\
\end{array}\right).
$$
By direct computation, we find
$$
195 \times M^{-1}=\left(\begin{array}{cccc}
1 & -5 & -8 & 25 \\
-8 & 1 & 25 & -5 \\
-5 & 25 & 1 & -8 \\
25 & -8 & -5 & 1 \\
\end{array}\right).
$$
Another example is $p=7$ and
$$
M=\left(\begin{array}{cccccc}
1 & 2 & 3 & 4 & 5 & 13 \\
4 & 1 & 5& 2 & 13 & 3 \\
5 & 3 & 1 & 13 & 4 & 2\\
2 & 4 & 13 & 1 & 3 & 5\\
3 & 13 & 2 & 5 & 1 & 4\\
13 & 5& 4 & 3 & 2 & 1\\
\end{array}\right).
$$
By direct computation, we find
$$
2^3 \times 7 \times 19 \times M^{-1}=\left(\begin{array}{cccccc}
15 & -9 &-13 &-25& -29 & 99 \\
-25 & 15 &-29 & -9 & 99& -13\\
-29 &-13 & 15 &99& -25 & -9\\
-9 &-25 & 99 & 15& -13& -29\\
-13 &99 & -9 &-29 & 15& -25\\
99 &-29 &-25 &-13 & -9 & 15\\
\end{array}\right).
$$
Note that if $M^{-1}$ exists and if we write its first column by $(v_1, v_2, \dots, v_{p-1})$ then the $ij$-th entry of $M^{-1}$ is $v_{\delta_j(i)}$.
When $p=5$, then the determinant is
$$
((a_1+a_4)^2-(a_2+a_3)^2)(a_1-a_4)^2(a_2-a_3)^2.
$$
Thus, at least we need more conditions like $a_i\neq a_{p-i}$.
|
2025-03-21T14:48:31.975560
| 2020-09-07T05:38:27 |
371063
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Derek Holt",
"Geoff Robinson",
"Noah Snyder",
"Siddhartha",
"https://mathoverflow.net/users/122414",
"https://mathoverflow.net/users/14450",
"https://mathoverflow.net/users/22",
"https://mathoverflow.net/users/35840"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632777",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371063"
}
|
Stack Exchange
|
Constructing a group of order $2187=3^7$
I am trying to look for the $2$-generated groups of order $3^7$ and class $4$ all whose upper central series quotients are elementary abelian of order 9 except the center which has order $3$.
A small check through GAP reveals there is a unique one which is a semi-direct product of $C_{81}$ and $C_{27}$, namely SmallGroup($2187,194$). I am trying to get a structural argument for this.
The second center is $2$-generated abelian of order $27$. Is it possible to construct the third center using the cohomology argument from this without knowing how the generators of the group behave?
Apologies, if the question is too easy.
${\mathrm{\bf{Revised~notes}}}$: The group SmallGroup($2187,194$) turns out to be powerful (Thanks to Derek!).
I take it that the cyclic group of order $81$ is the normal one. Do you mean that you want a structural explanation of why this is the ONLY example? Or a structural explanation of why this example fulfils your conditions?
I want a structural explanation of why this is the ONLY example.
Is your assertion about the structure of the second centre something that you know how to prove already, or are you just assuming that to be true? (It has to be abelian of order $27$, but is it clear that it is not elementary abelian?)
@Derek: the GAP code showed it and hence I wasn't suspicious about it. But you are right, I have a different and long reasoning that implies $Z_2(G)$ is $2$-generated abelian of order $27$.
I just tried some calculations with $\mathtt{SmallGroup}(2187,194)$, and I found that $G' < G^p$, which seems to contradict the first of your additional notes. (Or did you mean that it is not true that $G' \le G^p$, in which case why be so negative?)
@Derek: My apologies. I added the notes, but forgot to implement it in the program. A small check: IsPowerfulPGroup(SmallGroup(2187,194)) give yes. I need to revise the question to ask for a negative answer then.
I think I can see how to prove this now under the assumption that $G$ is powerful. I think the same approach would work without that assumption, but would involve eliminating more cases.
We are given that $G$ is a $2$-generated group, and that the upper central series of $G$ is $1=Z_0 < Z_1 < Z_2 < Z_3 < Z_4 = G$ with $|Z_1|=3$, $|Z_2|=27$, $|Z_3| = 243$, and $|Z_4|=|G|=2187$, with $Z_2/Z_1$, $Z_3/Z_2$ and $Z_4/Z_3$ elementary abelian
My approach is to identify the quotients $G/Z_i$ for $i=3,2,1,0$. I will just sketch the proof for now, and I can fill in details later if necessary.
We know that $G/Z_3 = C_p^2$ is elementary abelian. and it is not hard to see that $$G/Z_2 \cong \langle a,b \mid a^9=1, [[b,a],a] = [[b,a],b] = 1, b^3 = 1\ {\rm or}\ [b,a] \rangle.$$
Since the group with $b^3=1$ is not powerful, we can assume that $b^3 = [b,a]$ and in fact $G/Z_2 \cong \langle a,b \mid a^9=b^9=1, a^{-1}ba=b^4 \rangle.$
The hardest part is to identify $G/Z_1$, but using the facts that it is a powerful 2-generated group with centre is $Z_2/Z_1 \cong C_p^2$, it can be shown that $$G/Z_1 \cong \langle a,b \mid a^{27}=b^{27}=1, a^{-1}ba=b^4 \rangle.$$
The final step, showing that $G \cong \langle a,b \mid a^{27}=b^{81}=1, a^{-1}ba=b^4 \rangle$ is similar but easier.
${\bf Edit\!:}$ Since we would prefer to prove this without assuming that $G$ is powerful, we need to eliminate the possibility that $b^3=1$ in $G/Z_2$. So assume that $b^3=1$ in $G/Z_2$ and now replace $a,b$ by inverse images in $G/Z_1$.
Suppose that $a^{-1}ba = bt$ in $G/Z_1$. Then $a^{-1}b^3a = (bt)^3 = b^3t^3[b,t]^3 = b^3t^3$, because $[b,t] \in Z_2/Z_1$. So, since $b^3 \in Z_2/Z_1$, we have $t^3=1$.
But then $b^{-1}a^{-1}b = ta^{-1}$, so $b^{-1}a^{-3}b = t^3 a^{-3} = a^{-3}$, and so $a^3 \in Z_2/Z_1$, contrary to assumption, because $a$ has order 9 in $G/Z_2$.
So we have determined $G/Z_2$ up to isomorphism without assuming that $G$ is powerful.
A similar, but more complicated calculation, reveals that there is one other non-powerful (powerless?) possibility for $G/Z_1$ other than the one above, which is
$$\langle a,b,t \mid a^{9} = b^{27} = 1, a^{-1}ba=b^4t, t^3=[a,t]=[b,t]=1 \rangle,$$
where the generator $t$ is redundant. (This is $\mathtt{SmallGroup}(739,32)$, and the other (correct) option above is $\mathtt{SmallGroup}(739,22)$.
A similar argument to the one above shows that that this other option does not extend further to a group $G$ of order $2187$ with the required properties.
Although it is helpful to check things by computer, this can all be done by hand, and I can add further details if necessary. Of course whether you really gain much additional insight by doing it by hand rather than computer is an interesting question!
I think I just mislead you again. We are trying to prove that the assumptions implies $G$ must be powerful.
Some of those $Z_2$'s should be $Z_1$'s, right?
@Derek: Just a quick cross check. Writing ${\overline{G}} := G/Z$ we have a group of class $3$ order $729$ with upper central series quotients all elementary abelian of order $9$. Then ${\overline{G}}/{Z({\overline{G}})}$ is capable of order $81$. It seems that there are only two choices, as you wrote: SmallGroup($81,3$) and ($81,4$) depending on $[b,a] = 1$ or $[b,a] = b^3$. The Epicentre of ($81,3$) seems to contain a central element.
@NoahSnyder Yes I have attempted to correct the indexes
@Siddhartha Yes, there are just two choices for $G/Z_2$ (of order 81) and I rejected $\mathtt{SmallGroup}(81,3)$ because it is not powerful. But since you don't me to assume that $G$ is powerful, I will have to reconsider that case.
|
2025-03-21T14:48:31.976278
| 2020-09-07T08:21:14 |
371070
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Fabian Ruoff",
"Jason Starr",
"https://mathoverflow.net/users/13265",
"https://mathoverflow.net/users/164782",
"https://mathoverflow.net/users/519",
"naf"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632778",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371070"
}
|
Stack Exchange
|
Family over the coarse moduli space of curves
Let $k$ be an algebraically closed field. As the coarse moduli space of curves $M_g$ of genus $g$ over $k$ is not a fine moduli space, it does not have a universal family. But I am wondering if it has a family (proper and flat) such that the fiber over every point $[C]$ of $M_g$ is isomorphic to the curve $C$.
As a disclaimer: I am not that familiar with the language of stacks. As far as I understand the situation in this context, the stack $\mathcal{M}_g$ has an universal family $\mathcal{C}_g$. The corresponding coarse moduli space of $\mathcal{C}_g$ is $M_{g,1}$, so the coarse moduli space of curves with one marked point. The morphism $\pi \colon M_{g,1} \to M_g$ on the level of quasiprojective varieties is just forgetting about the marked point. This family has the property that the fiber over a point $[C]$ is isomorphic to $C$, at least if $C$ has no nontrivial automorphisms. In all other cases the fiber is isomorphic to $C/\operatorname{Aut}(C)$. Is it possible to get something better than that?
If you had such a family, then you'd get a section of the map $\mathcal{M}_g$ to $M_g$, and also a map from the total space of the family to $\mathcal{C}_g$.
There is the forgetful morphism from the coarse moduli space $M_{g,1}$ to $M_g$. However, that morphism is not everywhere flat.
@ulrich Is this something that should not exist?
My previous comment would imply that there is a morphism from any family (as in the question) to $M_{g,1}$ which on fibres is the quotient by the automorphism group. This morphism would be finite birational and this is not possible since $M_{g,1}$ is normal. (I am assuming that $g \geq 3$.)
To close up loose ends and for everyone finding this questions: Such a family does not exist in general. An argument for elliptic curves can be found in Robin Hartshorne Deformation Theory in Remark 26.3.1.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.