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2025-03-21T14:48:31.507920
| 2020-07-14T13:09:34 |
365608
|
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|
Stack Exchange
|
Binomial theorem for content polynomials of partitions
Let $\lambda$ be a partition, represented by a usual Young diagram in which $1\le i\le \ell(\lambda)$ labels the rows and, for each $i$, $1\le j\le \lambda_i$ labels the columns. For each box $\square$ in the diagram, $c(\square)=j-i$ is its content. The polynomial
$$ P_\lambda(x)=\prod_{\square\in\lambda}(x+c(\square))=\prod_{(i,j)\in\lambda}(x+j-i)$$
is the content polynomial.
I would like to know some family of coefficients $b^\lambda_{\mu\nu}$, as a function of three partitions, such that
$$P_\lambda(x+y)=\sum_{\mu,\nu\subset \lambda}b^\lambda_{\mu\nu}P_\mu(x)P_\nu(y).$$
(Notice that content polynomials are not linearly independent, so this equation alone is not enough to uniquely determine the coefficients. I ask for some family)
If you lift this to the level of symmetric functions then the structure constants are uniquely determined. Suppose $x$ denotes a set of $m$ variables and $y$ denotes a set of $n$ variables. Then you can start with the identity of schur polynomials
$$s_{\lambda}(x,y)=\sum_{\mu,\nu}c_{\mu,\nu}^{\lambda}s_{\mu}(x)s_{\nu}(y)$$
where $c_{\mu,\nu}^{\lambda}$ are the Littlewood-Richardson coefficients. Setting all the variables to $1$ gives
$$P_{\lambda}(m+n)=\sum_{\mu,\nu}b_{\mu,\nu}^{\lambda}P_{\mu}(m)P_{\nu}(n)$$
where $$b_{\mu,\nu}^{\lambda}=c_{\mu,\nu}^{\lambda}\frac{\prod_{\square\in\lambda}h_{\square}}{\prod_{\square\in\mu}h_{\square}\prod_{\square\in\nu}h_{\square}}$$
and the $h_{\square}$ represent the hook lengths of each partition.
It could be mentioned that the first identity is what defines the coproduct structure for the Hopf algebra of symmetric functions.
|
2025-03-21T14:48:31.508043
| 2020-07-14T13:19:01 |
365609
|
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"authors": [
"Alex Ravsky",
"Malkoun",
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|
Stack Exchange
|
Distance properties of the permutations of a set of points in a Euclidean space
We are given a set of $n$ distinct points $S=\{\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_n\}$ in a Euclidean space $\mathbb{R}^d$, where the distance between two points $\mathbf{x}_i,\mathbf{x}_j\in S$ is denoted by $d(i,j)$ for any $i,j\in[n]$, and $\max_{i,j\in [n]}d(i,j)=1$. Let $\mathcal{\Pi}$ be the set of all functions $\pi$ mapping each index of a point of $S$ to a distinct integer in $[n]$, namely a permutation on the indices of the points of $S$. We also define
$$R(\pi):=\sum_{\substack{(i,j,k)\in[n]: \\\pi(i)<\pi(j)<\pi(k)}} \left(1-d(i,j)\right)~,$$
$$\pi^*:=\arg\max_{\pi\in\mathcal{\Pi}} R(\pi)~.$$
Question: Can we prove the following conjecture? Does the answer depend on the number of dimensions of the Euclidean space?
For any pair of elements $\mathbf{x}_i,\mathbf{x}_j\in S$ such that $\pi^*(i)<\pi^*(j)$, the average value of $R(\pi)$ over all $\pi\in\mathcal{\Pi}$ such that $\pi(i)<\pi(j)$, is greater than or equal to the average value of $R(\pi')$ over all $\pi'\in\mathcal{\Pi}$ such that $\pi'(i)>\pi'(j)$.
It is an interesting question. I did not think carefully about it. Did you run some numerical calculations for instance?
Thank you. Not specifically for this problem. I run some numerical experiments for this related question (which is even more important for my research now): https://mathoverflow.net/q/365472/115803
Bu the way, about the problem of this webpage, I hope that two average values of the question cannot be arbitrarily close to each other (when $n\gg 1$, and even when $n\to\infty$).
I guess in the sum for $R(\pi)$ should be figured $k$.
Also functions $\pi$ mapping each point of $S$ to a distinct integer in ${1,\dots,n}$ are bijective maps from $S$ to ${1,\dots,n}$, whereas permutations of the elements of $S$ are bijective maps from $S$ to $S$.
Thank you Alex, I removed the sentence about permutations (but I would not modify the title, because the two concepts are tightly related and it can quickly provide information about the topics of this problem).
I could also define $\pi$ as a bijection between the indices of the vectors.
|
2025-03-21T14:48:31.508199
| 2020-07-14T13:23:56 |
365611
|
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|
Stack Exchange
|
About the metrizability of the space of Probability measures $\mathcal{P}(S)$
It is often proved in Books that the space of Probability measures $\mathcal{P}(S)$ on a Polish metric space $(S,\rho)$ endowed with the weak/narrow topology induced by declaring it to be be the coarsest topology on $\mathcal{P}(S)$, which makes the mappings
$$\mathcal{P}(S) \ni \mu \mapsto \int f d\mu \in \mathbb{R}$$
continuous for each bounded and continuous $f : S \rightarrow \mathbb{R}$, is metrizable. Two such metrics should be the Prokhorov metric $d_P$ and the Wasserstein metric $W_0$ of the bounded distance function $\min\{\rho,1\}$.
But the thing I do not understand is the following: It is often shown (For example in Villani, 2009, Optimal Transport) that $(\mu_n) \subset \mathcal{P}(S)$ converging weakly to some $\mu \in \mathcal{P}(S)$, that is,
$$\int f d\mu_n \rightarrow \int f d\mu$$
for each bounded and continuous $f : S \rightarrow \mathbb{R}$, is equivalent to $W_0(\mu_n,\mu) \rightarrow 0$, or $d_P(\mu_n,\mu)\rightarrow 0$. If we do not know a priori that the weak topology is metrizable, then we cannot conclude by the above, that the topology generated by $W_0$ or $d_P$ is exactly the weak topology. Or am I missing something?
How is the $W_0$ distance you mention defined? I only know the Wasserstein $p$-metric for $p \ge 1$.
I'm not really sure what Villani wrote in his monograph, but it is true that one needs to prove that the weak topology is induced by a distance, as a priori it could be another topology with the same converging sequences.
This is quite standard, though. The key point is to realize that it is sufficient to check the continuity of $\mu\mapsto\int f d\mu$ for a countable number of continuous and bounded functions. This is quite obvious if the space is compact (because in this case $C_b(S)$ is separable), but possible also for general Polish spaces.
The details can be found, e.g., in Section 8.3 of the second book by Bogachev "Measure Theory" or at the beginning of my book with Pasqualetto "Lectures on Nonsmooth Differential Geometry"
|
2025-03-21T14:48:31.508362
| 2020-07-14T15:24:56 |
365616
|
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"Gerhard Paseman",
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|
Stack Exchange
|
Is there a finite equational basis for the join of the commutative and associative equations?
I asked this on math stack exchange, but I was told to post it on mathoverflow. Consider the lattice of equational theories of a single binary operation $*$. The meet of the theory axiomatized by the commutative equation and theory axiomatized by the associative equation is the theory axiomatized by both of them. What about the join? Is there a finite equational basis for the join of those theories?
For this example, I don't know. I recall some work about chains of varieties, probably semi group varieties ,where every other member was not finitely based. I suspect nfb (and thus fb) is not well behaved under join. I don't recall the author names, but I would be unsurprised if one of them was Mark Sapir. Gerhard "Look Up Chains And NFB" Paseman, 2020.07.14.
@MattF. $((ab)c)(a(bc))=(a(bc))((ab)c)$ at least if you don't insist on "interesting".
Here's a purely algebraic (rough) formulation of the question: roughly: what are identities satisfied by all associative magmas, and by all commutative magmas? is there a finite number of identities generating all those?
|
2025-03-21T14:48:31.508500
| 2020-07-14T16:00:48 |
365619
|
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"Alexandre Eremenko",
"Carlo Beenakker",
"Tanya Vladi",
"Vova",
"YCor",
"Yemon Choi",
"fedja",
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"https://mathoverflow.net/users/1131",
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|
Stack Exchange
|
Looking for sufficient conditions for positive Fourier transforms
I am looking for some sufficient conditions for an even, continuous, nonnegative, non-increasing, non-convex function to be non-negative definite. In other words
$$
\int_0^\infty f(x)\cos(x\omega) \, dx\ge 0, \quad \omega \in \mathbb{R}.
$$
The function $f(x)=\exp(-x^{\alpha})[x^{\alpha} \log(x) + \delta/\alpha ]$, $\alpha\in(1,2)$, $\delta\ge 2$. Note that for $\alpha\in(0,1)$, $f(x)$ is convex.
I have tried complex detour numerically (as in Tuck, E. O.: On positivity of Fourier transforms, Bull. Austral. Math. Soc., 74 (2006) 133–
138) – did not work out. It seems the imaginary part always increases around $0$.
I have tried Polya type conditions () - no luck either.
For $\alpha\in(0,1)$ everything works out as the function is convex.
Thank you in advance for any hints or references!
You can find a list of necessary conditions in On the positivity of Fourier transforms.
@CarloBeenakker, thank you, good read, I am seeking sufficient conditions though as I believe the function I am interested in is positive definite - cannot prove it rigorously but have some hand waving arguments
positive definite for all $\delta$ ?
@CarloBeenakker, $\delta\ge 2$, I have edited the question, thank you
You did 11 edits in 2 days, please don't permanently edit your post. Edit 7 was changing "Sufficient" into "sufficient" and Edit 12 was the reverse edit, so it looks like artificially putting the post in the front list.
@YCor You give me too much credit, I just have OCD. In my list, the posts appear according to the time of submission, apologies.
Your last edit is the kind of thing which @YCor was counselling against. Please stop doing this - the question itself is fine, please just trust people to look at it as and when they find time
@YemonChoi, yes, I am trying, sorry
@mathworker21, someone else edited it to be capital, I have changed it back to small, then thought whoever did it is an administrator of some sort and may get offended - changed it back to capital and got accused of cheating the system;))). I do not mind if you change it back to small. Also it seems that people started to offer me necessary conditions - which are nice but not what asked for
please show me how to prove for $|\omega|>20$
With great pleasure. We shall just show that $F(y)=\int_0^\infty e^{-x^a}x^a\log x\cos(yx)\,dx>0$ for large enough $y>0$. Note that $\cos(yx)=\Re e^{iyx}$, so we have the real part of a contour integral from $0$ to $+\infty$ of $e^{-z^a}z^a\log z e^{iyz}\,dz$. The integrand oscillates like crazy on the line, so we would like to move the contour up to get less oscillation. Ideally we would like to have a curve $\Gamma$ parameterized by $z(t)$ so that $\Re \left[e^{-z^a}z^a\log z e^{iyz}z'(t) \right] \ge 0$ everywhere on $\Gamma$. If we can do it, it would be the end of the story. However, for the integral in question it is, clearly, impossible (without the extra $1$ or $2$, the Fourier transform has zero integral, so it cannot be positive everywhere), so we'll settle for less: the integral over the "head part" of the contour will have relatively large positive real part and the tail will be small.
We will use the same curve $\Gamma$ that is normally used to prove that the Fourier transform of $e^{-|x|^a}$ is non-negative, namely, the curve $z(x)=x+i\xi(x)$, $x>0$, where $\xi(x)$ satisfies $(x+i\xi)^a=g+iyx$, $\xi,g\in\mathbb R$. This will make $e^{-z^a}e^{iyz}$ real positive on $\Gamma$. Note that $\xi$ is well-defined and continuous in $x$ for $1\le a<2$ and $\xi(x)\asymp x^{2-a}$ at $+\infty$, so this change of the contour is legitimate.
On $\Gamma$, we have $e^{-z^a}e^{iyz}=e^{-H}$ where $H=g+y\xi$. Differentiating the identity defining $\Gamma$, we get $az^{a-1}(1+i\xi_x)=g_x+iy$, so
$$
H_x=g_x+y\xi_x=\Re [(g_x+iy)(1-i\xi_x)]=\Re[az^{a-1}|1+\xi_x|^2]>0
$$
for $z$ in the first quadrant, so $H$ is strictly increasing.
We now want to evaluate
$$
\Re\int_{\Gamma} e^{-H}z^a\log z\,dz=\int_{\Gamma} \Re\left[\frac{z^{a+1}}{a+1}\left(\log z-\frac 1{a+1}\right)\right]d(-e^{-H})\,.
$$
To do it, we will switch to polar coordinates $z=Re^{i\theta}$. Notice that the equation for $\Gamma$ becomes $R^a\sin(a\theta)=yR\cos\theta$. Since $\theta\mapsto \frac{\sin(a\theta)}{\cos\theta}$ increases from $0$ to $+\infty$ as $\theta$ runs from $0$ to $\pi/2$, the curve $\Gamma$ intersects every circumference centered at $0$ only once, so the radius $R$ is a legitimate parameter on $\Gamma$.
We also have $\Re [z^a\bar z]=\Re[(x-i\xi)(g+iyx)]=x(g+y\xi)=xH$, so
$$
H=R^a\frac{\cos((a-1)\theta)}{\cos\theta}
$$
Now come a couple of observations. The first one is that
$$
\sin(a\theta)=\sin((a-1)\theta)\cos\theta+\cos((a-1)\theta)\sin\theta
$$
and
$$
-\sin((a-1)\theta)\cos\theta+\cos((a-1)\theta)\sin\theta=\sin((2-a)\theta)>0
$$
for $\theta\in[0.\frac{\pi}2]$. Hence
$$
R^a\cos((a-1)\theta)\sin\theta\le R^a\sin(a\theta)=yR\cos\theta\le 2R^a\cos((a-1)\theta)\sin\theta\,.
$$
Juxtaposing this with the polar formula for $H$, we see that
$H\sin\theta\le yR\le 2H\sin\theta$ on $\Gamma$.
The second observation is that $\sin(a\theta)\le 1$, so for $R\in[0,1]$, we have $\cos\theta\le\frac 1y$, i.e., $\theta\in[\theta_0,\frac\pi 2]$ where $\theta_0=\arccos\frac 1y$ is quite close to $\frac\pi 2$ for large $y$.
Now we are ready to look at the real part of $\frac{z^{a+1}}{a+1}\left(\log z-\frac 1{a+1}\right)$ when $R\le 1$. It is
$$
\frac{R^{a+1}}{a+1}\left[\cos((a+1)\theta)\left(\log R-\frac 1{a+1}\right)-\theta\sin((a+1)\theta)\right]\,.
$$
Note that $(a+1)\theta\in [2\theta_0,\frac{3\pi}2)$, so as soon as $\theta_0\ge\pi/4$ ($y>\sqrt 2$), the cosine is negative. Thus we can ignore $\log R<0$, which leaves us with the expression
$$
-\cos((a+1)\theta)\frac 1{a+1}-\theta\sin((a+1)\theta)
$$
on $[\theta_0,\frac {\pi}2]$. Taking the derivative, we see that it is increasing in $\theta$, so $\theta=\theta_0$ is the worst case. Again, it can easily be verified that this expression is greater than, say, $c(a)=\frac 1{2(a+1)}$ if $\theta_0$ is close enough to $\frac\pi 2$. Taking into account the bound $H\le yR/\sin\theta_0=:H_1R$ on $[0,1]$ and noting that for $R=1$, we have $H=\Re[z^a]+y\xi\ge -1+y\sin\theta_0=:H_0$, we immediately conclude that
$$
\int_{\Gamma:R<1}\dots d(-e^{-H})\ge c(a)\int_0^{H_0/H_1}\frac {R^{a+1}}{a+1}H_1e^{-H_1R}\,dR
$$
For large $y$, we have $H_0\approx H_1\approx y$ (see the accurate estimates above), so the whole integral is about
$$
\frac{c(a)y}{a+1}\int_0^1 R^{a+1}e^{-yR}\,dR\approx \frac{c(a)}{a+1}y^{-a-1}\int_0^\infty r^{a+1}e^{-r}\,dr
$$
i.e., we have just power decay in $y$ here.
On the other hand, since $R\le 2H/y=R(H)$ all the way through, we have
$$
\left|\int_{\Gamma: R>1}\dots d(-e^{-H})\right|\le\frac 1{a+1}\int_{H_0}^\infty R(H)^{a+1}\sqrt{\left(\log R(H)+\frac 1{a+1}\right)^2+\frac{\pi^2}4} \,e^{-H}\,dH\,,
$$
which decays exponentially in $y$, so the whole integral is positive for large $y$. I leave the accurate estimates of the minimal $y$ for which it works to you. $y>20$ is certainly enough, but I won't be suprised if one can push it down to $y>10$.
I'm still struggling with small $y$... :-( Meanwhile, feel free to ask questions if something is unclear.
the proof is awesome, I am struggling to find a way to evaluate $y$ numerically. I believe it should be $y$ which makes the second integral positive. What is $H_0$?
@TanyaVladi Sorry, forgot to write: $H_0=-1+y\sin\theta_0$ (the lower bound for $H$ at the moment of escape from the unit circle).
got it, thank you. Just a follow up question, it seems using your method one can get a lower bound on $F(y)$, it is possible to estimate when the global minimum occur? the number of zeros of $F(y)$
I have run numerics. Basically I find points on the $\Gamma$ by solving $R^{\alpha-1}\sin(\alpha\theta)=\cos(\theta)y$ for fixed $y$ and $\alpha$- get pairs of $(x,\xi)$. As you can imagine $\xi$ is an increasing function of $x$. I have noticed that large $y$ do not work as real part of ${z^{\alpha+1}\over \alpha+1}$ ${\log(z)-{1\over \alpha+1}}$ is negative for large $\xi$. On the other hand tiny $y$s work perfectly as the real part is positive, What am I missing?
@TanyaVladi You are probably missing the logic of this proof: The real part for the large $y$ is negative once the curve leaves the unit disk and the second integral estimates the negative contribution, so it has to be subtracted. The positive contribution comes from the part of the curve in the unit disk, which is bounded from below by the first integral. The point is that for large $y$ the first (positive) integral decays in a power (in $y$) fashion while the second one decays exponentially, so the positive contribution wins. I'm getting a cutoff at around $y=8.5$ uniformly in $a$.
I think I got the logic, I am just trying to adjust the proof ti my original function. I actually do not see how you got the expression under square root. "On the other hand, since $R\le 2H/y=R(H)$ all the way through, we have"
$$
\left|\int_{\Gamma: R>1}\dots d(-e^{-H})\right|\le\frac 1{a+1}\int_{H_0}^\infty R(H)^{a+1}\sqrt{\left(\log R(H)+\frac 1{a+1}\right)^2+\frac{\pi^2}4} ,e^{-H},dH,,
$$ . Can you please drop a hint or two?
@TanyaVladi The expression under the square root is just the upper bound on $|\log z-\frac 1{a+1}|$ where $z$ is in the first quadrant with $1\le |z|\le R(H)$: the real part of $\log z$ is at most $\log R(H)$ and the imaginary part (the argument of $z$) is between $0$ and $\pi/2$, hence the sum of squares. I added $\frac 1{a+1}$ instead of subtracting for safety: $|a-b|\le A+b$ for $0\le a<A$, $b>0$ but not necessarily less than $|A-b|$. That's all.
@fedja, how do you get $8.5$? why did you remove the term $\delta \exp(-x^{\alpha})/\alpha$, it seems the term is needed to make sure the function is positive for smaller values (greater than 1 and apparently smaller than 8,5)
@Vova 1) Just by numerically comparing the integrals in the estimates. If you do it by hand with rough bounds, you get something like $12$. 2) Because I still cannot use it efficiently: even with that term (that is, indeed, needed for the full range, as I mentioned in the very beginning), I can prove only the same result (large $y$ only), so I just wanted to attract one's attention to the fact that in this regime it is not necessary.
@fedja thank you, can you estimated the global minimum?
@Vova Yes. but the estimate is not good enough to be beaten by $e^{-y^2/4}$. Of course, what I proved is enough to show that the FT is positive if $\delta$ is sufficiently large, but "sufficiently" is currently way above the required $\delta=2$.
@fedja, thank you for your comments. How about combining two approaches. The one from Tuck's paper requires convexity for the real part of the function and positive and decreasing imaginary part of the function, which is true (according to my numerical studies) for $≥1$ (can be smaller) and contour $+0.01$ for any $$. It seems with your approach the real part of the function is positive for any $$ and $\ge1$. I cannot formally prove that though. I do understand that $x=R\cos(\theta)$ and when $y$ is large $R>1$
@TanyaVladi "the real part of the function is positive for any $y$ and $x\ge 1$". Hmm... That is quite doubtful because, adjusting $y$, we can get any $z$ with fixed real part to lie on the curve. I'll look into it though my free time is limited :-)
@fedja, sorry I meant $x<1$. The idea is to split the range of $x$ into $[0,1]$ and $[1,\infty]$. You approach works for the first part and Tuck's for the second
@fedja, by the way my original problem/question was about showing that Fourier of $\exp(-|x|^{\alpha})/\alpha^2$, $x\in R$ decays as a function of $1<\alpha<2$, for any $\omega$. By taking derivative- I made a problem more complicated as it seems
This can be very complicated, so I will only give a reference which contains many highly non-trivial results of this sort:
MR0428382 J. V. Linnik and I. V. Ostrovskiĭ, Decomposition of random variables and vectors. Translated from the Russian. Translations of Mathematical Monographs, Vol. 48. American Mathematical Society, Providence, R. I., 1977.
Look especially in the chapter which is called Necessary conditions for $I_0$,
I think it is Chapter IV.
I have the book and checked it. Are you implying that my function is of Levy Hinchin representation? I do not see it, can you please drop another hint?
I only answered the first sentence in your message, not about this specific function. I do not claim that your function has a Levy Hinchin representation.
Change of contour should work (as I commented in your other post, it does work if you care only about $|\omega|>20$) but it seems rather delicate because the function has a zero in the first quadrant and treating it properly in the full range requires a lot of extra care. On the side note, asking slight variations of the same question several times in a row is usually frowned upon, though you seem lucky so far :-)
I am not an expert and do not know what extra care I need. I have done some numerical work and was discouraged by the results. I read the paper , but could not get the nice closed form analytical expression for Im or Re parts of a complex detour of the function. As for asking the same question- the question does not look difficult for a mathematician. I am surprised I did not get an answer so far - I blame myself for not being able to ask it correctly, hence I iterate.
@TanyaVladi It isn't easy if you want to get a proof in the full range and do not want to settle for anything less, as it seems to be the case (otherwise I would post my $|\omega|>20$ solution long ago). Your question has been crystal clear all the time. Just give people some time to figure it out. :-) Of course, there is no explicit formula for anything here but it is not what makes it tricky. The real issue is the unpleasant behavior of the argument of $1+z\log z$ in the upper half-plane and I'm still struggling with it in the case of small $\omega$.
@fedja, please show me how to prove for $|\omega|>20$. I can show it is positive for small $\omega$
@fedja, I found the following paper Luo, S. L. and Zhang, Z. M.: Estimating the first zero of a characteristic function,
C. R. Acad. Sci. Paris, Ser. I 338 (2004), 203–206. . . Basically I can find $\tau$, such that my transform is positive for all arguments less than that. It is, according to the paper, $\Delta(\alpha,p)=\pi/(2M_p(\alpha))^{1/p}$.$\Delta(\alpha,p)$ is small, around 1, but it is a good start
@TanyaVladi Done (large $\omega$)
How about the following reference T. Gneiting, Criteria of Po´lya type for radial positive definite functions, Proc. Am. Math. Soc. 129 (2001), 2309–2318. I have plotted the results for 6th derivative all numeric but it seems working,
Looks very interesting. So what particular $k,l,\alpha$ do you suggest?
|
2025-03-21T14:48:31.509536
| 2020-07-14T16:01:19 |
365620
|
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|
Stack Exchange
|
Adjoint operator
Assume that $$\left<f,g\right>_R=\Re \int_U f(z) \overline{g(z)} \, dx \, dy, f,g \in L^2(U),$$ where $U$ is the unit disk and assume that $A: L^2(U)\to L^2(U)$ is a real-linear operator. Assume also that $A^*$ is its adoint, with respect to $\left<f,g\right>_R$, that is $\left<Af,g\right>_R= \left<f,A^*g\right>_R$. My question is, whether in that case we have $\|A\|_{L^p\to L^p} = \|A^*\|_{L^q\to L^q}$, where $1/p+1/q=1$.
Yes, this is true, in the sense that if one side is finite then so is the other and they are equal. All you really need is to notice that $\|g\|_q = \sup \{{\rm Re}\int fg: \|f\|_p = 1\}$ (since you can multiply any $f$ by a scalar of modulus 1).
I'll show that $\|A^*\|_{L^q \to L^q} \leq \|A\|_{L^p\to L^p}$; the reverse inequality follows by symmetry. Choose $f$ and $g$ with $\|f\|_p = \|\bar{f}\|_p= \|g\|_q = \|\bar{g}\|_q =1$ and $$\langle Af, g\rangle = \langle f, A^*g\rangle = {\rm Re}\left(\int f\cdot \overline{A^*g}\right) \geq \|A^*\|_{L^q\to L^q} - \epsilon.$$ (By truncating, we can assume $f$ and $g$ both lie in $L^2(U)$.) That is, $\int Af\cdot\bar{g} \geq \|A^*\|_{L^q\to L^q} - \epsilon$, which shows that $\|A\|_{L^p\to L^p} \geq \|A^*\|_{L^q\to L^q} - \epsilon$.
|
2025-03-21T14:48:31.509650
| 2020-07-14T16:25:06 |
365621
|
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|
Stack Exchange
|
A nice pattern about Goldbach conjecture in French Wikipedia
In the following link: https://fr.m.wikipedia.org/wiki/Conjecture_de_Goldbach, one can see a nice pattern of pink and blue lines coming from each prime number, the intersection points thereof are circled. Call those circles on the pink line emanating from $p$ "the Goldbach $p$-dots". It is quite easy to figure out that there are $\pi(p)-1$ such $p$-dots, corresponding to the even numbers $2n$ between $p+3$ and $2p$ such that $2n-p$ is prime, hence to the primes between $3$ and $p$. Now if we sum up the number of $p$-dots for consecutive values of $p$ up to $x$, we get a lower bound for the sum of Goldbach decompositions of the even numbers not exceeding $2x$.
That is:
$$\sum_{2<p\leq x/2}\left(\pi(p)-1\right)\leq\sum_{n\leq x}G(2n)\leq\sum_{2<p<x}\left(\pi(p)-1)\right)$$.
While it doesn't prove Goldbach's conjecture by itself, it gives both a lower and an upper bound for the average growth rate of $G(2n)$, that is, the number of representations of $2n$ as the sum of two primes.
Would this and known bounds for $\pi(x)$ allow to prove that Goldbach's conjecture implies the extended Goldbach's conjecture derived semiheuristically by Hardy and Littlewood in 1923?
Edit: note that the LHS in the double inequality above is asymptotically equal to $\frac{\pi(x/2)^2}{2}\sim\frac{x^2}{8\log^{2}x}$ and thus $\frac{1}{x}\sum_{n\leq x}G(2n)\gtrsim\frac{x}{8\log^{2}x}$ so that the average value $\tilde{G}(2n)$ of $G(2n)$ fulfills $\tilde{G}(2n)\asymp\frac{n}{\log^{2}n}$.
What is the explicit statement of extended Goldbach by Hardy and Littlewood?
See https://en.m.wikipedia.org/wiki/Goldbach%27s_conjecture : $G(2n)\sim 2C_{2}\prod_{p>2}\frac{p-1}{p-2}\frac{2n}{\log^{2}2n}$, where $C_{2}=0.66016...$ is the twin prime constant.
|
2025-03-21T14:48:31.509803
| 2020-07-14T16:38:15 |
365623
|
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"Alexey Ustinov",
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|
Stack Exchange
|
Partial sums of signed binomial coefficients
I don't know if this is true or not but I want this to be true and so far I don't have any counterexample.
Let $i$ be odd. Do there exist coefficients $a_k \in \{0,1\}$ such that
$$\sum_{k=1}^{i-1} (-1)^k \binom{i}{k} a_k = 1\text{?}$$
Some related questions: https://mathoverflow.net/questions/74045/zero-sum-of-binomial-coefficients ; https://mathoverflow.net/questions/74191/what-is-the-degree-of-a-symmetric-boolean-function
No hope when i is a prime. Not much better when i is two more than a prime. Gerhard "Won't Find Any Prime Examples" Paseman, 2020.07.14.
Relevant question https://mathoverflow.net/q/216351/5712
Given that $i=2j+1$ is odd, $a_k$ and $a_{i-k}$ cancel out if both are equal. So you could equivalently ask for coefficients $b_k\in \{-1,0,1\}$ such that $$\sum_{k=1}^jb_k\binom{i}{k}=1.$$
I will ignore the cases of $k=0$ and $k>j$
If there is a prime $p$ such that $p|\binom
{i}{k}$ with few exceptions, then it may be possible to rule out a sum of the form you seek being any value congruent to $1 \bmod p.$
This eliminates the sum being $1 \bmod p$ for a prime power $i=p^e$, for any $i=3p$ and for all $i=5p$ except, perhaps, $i=35$ and $i=55$
If $i=p^e$ is a prime or prime-power then $p|\binom{i}{k}$ so the sum can't never be anything other than a multiple of $p.$
If $i=3p$ then $p|\binom{3p}{k}$ with the exception that $\binom{3p}{p} \equiv 3 \bmod p$ so the sum can only be $0,3$ or $p-3$ $\bmod p.$
For $i=5p$ we have $\binom{5p}{p} \equiv 5 \bmod p$ and $\binom{5p}{2p} \equiv 10 \bmod p$ so the sum can only be $0,\pm5,\pm 10,\pm 15$ $\bmod p$ For $p=7$ we do have $15 \equiv 1 \bmod p.$ And $p=11$ is not obviously ruled out.
However we do see that the only chance for $\sum_{k=1}^{27}b_k\binom{55}k=1$ is $b_{11}=0$,$b_{22}=-1$ and the other terms (which are all multiples of $11$) adding to $\binom{55}{22}+1.$ Looking $\bmod 5$ seems to force $b_5=0$ and $b_{25}=1.$
In general, for $i=qp$ with $q<p$ and both prime, $\binom{qp}{rp} \equiv \binom{q}{r} \bmod p$ This will eliminate all but a finite number of $p$ for any given $q$. And in the cases not immediately ruled, out the possibilities to search over are restricted.
Similar considerations rule out $i=7p$ for $p >7$ prime, with the possible exceptions of $p=11, 13, 17, 19, 29, 31, 41, 43$, the primes dividing $N-1,N$ or $N+1$ for $N=7,14,21,35,42,49,56,63$ , these being the numbers $N$ which can be formed from some or all of $7,21,35=\binom{7}{1},\binom{7}{2},\binom{7}{3}$ with addition and subtraction.
I don't immediately see an obstacle to $i=105.$
For $i = 35$, take $a_k = 1$ for $k = 11, 14, 15, 22, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34$, and $0$ otherwise.
Nice example. Given Gerhard's comments and some fiddling I did with small values, maybe the question should be about finding (characterizing?) $i$ for which this does work.
If my computations are correct, $35$ is the least.
That's a nice counterexample. Can we argue that once we have used some coefficients in a combination to sum to 1, then there does not exist any combination of the remaining terms that could sum to 0.
I believe 35 is the least for which it holds.
|
2025-03-21T14:48:31.510298
| 2020-07-14T16:56:06 |
365624
|
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|
Stack Exchange
|
A $q$-analogue of a characterization of polynomials by binomial coefficients
Considering the binomial coefficient $\binom{x}{m}$ as a polynomial in $x$, the span of $\binom{x}{0}, \binom{x}{1}, \ldots, \binom{x}{d}$ is exactly the polynomials of degree $\le d$. A closely related characterization is that this subspace is the kernel of $\Delta^{d+1}$, where $\Delta : \mathbb{C}[x] \rightarrow \mathbb{C}[x]$ is the difference operator defined by $(\Delta P)(x) = P(x) - P(x-1)$. Roughly stated, my question asks for a $q$-analogue of either of these results.
To make this more concrete, use the standard notation, so $[n]_q = 1+q+\cdots + q^{n-1}$, $[n]_q! = [n]_q[n-1]_q\ldots [1]_q$, and $\binom{n}{d}_q = \frac{[n]_q!}{[d]_q![n-d]_q!}$. Fix $N \in \mathbb{N}$ and for each $m \in \mathbb{N}_0$ let
$$u^{(m)}_q = \bigl( \binom{0}{m}_q, \binom{1}{m}_q, \ldots, \binom{N-1}{m}_q \bigr) \in \mathbb{C}[q]^{N}$$
By the first paragraph, for $P \in \mathbb{C}[x]$, we have
$$\bigl( P(0), P(1), \ldots, P(N-1) \bigr) \in \bigl\langle u^{(0)}_1, u^{(1)}_1, \ldots, u^{(d)}_1 \bigr\rangle$$
if and only if $\mathrm{deg} P \le d$. In a current research problem, it's useful that the same holds for $( P(N-1), \ldots, P(1), P(0))$; in fact the evaluation points $0$, $1, \ldots, N-1$ can be varied by an arbitrary affine transformation.
Is there an analogous characterization of the span $\langle u_q^{(0)}, u_q^{(1)}, \ldots, u_q^{(d)} \rangle$?
As a follow up, what transformations preserve this space? In particular, is it invariant under a $q$-analogue of the affine transformation just mentioned? Despite some thought I have not found any reasonable answer to these questions, but I find it hard to believe that there is nothing to be said.
We have
$$\binom{n}d_q=\frac{(q^n-1)\ldots(q^n-q^{d-1})}{(q^d-1)\ldots(q^d-q^{d-1})}=f_d(q^n)=g_d([n]_q)
$$
where $f_d$ and $g_d$ are polynomials of degree $d$ (depending on $q$ of course). Therefore $$\bigl( P([0]_q), P([1]_q), \ldots, P([N-1]_q) \bigr) \in \bigl\langle u^{(0)}_q, u^{(1)}_q, \ldots, u^{(d)}_q \bigr\rangle$$
if and only if $\deg P\leqslant d$.
See also: https://mathoverflow.net/questions/218696/q-integer-valued-polynomials
Thank you. I think the critical point is that the binomial coefficients $\binom{x}{d}$ are the minimal degree polynomials that interpolate the functions $0 \mapsto 0, 1 \mapsto 0, \ldots, d-1 \mapsto 0, d \mapsto 1$. Thus $(v_0, \ldots, v_{N-1})$ is in the span of $u_1^{(0)}, \ldots, u_1^{(d)}$ if and only if there is a polynomial $P$ of degree $\le d$ such that $P(j) = v_j$ for each $j$. This generalizes to the $q$-binomial coefficients replacing $j \mapsto 0$, $d \mapsto 1$ with $[j] \mapsto 0$ and $[d]_q \mapsto 1$ and $P(j) = v_j$ with $P([j]_q) = v_j$.
Hence if $P(ax+b) = Q(x)$ then $(P([0]_q), \ldots, P([N-1]_q)) \in \langle u_q^{(0)}, \ldots, u_q^{(d)} \rangle$ if and only if $(Q([0]_q), \ldots, Q([N-1]_q)) \in \langle u_q^{(0}), \ldots, u_q^{(d)} \rangle$. For binomial coefficients this is useful, because such affine changes of variable move the evaluation points in a 'natural way'. (E.g. they can be reversed.) But for $q$-binomial coefficients, it's not so useful: e.g, there is no affine transformation sending $[0]_1, [1]_q, [2]_q$ to $[2]_q, [1]_q, [0]_q$ when $q\not= 1$.
Anyway, thank you for the very clear answer, which I think is definitive and shows that the generalization (I now realise) I really wanted is not feasible.
|
2025-03-21T14:48:31.510545
| 2020-07-14T17:10:22 |
365626
|
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|
Stack Exchange
|
Prime gap distribution in residue classes and Goldbach-type conjectures
Update on 7/20/2020: It appears that conjecture A is not correct, you need more conditions for it to be true. See here (an answer to a previous MO question).
The general problem that I try to solve is this: if $S$ is an infinite set of positive integers, equidistributed in a sense defined here, and large enough as defined in the same post, then all large enough integers can be written as the sum of two elements of $S$. I call this conjecture A, and the purpose of my previous question (same link) was to find whether this is a conjecture, a known fact, or not very hard to prove.
Here I try to solve what I call conjecture B. Let $p_k$ be the $k$-th prime ($p_1 = 2$) and $q_k = (p_{k} + p_{k+1})/2 = p_{k} + g_{k}$ where $g_{k} =(p_{k+1}-p_{k})/2$ is the half-gap between $p_{k}$ and $p_{k+1}$. Let $S_1$ be the set of all the $q_k$'s, for $k=2,3,\cdots$. Is $S_1$ equidistributed in the same sense, that is equidistributed in all residue classes? For this to be true, it suffices to prove that the half-gaps are equidistributed in residue classes. There is an attempt to answer that question here, but it is not clear to me if the answer is yes, no, or unsure. What is your take on this?
Assuming conjectures A and B are true, then any large enough integer is the sum of two elements of $S_1$. Another interesting result is this: let $S_2$ be the set of all $\lfloor \alpha p_k\rfloor$ where the brackets represent the floor function, $k=1,2,\cdots$, and $\alpha > 0$ is an irrational number. Then any large enough integer is the sum of two elements of $S_2$.
The interesting thing about $S_2$ is that it is known to be equidistributed and furthermore, you can choose $\alpha=1+\epsilon$ with $\epsilon$ an irrational number as close to zero as you want, but NOT exactly zero. Since $\lfloor(1+\epsilon)p_k\rfloor = p_k + \lfloor \epsilon p_k\rfloor$, if conjecture A is true you have this result:
Any large enough integer $n$ can be written as $n=p + q + \lfloor \epsilon p\rfloor + \lfloor \epsilon q\rfloor$, with $p, q$ primes and
$\epsilon>0$ an irrational number as close to zero as you want (but not zero).
With $\epsilon=0$, this would be equivalent to Goldbach conjecture, but of course it does not work with $\epsilon=0$ since no odd integer $n$ is the sum of two primes, unless $n=p+2$ and $p$ is prime.
Two useful references
Provided by Andrew Granville, who also mentioned the following.
As to your question the answer is a little surprising and has been the subject of some recent publicity - there are two papers by Robert Lemke Oliver and Soundararajan who look at how often one has $p_n= a \bmod{q}$ and $p_{n+1} = b \bmod{q}$.
Turns out these counts are far from uniformly distributed though an analysis via the circle method reveals that they should be asymptotically the same, but there is a large secondary term which plays a significant role as far as one can ever hope to compute.
The distribution of consecutive prime biases and sums of sawtooth
random variables
Unexpected biases in the distribution of consecutive primes
Finally, I will try to prove that if $S$ is equidistributed in residue classes, then $S+S$ is also equidistributed. I posted this as a question on MSE, here.
Your conjectured result, if true, would be up to $O(1/\epsilon)$, no?
Yes, unfortunately the closer to zero you get, my obvious guess is that the exception set (numbers that can not be represented that way) even though finite, would be extremely large.
A first step to prove conjecture A might be this: prove that if $S$ is equidistributed, then $S+S={x,y$ with $x,y \in S}$ is also equidistributed.
Here I provide some insights about conjecture B. First, it is still a conjecture, and just like the paradox that I discussed here, it defies empirical evidence: the error term in the approximation involves $\log$ and $\log \log$ functions (see here) so you would need to use insanely large numbers to see convergence to uniform distribution in residue classes, for all moduli $m$. In particular, if you look "only" at the first million elements of $S_1$,
If $m>0$ is a multiple of $3$, then $q_k = 0, 3, 6,\cdots \bmod{m}$ far more often
than expected.
If $m>0$ is a multiple of $3$, then $q_k = 1 \bmod{m}$ far less often
than expected.
Yet if $m>2$ is prime, then the discrepancies tend to disappear much faster. In that case $q_k = r \bmod{m}$ much more frequently for $r=0$, and less frequently for $r=1,\cdots,m-1$. The case $r=0$ is the worst discrepancy. The table below summarizes the discrepancy at $r=0$ when $m$ is prime ($m=3, 5,\cdots, 23$):
A number such as $1.7037$ means that for the prime $m$ in question (in this case $m=3$) we have $q_k = 0 \bmod{m}$ about 1.7073 times more than expected, among the first million elements of $S_1$.
|
2025-03-21T14:48:31.511034
| 2020-07-14T19:01:49 |
365631
|
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|
Stack Exchange
|
What are some examples of proving that a thing exists by proving that the set of such things has positive measure?
Suppose we want to prove that among some collection of things, at least one
of them has some desirable property. Sometimes the easiest strategy is to
equip the collection of all things with a measure, then show that the set
of things with the desired property has positive measure. Examples of this strategy
appear in many parts of mathematics.
What is your favourite example of a proof of this type?
Here are some examples:
The probabilistic method in combinatorics As I understand it, a
typical pattern of argument is as follows. We have a set $X$ and want to
show that at least one element of $X$ has property $P$. We choose some
function $f: X \to \{0, 1, \ldots\}$ such that $f(x) = 0$ iff $x$ satisfies
$P$, and we choose a probability measure on $X$. Then we show that
with respect to that measure, $\mathbb{E}(f) < 1$. It follows that
$f^{-1}\{0\}$ has positive measure, and is therefore nonempty.
Real analysis One example is Banach's
proof
that any measurable function $f: \mathbb{R} \to \mathbb{R}$ satisfying
Cauchy's functional equation $f(x + y) = f(x) + f(y)$ is linear.
Sketch: it's enough to show that $f$ is continuous at $0$, since then
it follows from additivity that $f$ is continuous everywhere, which makes
it easy. To show continuity at $0$, let $\varepsilon > 0$. An
argument using Lusin's theorem shows that for all sufficiently small
$x$, the set $\{y: |f(x + y) - f(y)| < \varepsilon\}$ has positive
Lebesgue measure. In particular, it's nonempty, and additivity then
gives $|f(x)| < \varepsilon$.
Another example is the existence of real numbers that are
normal (i.e. normal to every base).
It was shown that almost all real numbers have this property
well before any specific number was shown to be normal.
Set theory Here I take ultrafilters to be the notion of measure, an
ultrafilter on a set $X$ being a finitely additive $\{0, 1\}$-valued
probability measure defined on the full $\sigma$-algebra $P(X)$. Some
existence proofs work by proving that the subset of elements with the
desired property has measure $1$ in the ultrafilter, and is therefore nonempty.
One example is a proof that for every measurable cardinal
$\kappa$, there exists some inaccessible cardinal strictly smaller than
it. Sketch: take a $\kappa$-complete ultrafilter on $\kappa$. Make an inspired choice of function $\kappa \to \{\text{cardinals } <
\kappa \}$. Push the ultrafilter forwards along this function to give
an ultrafilter on $\{\text{cardinals } < \kappa\}$. Then prove that the set
of inaccessible cardinals $< \kappa$ belongs to that ultrafilter ("has
measure $1$") and conclude that, in particular, it's nonempty.
(Although it has a similar flavour, I would not include in this list the cardinal arithmetic proof of the
existence of transcendental real numbers, for two reasons. First,
there's no measure in sight. Second -- contrary to
popular belief -- this argument leads to an explicit construction
of a transcendental number, whereas the other arguments on this list
do not explicitly construct a thing with the desired properties.)
(Mathematicians being mathematicians, someone will probably observe that
any existence proof can be presented as a proof in which the set of things
with the required property has positive measure. Once you've got a thing
with the property, just take the Dirac delta on it. But obviously I'm
after less trivial examples.)
PS I'm aware of the earlier question On proving that a certain set is
not empty by proving that it is actually
large. That has some good
answers, a couple of which could also be answers to my question. But my
question is specifically focused on positive measure, and excludes
things like the transcendental number argument or the Baire category
theorem discussed there.
You got the argument for the probabilistic method in combinatorics the wrong way around: $\mathbb{E}(f)<1$ does not imply $f^{-1}(0) \neq \emptyset$. The argument works the other way: One shows $\mathbb{E}(f)>0$ and infers that there must be at least one point $x\in X$ with $f(x)>0$.
Not sure what you mean, Johannes. If $\mathbb{E}(f) < 1$ then there's some $x$ such that $f(x) < 1$. Since $f$ is valued in $\mathbb{N}$, that's equivalent to $f(x) = 0$. Of course, one can make similar arguments with $\mathbb{E}(f) < a$ or $\mathbb{E}(f) > a$ for any constant $a$, as you've done with $a = 0$.
I feel like basically all Brownian motion sample path properties are of this form. Given any one of them, it's usually not so obvious how you would construct even one continuous function with that property (nowhere differentiable, uncountable level sets, nowhere increasing, etc, etc); yet "almost every" continuous function enjoys all of them simultaneously.
The proof that there exist normal numbers can also be made into a construction of a normal number. See e.g. my answer here.
@TomLeinster Oh right, I overlooked that the function is supposed to be $\mathbb{N}$-valued and thought of real-valued functions instead.
https://en.wikipedia.org/wiki/Shannon%27s_source_coding_theorem#Proof:_Source_coding_theorem
One of the most spectacular examples (at least in functional analysis) is Gluskin’s results that the Banach Mazur compactum, i.e., the family of $n$ dimensional Banach spaces, which is a compact space under a natural metric, is as bad as can be. He did this by showing that there is a suitable measure so that pairs of randomly chosen polytopes generate spaces with large Banach Mazur distances on a set of positive measure. A more precise statement can be found in the (in this case reliable) wikipedia article “Banch Mazur compactum”.
Szemerédi's theorem asserts that every set $A$ of integers of positive upper density (thus $\limsup_{N \to \infty} \frac{|A \cap [-N,N]|}{|[-N,N]|} > 0$) contains arbitrarily long arithmetic progressions. One of the shortest (but not the most elementary) proofs of this remarkably deep theorem deduces it from a result in ergodic theory:
Furstenberg recurrence theorem: Let $E$ be a subset of a probability space $(X,\mu)$ of positive measure, and let $T: X \to X$ be an invertible measure-preserving shift. Then for any $k \geq 1$ there exists a positive integer $n$ such that $E \cap T^n E \cap T^{2n} E \cap \dots \cap T^{(k-1) n} E$ has positive measure.
The case $k=1$ is trivial, and the case $k=2$ is the classical Poincare recurrence theorem. The general case was established in
Furstenberg, Harry, Ergodic behavior of diagonal measures and a theorem of Szemeredi on arithmetic progressions, J. Anal. Math. 31, 204-256 (1977). ZBL0347.28016.
Roughly speaking the deduction of Szemerédi's theorem from Furstenberg's theorem is as follows. By hypothesis, there is a sequence $N_j \to \infty$ such that $\frac{|A \cap [-N_j,N_j]|}{|[-N_j,N_j]|}$ converges to a positive limit. One can define a generalised density of subsets $B \subset {\bf Z}$ by the formula $\mu(B) := \tilde \lim_{j \to \infty} \frac{|B \cap [-N_j,N_j]|}{|[-N_j,N_j]|}$ where $\tilde \lim$ is an extension of the limit functional $\lim$ to bounded sequences (this can be constructed using the Hahn-Banach theorem or using an ultrafilter). Morally speaking, this turns the integers ${\bf Z}$ into a probability space $({\bf Z},\mu)$ in which $A$ has positive measure and the shift $T: n \mapsto n-1$ is measure-preserving. Then by the Furstenberg recurrence theorem, for every $k$, there is a positive integer $n$ such that $A \cap T^n A \cap \dots \cap T^{(k-1) n} A$ has positive measure, hence non-empty, hence $A$ contains arbitrarily long arithmetic progressions.
(I cheated a little because $\mu$ is only a finitely additive measure rather than countably additive, but one can massage the finitely additive probability space $({\bf Z},\mu)$ constructed here into a countably additive model $(X, \tilde \mu)$ by a little bit of measure-theoretic trickery which I will not detail here.)
a curiosity: is it really necessary for this argument to work to have at disposal the (weak version of - but still stronger than Countable Dependent) Axiom of Choice? I ask because I don't think anything more than DC is needed to prove both Furstenberg's theorem and Szemerédi's one (but I might be wrong here), thus I find curious that some `serious choice' is needed to bridge them
There is another way to proceed with the derivation by constructively building some finite probability measures on a Cantor space that approximately model the set $A$ and then extracting a weakly convergent subsequence. I think this has a chance of being done completely in a choice-free manner (countable iterations of Bolzano-Weierstrass plus a diagonalization argument). Not an expert on these questions though. [And now one can go further and ask if one can make this argument intuitionistic, my guess is no...]
Actually I now remember that I looked into this question back in 2005 and found a version of the Furstenberg correspondence principle that does not require the axiom of choice: https://www.math.ucla.edu/~tao/preprints/Expository/limiting.pdf
Very interesting, thank you very much
@NicolaGigli: Doesn't non-dependence on AC follow from $Π^1_1$-absoluteness?
@user21820: I don't really know, it is not my field of expertise. I've just seen the use of Hahn-Banach on a non separable space in the argument above and wondered whether more than DC was truly needed to carry it out. It might be that one could answer `no' by some general principle of which I known nothing about (but I'd like to know more: can you explain or point me to some relevant literature?)
@NicolaGigli: I am not a set theory expert, but from what I know AC can be eliminated from any ZFC proof of any theorem that is $Σ^1_2$ or $Π^1_2$, by simply applying Shoenfield's absoluteness theorem. In particular, the statement $Q$ here is equivalent to a $Π^1_1$ sentence that says "$∀A{⊆}N\ $ $( \ ∃p,q{∈}N\ ∀k{∈}N\ ∃m,c{∈}N\ ∃t{∈}(A⋂[-m,m])^c ( \ p,q>0 ∧ c·q > (2·m+1)·p \ )$ $⇒ ∀c{∈}N\ ∃t{∈}A^c\ ( \ \text{$t$ is an AP} \ ) \ )$" where $N$ is the naturals.
@user21820: thank you, I'll think about this. Still, I think it is interesting to have a more direct path like the one in Terence Tao's note (whether this can be regarded as an unpacking of Shoenfield's theorem in this particular circumstance, I have no idea)
@NicolaGigli: Sorry there was a typo; the "$∃m,c{∈}N$" should be "$∃m{\in}N_{>k}\ ∃c{∈}N$". It is unlikely that Terry's note is an unpacking of the full Shoenfield absoluteness, since Shoenfield did it for sentences higher in the analytical hierarchy, whereas here we only need it for a $Π^1_1$ sentence. I believe there is a much simpler proof for this special case, but haven't really thought about it.
@user21820: "It is unlikely that Terry's note is an unpacking ..." yeah, that was my bet as well
Sard's theorem implies that the measure of the set of critical points of a smooth function $f:M_1\to M_2$ between smooth manifolds has measure zero. Hence the preimage $f^{-1}(x)$ of almost every point in $M_2$ is a smooth submanifold. This can be used, for example, to prove the existence of Morse functions. Following Milnor's Morse Theory, Section 6, one can embed $M$ into $\mathbb{R}^n$. Then for almost all points in $\mathbb{R}^n$, the distance map is a Morse function. This may be seen by applying Sard's theorem to the normal bundle. The set of focal points has measure zero, and corresponds to the points at which the distance function is degenerate.
The Chevalley-Warning theorem asserts that if a system of polynomial equations in $r$ variables over a finite field of characteristic $p$ has total degree less than $r$, then the number of solutions to this system is a multiple of $p$.
An immediate corollary of this is Chevalley's theorem: if such a system of polynomials has a "trivial" solution (often this is the origin $(0,\dots,0)$), then it must necessarily have a non-trivial solution as well. This is often applied for instance as part of the "polynomial method" in combinatorics.
Thanks, but in the spirit of the game: where's the measure here? Did you just mean counting measure?
Yeah, counting measure, applied to the non-trivial solutions. (Arguably one should be measuring here using the p-adic valuation rather than the Archimedean valuation, though.) There are some proofs of at least one of the Sylow theorems in a similar spirit, though I don't remember the details offhand.
For the Sylow theorem. Let $G$ act on the set $X$ all subsets of size $p^k$ (where $|G|=p^k.a$) by right multiplication (or left if you prefer). Since $|X|$ is not divisible by $p$, there is an orbit of this action of size not divisible by $p$. Thus its stabilizer contains a multiple of $p^k$ elements in it. Either induction or an easy argument shows that the stabilizer has order exactly $p^k$.
In the same spirit, a p group acting on a finite vector space over F_p always has a fixed point. The orbits are all sizes power of p. There is an orbit of size 1 corresponding to 0 so there must be at least p-1 other orbits of size 1.
@TerryTao could u give me your opinion about my example here as an answer to the titled question, Thanks for any comment
Kahn and Markovic showed the existence of immersed essential surfaces in closed hyperbolic 3-manifolds. The idea was to construct many immersed pants in the manifold using the frame flow. From the exponential mixing of the frame flow, they showed that the cuffs of the pants were equidistributed in a sufficiently uniform way so that they could use Hall's marriage theorem to pair the cuffs in a way that created a closed nearly geodesic (and hence essential) surface. They used similar ideas to resolve the Ehrenpreis conjecture, although the proof was more subtle since they couldn't use Hall's marriage theorem.
The proofs of existence of expanders by Barzdin - Kolmogorov and Pinsker,
and (somewhat related)
Gromov's proof of the existence of groups with no coarse embedding into a Hilbert space.
A very famous and important theorem in the theory of metric embeddings is known as "Assouad's Embedding Theorem". It concerns doubling metric spaces: metric spaces for which there is a constant $D$ such that every ball can be covered by $D$ balls of half the radius.
Theorem (Assouad, 1983): For every $\epsilon\in (0,1)$ and $D>0$, there are constants $L$ and $N$ such that if $(X,d)$ is doubling with constant $D$, then the metric space $(X,d^\epsilon)$ admits an $L$-bi-Lipschitz embedding into $\mathbb{R}^N$.
This theorem is widely used throughout metric geometry and analysis on metric spaces. (See, e.g., here or here.)
An $L$-bi-Lipschitz embedding is simply an embedding that preserves all distances up to factor $L$. It's easy to see that the doubling condition is necessary for this theorem to hold. Moreover, there are known doubling metric spaces (the Heisenberg group for one) that are doubling but do not admit a bi-Lipschitz embedding into any Euclidean space, so one cannot allow $\epsilon=1$ in Assouad's theorem.
This means, of course, that the constants $L$ and $N$ must blow up as $\epsilon\rightarrow 1$, and this is reflected Assouad's proof.
Except, that's not quite true. In a really surprising construction, Naor and Neiman showed in 2012 that the dimension $N$ in Assouad's theorem can be chosen independent of the ``snowflake'' parameter $\epsilon$ as $\epsilon\rightarrow 1$. (The distortion $L$ must necessarily blow up in general.) In other words, one need not use too many dimensions for the embedding, no matter how close $\epsilon$ gets to $1$. I believe this shocked many people.
The construction of Naor and Neiman is probabilistic: they construct a random Lipschitz map from $(X,d^\epsilon)$ into $\mathbb{R}^N$, and show that it is bi-Lipschitz with positive probability. The proof is also a nice application to geometry of the Lovasz Local Lemma.
Assouad's paper: http://www.numdam.org/article/BSMF_1983__111__429_0.pdf
Naor-Neiman's paper: https://www.cs.bgu.ac.il/~neimano/Naor-Neiman.pdf
In general, the probabilistic method of Erdos follows exactly this philosophy: prove that an object with a certain property of number theoretic interest exists by showing that the probability a random set satisfies the desired property with positive probability (usually the probability is one!)
Example: a subset $S \subset \mathbb{N}$ is an asymptotic additive basis of order $k$ if there exists $N_0 > 0$ such that for all $N > N_0$, there exists $x_1, \cdots, x_k \in S$ (not necessarily distinct) such that $N = x_1 + \cdots + x_k$. In other words, every sufficiently large positive integer is the sum of $k$ elements of $S$ (with possible repetition).
If we define $r_S^k(n) = \# \{(x_1, \cdots, x_k) \in S^k : n = x_1 + \cdots + x_k\}$ to be the representation function of order $k$ with respect to $S$, then the average size of $r_S^k(n)$ is a measure of how "optimal" the set $S$ as an additive basis. For instance it is known that the set $\mathcal{S}$ of square integer is an additive basis of order $4$ (Lagrange's theorem), but it is hardly optimal since $r_\mathcal{S}^4(n) \gg n$ for all $n$. How small can $r_S^k(n)$ be on average provided that it is positive for all sufficiently large $n$?
Erdos and Fuchs gave a "lower bound" for this average: $r_S^k(n)$ cannot be constant on average. Further, Erdos and Turan made the following conjecture: if $S$ is an asymptotic additive basis of order $k$, then $\liminf_{n \rightarrow \infty} r_S^k(n) = \infty$.
Erdos further refined this conjecture to assert that the lower bound ought to be of order $\log n$. To show that such optimal additive bases exist he used the probabilistic method. The case $k = 2$ is due to Erdos and the general case due to Erdos and Tetali.
Universality of Riemann zeta function , Which related to the approximation of every Holomorphic function $f(z)$ by Riemann zeta function in the strip .
Corollary: Let $K_0$ be a compact set in the right half of the critical stripe $1/2< \Re z<1$.
Let $f$ be a continuous function on $K_0$, which is holomorphic on an open set containing $K_0$ and does not have any zeros in $K_0$ . For every $\epsilon_0>0$, we have that the limit (lower density )
$$ \inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \Big (\{ t\in[0,T]: \max\limits_{z \in K_0} \left| {\zeta(z+it) -f(z) )}\right| < \epsilon_0\Big\}) $$
is positive for $\lambda$ being the Lebesgue measure.
That's a great example! But I was wondering what L was and checked the linked wiki page and learned that the statement on universality there is purely in terms of $ \zeta$ (and without a log). So is there a reason why you quoted this more complicated statement?
@Dirk,yes am going to edit it, this is a particular case of Universality of Riemann zeta function, Thanks for your attention
@PierrePC, yes thanks for your attention it fixed now
Lubotzky, Maher, and Wu showed for any $n\in \mathbb{Z}, g\in \mathbb{N}$ the existence of homology 3-spheres of Heegaard genus $g$ and Casson invariant $n$ via a probabilistic argument. The idea is to take an appropriate subgroup of the Torelli subgroup of the mapping class group of a genus $g$ surface, and modify a Heegaard splitting of $S^3$ of genus $g$ by a random walk on this subgroup. On this subgroup, the Casson invariant is realized by a homomorphism to $\mathbb{Z}$. Since random walks are recurrent, each integer is realized as a Casson invariant infinitely often. And they show that with probability tending to 1, the Heegaard genus is $g$. Hence there exists manifolds with the desired invariants.
An example from pre measure theory real analysis is in the proof of the Bounded Convergence Theorem for Riemann integrals in [1, §22.14, p288]. Here the author omits a portion of the proof, and in his other book [2, §7.2.5, p304] the proof is omitted entirely and described as 'quite delicate'. The proof required is of the following : if $\delta > 0$ and if $(E_n)$ is a sequence of sets in $[0, 1]$ with each $E_n$ containing a finite number of non-overlapping closed intervals with a total length $\geq \delta$, then there exists a point belonging to infinitely many of the $E_n$. In other words we need to show the set $E = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} E_m$ of all such points is non-empty and we can do this by showing it contains a set of positive measure : denote by $F_n$ the union of the above intervals within $E_n$ (so $F_n$ is a Borel set), and let $F = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} F_m \subseteq E$, then by [3, §2.60, p44], since this is a decreasing intersection, we have $|F| = \lim_{n \rightarrow \infty} |\bigcup_{m=n}^{\infty} F_m|$, where $\forall\ n\ |\bigcup_{m=n}^{\infty} F_m| \geq |F_n| \geq \delta$ (noting $F_n$ equals a finite union of disjoint intervals with a total length $\geq \delta$), so that $|F| \geq \delta > 0$ and hence $F \neq \emptyset$ and thus $E \neq \emptyset$, as required. The author then uses the above theorem to prove the Monotone Convergence Theorem and Dominated Convergence Theorem for Riemann integrals, [1, §22.15, p289], [1, §25.21, p359], [4, App A], and the latter of these facilitates the proof of Stirling's Formula in [5].
Another example is the problem stated in this answer to Theorems that are 'obvious' but hard to prove, as an example of a problem that may seem obvious but which requires a non-trivial proof : if intervals $I_n \subseteq [0, 1]\ \forall\ n$ and $\sum_{n=1}^{\infty} l(I_n) < 1$ then $\bigcup_{n=1}^{\infty} I_n \subsetneq [0, 1]$, so we seek a point in $[0, 1]$ which has the property of not being in any $I_n$. Noting all the sets below are Borel sets we have :
\begin{eqnarray*}
|[0, 1]| & = & |\bigcup_{n=1}^{\infty} I_n| + |[0, 1] \setminus \bigcup_{n=1}^{\infty} I_n|, \hspace{2em} \mbox{using countable additivity} \\
\Rightarrow |[0, 1] \setminus \bigcup_{n=1}^{\infty} I_n| & = & 1 - |\bigcup_{n=1}^{\infty} I_n| \\
\Rightarrow |[0, 1] \setminus \bigcup_{n=1}^{\infty} I_n| & > & 0
\end{eqnarray*}
since by countable subadditivity $|\bigcup_{n=1}^{\infty} I_n| \leq \sum_{n=1}^{\infty} |I_n| = \sum_{n=1}^{\infty} l(I_n) < 1$. Thus the set of all the points of interest, ie $[0, 1] \setminus \bigcup_{n=1}^{\infty} I_n$, is non-empty as required. The proof is not 'hard' but does require the non-trivial concept of the outer measure on $\mathbb{R}$.
[1] Robert G. Bartle (1964), The Elements of Real Analysis, John Wiley & Sons.
[2] Robert G. Bartle & Donald R. Sherbert (1982), Introduction to Real Analysis, John Wiley & Sons.
[3] Sheldon Axler (2020), Measure, Integration & Real Analysis, Springer Graduate Texts in Mathematics, https://measure.axler.net/.
[4] Ross Ure Anderson (2023), Intervals and Outer Measure on R
, https://ia601904.us.archive.org/19/items/intervals-and-outer-measure-on-r/IntervalsAndOuterMeasureOnR.pdf
[5] Keith Conrad, Stirling's Formula, https://kconrad.math.uconn.edu/blurbs/analysis/stirling.pdf (mirror : https://archive.org/details/stirlings-formula-keith-conrad)
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2025-03-21T14:48:31.512556
| 2020-07-14T19:43:57 |
365633
|
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|
Stack Exchange
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Picard scheme of family of quartic surfaces
Recall that a quartic surface in $\mathbb{P}^3_\mathbb{C}$ has $N = 35$ coefficients. Let $U$ be the open subset of $\mathbb{P}^{N-1}$ parametrising smooth quartic surfaces and let $Q \to U$ be the corresponding family of all smooth quartic surfaces. Let $Pic_{Q/U}$ denote the Picard scheme of this family.
What is $Pic_{Q/U}$?
Picard groups of quartic surfaces can behave quite erratically, but my guess is that $Pic_{Q/U} \cong \mathbb{Z}$ is a constant group scheme, generated by $\mathcal{O}(1)$. It would be nice to have confirmation of this, ideally with a proof.
Formation of the Picard scheme commutes with base change, so the fibers of $Pic_{Q/U} \to U$ are complicated.
Thanks this makes sense. Still, I'm struggling to come up with a flat map $V \to U$ with $V$ connected such that $Pic_{Q/U}(V) \neq \mathbb{Z}$, since the genetic quartic surface has Picard group $\mathbb{Z}$.
There is no such flat map; the special fibers of $Pic_{Q/U} \to U$ are concentrated on proper subvarieties of $U$.
Yes I finally get it now. I think I was getting confused between the small and big fppf site; now it all makes sense, thanks!
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2025-03-21T14:48:31.512666
| 2020-07-14T20:00:54 |
365634
|
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|
Stack Exchange
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Quantitative bounds on convergence of Bayesian posterior
Let $Y$ be a random variable in $[0,1]$, and let $X_1, X_2, \ldots$ be a sequence of random variables in $[0,1]$. Suppose that the $X_i$'s are conditionally i.i.d given $Y$ ; in other words, I'd like to think of each $X_i$ as a "measurement" or "signal" regarding $Y$, and the signals are i.i.d.
Let $Z_k = Y | X_1,\ldots,X_k$ be the posterior on $Y$ given the first $k$ observations. I believe that the Bernstein von-Mises theorem implies that the sequence $Z_k$ converges to some limit posterior as $k$ grows large. However, BVM does not provide any quantitative guarantees on the rate of convergence as far as I'm aware.
I'm interested in quantitative statements of the form: "$Z_k$ is close to $Z_{k-1}$", for some suitable choice of distance metric on distributions. More precisely, what I'd love to know is whether, for some suitable distance function $d(.,.)$, there exists a bound of the form $d(Z_k, Z_{k-1}) \leq f(k)$ for some decreasing function $f$. Note that I'm asking that the bound on $d(Z_k,Z_{k-1})$ not depend on the particulars of $Y$ and the $X_i$'s, which may or may not be asking too much... If that is asking too much, bounds on $d(Z_k,Z_{k-1})$ which depend as minimally as possible on $Y$ and the $X_i$'s would be appreciated as well.
I've seen a few somewhat similar questions on Mathoverflow and stackexchange, but didn't find anything that really answered this question. Any literature references would be appreciated.
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2025-03-21T14:48:31.512801
| 2020-07-14T20:24:13 |
365637
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Stack Exchange
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Compute $ \int_{0}^{+\infty} \left( \frac{\ln(x)}{e^x}\right)^2 dx $
How can I compute this integral?
$$
\int_{0}^{+\infty} \left( \frac{\ln(x)}{e^x}\right)^2 dx
$$
What do you mean by compute? Do you want a closed-form formula?
$\int_{0}^{\infty} \bigl( \frac{\ln x}{e^x}\bigr)^2 dx=\frac{1}{12}\pi ^2+\frac{1}{2} (\gamma_{\rm Euler} +\ln 2)^2$
I don't need the result but rather the method, I used Integration by parts
but had problems with $ \int_{0}^{\infty} \frac{e^{-2x}}{x} ln(x) $
@mathouv, look my below linked answer, I already asked this question yesterday
Let
$$f(a):=\int_0^\infty x^{a-1}e^{-2x}\,dx.$$
Then
$$f''(1)=\int_0^\infty \ln^2x\,e^{-2x}\,dx,$$
which is the integral in question.
On the other hand, $f(a)=2^{-a}\,\Gamma(a)$, and hence
the integral in question is
$$f''(1)=\frac{\ln^2 2}2 - \Gamma'(1)\ln2
+ \Gamma''(1)/2
=\frac{\pi^2}{12}+ \frac{(\gamma +\ln2)^2}2=1.6293\dots,$$
where $\gamma=0.57721\dots$ is the Euler gamma constant.
(The second equality in the latter display follows because $\Gamma'(1)=-\gamma$ and $\Gamma''(1)=\gamma^2+\pi^2/6$. In turn, the latter two equalities can be obtained using the last two displays in Section Recurrence relation.)
$f(a)$ is a Mellin transform, right? Is this often used to compute seemingly tricky integrals?
@BenCrowell : Yes, of course.
@SylvainJULIEN : Yes, $f$ is the Mellin transform of $e^{-2\cdot}$. I am very far from an expert in the Mellin transform, though.
Hint: you may try $n=2$ and follow in general this :\begin{align} \int_0^\infty \left(\frac{\log(x)}{e^x}\right)^n\,dx&=\int_0^\infty e^{-nx}\log^n(x)\,dx\\\\ &=\frac1n\sum_{k=0}^n\binom{n}{k}\log^{n-k}(n)\int_0^\infty e^{-x}\log^n(x)\,dx\\\\ &=\frac1n\sum_{k=0}^n\binom{n}{k}\log^{n-k}(n) \left.\left(\frac{d^n \Gamma(x+1)}{dx^n}\right)\right|_{x=0} \end{align}
For more information check answers of my question here
For more clarification according to the answer of @Nikunj, try $n=2$ using this steps in general: Let $$I(a) = \int_0^\infty e^{-nx}x^a\,dx$$
$$\implies \frac{d^nI(a)}{da^n} = \int_0^\infty e^{-nx}x^a(\ln x)^n\,dx$$
Put $nx \rightarrow v$ in the first integral to get:
$$I(a) = \frac1{n^{1+a}}\int_0^\infty e^{-v}v^a\,dv$$
$$\implies I(a) = \frac{\Gamma(1+a)}{n^{1+a}}$$
Now $$\implies \frac{d^nI(a)}{da^n}\bigg|_{a=0} = \frac{d^n}{da^n}\left(\frac{\Gamma(1+a)}{n^{1+a}}\right)\bigg|_{a=0}$$
Which evaluates to:
$$\frac1{n}\sum_{k=0}^n(-1)^k\binom{n}{k}\Gamma^{(n-k)}(1+a)\ln^k(n)\bigg|_{a=0}$$
Where $\Gamma^{(n-k)}(1+a)$ is the $(n-k)$th derivative of the Gamma function.
This doesn't seem to work for $n = 2$. Do you mean to have $\displaystyle \frac1n\sum_{k=0}^n(-1)^k\binom{n}{k}\log^{n-k}(n)\int_0^\infty e^{-x}\log^k(x),dx$?
@ChipHurst, I had a wrong linked question, it fixed now , I already asked this question yesterday here
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2025-03-21T14:48:31.513122
| 2020-07-14T21:23:55 |
365644
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|
Stack Exchange
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Character degrees in induced blocks
Let $G$ be a finite group and $U\leq H\leq G$ a chain of subgroups.
Presume that $p$ is a prime dividing the order of $U$.
Suppose that $b_1$ is a $p$-block of $U$ and $b_2$ a $p$-block which is conjugated to $b_1$ via an element $g\in G$. Assume that both $B_1:=b_1^H$ and $B_2:=b_2^H$ are defined in the sense of Brauer.
If the character degrees of $\text{Irr}(b_1)$ and of $\text{Irr}(B_1)$ are known, what can be derived about the character degrees of $\text{Irr}(B_2)$ ?
This is probably the case, but are you assuming that $g$ normalizes $H$?
@DavidCraven : If $g$ normalizes $H$, isn't it clearly the case that $B_{1}$ and $B_{2}$ have the same character degrees?
@GeoffRobinson That's what I thought. But if $g$ doesn't normalize $H$ surely you have very little control over the induced block?
@DavidCraven : I think so, which is why I hadn't attempted an answer
Alas, I cannot suppose that $g$ normalizes $H$. :-(
I think it's not even going to be true in general that $B_2$ has the same number of characters as $B_1$. @GeoffRobinson do you agree?
@DavidCraven: I can see no reason why they should have the same number of characters.
Ok. Thank you very much.
|
2025-03-21T14:48:31.513262
| 2020-07-14T22:00:32 |
365646
|
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"Alex Kruckman",
"Emil Jeřábek",
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|
Stack Exchange
|
Is an abelian group of bounded exponent $\aleph_0$-categorical
For an abelian torsion group of finite exponent, i.e. there is an integer $n$ such that $g^n=1$ for all $g\in G$, its theory appears to be $\aleph_0$-categorical by the theorem of Engeler, Ryll-Nardzewski and Svenonius. I want to confirm this fact.
Just to clarify: for you, "totient group" is a synonym for "finite abelian group"? The complete theory of any finite structure has only one model up to isomorphism. In particular, it has no infinite models. So no, it is not $\aleph_0$-categorical.
@AlexKruckman I added an answer assuming "finite order" meant "finite/bounded exponent". But perhaps it would have been good to clarify the terminology first since of course what you've said is correct!
@GabeConant Yes, I understood "finite order" to mean "finite cardinality", but I'm sure you're right that the condition I would call "finite exponent" is what the OP intended.
Why do you edit without addressing the comments? what do you mean by "finite order"?
Perhaps “totient” here is a mangling of “torsion”? (That’s a weaker property, but related.)
Yes, it is a typo and should be torsion group. Finite order means bounded order.
@hermes (1) the definition of "torsion group" is a group in which every element has bounded order. (2) the definition of "group of finite order" is just "finite group". (3) a group in which elements have finite bounded order is called "group of finite exponent".
@YCor, I modified it. One thing I am confused is that, "finite group" means a group of finite size and thus each element of it should be of finite order, right?
To clarify: (1) I've only used "bounded order" to stay consistent with Rosenstein. "Bounded exponent" or "finite exponent" is the more standard terminology. (2) Finite groups are special cases of groups of bounded exponent. (3) Groups of bounded exponent are special cases of torsion groups, but these are not the same. A torsion group is defined as one in which every element has finite order, but there need not be a uniform finite bound on the orders. For example: $\oplus_n \mathbb{Z}/n\mathbb{Z}$ is a torsion group that does not have bounded exponent. (4) A torsion group need not be abelian.
But if the order of elements in a group is not uniformly bounded, i.e. the order can be arbitrarily large, then it is (generally) not considered as a group of finite order.
It is correct that such a group is not considered to be a group of bounded order, but it could still be a torsion group. I say all of this because you have written "A torsion group is an abelian group of bounded order." But torsion groups need not have bounded order (or be abelian).
I've rolled back your recent edit, in which you wrote "An abelian torsion group is an abelian group in which each element has finite order, i.e. there is an integer $n$ such that $g^n=1$ for all $g\in G$." The condition "$G$ is torsion" means "for all $g\in G$, there is an integer $n$ such that $g^n = 1$." There are abelian torsion groups which are not $\aleph_0$-categorical, like $\mathbb{Q}/\mathbb{Z}$. On the other hand "$G$ has finite exponent" means "there is an integer $n$ such that for all $g\in G$, $g^n = 1$." This is the type of group Gabe's answer addresses.
@hermes The question is fine as it is, and it has a good accepted answer. There's no need to change it. Looking at your profile, I don't see that you're suspended, so you should be able to ask questions now. Maybe you were previously suspended - if you were, it was for some rule violation, not because someone thought this was a bad question.
I have to pay close attention to questions with lower upvotes. Since this question has one upvote and one downvote, I guess it is considered a bad question. I am still banned to ask questions on this site which can not be seen from the profile. I do not have any rule violation and so the ban must be due to some questions I have asked.
This is a theorem of Rosenstein from the paper $\aleph_0$-categoricity of groups. But note that he uses the terminology bounded order rather than finite exponent.
I say that a group is $\aleph_0$-categorical if its complete theory in the language of groups is $\aleph_0$-categorical (i.e., has a unique countable model).
It is not hard to show that an $\aleph_0$-categorical group has finite exponent. This is also Theorem 1 of Rosenstein's paper. Theorem 2 of his paper provides a converse for abelian groups. So we have:
Theorem. (Rosenstein 1971)
An infinite abelian group is $\aleph_0$-categorical if and only if it has finite exponent.
Appendix. The discussion below has inspired me to add some discussion of the proof of theorems, and some other remarks. The theorem above comes from two results in Rosenstein's paper.
Theorem 1. An $\aleph_0$-categorical group has finite exponent.
Proof. Rosenstein uses the following fact, due to Ryll-Nardzewski/Enegler/Svenonius independently.
A countably infinite structure $M$ is $\aleph_0$-categorical if and only if for all $m>0$, the action of $\operatorname{Aut}(M)$ on $M^m$ has finitely many orbits.
Now suppose $\operatorname{Th}(G)$ is $\aleph_0$-categorical. We may assume $G$ is countable. Since elements of distinct orders are in distinct orbits of the action of $\operatorname{Aut}(G)$ on $G$, there is a uniform bound on the orders of torsion elements in $G$. Moreover, $G$ has no elements of infinite order since, if $g$ were such then $(g,g^n)$ for varying $n$ would be in distinct orbits in the action on $G^2$.
Theorem 2. An infinite abelian group of finite exponent is $\aleph_0$-categorical.
See Rosenstein's paper. The main tool is a structure theorem for abelian groups of finite exponent as direct sums of cyclic groups. This is apparently called Prufer's First Theorem.
Some remarks:
By YCor's comment about $\mathbb{Q}^{(\omega)}$, one cannot prove Theorem 1 by considering only orbits of singletons. ($\mathbb{Q}$ works also.)
It is easier to prove that an $\aleph_0$-categorical torsion group $G$ has finite exponent. Indeed, if not then by compactness/DLS there is a countable model of $\operatorname{Th}(G)$ with an element of infinite order, which cannot be $G$. But if $G$ is not a torsion group, I don't see a quick way to avoid Ryll-Nardzewski or Omitting Types of some kind (although one can substitute other facts that use these results, e.g., in an $\aleph_0$-categorical structure, the algebraic closure of a finite set is finite).
An $\aleph_0$-categorical group need not be abelian. An example is the countably infinite extraspecial $p$-group (see Definition 5.15 here). On the other hand, there are many results in model theory along the lines of "$\aleph_0$-categorical plus some model-theoretic property" implies some abelian-like structure. For example, Bauer, Cherlin, and Macintyre showed that an $\aleph_0$-categorical superstable group is abelian-by-finite.
An infinite group of finite exponent need not be $\aleph_0$-categorical. For example, take an infinite finitely generated group of finite exponent (like a Tarski monster). Indeed, by the more general result about algebraic closure stated above, if $G$ is $\aleph_0$-categorical then any finite subset of $G$ generates a finite subgroup.
In fact, a google search for "totient group" turns up this question as the only usage of that phrase in a mathematical context.
@Alex On the page 2 "badlands" of Google searches I found it used in this paper. I think that there author actually does mean the group of units mod $n$, which supports your interpretation. I'm now also quite intrigued by this.
Haha. I found that paper too, but I considered "the totient group of $n$" to be a different usage than "totient group" as a general concept (certainly not every finite abelian group is the group of units mod $n$ for some $n$).
Actually, that every abelian group of bounded exponent is $\aleph_0$-categorical is pretty easy (and probably straightforward from Szmielew); I guess the bulk of Rosenstein's theorem is the reverse direction.
@YCor His proof of Thm2 is definitely longer, for whatever that's worth. But yes I agree. Anyway, in light of all these comments I'm definitely less sure what the original intention of the question was. Hopefully it will be clarified.
For the other direction: If $G$ has elements of arbitrarily large finite order then there are infinitely many orbits of singletons. If $G$ has an element $g$ of infinite order then $(g,g^n)$ for varying $n$ yield infinitely many orbits of pairs.
I thought categoricity was a property of theories, not structures. I'm guessing that people are saying "$\aleph_0$-categorical sructure" when then mean "structure whose elementary theory is $\aleph_0$-categorical".
@GabeConant what's an orbit of singleton? for the group $\mathbf{Q}^{(\omega)}$ the automorphism group acts transitively on nonzero elements.
@bof what's the meaning for a noncomplete theory to be $\aleph_0$-categorical? (I didn't find the answer)
@YCor I never studied logic seriously so my knowledge of the subject is rather scattershot, but I think I know that one. A theory $T$ is $\kappa$-categorical if all models of $T$ of cardinality $\kappa$ are isomorphic. Examples: theory of plain sets with no additional structure; theory of dense linear orders with (or without) top and bottom elements; theory of atomless Boolean algebras. Hmm. I guess the theory of the random infinite graph is another one.
@YCor I’m talking about the action of $Aut(G)$ on $G^m$ for varying $m$. $\aleph_0$-categoricity is equivalent to those actions having finitely many orbits for all $m$. “Singleton” and “pair” refers to $m=1$ and $m=2$. In the first case elements of different orders must be in different orbits. One can also think of $m$-types over $\emptyset$ rather than orbits of $m$-tuples.
@bof Yes you are correct it’s a property of the theory. But it’s common for people to say that a structure has a property when they mean its complete theory does. E.g. “$M$ is stable” means “$Th(M)$ is stable.”
@Emil Your argument is doesn't cover the case that $G$ is torsion-free.
Fair enough. --
@bof, $\aleph_0$-categorical structure is defined as a structure of size $\aleph_0$ whose theory is $\aleph_0$-categorical .
@hermes According to that definition, isn't the answer to your question trivially "no", since an Abelian group of bounded exponent may be uncountable or finite? Did you mean to say "is a countably infinite Abelian group of bounded exponent $\aleph_0$-categorical?
@bof, I mean it is countable because $ℵ_0$-categorical is countable.
I don't think Theorem B is correct as stated: $G=C_4\times C_2^{(\omega)}$ and $H=C_4^2\times C_2^{(\omega)}$ both have infinitely many elements of orders 2 and 4, but they are not isomorphic: $G/G_2\simeq C_2$ and $H/H_2\simeq C_2^2$, where $G_2={g\in G:g^2=1}$.
FWIW, the number (finite or infinite) $n$ of copies of $C_{p^k}$ in the decomposition of $G$ to prime power cyclic groups is given by $G_{p^k}/(G_{p^{k-1}}(G^p\cap G_{p^k}))\simeq C_p^{(n)}$ if I didn't make a mistake.
Thanks @Emil I will make some edits
|
2025-03-21T14:48:31.514017
| 2020-07-14T22:04:36 |
365647
|
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|
Stack Exchange
|
Information theory for uncountably infinite-dimensional continuous random variable
I'm exploring the possibility to apply information theory on an uncountably infinite-dimensional scenario. I found the concept of generalized entropy for continuous random variables defined on finite-dimensional Euclidean space. I wonder whether there are similar concepts for uncountably infinite-dimensional cases. For instance, if I have a distribution on $C([0, 1])$, which is the space of continuous function on $[0, 1]$, then how can I quantify the entropy of this distribution?
Your link does not lead to any content.
@IosifPinelis I've updated the link. Thanks for pointing out this!
You can do this exactly in the same way, except that the right notion is that of relative entropy and that you need a reference measure. Let me explain: on an abstract measurable space $(\Omega,\Sigma)$ choose any reference probability measure $R$. The relative entropy of an arbitrary probability measure $P\in\mathcal P(\Omega)$ with respect to $R$ is then simply
$$
H(P|R):=\int_{\Omega}\frac{dP}{dR}(\omega)\log\left(\frac{dP}{dR}(\omega)\right) \,d R(\omega),
$$
with the convention is that $0\log 0=0$, that $\frac{dP}{dR}$ denotes the Radon-Nykodim density of $P$ with respect to $R$ if the absolute continuity $P\ll R$ holds, and that $H(P|R):=+\infty$ whenever $P$ is not absolutely continuous w.r.t. $R$.
The fact that $R$ is a probability can obviously be relaxed, it can actually be unbounded (but is must still be nonnegative, of course).
Clearly, that $\Omega$ be finite- or infinite-dimensional plays no distinguised role whatsoever in this abstract definition. In my humble opinion people are often misled because in finite dimensions there is a "canonical" reference measure, which is the Lebesgue one $R=dx$. So people often don't realize that $H(\rho)=\int_\Omega \rho(x)\log\rho(x)\,dx$ is actually a relative entropy $H(\rho|dx)$, with a slight abuse of notations that the probability measure $\rho$ and its density $\rho(x)$ w.r.t the Lebesgue measure are identified.
In your particular example $\Omega=C([0,1])$ one possible and usual reference measure is the law of a Brownian motion. The resulting entropy then plays a role sometimes in Girsanov theory and optimal transport, see e.g. this paper or that paper.
Hi Leo, thanks for the answer here! While I'm digesting the references, I have some quick questions here. 1) Do we have any convenient way to compute $H(P|R)$? As far as I have concerned, the Radon-Nykodim density is not easy to compute... 2) Is there a good choice of $P$ in infinite-dimensional case? I know that a uniform distribution doesn't exist in infinite-dimensional case... 3) If I want to extend this notion of relative entropy to a space of random variables (e.g., I would like to choose an r.v. under some probability distribution), then what reference measure would be available?
is not an easy task in whole generality, but for example when $R$ is the (law of) a Brownian motion then the RN density can be computed using Girsanov theory, as I mentioned. 2) not really, at least not a canonical one. Again, the Wiener measure is a reasonable choice, but I guess it depends on the applications you have in mind. 3) same as 2), I guess it depends on your applications, but perhpas check out Sanov's theorem for a starting point?
|
2025-03-21T14:48:31.514292
| 2020-07-14T22:05:44 |
365648
|
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"Alessandro Codenotti",
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|
Stack Exchange
|
Is the Hilbert cube the countable union of punctiform spaces?
Recall that a (separable) metric space is called punctiform, if all its compact subspaces are zero-dimensional. While "natural" spaces would seem to be punctiform if they already themselves zero-dimensional, there are even infinite dimensional punctiform spaces. The constructions I have seen however are still yielding spaces that "feel sparse" to me.
The Hilbert cube $[0,1]^\omega$ is large in the sense that it is not a countable union of zero-dimensional spaces. What I am now wondering is whether we can write the Hilbert cube as a countable union of punctiform spaces. Note that I do not want to impose any complexity constraints on the pieces.
If the answer should be "yes", I'd be very interested in understanding the structure of the punctiform spaces involved.
Had the answer been "no", this would have answered an open problem in computability theory, see Question 5 on Page 99 (v1) here: https://arxiv.org/abs/1904.04107
I don't know how updated this list is, but it seems to be open (question 4)
@AlessandroCodenotti The question arises out of a joint paper with Kihara and Ng in computability theory. So we don't know, and we know that some others don't know either, but maybe someone here knows, or has an idea how to solve this.
this may be of interest https://mathoverflow.net/a/357441/53155
The Hilbert cube can be written as the union of two punctiform spaces. Just take any Bernstein set $X\subset[0,1]^\omega$ and observe that compact subsets in $X$ and $Y=[0,1]^\omega\setminus X$ are at most countable. So, $X$ and $Y$ are punctiform spaces and $X\cup Y=[0,1]^\omega$.
A less trivial fact says that the Hilbert cube cannot be written as the countable union of hereditarily disconnected sets; for the proof of this fact, see Main Lemma in this paper. A topological space $X$ is called hereditarily disconnected if all connected subspaces in $X$ are singletons.
I suspect that your result on the hereditarily disconnected spaces is what I actually needed in the first place, thanks a lot for bringing it my attention!
@Arno You are welcome. I looked at the questions of Kihara (https://math.berkeley.edu/~kihara/questions.html). Is his 5th question still open?
To my knowledge, Question 5 is still open (and in fact is the motivation for this question). The fascinating thing about it is that it is equivalent to an open problem in computability theory concerning enumeration degrees. The latter question is Question 5 here https://arxiv.org/abs/1904.04107 (Page 99)
@Arno Thank you for the link to your paper. What is known about metrizable subspaces of $(\omega_{cof})^\omega$? All of them are punctiform. Does $(\omega_{cof})^\omega$ contain a non-trivial metrizable connected subspace? Is the Erdos space (it is totally disconnected but not zero-dimensional) embeddable into $(\omega_{cof})^\omega$? Is each metrizable subspace of $(\omega_{fin})^\omega$ zero-dimensional?
We still need to carefully check the argument, but the Erdos space should embed into $(\omega_{\mathrm{cof}})^\omega$.
@Arno Wow! Very exciting news!
|
2025-03-21T14:48:31.514525
| 2020-07-14T22:29:44 |
365650
|
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|
Stack Exchange
|
Using Paley-Wiener Theorem to prove the decay of $G(x-y)$
This question is related to my previous one, where I was looking for some help to prove the decay of the lattice Green function:
\begin{eqnarray}
G(x-y) = \int_{[-\pi,\pi]^{d}}\frac{d^{d}k}{(2\pi)^{d}}\frac{e^{ik\cdot (x-y)}}{\gamma(k)+p} \quad (x,y \in \mathbb{Z}^{d}) \label{1}\tag{1}
\end{eqnarray}
where $p>0$, $k\cdot (x-y) := k_{1}(x_{1}-y_{1}) +\cdots+k_{d}(x_{d}-y_{d})$ and:
$$\gamma(k) := 2\sum_{j=1}^{d}(1-\cos(k_{j})). $$
As mentioned in the previous question, this decay should be like:
$$G(x-y) \le Ke^{\alpha ||x-y||_{\infty}}$$
where $||x-y||_{\infty} := \max_{i}\{|x_{i}-y_{i}|\}$. As cited in the comments of the previous question, we can address this problem by using contour integrals or representation as a Neumann series. However, in this article (page 183) the authors propose a very interesting way to prove this result by using the Paley-Wiener Theorem. I'd really like to understand it but they just briefly mention the idea and I'm having trouble working out the details and I'm having trouble making the Paley-Wiener Theorem fit in this context. Shouldn't my function $G$ be an entire function of $n$ complex variables given by the Fourier transform of a $C_{0}^{\infty}(\mathbb{R}^{d})$ function? It does not seem the case to me. Any help would be appreciated!
My guess would be that this is the same as what Abdelmalek Abdesselam wrote in his comment to the other question — so strictly speaking not really an application of the Paley–Wiener theorem, but something to some extent similar. (And yes, $G$ is an entire function, once you define it for by the same formula for all $x \in \mathbb{C}^d$.)
That said, the easiest way to get such a crude upper bound is perhaps as follows: If $p = \cosh \alpha - 2$, then the function $F(x) = \exp(-\alpha|x|_\infty)$ is discrete $p$-superharmonic, in the sense that $H(x) := \Delta F(x) - p F(x) \leqslant 0$ for all $x \in \mathbb{Z}^d$. Therefore, $F$ is the discrete convolution of $G$ and $-H(x)$ (more precisely: $F + G*H$ is discrete harmonic and goes to zero at infiinity, and hence it is constant zero)). In particular, $F(x) \geqslant (-H(0))G(x)$, that is, $G(x) \leqslant (-H(0))^{-1} F(x)$.
Oh, this seems a nice approach but I'm not very familiar with some of these objects. If $F(x)$ is $p$-superharmonic does it follow that it is the discrete convolution of $G$ and $-H$? Is this a theorem? Do you have any reference on the precise statement?
All you need to know is that $(-\Delta + p) G(x) = \delta_0(x)$ (which follows by looking at the Fourier transform, for example). Then we find that $(-\Delta + p) (G * H) = H = (\Delta - p) F)$, and consequently $(-\Delta - p)(F + GH) = 0$. Now $F + GH$ goes to zero at infinity (requires a proof!), and — if non-zero — it attains a positive maximum or a negative minimum somewhere. However, if $F(x_0) + GH(x_0) > 0$ is a maximum, then $\Delta (F + GH)(x_0) < 0$, and so, in particular, $0 > (\Delta - p)(F + GH)(x_0) = 0$, a contradiction. Thus, $F + GH = 0$, as desired.
Amazing! I will try to work out the details! Thanks a lot!
I think I got it! Just one thing: (it is probably easy to see but) why does $F(x) \ge (-H(0))G(x)$?
That is indeed quite simple: $H(x) \leqslant 0$, and hence $F(x) = (-H) * G(x) = \sum (-H(y)) G(x - y) \ge -H(0) G(x - 0)$.
|
2025-03-21T14:48:31.514784
| 2020-07-14T22:55:49 |
365652
|
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|
Stack Exchange
|
Existence of the differential entropy for infinitely divisible laws
Let $X$ be an absolutely continuous (i.e. its law is absolutely continuous with respect to the Lebesgue measure) random variable with probability density $p$. Its differential entropy is given by
$$h(X) = - \int_{\mathbb{R}} p(x) \log p(x) \mathrm{d} x$$
with the convention $0 \log 0 = 0$, as soon as the integral is absolutely convergent.
A random variable is infinitely divisible if, for any $n \geq 1$, $X$ can be decomposed as the sum of $n$ i.i.d. random variables.
Question: Are there infinitely divisible and absolutely continuous random variables for which the differential entropy does not exist?
Comment: It is possible to construct random variables for which the differential entropy does not exist. The constructions I could find are however handcrafted to make the differential entropy undefined. Since infinitely divisible random variables have a strong structure, I am wondering what can be said in this case.
It is moreover possible to find simple conditions so that the differential entropy is well-defined, for instance if $X$ admits some positive moments and $p$ is a bounded probability density. The former condition is however not always true for infinitely divisible laws, and I have no idea for the latter.
Any help would be appreciated.
For real $t>0$, let
\begin{equation}
p_t:=e^{-t}e^{*tf}*g_t:=e^{-t}\sum_{n=0}^\infty\frac{t^n f^{*n}}{n!}*g_t, \tag{0}
\end{equation}
where $f$ is the (bounded by $c:=1/e$) pdf given by
\begin{equation}
f(x)=\frac{1\{x\ge e\}}{x\ln^2 x}, \tag{0.5}
\end{equation}
$f^{*n}:=f*\cdots*f$ ($n$ times, with $f^{*0}$ defined as the Dirac delta function at $0$), and $g_t$ is the normal pdf with mean $0$ and variance $t$. Then $p_s*p_t=p_{s+t}$ for all real $s,t>0$. So, $$p:=p_1$$ is the pdf of an infinitely divisible distribution. Moreover, similarly to the main result in this paper, we have
\begin{equation}
p(x)\sim f(x) \tag{1}
\end{equation}
(the convergence everywhere here is as $x\to\infty$),
whence
\begin{equation}
p(x)\ln p(x)\sim-\frac1{x\ln x}, \tag{1.5}
\end{equation}
so that the differential entropy does not exist.
Since the proof of (1) is a bit involved, let us make do with something weaker than (1), which however can be proved quickly. Indeed, note first here that, by (0) and (0.5), for $g:=g_1$ and all real $x$
\begin{equation}
p(x)\ge e^{-1}(f*g)(x)\ge\frac1e\,\int_{-1}^1f(x-y)g(y)\,dy\sim\frac c{x\ln^2 x}=:q(x), \tag{2}
\end{equation}
where $c:=\frac1e\,\int_{-1}^1g(y)\,dy\in(0,\infty)$.
On the other hand, again by (0),
\begin{equation}
p(x)=\frac1e\,\int_{-\infty}^\infty g(x-y)\,(e^{*f})(y)\,dy\to0, \tag{3}
\end{equation}
by dominated convergence.
Now note that the function $u\mapsto-u\ln u$ is positive and increasing in a right neighborhood of $0$. Hence, by (2) and (3), for all large enough $x>0$
\begin{equation*}
-p(x)\ln p(x)\ge-\frac{q(x)}2\,\ln\frac{q(x)}2\sim\frac c{2x\ln x}
\end{equation*}
(cf. (1.5)).
So, the differential entropy does not exist.
Nice! I will definitely keep this way of constructing infinitely divisible laws with given asymptotic behaviors for the pdf. Thanks a lot, this solves my question.
@Goulifet : I am glad this was of help.
@IosifPinelis One can just take the exponential semigroup of any infinite differential entropy measure on $\mathbb R$ without involving the Gaussian semigroup. On the other hand, one can also take the exponential semigroup of an infinite entropy measure on $\mathbb Z$ and then convolve it with the Gaussian semigroup.
@RW : What I used here was indeed the $$-exponentiation of a finite measure $\mu$. However, one cannot do here without the Gaussian component (even if $\mu$ is absolutely continuous), because otherwise the resulting measure $e^{\mu}$ would not be absolutely continuous. Anyhow, the main problem here is to prove (1) or its adequately relaxed counterpart.
@ IosifPinelis Sorry - forgot about the 0 order term in the exponent, which gives yet another reason to deal with a distribution on $\mathbb Z$, and then to smooth it. However, I believe that it can be done without using any explicit estimates, just from infinitude of the entropy.
@IosifPinelis, your construction creates an infinitely random random variable $X$ such that all the positive moments are infinite. Do you believe it is possible to construct an example with a finite positive moment $\mathbb{E}[|X|^p] < \infty$ for some $p>0$?
@Goulifet : No, this is not possible. Indeed, for any real $r>0$, if the $r$th moment of the distribution with a pdf $p$ is finite, then the differential entropy of $p$ is also finite; cf. e.g. https://en.wikipedia.org/wiki/Differential_entropy#Proof . If you need further details, you may want to ask this in a separate post.
@IosifPinelis I know this result for finite variance random variables using the KL with the Gaussian, but I am not aware of a general result with no additional assumption on $p$. You are right then, this may deserve a new post.
@IosifPinelis I am using your answer in a paper to illustrate the existence of infinite divisible laws with properties close to the one you proved. Currently, I am referring to the Mathoverflow link. Would you agree to be mentioned in the acknowledgement of the paper for providing the argument?
@Goulifet : Yes, that would be fine.
@IosifPinelis Perfect, thanks.
|
2025-03-21T14:48:31.515285
| 2020-07-14T23:36:53 |
365654
|
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|
Stack Exchange
|
What is the relation between different generalizations of linear programming?
Linear programming subsumed by each of
Semidefinite programming (SDP)
Convex programming (CXP)
SOS programming (SSP)
Is there any relation between each pair in the three?
Are all three equivalent in expressive capability?
$LP\subset SDP\subset SSP\subset CXP$. Also, you can encode SSP as SDP by lifting. These slides are relevant: http://www.mit.edu/~parrilo/pubs/talkfiles/Eckman.pdf
So it is rather $LP\subset_{poly} SDP={poly}SSP\subset{poly} CXP$? By subscript $poly$ I mean you are blowing up the program only polynomially. So the inclusions are upto polynomial blowup?
|
2025-03-21T14:48:31.515365
| 2020-07-15T00:47:02 |
365656
|
{
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"Nate Gallup",
"Tim Campion",
"Zhen Lin",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365656"
}
|
Stack Exchange
|
Closure of the product of subfunctors
Background:
Let $X: \textbf{CRing} \to \textbf{Set}$ be a presheaf on the category of affine schemes and $Z \subseteq X$ a subfunctor. One defines $Z$ to be closed if for every ring $A$ and every morphism $f: \text{Hom}(A , -) \to X$ the inverse image $f^{-1}(Z)$ is of the form $R \mapsto \{ \varphi : A \to R | \varphi(I) = 0 \}$ for some ideal $I \subseteq A$.
The intersection of subfunctors is defined naively, as is the closure (denoted by $\overline{Z}$) of a subfunctor $Z \subseteq X$ (it is the intersection of all closed subfunctors of $X$ containing $Z$).
If $Y$ is another presheaf, the product of $X$ and $Y$ is also defined naively.
Context: In section 1.14 of Jens Jantzen's great book "Representations of Algebraic Groups", the following is stated: If $X$ and $Y$ are presheaves which are schemes over a noetherian ring $k$ and $Z \subseteq X$ is a subscheme, and if $Z, X$ are algebraic and $Y$ is flat, then $\overline{Z \times Y} = \overline{Z} \times Y$. For the proof, he references Demazure-Gabriel I, section 2, 4.14 (although in my copy of Bell's translation this reference unfortunately doesn't exist).
Actual Question:
Is this true for general presheaves? I.e. if $X$ and $Y$ are presheaves and $Z \subseteq X$ is a subfunctor, is it true that $\overline{Z \times Y} = \overline{Z} \times Y$? I worry that it isn't because of the conditions in Jantzen stated above, but I haven't been able to decide either way. (Also side question: does anyone know the correct reference in the translation?)
This is not true even for affine schemes. Let $k = \mathbb{Z}$, let $X = \operatorname{Spec} \mathbb{Z}$, let $Y = \operatorname{Spec} \mathbb{F}_p$, and let $Z \cong \operatorname{Spec} \mathbb{Z} [ p^{-1}]$. The closure of $Z$ in $X$ is $X$ itself, but $Z \times Y \cong \operatorname{Spec} \{ 0 \}$, which is already closed in $X \times Y$. Of course, $Y$ is not flat.
It's good to see you around, Zhen Lin!
Thanks so much for your answer!! (Even though I was hoping the opposite would be true ;). ) This is a great example! I'm actually secretly interested in the special case when $G$ is a $k$-group functor acting on a $k$-functor $X$. If $Y \subseteq X$ is a subfunctor and the action of $G$ on $X$ restricts to an action of $G$ on $Y$, do you happen to know if $G$ also acts on the closure $\overline{Y} \subseteq X$?
I think you want $G$ to be at least flat over $k$ there. The analogous fact for continuous group actions has an abstract nonsense proof using the fact that a group action $G \times X \to X$ is automatically an open map and the fact that pullback along open maps preserves denseness. But for schemes even the trivial action $G \times 1 \to 1$ might not be open.
Thanks again for your comment! This has been extremely helpful. If you don't mind my bombarding you with a few more questions... What do you mean by pullback along the group action preserves denseness? Just that the inverse image of $Y$ under $G \times X \to X$ (which is just $G \times Y$) is dense in the inverse image of $\overline{Y}$?
Yes, precisely. Of course, the preimage of $\overline{Y}$ is closed in $G \times X$, so the preimage of $Y$ being a dense subset means that the closure is the preimage of $\overline{Y}$, as desired. Incidentally, I think it is misleading to think of the preimage of $Y$ by the group action as $G \times Y$ – although it is true when $Y$ is closed under the group action, if it is not, there will be twisting by the group action.
|
2025-03-21T14:48:31.515631
| 2020-07-15T04:28:05 |
365665
|
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"Geoff Robinson",
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"url": "https://mathoverflow.net/questions/365665"
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|
Stack Exchange
|
Literature and history for: lifting matrix units modulo various kinds of ideal
This is not so much a mathematics question as a cross between a "history of mathematics" question and a reference request.
My PhD student has been working on some problems concerning subhomogeneous Banach algebras, and at one point this requires lifting matrix units from $A/J$ to $A$, where $A$ is a Banach algebra and $J$ a closed ideal with certain properties. In our setting, the ideal $J$ is not contained in the Jacobson radical (indeed, $A$ has trivial Jacobson radical), but $J$ does have some special features that allow one to adapt various proofs from the ``$J$ is radical'' setting.
The arguments that my student is adapting come from Rowen's book on Ring Theory, but I wondered where one could find the initial literature about such lifting theorems. The proof in Rowen's book is fairly direct but it has the effect of making the calculations seem "pulled out of a hat", rather than the consequence of some bigger conceptual picture; and we would like to have some idea of the bigger picture, both for general acknowledgement of the literature and for possible inspiration in the Banach-algebraic setting.
Here are some specific questions. My hope is that some members of the MO community who specialise in topics adjacent to noncommutative ring theory will have seen appropriate articles or textbook expositions.
We assume characteristic zero throughout (and all algebras are associative).
Q1. If $A$ is a finite-dimensional algebra and $J$ is its Jacobson radical then the quotient map $q: A\to A/J$ has a right inverse homomorphism $\sigma: A/J \to A$ (Wedderburn decomposition). In particular, matrix units in $A/J$ will lift to matrix units in $A$. Did Wedderburn's original paper make this observation, either as a corollary or as an ingredient of the decomposition theorem? Are there other pre-1950 sources which explicitly link the lifting of matrix units modulo the radical to Wedderburn decomposition?
Q2. If $A$ and $J$ are as in Q1 and one knows in advance that $A/J$ is a sum of matrix algebras, then one can use an iterative argument based on vanishing Hochschild cohomology groups $H^2(A/J, \underline{\quad})$ to construct the splitting map $\sigma$. Has this approach been extended to provide lifting of matrix units in settings where $A$ is infinite-dimensional and $J$ is no longer a nil ideal?
Q3. Following on from Q2, are there some standard references for lifting results in settings where $J$ is not (contained in) the Jacobson radical?
Q4. It seems much easier to find literature on lifting orthogonal idempotents modulo ideals that satisfy various properties. Am I missing some "obvious" reason why the problem of lifting matrix units reduces to this problem? (In Rowen's book, given certain technical conditions on $J$, the first step in lifting matrix units modulo $J$ is to lift the diagonal entries to orthogonal idempotents; is this a matter of technical convenience, or done to follow some deeper principle?)
Have you looked at the work of Malcev?
|
2025-03-21T14:48:31.515872
| 2020-07-15T06:02:02 |
365670
|
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|
Stack Exchange
|
Ratner's orbit closure for a unipotent semigroup
For Ratner's orbit closure theorem, one may refer to the following Wikipedia page.
Let $\{u_t\mathrel: t\in \mathbb{R} \}\subset G$ be a unipotent one-parameter subgroup of a connected Lie group. Let $\Gamma$ be a lattice of $G$. From the theorem, we know that the closure of every full orbit $\{x\cdot u_t\mathrel: t\in \mathbb{R}\}$ is dense in the orbit $xS\subset \Gamma\backslash G$ of a certain subgroup $S$ of $G$.
Do we also have the density of the semigroup orbit $\{x\cdot u_t\mathrel: t\geq 0\}$ in certain orbits of subgroups? Is $\{x\cdot u_t\mathrel: t\geq 0\}$ just dense in $xS$ as well?
I guess, one should look at the proof to see whether it uses negative times.
An easier special case (with the full Borel group rather than just U) is discussed in Section 2.2 of http://perso.ens-lyon.fr/ghys/articles/dynamiqueflots.pdf . In the cocompact case, density is reduced to density of $\Gamma$ in $G/B$, so one may look at $G/B^+$ and wonder whether density of $\Gamma$ still holds.
$\DeclareMathOperator\supp{supp}$The theorem holds for semigroups as well (well, in the finite volume setting! in the infinite volume setting there are subtleties between two-sided and one-sided averages).
Assume $\mu$ is an $S$-invariant and ergodic probability measure, where $S$ is a semigroup inside a one-parameter unipotent subgroup $U$.
We will show that $\supp(\mu)$ contains a full $U=\langle S\rangle$ orbit, that's enough.
Pick some $x\in \supp(\mu)$.
Consider $\overline{S\cdot x}=P$. For a generic point $x$, $P=\supp(\mu)$ by the ergodic theorem.
Notice that $S\cdot P\subset P$, moreover $S^{2}\cdot P=S\cdot P$, so this is an $S$-invariant subset. By ergodicity $\mu(S\cdot P)=1$, but we also have that $\mu(P)=1$ as well, so $P=S\cdot P$ up to a measure zero set, or in other words $S^{-1}\cdot P=P$ as well, hence $P$ is invariant under $S\cup S^{-1}$, now you may apply the orbit closure theorem.
P.S. you might want to consider the paper by Nimish Shah - "Invariant measures and orbit closures on homogeneous spaces for actions of subgroups generated by unipotent elements" (MSN) in Lie groups and ergodic theory, where he discusses the move from Ratner's theorem to discrete subgroup actions!
I did some light proofreading, which I hope is OK. One thing that I didn't change is that you refer to a semigroup $S$ inside a unipotent group $U$, and later say $U = \langle S\rangle$. I think you meant to replace $U$ by the group generated by $S$, but I wasn't sure if this was instead an additional hypothesis.
@LSpice, thanks, I meant $S$ is the positive semigroup inside $U$, which is a one-parameter unipotent group.
@LSpice, he meant the full-positive semigroup. Not a discrete one. Dealing with a discrete subgroup/group situation is a bit more complicated (and needs a suspension construction, as in Shah's article I have mentioned).
I don't think I mentioned discreteness, and I didn't have it in mind, so I'm not sure to what that is responding (although certainly I believe you that there is more complication involved). Your first comment is what I meant.
@LSpice, it is clear that for $\mathbb{R}$, the group generated by the positives (or say any non-discrete semigroup) is the whole group.
I suspected that you might have considered a discrete semigroup like $\mathbb{N}$.
Since we're on the same page, there's no point fussing about it; but the original language just said that $S$ was a semigroup inside a unipotent group $U$. This allows, for example, the possibility of $S$ being the semigroup $\begin{pmatrix} 1 & 0 & \mathbb R_{> 0} \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}$ inside $U = \begin{pmatrix} 1 & * & * \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}$, in which case $S$ does not generate $U$. Your new language clarifies that that's not what you mean, so all is well.
|
2025-03-21T14:48:31.516395
| 2020-07-15T07:28:14 |
365674
|
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"Qiaochu Yuan",
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|
Stack Exchange
|
Smallest family of subsets that generates the discrete topology
If $X$ is a finite set, what is the smallest (in cardinality) family of open subsets $\mathcal U\subseteq 2^X$ such that $\mathcal U$ generates the discrete topology, i.e. if $\mathcal U\subseteq \tau\subseteq 2^X$ and $\tau$ is a topology, then $\tau=2^X$?
Let $\mathcal{U}=\{A_1,\ldots,A_k\}$. Then for any element $x\in X$ there should exist a set $I(x)\subset \{1,\ldots,k\}$ such that $\cap_{i\in I(x)} A_i=\{x\}$. Note that $I(x)$ is not contained in $I(y)$ for $x\ne y$. Therefore $|X|\leqslant \binom{k}{\lfloor k/2\rfloor}$ by Sperner's theorem. On the other hand, if $|X|\leqslant \binom{k}{\lfloor k/2\rfloor}$, we may construct an injection $f$ from $X$ to $\lfloor k/2\rfloor$-subsets of $\{1,\ldots,k\}$ and define $A_i=\{x:i\in f(x)\}$. Then $\cap_{i\in f(x)} A_i=\{x\}$.
So the answer is the minimal $k$ for which $|X|\leqslant \binom{k}{\lfloor k/2\rfloor}$.
Here is a construction that is within a constant factor of optimal.
You can find such an $\cal U$ containing $2 \lceil \log_2 |X|\rceil $ sets: identify $X$ with a subset of $\{0,1\}^{\lceil \log_2 |X|\rceil}$ and take $U_i$ to be the set of all elements whose $i$-th coordinate is $0$, and $V_i$ to be the set of all elements whose $i$-th coordinate is $1$. Then each singleton is an intersection of appropriate sets $U_i$ and $V_i$, so the generated topology includes all singletons and thus is discrete.
On the other hand, we cannot do much better than that: if $\cal U$ contains fewer than $\log_2 |X|$ sets, then there are two points contained in exactly the same sets of $\cal U$ (and clearly these points cannot be distinguished by the resulting topology). To find such a pair, let $U_1,U_2,\dots$ be an enumeration of the elements of $\cal U$. Let $X_1 = X$ and inductively let $X_{i+1}$ be the larger set of $X_i \cap U_i$ and $X_i \setminus U_i$. Note that in every step we keep at least half of the elements, hence the last $X_i$ contains at least two elements.
Connecting this up with Fedor's answer we have ${2n \choose n} \approx \frac{4^n}{\sqrt{\pi n}}$ so the optimal $\mathcal{U}$ Fedor constructs has just a bit more than $\log_2 |X|$ sets, something like $\log_2 |X| + \frac{1}{2} \log_2 \log_2 |X|$ ish?
|
2025-03-21T14:48:31.516563
| 2020-07-15T07:31:56 |
365675
|
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"Florian Lehner",
"Gerhard Paseman",
"M. Winter",
"Nandakumar R",
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|
Stack Exchange
|
On folding a polygonal sheet
Consider a polygonal sheet $P$ of area $A$ with $N$ vertices (it material is not stretchable or tearable). Let $n$ be a positive integer >=2.
Question: Let $P$ lie on a flat plane. We need to fold up $P$ so that it now occupies only an area $A/n$ of the plane. It is also needed that the folding is as uniform as possible - ie the number of layers of the material above any given point should be as close to $n$ as possible. We need an algorithm that does it and an estimate of its complexity.
Example: If $P$ is a rectangle of area $A$ and $n$ is an integer, it is easy to see that we can fold it to an area $A/n$ such that it is exactly $n$ layers thick throughout - the 'creases' could simply be $n-1$ equally spaced parallel lines. It appears that no other shape of $P$ has this property of 'perfectly uniform foldability'. Which is the shape(s) of $P$ that causes greatest variation in the number of layers for a given $n$?
Further possibilities: One can further ask: Minimize the perimeter of the area $A/n$ region that is covered by the folded polygon. Alternatively, We could require $P$ to be folded as uniformly as possible so that it can be packed into a rectangular or square box of some specified dimensions - and area not necessarily equal to $A/n$ with n an integer.
Do you assume that $P$ is convex?
Yes A is the area. Made the necessary edit. Thanks. As for P being convex, not necessary.
For every $k\ge 2$ there is a non-rectangular shape allowing a uniform folding for all $n$ that are multiplies of $k$:
The reason is that you can uniformly fold it into a rectangle with $k$ layers.
If you are looking for convex shapes, then $k=2$ above is convex.
Here is a non-rectangular convex shape admitting a uniform folding with three layers.
If you want to make it rectangular you will get six layers, but then you can proceed in all multiples of 6.
More generally, every regular $n$-gon admits a $2n$-layer folding. And it can be further made into a rectangular $4n$-layer folding (and then every multiple thereof).
The differently colored creases in the pentagon are to be understood as alternatingly folded upwards and downwards.
Or even better, for every $n$ there is a non-rectangular convex shape admitting an $n$-layered folding, or a $2n$-layerd rectangular folding (and then every multiple thereof).
So an interesting question might be whether for every $n$ there is a non-rectangular convex shape admitting a rectangular $n$-layer folding.
The $(k=3)$-shape folds uniformly into a rectangle with $n$ layers for every $n \geq 2$, not just multiples of $3$. First fold the triangles in. If $n$ is even, then fold the right half along $n/2-1$ creases and the left half along $n-1$ creases; if $n$ is odd, then fold the right half along $(n-1)/2-1$ creases, fold the left half "over it" (i.e. fold in the middle) and fold the "remaining part" (i.e. the part that does not cover the folded right half) of the left half $n-1$ times.
Then there is the carpet roll approximation. Convert the sheet into a tube of small diameter and squash it. It won't be perfect, but for large n the thicknesses may not vary much. Gerhard "Exercise: Which Shapes Vary Much?" Paseman, 2020.07.15.
Those are quite nice examples! And the naïve guess would be that the answer to your parting question of whether there is some non-rectangular convex shape that admits a uniform n-layer folding for any n might be "no" even if the final folded object is allowed to be non-rectangular. And somehow all smart foldings seem to involve a rectangle at some stage.
@NandakumarR Do you mind whether I ask this particular instance (the last paragraph of my asnwer) as a separate MO question? You question seems to be also focused on the algorithmic aspect which my asnwer does not touch at all.
@M.Winter, Sure... Indeed, happy if the question acquires dimensions I didn't anticipate!
A very late clarification, Sorry! @M.Winter, Please go ahead if you so wish.
|
2025-03-21T14:48:31.516863
| 2020-07-15T08:24:23 |
365677
|
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|
Stack Exchange
|
Period domain closure and mixed Hodge structures
The moduli space of Hodge structures is the period domain
$$D\ = \ \coprod_{V,\psi} \text{Hom}_{\mathbf{R}\text{ alg.gp.}}(\mathbf{C}^*,G_\mathbf{R})/G_\mathbf{Z}$$
where $G\subseteq \text{GL}(V)$ are the linear maps preserving a given (skew)symmetric form $\psi$ on vector space $V$. There is a foliation on $D$ and a horizontal map
$$X \ \longrightarrow \ D$$
is the same thing as a variation of Hodge structures on $X$.
There are two other similar objects:
The compactification $\overline{D}$ of $D$: when the period domain is a variety, this exists by a theorem of Nagata.
The period domain $D_M$ of mixed Hodge structures.
Question A: what is the relation between 1 and 2? What are they both for the component $\mathbf{H}^g/\text{Sp}(2g,\mathbf{Z})$ of $D$?
My guess is that $\overline{D}\subseteq D_M$ because of limiting mixed Hodge stuctures. If they are not the same, then
Question B: what is $\overline{D}_M$?
The period domain is not a moduli space for Hodge structures, in the sense that there is no universal variation of Hodge structure on $D$. Rather, there is a certain foliation on $D$ and horizontal maps $X\to D$ correspond to VHS on $X$. This is Griffiths transversality.
Also, the period domain is not a variety in general. In situations where Griffiths transversality is automatic (e.g. weight 1 HS), then indeed the period domain is a moduli space for such VHS and is a variety (a Shimura variety).
@PiotrAchinger Thank you, I've edited my question accordingly.
|
2025-03-21T14:48:31.516995
| 2020-07-15T09:29:32 |
365680
|
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"Daniele",
"Fedor Petrov",
"https://mathoverflow.net/users/161126",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365680"
}
|
Stack Exchange
|
Expressing the elementary symmetric polynomials in the $(x_i-x_j)^2$ variables in term of the elementary polynomials in the $x_k$ variables
Let $n>1$ be an integer and let $P$ be a multivariate symmetric polynomial in $n(n-1)/2$ variables.
Let us define the multivariate polynomial $Q_P$ in n variables as:
$Q_P(x_1,...,x_n)=P\left(\{(x_i-x_j)^2\}_{i>j}\right)$
(The order of the variablesing inside $P$ shouldn't matter thanks to its symmetry).
It turns out that $Q_P$ is a symmetric polynomial, and so I should be able to express it in term of the elementary symmetric polynomials in n variables.
In the particular case of $P$ being itself a elementary symmetric polynomial in $n(n-1)/2$ variables, is there an expression for the decomposition of $Q_P$ in term of elementary symmetric polynomials in n variables?
The only vaguely related thing I found is the expression of the subdiscriminants as determinants of symmetric matrices with entries equals to power sums of the $x_i$ variables (and the power sums can be related to the elementary symmetric polynomials) (see Basu: Algorithms in Real Algebraic Geometry)
but subdiscriminants don't seem to form a basis for the space of symmetric polynomials in the $n(n-1)/2$ variables (but maybe they can be completed to form one? Or the determinant technique can be adapted?)
It is not hard to express the power sums via power sums. Then you may use Newton identities to relate power sums and elementary symmetric polynomials.
If I understand correctly the strategy is to consider the case for $P$ power sum in the $(x_i-x_j)^2$ variables and express $Q_P$ as power sums in the $x_k$ variables and then use Newton identities to re express everything in term of elementary symmetric polynomial? While I still haven't found explicitly a formula it looks like (using a CAS and the integer sequence site) that effectively the relationship is simple (and can be expressed through binomial coefficients). Thanks
Yes, the strategy is this. You start with $\sum_{i<j} (x_i-x_j)^{2k}=\frac12\sum_{i,j} (x_i-x_j)^{2k}=\frac12 \sum_{s=0}^{2k}(-1)^s \binom{2k}s (\sum x_i^s)(\sum x_j^{2k-s})$.
|
2025-03-21T14:48:31.517153
| 2020-07-15T09:53:12 |
365681
|
{
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"authors": [
"Igor Khavkine",
"Martin Hairer",
"fsp-b",
"https://mathoverflow.net/users/160714",
"https://mathoverflow.net/users/2622",
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"sort": "votes",
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}
|
Stack Exchange
|
Have you seen this PDE before?
Consider the second-order nonlinear PDE
$$(\partial_x\partial_y\varphi)\cdot\varphi = \partial_x\varphi\,\partial_y\varphi.$$
This PDE is solved by all ('separable') functions $\varphi\in C^2(\Omega)$ of the form $\varphi(x,y) = \ell(x)\cdot r(y)$ (any $\Omega\subseteq\mathbb{R}^2$ open).
I'd like to know whether each (classical) solution of the PDE is of this form.
Are you aware of any 'uniqueness' results for the above PDE which might (dis)prove this?
This equation is equivalent to $\partial_x \partial_y \log(\varphi) = 0$. This is the 2-dimensional wave equation for $\log(\varphi)$ in "null" coordinates.
That settles it, thanks. @IgorKhavkine
@fsp-b Not completely: it shows that it is true for positive solutions, but since the equation has compactly supported solutions, you can combine them to form solutions that aren't (globally) separable.
Many thanks for your comment, @MartinHairer I in fact happen to be interested in positive solutions only, so for me, this in fact gives what I needed.
|
2025-03-21T14:48:31.517263
| 2020-07-15T10:02:20 |
365683
|
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|
Stack Exchange
|
On the speed of divergence of the converse of the Strong law of large numbers
By the converse of the strong law of large numbers, we know that, given a sequence of i.i.d random variables $X_1,X_2,\dots$ such that $\mathbb{P}(X_1 \ge 0)=1$ and $\mathbb{E}X_1= \infty$,
then I have
$$
S_N:=\frac{1}{N}\sum_{i=1}^N X_i \longrightarrow \infty \quad \mathbb{P}\textit{-a.s}.
$$
I suppose that, just as in the case of the strong law of large numbers (without further assumptions on the moments of such random variables), we don't have a a-priori bound for the speed of divergence. My question is: given the law $\mathbb{P}_X$ of $X_1$, is there a well defined deterministic and diverging sequence $a_N=a_N(\mathbb{P}_X)$ and two positive but finite random variables $c$ and $C$ such that
$$
\mathbb{P}\Big(\limsup_{N\to \infty} [S_N \ge C\cdot a_N] \Big)=0
$$
and
$$
\mathbb{P}\Big(\limsup_{N\to \infty} [S_N \le c\cdot a_N] \Big)=0.
$$
Is it possible to get bounds for $a_N$ in terms of the law $\mathbb{P}_X$?
Sorry, just fixed the n/N situation. Also, it was supposed to be limsup, not liminf. But I am referring to the limsup of events, putting in words I want to prove that $[S_N > C\cdot a_N]$ only happen a finite (but random) ammount of times almost surely.
Such a sequence $a_n$ does not exist even for a well studied example like returns to the origin of simple random walk in one dimension. If $X_i$ denotes the number of steps from the $i-1$ time the walk returned to the origin to the $i$'th time, then $X_i$ are i.i.d. and their sum $S_n$ is the number of steps until the $n$'th return time to the origin. In [1] $S_n$ is denoted by $\rho_n$ and we will follow this. Taking $\epsilon=1$ in Theorem 11.6 on page 119 in [1] we find that
$$\rho_n>n^2 \log (n)$$ infinitely often almost surely, yet
$$\rho_n<\frac{n^2}{\log \log n}$$
infinitely often almost surely. More precise information is in Theorem 11.5 there. To exclude "wild" sequences $a_n$, just separate cases, using Theorem 9.11 in [1] that yields for all $x \in (0,\infty)$,
$$\lim_{k \to \infty} P(\rho_k <k^2 x) = f(x) , $$
where $f(x) \in (0,1)$ is explicitly given.
Case 1: If $a_n \le n^2$ infinitely often, then for any constant $C \in (0,\infty)$ we will have $P(\limsup [\rho_n>Ca_n]) \ge 1-f(C) $ by Fatou's lemma so
$P(\limsup [\rho_n>Ca_n]) =1 $ by the Hewitt-Savage zero-one law.
Case 2: If $a_n \le n^2$ finitely often, then a similar argument shows that $P(\limsup [\rho_n \le c a_n]) =1 $ for all $c \in (0,\infty)$.
The situation is more extreme for return times of two dimensional simple random walk, for which see Theorem 20.5 page 218 in [1].
[1] P. Revesz, Random Walk in Random and Non-Random Environments, World Scientific
Publ., Second edition (2005). https://www.worldscientific.com/worldscibooks/10.1142/5847
|
2025-03-21T14:48:31.517464
| 2020-07-15T10:24:55 |
365684
|
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"authors": [
"Andre of Astora",
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|
Stack Exchange
|
Condition such that the fibres of a polynomial map $p :\mathbb{C}^n\rightarrow \mathbb{C}^n$ are finite
I was told that if $A$ is the subring of $\mathbb{C}[x_1,\ldots, x_n]$ generated by the polynomials $p_1(x_1,\ldots, x_n),\ldots, p_1(x_1,\ldots, x_n)$, then the preimage $p^{-1}(c)$ via the map $p = (p_1,\ldots, p_n):\mathbb{C}^n\rightarrow \mathbb{C}^n$ is finite for all $c\in \mathbb{C}^n$ if the ring $\mathbb{C}[x_1,\ldots, x_n]$ is integral and flat over the subring $A$. Does anyone know where I can find a reference (preferably a textbook) that has this or an equivalent statement?
Is this not essentially Noether normalization?
$\def\CC{\mathbb{C}}$User "anon" points out to me that this is Proposition 8.28 in Milne's notes; see also Example 8.36 for a quasi-finite map $\CC^2 \to \CC^2$ which is not finite. The rest of my answer is probably not as useful now that there is a good reference, but I'll leave it.
Here is the right statement:
Theorem Let $X$ and $Y$ be affine varieties over an algebraically closed field $k$, with corresponding coordinate rings $A$ and $B$. Let $\pi : Y \to X$ be a map and let $\pi^{\ast} : A \to B$ be the corresponding map of rings. If $B$ is a finitely generated as an $A$-module, then $\pi^{-1}(x)$ is finite for all $x \in X$.
In your case, $X = Y = \CC^n$ and $A=B = \CC[x_1, \ldots, x_n]$. You phrase your hypothesis as $B$ is integral over $\pi^{\ast} A$ but, since $B$ is a finitely generated $\CC$-algebra, that is the same as asking that $B$ is finitely generated as an $A$-module. Also, the adjective "flat" isn't needed, and actually follows from finiteness in your case by the Miracle Flatness Theorem.
Proof Let $x \in X$ and let $\mathfrak{m}_x$ be the corresponding maximal ideal of $A$. Then $\pi^{-1}(x)$ corresponds to the radical ideal $\sqrt{B \pi(\mathfrak{m}_x)}$ so we want to show that $B/\sqrt{B \pi(\mathfrak{m}_x)}$ is a finite dimensional $k$-algebra. It is enough to show that $\dim_k B/B\pi(\mathfrak{m}_x)$ is finite, since $B/\sqrt{B \pi(\mathfrak{m}_x)}$ is a quotient of $B/B\pi(\mathfrak{m}_x)$. But $B/B\pi(\mathfrak{m}_x) \cong B \otimes_A A/\mathfrak{m}_x$. Since $B$ is a finite $A$-module, $B \otimes_A A/\mathfrak{m}_x$ must be a finite $A/\mathfrak{m}_x$ module, and $A/\mathfrak{m}_x$ is just $k$. $\square$.
Here is where this can be found in some other books: Shaverevich introduces finite maps in Section I.5.3, but doesn't show that they have finite fibers until Section II.6.3 (Theorem 3) and then only under the hypothesis that $X$ is normal. Milne introduces the words "finite" (meaning $B$ is finitely generated as an $A$-module) and "quasi-finite" (meaning the fibers of $\pi$ are finite) in Definition 2.39, and proves the result we want as Proposition 8.28; as noted above. In Hartshorne, this is Exercise 3.5.(a) in Section II.3. Vakil makes this Important Exercise 7.3.K. When I taught Algebraic Geometry, I got to this one month in, see the notes for October 8.
Well, there is EGA II.6.1.7, but this is probably not at the level you want to stay...
Milne's notes 8.28?
@anon Thanks! That's probably the best choice. I should have known to look harder when it appeared that Milne omitted a basic fact. Also nice to note that Milne's Example 8.36 gives a quasi-finite map $\mathbb{C}^2 \to \mathbb{C}^2$ which is not finite.
Thank you for the references. Your notes give me the exact statement I was looking for.
|
2025-03-21T14:48:31.517858
| 2020-07-15T10:39:27 |
365685
|
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|
Stack Exchange
|
Name for a class of almost symplectic manifolds
A $2n$-dimensional manifold $M$ is said to be almost symplectic if it possesses a non-degenerate two-form $\omega \in \Omega^2(M)$. Equivalently, an almost symplectic structure is a $G$-subbundle $P \subset F(M)$ of the frame bundle where $G < GL(2n,\mathbb{R})$ is isomorphic to the symplectic group $Sp(2n,\mathbb{R})$.
The intrinsic torsion of such a $G$-structure is captured by the three-form $d\omega \in \Omega^3(M)$. The bundle $\wedge^3 T^*M$ breaks up into the Whitney sum of two $G$-stable sub-bundles corresponding to the $\omega$-traceless 3-forms and their $\omega$-perpendicular complement. This therefore gives rise to four types of almost symplectic manifolds:
symplectic, where $d\omega = 0$
locally conformal symplectic, where $d\omega = \omega \wedge \varphi$ for some one-form $\varphi$ which is closed and hence $\varphi = df$ locally, allowing us to construct a local symplectic form $e^{-f}\omega$.
name?, where the volume form $\omega^n$ is left invariant by the hamiltonian vector fields $X_f = \omega^\sharp(df)$
generic, where $d\omega$ is none of the above.
My question is whether there is an accepted name for the third type. I would also appreciate a link to where this classification was made explicit for the first time.
Thank you.
Edit As Robert Bryant pointed out below, the condition name? is actually $d\omega^{n-1} = 0$. I will leave the question unmodified, except for this.
Hi Jose! I thought this classification appeared for the first time in Gray and Hervella's paper on almost-Hermitian manifolds, section 8. Do you believe there is an earlier source?
Hi Paul! That's embarrassing... I studied that paper a long time ago, but all I remembered from it were the almost hermitian classes. I totally forgot that they also discussed almost symplectic. Thanks for reminding me! I'd still like to know if there is an accepted name for the third class.
I'm a bit confused by your question, because I believe that, if one defines an $\omega$-Hamiltonian vector field to be a vector field of the form $X_f = \omega^\#(\mathrm{d}f)$ where $f$ is a (smooth) function on $M$, then $\omega^{n}$ is always invariant under the flow of $X_f$.
To see this, recall that, when $n>1$, if $\omega$ is a non-degenerate $2$-form on $M^{2n}$, its exterior derivative can be written uniquely in the form
$$
\mathrm{d}\omega = \phi\wedge\omega + \psi
$$
where $\phi$ is a $1$-form and $\psi\in\Omega^3(M)$ is $\omega$-primitive, i.e., $\omega^{n-2}\wedge\psi = 0$.
Meanwhile, by Cartan's formula for the Lie derivative with respect to $X_f$, we have, since $\iota(X_f)\omega = -\mathrm{d}f $ (where $\iota(X)$ denotes interior product with $X$),
$$
\begin{align}
\mathcal{L}_{X_f}\omega^n &= n\,\omega^{n-1}\wedge \mathcal{L}_{X_f}\omega
= n\,\omega^{n-1}\wedge\bigl(\iota(X_f)(\mathrm{d}\omega) + \mathrm{d}(\iota(X_f)\omega)\bigr)\\
&=n\,\omega^{n-1}\wedge\bigl(\iota(X_f)(\phi\wedge\omega + \psi) + \mathrm{d}(-\mathrm{d}f))\bigr)\\
&=n\,\omega^{n-1}\wedge\bigl(\phi(X_f)\wedge\omega + \phi\wedge\mathrm{d}f+ \iota(X_f)\,\psi)\bigr)\\
&= n\,\omega^{n-1}\wedge\bigl(\iota(X_f)\,\psi\bigr) =0
\end{align}
$$
since $\omega^{n-2}\wedge\psi=0$ implies
$$
0 = \iota(X_f)(\omega^{n-1}\wedge\psi) = (n{-}1)\,\omega^{n-2}\wedge(-\mathrm{d}f)\wedge\psi + \omega^{n-1}\wedge\bigl(\iota(X_f)\,\psi\bigr).
$$
Instead of the 'Hamiltonian flow invariance' criterion you propose for name? (by which, I think, you are trying to capture the condition $\phi=0$), you should instead just require that $\omega^{n-1}$ be closed. (This condition is sometimes known as 'balanced' in the literature.)
By the way, your second type "locally conformally symplectic" is only apt if $n>2$. When $n=2$, you do not automatically get that $\phi$ is closed from the condition $\mathrm{d}\omega = \phi\wedge\omega$. (In fact, it is generically not true, even though $\psi$ does vanish identically when $n=2$.)
Thanks, Robert. Indeed, my calculation had shown that the condition of vanishing $\psi$ was $d\omega \wedge \omega^{n-2} = 0$, which is closure of $\omega^{n-1}$. For some reason I wedged with $df$ and then interpreted it as $X_f$ preserving the volume form, but did not realise that this was always true. Thanks a lot! Also for the name.
@JoséFigueroa-O'Farrill: I think you meant 'condition of vanishing $\phi$' (not $\psi$) in your comment.
Ah, yes, sorry. I meant to write $\varphi$ as in my calculation.
|
2025-03-21T14:48:31.518164
| 2020-07-15T12:59:02 |
365692
|
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"authors": [
"Gerald Edgar",
"Loïc Teyssier",
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|
Stack Exchange
|
Is it possible to express $\int_{0+\epsilon}^{1-\epsilon}\left(\sqrt{1-x^2}^{\sqrt{1-x^2}^{\cdots}}\right) dx$ in elementary functions?
let $\epsilon >0$, I tried to evaluate $\int_{0+\epsilon}^{1-\epsilon}\left(\sqrt{1-x^2}^{\sqrt{1-x^2}^{\cdots}}\right) dx$ , using the fact $x= \cos t$ yield to have integrand using $\sin $ function seems is not easy to get such closed form by this variable change , For one iteration by means $\int_{0}^{1}\left(\sqrt{1-x^2}^{\sqrt{1-x^2}}\right) dx$ we have the integrand converge approximately to $\frac{\sqrt{3}}{{2}}$, For some odd iterations we have $l=0.89..$ and for even iterations we have $l=0.9..$ , Now if we fixe $\epsilon$ at at some small value for example $\epsilon=0.05$ such that $x$ lie at a least between $(0+\epsilon,0.99782-\epsilon)$ to get convergence , My question here is : Is it possible to express the titled integral in elementary functions ?
Note:The Copy of this question is posted yesterday here in MSE, And I bielive that integrand has a closed form because the integrand is of the trigonometric form
I guess $l$ is the first integral with an infinite sequence of powers.
Note: the continued exponential $u^{u^{\dots}}$ does not converge for $u$ near $0$. In your case, the integrand does not converge for $x$ near $1$.
@GeraldEdgar , I have edited it to assure convergence look it and thanks
I think this is
$$
-\int_{0+\epsilon}^{1-\epsilon}
{\frac {2\;{\rm W} \left(-\frac12\,\ln \left( 1-{x
}^{2} \right) \right)}{\ln \left( 1-{x}^{2} \right) }}\,{\rm d}x
$$
not elementary.
Could u explain a little bit how you got this form ?
|
2025-03-21T14:48:31.518311
| 2020-07-15T13:47:09 |
365696
|
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|
Stack Exchange
|
Riesz-Thorin interpolation theorem
Does the Riesz-Thorin interpolation theorem hold for real-linear operators?
The standard proof uses complex linearity, but maybe another proof avoids the assumption of complex-linearity?
What is the problem. You can complexify a real-linear operator. Perhaps the non-obvious (though true) fact is that the complexification does not change the $L^p$-norm.
The norm of the complexification can change if the operator maps $L^p$ to $L^q$ with $q<p$. When such a situation occurs one needs a constant (2 suffices) in the real Riesz-Thorin, I guess.
@Denis Serre. Assume that $Tf(z)= \int_U (A(z,w) f(z)+B(z,w) \overline{f(z)})dudv, w=u+iv, z=x+i y$, $U$ is the unit disk. This operator is real-linear, but the function $f:U\to U$ need not be real. So what is the complexification of this operator. $A$ and $B$ are certain kernels?
@Lira. You should not focus on the fact that $f$ is a complex-valued function. Your operator is real-linear, so just write it in terms of a pair of unctions $(g,h)$ (real and imaginary parts). Then you may complexify :).
|
2025-03-21T14:48:31.518423
| 2020-07-15T14:02:46 |
365697
|
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"Libli",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365697"
}
|
Stack Exchange
|
Spaces intersecting a plane non-trivially in $G(3,6)$
I want to understand the Schubert variety $\Sigma\subseteq G(3,6)$ representing 3-dim subspaces intersecting a given 2-dim subspace non-trivially. Is it smooth? How to describe $det(T_{\Sigma})$?
if you denote by $A$ the given two dimensional subspace, then the locus you are looking for is the degenracy locus of $\mathcal{R} \longrightarrow \mathbb{C}^6/A \otimes \mathcal{O}_{G(3,6)}$, where $\mathcal{R}$ is the tautological bundle on $G(3,6)$. It should be of codimension 2 in $G(3,6)$. is not smooth and its singular locus is the strata ${L \in G(3,6), \ L \supset A }$. There is a huge litterature on degeneracy loci
|
2025-03-21T14:48:31.518507
| 2020-07-15T14:14:50 |
365698
|
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"Jamie Gabe",
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"url": "https://mathoverflow.net/questions/365698"
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|
Stack Exchange
|
Ideals of maximal tensor product of $C^{\ast}$-algebras
Let $A$ and $B$ be $C^{\ast}$-algebras. It is well known that maximal tensor product of simple $C^{\ast}$-algebras need not be simple. So basically the ideal structure of $A\otimes_{max}B$ does not really depends on ideal structure of $A$ and of $B$.
What is known about ideals of $A\otimes_{max}B$ .
Any references?
You've asked a very similar question before. The question is so broad that it is very hard to think of a meaningful answer. If there is something specific you want to know, ask that.
As it stands it sounds a bit like "somebody prove some things for me, I don't know what, you'll have to figure that out"...
@NikWeaver: I apologise for the broad question. I am mainly interested in classification of (closed) ideals of max tensor product(if there's any). Also, I'm interested in questions like whether every ideal has elementary tensor or not? As, I have not read anything about ideals of max tensor product so I'm mainly looking for a paper or book which deals with the same topic!
I don't know the answers, but I'd be very surprised if there is any meaningful classification of ideals.
@MathLover What is the reason you want to know about ideals rather than representations?
Clearly the kernel $\ker (A \otimes_{\max{}} B \to A\otimes_{\min{}} B)$ contains no non-zero elementary tensors. More generally, a two-sided closed ideal in $A\otimes_{\max{}} B$ contains an elementary tensor if and only if it has non-zero image in $A\otimes_{\min{}} B$. This can be shown by tweaking the proof of Kirchberg's Slice Lemma (Lemma 2.15 in Blanchard, Kirchberg "Non-simple purely infinite $C^\ast$-algebras: the Hausdorff case").
@RobertFurber: Are you saying that every ideal would be kernel of some irreducible representations? I am not sure about it. I would love to know if there is anything related to this. Would love to know references.
@JamieGabe: I'll have a look at that paper.
@MathLover I did not intend such a statement. Rather, the kernels of irreducible representations are known as primitive ideals. If a C$^\ast$-algebra is type I, meaning that in every representation the bicommutant of its image is a type I von Neumann algebra, then the map from irreducible representations to primitive ideals is a bijection (because it's an injection). For type I C$^*$-algebras the maximal and minimal tensors coincide, and you may get a good answer to your question, if restricted to primitive ideals. Some information on this is in Dixmier's book on C$^\ast$-algebras.
@MathLover Of course, I meant "isomorphism classes of irreducible representations" in my previous comment. The statement I was referring to is the theorem at the start of Chapter 9 of Dixmier's C$^\ast$-algebras.
|
2025-03-21T14:48:31.518702
| 2020-07-15T14:44:34 |
365701
|
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|
Stack Exchange
|
(Big) Category O for rational Cherednik algebras
Let $H_{c}$, for simplicity, be the rational Cherednik algebra with parameter $t=1$, with triangular strucuture $\mathbb{C}[h] \otimes \mathbb{C} W \otimes \mathbb{C}[h^*]$, and $(W,h)$ the defining complex representation.
Consider the full subcategory category $O_c$ of finitely generated $H_c$-modules $M$ such that $h \subset \mathbb{C}[h^*]$ acts locally finitely on $M$. (cf. Gordon, section 5.3 in https://arxiv.org/pdf/0712.1568.pdf)
We have $O_c$=$\bigoplus_{\lambda \in h^*/W} \mathcal{O}_c(\lambda)$, where $\mathcal{O}_c(\lambda)$ is the full subcategory of $O_c$, consisting of modules $M$ where each $P \in \mathbb{C}[h^*]^W$ is such that $P-P(\lambda)$ acts locally nilpotently. $\mathcal{O}_c(0)$ is the usual category $\mathcal{O}_c$ considered by a number of people.
$\mathcal{O}_c$ is quite well understood: it is a highest weight category in the sense of Cline-Parshall-Scott; we have analogues of Verma modules of Lie theory, BGG reciprocity, KZ-functor, etc (cf. Guay Projective modules in the category O for the Cherednik algebra; Ginzburg et al. On the category O for rational Cherednik algebras).
Question What is the current understanding of the bigger category $O_c$?
Works that discuss this bigger category that I know of are the aforementioned work of Ginzburg et al., and also to a certain extent Ginzburg, On primitive ideals, and Bezrukavnikov and Etingof, Parabolic induction and restriction functors for rational Cherednik algebras.
References or any discussion you might have on this topic would be greatly appreciated.
I finish with:
Small question If $\mathcal{O}_c(0)$ is a semisimple abelian category, is also $\mathcal{O}_c(\lambda)$ a semisimple category for every $\lambda$ as well?
|
2025-03-21T14:48:31.518844
| 2020-07-15T14:49:11 |
365702
|
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|
Stack Exchange
|
Involutions on Hopf algebra crossed products
I am interested in a (cocycle) crossed product of a Hopf algebra $H$ with a (twisted) $H$-module algebra $A$, often denoted $H\#_\sigma A$, where $\sigma$ is the associated cocycle. This construction was introduced by, I think, Doi and Takeuchi and Blattner, Cohen and Montgomery; and also studied by Majid in the context when $A$ is also Hopf, and $H\#_\sigma A$ is a bialgebra (so a bicrossed product).
I am interested in the case when $H$ is a Hopf $*$-algebra and $A$ is a $*$-algebra, and what conditions we might place on $\sigma$ to obtain a "natural" $*$-algebra structure on $H\#_\sigma A$. This problem is briefly considered by Andruskiewitsch; see Section 3.2.
Are there other sources which consider $H\#_\sigma A$ with a $*$-structure; what were the first papers to do this?
When considering $*$-structures, it is of course natural to start to look at $C^*$-algebras and von Neumann algebras, and analogues of cocycle crossed products have been considered for e.g. Kac algebras and locally compact quantum groups. I am here interested only in the purely algebraic framework.
|
2025-03-21T14:48:31.518942
| 2020-07-15T14:59:46 |
365703
|
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|
Stack Exchange
|
Secondary operations in $H^*P$
I would be very grateful for any answer to this or pointing at a reference.
Let $P$ be the infinite dimensional real projective space and $\Phi_{j,j}$ be the Adams operation based on the Adem relation for $Sq^{2^j}Sq^{2^j}$. I wonder what are the known examples of computing the action of these operations on $H^*P$. More specifically, I would like to know the action of this operation on $H^{2^{j+1}}P$. From the Hopf invariant one result, I can read off that the action of $\Phi_{j,j}$ for small values of $j$ is trivial. But, how can I understand the other ones?
Maybe try John Harper's book Secondary Cohomology Operations pg. 147 onwards. Especially Pr. 5.3.5.
@Tyrone Thank you for reminding of the book. I will take a look!
@Tyrone I like to add that the aforementioned Proposition in your comment is about evaluation of $\phi_{0,j}$. But, thanks for reminding of this and of the book.
|
2025-03-21T14:48:31.519041
| 2020-07-15T15:58:17 |
365706
|
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"Erel Segal-Halevi",
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|
Stack Exchange
|
Perfect matching in hypergraphs: tripartite, regular and unbalanced
In a balanced bipartite graph - where both sides have the same size - a sufficient condition for the existence of a perfect matching is that the graph is regular - all vertices have the same degree.
Moreover, if the graph is unbalanced - one side is shorter than the other - it is still sufficient that the graph is 'regular' in the sense that, in each side, all vertices have the same degree. In this case, there exists a matching that is 'perfect' in that it covers all vertices in the shorter side. For example, if $G = (X\cup Y,E)$ and $|X|<|Y|$ and the degree of each vertex in $X$ is $|E|/|X|$ and the degree of each vertex in $Y$is $|E|/|Y|$, then the graph has a matching of size $|X|$.
For balanced tripartite hypergraphs, the above theorem does not hold. Zur Luria shows an example with 3 vertices in each side. Here is an even smaller example, with 2 vertices in each side. The sides are {1,2}, {a,b}, {A,B} and the edges {1,a,A}, {1,b,B}, {2,b,A}, {2,a,B}; this graph is 2-regular but does not have a perfect matching.
However, a perfect matching does exist if we make the hypergraph slightly unbalanced: in particular, if $H = (X\cup Y\cup Z,E)$ and $|X|=|Y|=2$ and $|Z|=3$, the degree of each vertex in $X \cup Y$ is $|E|/2$ and the degree of each vertex in $Z$ is $|E|/3$, then the hypergraph has a matching of size 2. I proved this by checking all cases (there are not many of them).
Does this idea work for larger hypergraphs? Is there a function $f(n)$ such that, if $|X|=|Y|=n$ and $|Z| = f(n)$, and all vertices in each side have the same degree, then the hypergraph has a matching of size $n$?
MY RESULTS SO FAR:
(*) There are many results regarding Perfect matching in high-degree hypergraphs. Some of these results are for $r$-partite hypergraphs, but only for balanced ones - they assume that all $r$ parts have the same size.
(*) There are many results regarding rainbow matchings. A rainbow matching in a bipartite graph is equivalent to a matching in a tripartite hypergraph. A theorem of Aharoni and Berger (2009) implies that, if $|Z|=2n-1$ and the set of neighbors of each $z\in Z$ is a matching of size $n$, then the hypergraph has a perfect matching. In our case the sets of neighbors need not be matchings - the only constraint is on the degree of each vertex.
(*) If such an $f$ exists, then $f(n)\geq 2n-1$. Consider a tripartite graph with $X = \{x_1,\ldots,x_n\}$ and $Y = \{y_1,\ldots,y_n\}$ and $Z = \{z_1,\ldots,z_{2n-2}\}$ and the following edges:
$(x_1, y_1, z_i), (x_2, y_2, z_i),\ldots, (x_n, y_n, z_i)$ for $1 \leq i \leq n-1$;
$(x_1, y_2, z_i), (x_2, y_3, z_i),\ldots, (x_n, y_1, z_i)$ for $n\leq i\leq 2n-2$.
All in all, there are $n(2n-2)$ edges, the degree of each vertex in $X\cup Y$ is $2n-2$ and the degree of each vertex in $Z$ is $n$, so the hypergraph is regular. But the largest matching it contains is of size $n-1$. For example, the matching $(x_i,y_i,z_i)$ for $1 \leq i \leq n-1$ is a matching, but a larger matching does not exist.
Is $f(n) = 2n-1$ sufficient?
When you write if ... all vertices in each side have the same degree, then the graph has a matching, do you really mean graph or, maybe, hypergraph?
@Seva yes, I meant hypergraph. fixed
|
2025-03-21T14:48:31.519270
| 2020-07-15T16:22:55 |
365709
|
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|
Stack Exchange
|
Solution for $AX+XA^T+XBX=C $ where $X$, $B$ and $C$ are symmetric
Is there a solution for $AX+XA^T+XBX=C$ where $X$, $B$ and $C$ are symmetric?
That equation is called a (continuous-time) algebraic Riccati equation, and there is ample literature on when they are solvable; just look for this search term. For instance, the book Algebraic Riccati equations by Lancaster and Rodman, or Numerical solution of AREs by Bini, Iannazzo, Meini.
In the generic case there is a finite number of solutions that grows exponentially with the dimension $n$: think about the scalar case, for instance, which already has 2 (possibly complex) solutions if $b \neq 0$. So usually the first point is understanding which solution you need. Typically one seeks the so-called stabilizing solution (or the antistabilizing one), i.e., the one such that $A^T+BX$ has eigenvalue in the left (resp. right) half-plane.
|
2025-03-21T14:48:31.519355
| 2020-07-15T16:29:13 |
365710
|
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|
Stack Exchange
|
On the Gelfand-Kirillov Conjecture
The base field $k$ is of zero characteristic.
Notation: $A_{n,s}(k):= A_n(k(x_1,\ldots,x_s))$, the Weyl agebra over a purely transcedental extension of the base field; $F_{n,s}(k)$, the Weyl field, is the skew field of fractions of $A_{n,s}(k)$.
We have the following classical problem which was formulated by I. Gelfand and A. Kirillov in 1966 (in the same wounderful paper where the beloved GK-dimension was introduced):
Let $\mathfrak{g}$ be a finite dimensional algebraic Lie algebra over an algebraically closed field of zero characteristc. When the skew field of fractions of $U(\mathfrak{g})$ is isomorphic to a Weyl field $F_{n,s}(k)$?
Widely known as the Gelfand-Kirillov Conjecture, it was eventually shown to be false in general, in [ Alev, Jacques; Ooms, Alfons; Van den Bergh, Michel A class of counterexamples to the Gelʹfand-Kirillov conjecture, 1996 ]
An interesting contribution was made by Premet in 2010 in [ Modular Lie algebras and the Gelfand-Kirillov conjecture ]. For $\mathfrak{g}$ a simple Lie algebra, it was known that Gelfand-Kirillov problem had a positive solution for type $A$ algebras since the seminal work of 1966, but the for other simple Lie algebras nothing was known until Premet showed in the above that Gelfand-Kirillov Problem is false for simple Lie algebras of type $B,D,F,E$.
The remaining simple Lie algebras are quite elusive. The last work I know of discussing them is [ Anderson, Dave; Florence, Mathieu; Reichstein, Zinovy The Lie algebra of type G2 is rational over its quotient by the adjoint action, 2013 ].
A couple of years ago, I've met Alev personally, and according to him, nothing is known about the skew field of fractions of $U(\mathfrak{g})$, when $\mathfrak{g}$ is of type $C,G$.
Has the understanding of these skew field of fractions been clarified by some recent work?
In a certain sense, Gelfand-Kirillov Conjecture fails by a small margin. Namely, in their second seminal work of 1969 [ The structure of the Lie field connected with a split semisimple Lie algebra ] the authors have introduced the following Conjecture
Let $D(\mathfrak{g})$ be the skew field of fractions of $U(\mathfrak{g})$, $\mathfrak{g}$ an algebraic Lie algebra as above. Let $Z$ be its center. Then there exists a finite field extension of $K$ of $Z$, purely transcedental over $k$, such that the skew field of fractions of $D(\mathfrak{g}) \otimes_Z K$ is a Weyl field $F_{n,s}(k)$.
In their work of 1969, it was shown that this Modified Gelfand-Kirillov Conjecture is true for all semisimple Lie $\mathfrak{g}$. In [ Alev et. al. op. cit ] the authors say they are quite confident that this modified Conjecture is true.
I've not known on any advance on it in the last decades. In the same conversation with Alev a couple of years ago, he pointed that there is a connection between Gelfand and Kirillov 1969 approach to this Conjecture and his work with F. Dumas on the so called Noncommutative Noether's Problem ( [ Opérateurs différentiels invariants et problème de Noether, 2006 ]), but I was unable to figure out precisely the connection.
Anyway,
Is there any recent progress on the modified Gelfand-Kirillov Conjecture?
|
2025-03-21T14:48:31.519677
| 2020-07-15T16:37:52 |
365712
|
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|
Stack Exchange
|
When $\Sigma^{\infty}Y^{\wedge}_p\simeq (\Sigma^{\infty} Y)^{\wedge}_p$?
When studying the stable homotopy of $BG^{\wedge}_p$, with $G$ a finite group, authors know that this abuse of notation is not dangerous because $\Sigma^{\infty}BG^{\wedge}_p$ and $(\Sigma^{\infty}BG)^{\wedge}_p$ are stable homotopy equivalent. However, that abuse is also common when $G$ is a compact Lie group, I do not know any reference that justifies this abuse unlike the finite group case (in this case $\Sigma^{\infty}BG^{\wedge}_p$ and $(\Sigma^{\infty}BG)^{\wedge}_p$ are not always stable homotopy equivalent).
What is more, some result such as the following from a Chun-Nip Lee's article:
Let $Y_1, Y_2$ be connected CW complexes such that both $H^{\ast}(Y_1, \mathbb{F}_p )$
and $H^{\ast}(Y_2, \mathbb{F}_p )$ are finitely generated as graded rings. Then the image of the ring
homomorphism
$\{(Y_1)^{\wedge}_p,(Y_2)^{\wedge}_p\}\rightarrow\mathrm{Hom}_{\mathbb{F}_p} (\widetilde{H}^{\ast}(Y_1, \mathbb{F}_p ), \widetilde{H}^{\ast}(Y_1, \mathbb{F}_p ))$
is finite.
What did the author mean by $(Y_i)^{\wedge}_p$?. In general, is there a criterion to know whether $\Sigma^{\infty}Y^{\wedge}_p\simeq (\Sigma^{\infty} Y)^{\wedge}_p$ or not?
See https://arxiv.org/abs/1712.07633
@skd Thanks, that article proves that $\Sigma^{\infty} (BS^{1})^{\wedge}_p$ is not $p$-complete, what made me think that $BG^{\wedge p}$ in the stable homotopy context means $(\Sigma^{\infty} BG)^{\wedge p}$, although it seems counterintuitive to me.
@VictorTC Unfortunately those kinds of notations are wildly inconsistent from paper to paper (especially when you go to somewhat old ones). There's not much to do but be extra careful.
When in doubt interpret $\Sigma^\infty(X_p)$ as $(\Sigma^\infty X)_p$. Or, rather, interpret it as $(\Sigma^\infty X_p)_p$, which is equivalent to $(\Sigma^\infty X)_p$. So this is a functor that you can apply to $p$-complete spaces. (How generally are they equivalent? If you're using $\mathbb Z/p$-localization, it's easy to see that they're always equivalent. If you're using $p$-completion, I'm not sure if it's always equivalent, but usually you're using it because in that case it's equivalent to $\mathbb Z/p$-localization.)
|
2025-03-21T14:48:31.519833
| 2020-07-15T16:56:02 |
365715
|
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"url": "https://mathoverflow.net/questions/365715"
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|
Stack Exchange
|
The contraction principle in quasi metric spaces
I am researching contractive mappings and I need the article of I. A. Bakhtin "The contraction principle in quasi metric spaces"(1989) or at least part where explanation is given for introduction of quasi metric spaces (in contemporary literature they are called b-metric spaces) as it is a pioneering work in the field but impossible to find.
Is there anyone who has read this paper or knows the answer to the question?
Send me your address email .
Im interests research in b-metric space
|
2025-03-21T14:48:31.519908
| 2020-07-15T16:57:14 |
365716
|
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|
Stack Exchange
|
How to solve this conditional recurrence relation?(two variable and conditions)
I am trying to solve the following recurrence relation
$4 \leq n \;\; \; \; \; \;$ and $\; \; \; \; 2\leq i \leq \lfloor{\frac{n}{2}}\rfloor$
$F(2i,n)=$
$\begin{cases}
\frac{1}{2(2i)-5}F(2i-2,2i-1),& \text{if } n=2i\text{, } i\geq3\\
\frac{n}{2n-5}F(4,n-1)=\frac{n!*3}{4!(2n-5)!!},& \text{if } i=2 \text{, } n\geq5\\
\frac{2i+n-4}{2n-5}F(2i,n-1)+\frac{n-2i+1}{2n-5}F(2i-2,n-1),& \text{otherwise}
\end{cases}$
$F(4,4)=1$
I don't know how to solve a conditional recurrence relation, and I didn't find anything useful about it. any sugestion would be really helpful.
What I did up to now, as the first two conditions and also f(4,4) are initial and special case of the third condition, I only tried to solve the third one without any initial condition, I used generating function. But now I don’t know how to consider the other conditions and even whether the idea of solving the only third one alone was a good idea or not.
What does it mean to solve it? A closed-form formula seems a priori unlikely, unless you have some reason to think it exists. (Also, is this an exercise? If so, then it doesn't belong here. If not, then how does such a recurrence arise?)
Thank you so much for your comment, still I am trying to get the closed formula, it is part of a project, the recurrence come from calculating a conditional expectation, we need to get the closed form for further analysis. My problem is with condition function here, how to consider the changing initial condition, the initial of one variable depends on the other variable, it seems like there is something we can do to spin this.
I strongly believe in the second case, it was meant to be "if $i=2$" - I took a liberty to edit it.
Under the assumption that $F(2i,n) = 0$ when $i<4$ or $2i>n$, the first and second cases in the recurrence become partial cases of the third one. Hence, the recurrence reduces to
$$F(2i,n) =
\begin{cases}
1, & \text{if } i=2,\ n=4;\\
0, & \text{if } i<2\text{ or } 2i>n;\\
\frac{2i+n-4}{2n-5}F(2i,n-1)+\frac{n-2i+1}{2n-5}F(2i-2,n-1) & \text{otherwise}.
\end{cases}
$$
Consider the generating function
$$f(x,y) := \sum_{i\geq 2} \sum_{n\geq 2i} F(2i,n) x^{n+2i-8} y^{n-2i+1}.$$
Then the first two cases imply $f(x,0) = 0$ and $f(0,y)=y$.
Now, the third case translates into a linear PDE:
$$\frac{\partial}{\partial x} (x^2\cdot f(x,y)) + xy\frac{\partial}{\partial y} f(x,y) = 3xy + \frac{y}{x^2}\frac{\partial}{\partial x} (x^5\cdot f(x,y)) + x^4y\frac{\partial}{\partial y} (y^{-1}\cdot f(x,y)),$$
which simplifies to
$$(x^2y^2-xy)f_x + (x^3y - y^2)f_y = (2y+x^3-5xy^2)f - 3y^2.$$
Thank you very much.
|
2025-03-21T14:48:31.520112
| 2020-07-15T17:33:09 |
365718
|
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|
Stack Exchange
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A global mathematics library
Just for personal interest, I am not (yet) professionally involved in it. My question is about the state of arts in digitalization of mathematics and to what extent it is possible and reasonable.
There are different levels of digitalizations:
OCR scan all historical mathematical texts
organize metadata of references and authors (e.g. as graphs)
extract mathematical objects (like theorems, definitions, etc)
extract proofs and ideas
formalize mathematics such that it can be completely checked by theorem provers
The main effort I found is
https://imkt.org/
Steps 3/4 and 5 may be of independent interest and should be understood more parallel than chronological. Point 5 is more interesting in having (error-free) formalized math. It should also be allowed to choose different foundations of mathematics and the possibility to switch in between them. Point 3/4 is more interesting for a researcher that wants all references for a definition, a theorem, a keyword. It would be a wonderful source for doing data analysis of mathematical knowledge (historical, social, semantical, etc). In contrast to 5 it can contain errors and speculations. The main interest is in identifying and referencing mathematical objects over all the produced texts in history of math.
My question is:
The goal of https://imkt.org/ is huge but when looking at its first projects it looks (sry) a little bit disappointing. The main focus (also of other literature that I scanned through) lies on connecting the existing databases and languages, i.e. of theorem provers, computer algebra systems (and maybe wikis). I understand that different applications in math demand different systems (e.g. integer series http://oeis.org/ should also be part of it?) Can or shouldn't there be one system that contains everything that can be accessed (and is stored not just referenced!) through the same system? Are my dreams of such a system over the top?
One of the largest issues is the copyright of the big publishers.
More and more is going in direction open math. Until then it is
unclear to what extent a library can be complete (gaps are somehow
missing the point of this system).
The other issue is the efficiency in producing the content extraction
and to advance it by advertising the advantages of such a library to
mathematicians such that it will move itself at some point.
There have been many efforts in the past that were abandoned again or here for years (like Mizar) but far from being known and used in daily mathematics.
You might look at the IMU and their efforts on WDML. Part of the idea is to archive sites like MathOverflow which show the (relatively informalized) development of mathematics in this century. Gerhard "Says 'Informalized' Is A Word" Paseman, 2020.07.15.
So far I just found articles either analyzing existing systems or abstractly sophisticating about properties that a perfect library should have. Do you know about a recent concrete effort of building a perfect system? And can you give me a link for that IMU effort? Sounds interesting.
IMU is international Mathematics Union. WDML is World Digital Mathematics Library. I think you can find the links as well or better than I can. Gerhard "Ingrid Posted About It Here" Paseman, 2020.07.15.
There is slow progress towards point 5 (see various libraries that have grown around systems like Coq and Mizar), but at the moment nothing too high can grow on this field since most proof languages are too unstable and don't feel like "the right thing" (so any work done now will probably have to be painstakingly adapted in 10 years if not earlier). Most of the time, these libraries are written by humans, typically inspired by existing literature but in no way just straightforwardly translating it. Maybe AI could do this in 20 years, but even that's far from a given.
I'm not sure whether I understand point 3 correctly: If I get it right, you would like to have a database which contains theorems, definitions, etc., and which links to articles books, etc, where they are used. I am, however, under the impression that this idea underestimates the diversity of mathematical literature, even within very specific sub-areas: For instance: What precisely is the statement of Banach's fixed point theorem? What is "the" Perron-Frobenius theorem in matrix analysis? What does the existence and uniqueness theorem for ODEs say? And these are only rather mild examples...
I am aware that there is sometimes no unique formulation of a theorem. That is not a problem. A perfect system/database would try to just capture exactly in a machine-readable form how a textbook states a theorem. Using some ML one then aggregates similar statements (the "same") theorems and displays them in a frontend view. Maybe listing the most stated version first and referring to other formulations etc.
First one needs a good strategy how to store statements. One this is defined the aggregating process and linking by ML and the user friendly search engine frontend can be independent and more and more improved over the years.
"Same theorem" is much more of a problem than you think. Let me point out one particularly problematic situation. The foundations of algebraic geometry were completely redone in the 1960s by Grothedieck and others, to the extent that modern algebraic geometers find it very hard to read the standard text of Hodge and Pedoe from the 1950s. A much broader class of objects was introduced, and the objects previously studied were redefined in a different way. (continued)
Some old theorems generalized to (at least some of) the new class of objects by doing the 'obvious' translation to the new language; some generalized but required genuinely new proofs; some remained true only for the 'classical' objects (redefined). In which cases do we have the "same theorem"?
None of your five levels of digitalization are at all within reach. Let's take just the first one. As you say, copyright is a huge obstacle. Consider what happened when the same idea was attempted by Google (resulting in Google Books), as explained wonderfully in this article in the Atlantic. Larry Page (as in, PageRank) set up a tremendous effort to get books from libraries and scan them en masse with OCR technology to make them searchable.
After doing this for some time, publishers filed suit against Google. Google hired a massive legal team because it cared so much about creating such a library. Google managed to bring the publishers onboard with the idea, pointing out that it would allow publishers to sell out-of-print works. Unfortunately, this raised another issue, like a sad game of whack-a-mole, and the Department of Justice got involved to prevent Google from having a de-facto monopoly on out-of-print books. In the end, the Judge prevented the settlement that Google and the publishing companies had agreed to.
The same things would happen in the math world. The point is: you have a bunch of authors who died a long time ago and can't give consent to update their old publishing contracts. You have publishers who won't let go of content unless they can profit on it. And you don't have any organization large enough to be the central clearing house to connect customers who want to read these works with the places they can view and buy them. The problem only gets harder every year, because we produce more and more mathematics all the time (and OCR for math is not great, throwing a wrench into your third idea). My opinion: if Google couldn't pull together a global library, despite all the effort they put into this problem, there's no hope for the rest of us.
So end run around the old system: publish newly created mathematics (and synopses and surveys of older math) under a free-to-distribute license. Just because it was done the old-fashioned way does not mean it has to stay that way. Gerhard "Get Rid Of Tradition,... Tradition!" Paseman, 2020.07.15.
Nice answer and also nice cited article. And I also partly agree with your conclusions: if the problem is dealt only on a pure economic interest basis, if Google has not succeeded, then no one will. However I hope that there will be more space for science and a little less (but reasonably sufficient) space for business. After all, the Taj-Mahal, St. Peter's Basilica, and almost everything it made by men that is worth to see, has not been built with the aim of making money about it.
De-facto, Library Genesis is the library everyone uses these days as a first source for mathematics books; physical libraries are consulted as a last resort, for the rare books that are not (yet) on Library Genesis. But there is a huge gap between having a scan and having a readable "true PDF" or TeX file. MathPix is a promising OCR tool for mathematical formulas, but it has its limits, and typesetting can be very diverse. (Not to mention that lots of scans aren't good enough to leave OCR to machines.)
@darijgrinberg, I'm not sure that LibGen is the preferred resource over library books for everyone. A, uh, friend of mine definitely consults it whenever needed to stock an electronic library (say, for books to take along on travel), but, pre-pandemic, I would almost always go to the physical library to get a book (not an article). It's just too hard to browse, properly to bookmark, to compare side-by-side, etc. in an electronic reference; the electronic reading experience remains sub-par for me.
@LSpice your friend can print out the relevant pages
@erz Many books cross-reference themselves across hundreds of pages constantly. There's a reason books are books, and not sequences of articles.
@KevinArlin one can print the whole book. Also, I think it is acceptable to print out only the relevant pages and read them from the paper, and once a reference to some early place pops out look it up on the screen
@LSpice: Having to crawl or climb on a chair to reach the appropriate shelves of a library isn't always fun either :) And I have to say, after seeing publisher after publisher lose the art of bookbinding, that there are many books I'd prefer to print out myself.
@darijgrinberg, I definitely have nothing against those who prefer e-books, and am glad that they're there when I need them (like in the US now—I'm definitely not going in our library, even when it opens). I just don't agree that it's everyone's first choice.
What if there will be ONE country in the word which openly declare that they do not accept publisher rights, patents, etc., and by the law of this country everyone has the right to scan and put online any information they like? Then this country could create and maintain such database. If would be illegal in other countries, but at least it would exist.
On my opinion, 2-5 are non-realistic (and hardly needed), while 1 is indeed desirable and important. To be sure, a substantial progress was made in the recent years towards 1. But it still remains a remote goal.
David White mentioned the main obstacle: the existing shameful copyright system. But there are several other obstacles. The number of math periodicals is really enormous, many were owned by universities and various societies in the past, and there is no real will to digitalize all these obscure publications. There are many old publications of various societies in most European countries which are still not digitalized. Big companies probably see no perspective of profit in this (and they are probably right), and volunteers have more important things to do.
As I understand copyright does not apply to 18-19 centuries math journals.
Still, there is little interest in their digitalization. Few big journals
which still continue, digitalized the old issues (and claim copyright for this digitalization). But no one seems to be interested in those journals which do not continue.
In many cases, even when a digitalization is available, the quality is too poor to read or print, or search. The old volumes of Comptes rendus are available, but try to find anything in them. Sometimes they show you "digitalized" pages on the screen but do not allow to download the pdf file of the article. So you are supposed to print a large paper one page at a time, if allowed to print at all.
The result is easy to predict: with the current demise of the usual libraries, most this mathematics will be simply lost. And no realistic remedy is visible. Well, this is not the first time that such thing happens: by various estimates about 90% of Hellenistic mathematics and science were lost. (And practically 100% of Greek pre-Hellenistic).
Of course one can argue that "the most important" works are preserved. For example Euclid is preserved but all his predecessors are not. Perhaps there is some truth in this. (Imagine that of all 20 century mathematics only Bourbaki is preserved:-)
I”m not sure I agree on your 18-19 centuries assessment. Most everything to 1850 or 1900 is scanned and available in archives like numdam, gallica, hathitrust, biodiversity, archive.org that seem safe even if Google books were to shut down. After 1950 or so is another matter with, anyway, a much reduced signal/noise.
Most German "provincial" journals (which have names beginning with Sitzungsberichte, Abhandlungen, etc. are not digitalized, even the famous one published in Gottingen). And this is only one example. The famous Russian journal Doklady of the Academy is not digitalized. We only have translations for few years when it was translated.
Well, the link leads to a list where everything from Göttingen seems available (GDZ). Likewise Leipzig (babel, wikisource).
The links you provided even do not contain the word "Göttingen".
My first comment, linked word “available”. Search the list there for “Göttingen”. One of the links seems broken but here it is again, fixed.
Here is the complete title of the journal I was never able to find: Nachrichten von der Königlichen Gesellschaft der Wissenschaften und der Georg-Augusts-Universität zu Göttingen (and still cannot).
(1), (2), (3), (4),...
You say Doklady is not digitized. It is available for the period 1957-1995 at http://www.mathnet.ru/php/archive.phtml?jrnid=dan&wshow=contents&viewarchiveID=8&option_lang=eng
Why isn't there an effort to survey this old mathematics to prevent it from vanishing? I don't think the texts as such are all that valuable, it's the ideas that are there.
@erz: I can give you one reason: reading old mathematics is difficult and requires highest qualification. People who have it prefer to spend their time more productively: by proving their own theorems:-)
@KConrad: this appeared recently, thanks for noticing. Still the first 111 volumes are not available. Hope they will do all volumes it eventually.
Earlier volumes of Doklady back to 1933 are listed online at https://catalog.hathitrust.org/Record/000053018 but the volumes can only be searched by keyword.
There's a lot of work lately on formalizing large parts of classical mathematics using proof assistants, see for example the Lean prover and the accompanying library mathlib. You can see an overview of what has been formalized here; it includes a lot of classical abstract algebra and real analysis, although there's also a lot of undergraduate math they haven't formalized yet which they list here. If you think that's all "trivial", there's ongoing work to formalize the proof of the independence of the continuum hypothesis, the proof that it's possible to evert the sphere, etc.
An interesting facet of this project is that all the proofs are kept under a version control system and you can see the entire history of every proof. It's common for the initial proof of some result to be very long and verbose, only for people to find ways to shorten it or make it more elegant later -- for example, von Neumann's proof of the Radon-Nikodym theorem. Where before you might have had to trace through the citations of loads of historical papers that your library might not even have, now you can see exactly how this process played out by just doing git blame.
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2025-03-21T14:48:31.521344
| 2020-07-15T19:09:23 |
365725
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Stack Exchange
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Examples of $\aleph_0$-categorical nonhomogeneous structures
Macpherson in a survey of homogeneous structures, states that there are many $\aleph_0$-categorical structures which are not homogeneous. I'd like to know more of such examples.
Edit: The homogeneity mentioned here is the ultrahomogeneity that is defined as every isomorphism between two finite substructures of a structure $M$ can be extended to an automorphism of $M$. There is another homogeneity known as $ℵ_0$-homogeneity that is defined as two $n$-tuples with the same type in $M^n$ must be on the same orbit of $\rm Aut$$(M)$.
$\aleph_0$-categorical structures need not be ultrahomogeneous, but are always $\aleph_0$-homogeneous. So $\aleph_0$-homogeneous is a weaker notation than ultrahomogeneous in general. These two types of homogeneity become equivalent if and only if $\rm Th$$(M)$ has quantifier elimination.
Here is my favorite example.
Theorem. Fix $n\geq 1$. Then there is a unique (up to isometry) countable metric space $(M_n,d)$ satisfying the following properties:
$d(x,y)\in\{0,1,2,\ldots,n\}$ for all $x,y\in M_n$
Any finite metric space with distances in $\{0,1,2,\ldots,n\}$ embeds as a subspace of $M_n$.
Any partial isometry between two finite subspaces of $M_n$ extends to a total isometry of $M_n$.
I don't really know who to attribute this to (perhaps Casanovas & Wagner, or Delhomme, Laflamme, Pouzet, Sauer). The point is that the class of finite metric spaces with distances in $\{0,1,\ldots,n\}$ is a Fraisse class in an appropriate relational language, and so $M_n$ is the Fraisse limit. In particular, for all $k\leq n$, add a binary relation $d_k(x,y)$ interpreted as "$d(x,y)\leq k$". In this language, $M_n$ is homogeneous. But....
View $M_n$ as a graph under just the relation $d_1(x,y)$. Then $M_n$ is still $\aleph_0$-categorical because the metric is "definable" from the graph language using existential quantifiers. (Specifically, the defining properties of $M_n$ force the metric $d$ to coincide with the "path metric" given by distance $1$. E.g, $d(x,y)\leq 2$ iff $\exists z(d_1(x,z)\wedge d_1(z,y))$.) However, in the graph language, $M_n$ is homogeneous if and only if $n=1$ or $n=2$. Indeed, if $n\geq 3$ then we can find points $a,b,c,d\in M_n$ such that $d(a,b)=2$ and $d(c,d)=3$. In the graph language, the finite substructures $\{a,b\}$ and $\{c,d\}$ are isomorphic. But an automorphism of $M_n$ has to respect the metric, and so there is no automorphism sending $(a,b)$ to $(c,d)$.
The point, of course, is that by looking only at the distance $1$ relation, we lose quantifier elimination.
Note, on the other hand, $M_1$ is a countably infinite complete graph and $M_2$ is the countable Rado graph, both of which are homogeneous as graphs.
By the way, $M_n$ is an example of what is called a metrically homogeneous graph. For more on these, see Cherlin's work on the classification program for these graphs.
Macpherson states that for $ℵ_0$-categorical structures, they are homogeneous iff they have quantifier elimination. Why is that? Intuitively, $ℵ_0$-categorical structures along can have quantifier elimination.
@hermes When Macpherson says a countable structure $M$ in a relational language is homogeneous, he means that every isomorphism between finite substructures extends to an automorphism of $M$. It is easy to see that this is equivalent to the following condition: if $\overline{a}$ and $\overline{b}$ are finite tuples with the same quantifier-free type, then there is an automorphism $\sigma\in \mathrm{Aut}(M)$ such that $\sigma(\overline{a}) = \overline{b}$. In other sources (e.g. Hodges), this condition is sometimes called "ultrahomogeneity".
It is also a theorem that every $\aleph_0$-categorical structure $M$ is $\aleph_0$-homogeneous, which means that if $\overline{a}$ and $\overline{b}$ are finite tuples with the same type, then there is an automorphism $\sigma\in \mathrm{Aut}(M)$ such that $\sigma(\overline{a}) = \overline{b}$ (this holds because $M$ is atomic, and also because $M$ is saturated). These two conditions, ultahomogeneity and $\aleph_0$-homogeneity, are equivalent if and only if the quantifier-free type of a tuple determines its type, i.e. exactly when $\text{Th}(M)$ has quantifier elimination.
@hermes No. To reiterate Alex's main points: for Macpherson, "homogeneous" means "ultrahomogeneous" and $\aleph_0$-categorical structures need not be ultrahomogeneous. $\aleph_0$-categorical structures are always $\aleph_0$-homogeneous, which is a weaker notation than ultrahomogeneous in general. They become equivalent in the presence of QE.
How about: dense linear order with endpoints.
It's $\aleph_0$-categorical by the same proof as for the case without endpoints.
It's not homogeneous because of the endpoints.
Looks like one can do something similar from any $\aleph_0$-categorical theory by adding finitely many constants.
@YCor well, in my example it's important that the endpoints are not named by constants, so I'm not sure about that
In which way is it important for the question? one could consider an poset with two constants in which the order is total and dense without endpoints outside the constants, and the constants are incomparable to the other elements.
@YCor ah okay, yes I think that works. Only these would not be "constants" in the usual sense; they will not be interpretations of constant symbols in the first-order language.
@YCor More generally, you can do something similar with any two distinct $\aleph_0$-categorical theories $T$ and $T'$. Define a new theory whose models are disjoint unions of a model of $T$ and a model of $T'$ (using disjoint vocabularies). Your observation is the case where $T'$ is the theory of a structureless finite set of specified size.
@hermes it's the back and forth argument + the observation that endpoints are sent to endpoints by isomorphism
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2025-03-21T14:48:31.521743
| 2020-07-15T19:15:30 |
365726
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Stack Exchange
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Continuity of conformal grafting, wrt the (infinite type) surface
Say I have two closed hyperbolic surfaces $X,Y$ and a smooth, $(1+\epsilon)$-bilipschitz map $f : X \to Y$ for some small $\epsilon$. Pick a simple closed curve $c \subset X$, and let $X',Y'$ be the results of `length t conformal grafting' along $c$ and $f(c)$, respectively. (Namely, we cut $X$ along $c$, patch in a Euclidean annulus of length $t$, and then uniformize to get another hyperbolic surface, and we do something similar on $Y$ to get $Y'$.) Then $X',Y'$ are also metrically close, since you can use $f$ to construct a quasiconformal map from one to the other that has small dilatation, and nearly conformal hyperbolic surfaces are nearly isometric. (Essentially since you can take limits and use that conformal maps $D \to D$ are hyperbolic isometries.)
So what if I now have infinite type surfaces? Let's say $(X,x),(Y,y)$ are two pointed hyperbolic surfaces (possibly infinite type) that are close, meaning that for some large $R$, there are subsets $A\subset X$ and $B\subset Y$ containing the $R$-balls around $x$ and $y$, and a $(1+\epsilon)$-bilipschitz map $f : A \to B$ with $f(x)=y$, for some small $\epsilon$. Say I do length $t$ conformal grafting on some curve $c$ that's close to $x$, and on its $f$-image in $Y$. Are the resulting surfaces $X',Y'$ close, in the sense that large hyperbolic neighborhoods around their basepoints are close? You can construct quasiconformal maps between $A \cup \text{ (annulus)}$ and $B \cup \text{(annulus)}$ using $f$, just like before, but the problem I have is saying that the these sets contain large radius hyperbolic balls around the basepoints after uniformization.
Any thoughts on this from the grafting gurus? Thanks in advance.
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2025-03-21T14:48:31.522002
| 2020-07-15T20:37:25 |
365733
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Stack Exchange
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Elementary proof for $n^2>p>n$ for all $n>1$
Is there any elementary way of proving that for all natural numbers $n>1$ there exists a prime $p$ such that $n<p<n^2$. And I mean elementary, not using the Prime Number Theorem or Bertrand's Postulate.
Welcome to MathOverflow SE. FYI, on the Math SE site, there's There is a prime between $n$ and $n^2$, without Bertrand. On this site, there's a closely related one of What is the shortest proof of the existence of a prime between $p$ and $p^2$ ? other examples?.
Erdos has a nice proof in his paper on the Sylvester Schur theorem from 1934. I will sketch part of it here.
Look at the prime factors of the binomial coefficient $B = \binom{m+n}{n}$. An elementary argument says that the prime $p$ divides $B$ to a power that is less than $m+n$. Therefore $B$ has all its prime factors less than $n$ multiply to a number $C$ that is less than $(m+n)^{\pi(n)}$.
Now when $n > 8$ and $m+n \geq n^2$, $B$ gets bigger than the $n$-smooth part of $B$. This means there is a prime bigger than $n$ dividing a number in $(m,m+n]$. Therefore there is a prime in $(n,n^2]$ for $n > 8$. The remaining cases are left to the reader.
For more elementary arguments giving a prime bigger than $n$ dividing $B$ when $m \geq n$, read the 1934 paper. For those interested, I am reworking the arguments to make it more general and still elementary. I invite you to critique a rough draft.
Gerhard "That Simple Enough For You?" Paseman, 2020.07.15.
FYI, this is quite similar to the Math SE's answer to There is a prime between $n$ and $n^2$, without Bertrand.
It does not differ much from the proof of Bertrand postulate.
Erdos has a little more to the sketch than what I present here. However, the whole paper is elementary, and proves SS for all n greater than 3000. My understanding is that there is more to proving Bertrand, and seems like much more to me. Gerhard "Lots Of Little Cases Left" Paseman, 2020.07.15.
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2025-03-21T14:48:31.522524
| 2020-07-15T20:45:18 |
365734
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Stack Exchange
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Classification of finite crossed modules
A finite crossed module is a 4-tuple
$$(G_1,G_2,\delta: G_2 \to G_1, \alpha: G_1 \to Aut(G_2))$$
satisfying certain compatible conditions, where the $G_i$ are finite groups and the maps are group homomorphisms.
Given two arbitrary finite groups $G_1, G_2$, are all finite crossed modules with the same first two entries known?
Given a group homomorphism $\delta$ from $G_2$ to $G_1$, can we describe the corresponding set of equivalent crossed modules from $H^3(K,A)$, where
$$ A = ker(\delta), K = cok(\delta)? $$
If unfortunately both cases are wildly unknown.. are there some situations in which the classification problem has been settled?
You could look at
HOMOTOPY 2-TYPES OF LOW ORDER
GRAHAM ELLIS AND LE VAN LUYEN
in the J Experimental Math,
@RonnieBrown It's very helpful. With both groups fixed, up to quasi-isomorphism the structure only depends on the Postnikov invariant. How about "up to isomorphism"? In section 3 the author implemented an algorithm, which seems to be brute forcing all possibilities.. To summarize.. is there a more conceptual way to classify finite Cat1Group/CrossedModules structures up to isomorphism?
As you have discovered, the method in the GAP package HAP constructs all cat1-groups G->G, and provides a library of all cat1-group structures on groups of order up to 255 (with some exceptions). The XMod package also has an operation AllCat1GroupsUpToIsomorphism(G). But it is not clear how to implement an operation AllXModsUpToIsomorphism(G1,G2) - AllHomomorphisms(G1,G2) is available, in GAP, but then all possible actions would be required. Would such a function be useful? For a crossed module X0, the XMod package provides an operation KernelCokernelXMod(X0) which constructs your A->K. None of this, of course, involves a conceptual approach.
The questioner could also find the following helpful on the use of crossed modules. 1. "Groupoids and crossed objects in algebraic topology"Homology, Homotopy Applications, 1(1999) 1-78, and the book "Nonabelian Algebrqic Topology, both downloadable from my wen site. The uses suggest the kind of classification one seeks. .
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2025-03-21T14:48:31.522694
| 2020-07-15T21:06:00 |
365735
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}
|
Stack Exchange
|
Affine scheme as algebraic space
We working in the following with Knutson's definition of an algebraic space
(ie via equivalence relation; there is also another equivalent def via
sheaves but let us work here with the following one):
An algebraic space $X$ comprises a scheme $U$ and a closed subscheme
$R \subset U \times U$ satisfying the following two conditions:
$R$ is an equivalence relation as a subset $U \times U$;
the two projections $p_i: R \to U$ onto each factor are étale.
Knutson adds an extra condition that the diagonal map is quasi-compact.
A couple of notes on used notations: the equivalence realtion $R
\subset U \times R$ is considered as
categoretical equivalence relation (also called "internal relation"),
that means that for all
$T \in (Sch)$ the set $Hom(T,R) \subset Hom(T, U \times U)=
Hom(T,U) \times Hom(T,U)$ is the equivalence relation in usual sense.
Question: How one can see that an "usual" scheme $U$ is an
algebraic space in the sense above? Assume wlog $U$ affine. The
crucial task is to find an equivalence relation $R \subset U \times U$
corresponding to $U$ such that projections $p_i: R \to U$ are etale.
The most natural choice seems to me the image with respect the
diagonal map $\Delta: U \to U \times U$, ie $R:= \Delta(U)$.
$\Delta$ is always an immersion and thus
$\Delta(U)$ is always a locally closed subscheme of $U \times U$.
If we take this choice for $R$, why $p_i: R \to U$ are etale?
Or is it conventional to take another choice for $R$? eg the
closure of the image? if yes, why?
The restriction of each of the projections $U \times U \rightarrow U$ to $\Delta(U)$ gives an isomrphism of $\Delta(U)$ to $U$, so it is in particular etale.
@AlexI: yes I see. More precisely the restricted $p \vert _{\Delta(U)}: \Delta(U) \to U$ must be an isomorphism because it's nothing but the composition of isomorphisms $\Delta^{-1}: \Delta(U) \to U$ and $id_U= p \circ \Delta: U \xrightarrow{\Delta} U \times U \xrightarrow{\text{p}} U$.
So indeed $R:=\Delta(U)$ is the right choice and not the schematic closure $\overline{\Delta(U)} \subset_c U \times U$, I think.
A curious nitpick beside: Do you know if, if we assume that $\overline{\Delta(U)} \not \cong \Delta(U)$ (eg $U$ not separated). Is the restricted projection map $p \vert _{\overline{\Delta(U)}}:\overline{\Delta(U)} \subset U \times U \to U$ nevertheless etale, or is it in general wrong?
Look at the example where $U$ is the affine line with doubled origin. The closure of the diagonal is then an affine line with four points "at the origin". Each of the projection maps $\overline{\Delta{U}} \rightarrow U$ is an isomorphism outside origin and over each of the origins of $U$ there lie two of the four points of $\overline{\Delta{U}}$. This map is an isomorphism locally on the source, hence etale. See also https://math.stackexchange.com/questions/1438886/are-etale-morphisms-separated
I understand why in your example $p: \overline{\Delta(U)} \to U$ is also etale but
I'm not sure if this example covers all pathological phenomena can occure
when passing from $\Delta(U)$ to $\overline{\Delta(U)}$ as subscheme of
$U \times U$. Does your example prove a general argument why for arbitrary $U$
the projection $p: \overline{\Delta(U)} \to U$ should be etale?
|
2025-03-21T14:48:31.522948
| 2020-07-15T22:43:03 |
365743
|
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}
|
Stack Exchange
|
Explicit construction of Fubini Study Metric
I have a question about a remark on Fubini Study metric on $\mathbb{CP}^n$
from Notes on canonical Kähler metrics
on page 8 is remarked (Example 2.12 4.):
Fix a Hermitian innerproduct on $\mathbb{C}^{n+1}$.
Then $\mathbb{CP}^n$ inherits a unique
(up to scale) $U(n+1)$-invariant Riemannian metric, called the Fubini–
Study metric. It is Kähler, as can be seen in various ways. [...]
My question is how Fubini–Study metric $f_{FS}$ concretly
arise as inherited
metric from standard Hermitian product $\langle-,-\rangle_h$ on $\mathbb{CP}^{n+1}$?
It suggests that to obtain $f_{FS}$ we consider the canonical projection
$\mathbb{C}^{n+1} \backslash \{0\} \to \mathbb{CP}^n$ where
$\mathbb{C}^{n+1} \backslash \{0\}$ is endowed still with restricted metric
$\langle-,-\rangle_h$ and $f_{FS}$ on is induced from $\langle-,-\rangle_h$ in certain canonical
way.
Could somebody give a sketch how this explicit construction of
$f_{FS}$ from standard metric $\langle-,-\rangle_h$ works?
More precisely the problem is that in almost
every source
the Fubini Study metric beeing introduced by a
"let define the FS as ..." and then one verifiy that the defined metric is
$U(n+1)$-invariant Riemannian and so on (see e.g. Here or the classical literature by Griffith/Harris or R. Wells)
But the quoted sentence above suggest that there is a way to
construct the Fubini-Study metric from
$\langle-,-\rangle_h$ on $\mathbb{C}^{n+1}$ explicitly / canonically.
That's what I understand
by "inherited". And I'm asking where I can find source
on Fubini-Study metric where it is constructed from $\langle-,-\rangle_h$$
and not simply defined as something "fallen from heaven" &
has desired properties.
The most "natural" approach I found was by defining
$$f_{FS}:= \bar\partial\partial \log <Z,Z>=
\bar\partial\partial \log (1+z^2)$$
where $Z =[1:z_1:....z_n] \in U_0 :=\{Z \in \mathbb{CP}^n \ \vert
z_0 \neq 0 \} \subset \mathbb{CP}^n$.
Although this construction indeed only depends on $\langle-,-\rangle_h$
this also not persuades me completly. The
$\bar\partial\partial \log \langle Z,Z \rangle_h$ shape looks too artificially; i.e. more like a
"sophisticated guess" than something "canonical".
I expected to find something like a general machinery for it like: when $M$ is
a Riemannian/Hermitian manifold endowed with metric $h$ and suppose
a group acts on $M$ nice enough such that $M/G$ is also a manifold.
Then there is a standard method to obtain a metric $\bar{h}$
on $M/G$ inherited from $h$ on $M$. And in our example the
$\bar{h}$ is nothing but the Fubini Study.
Does there exist such procedure in cases for $M$ & group action by $G$ nice enough? Or do I looking for something that not exist and the definition $f_{FS}= \bar\partial\partial \log \langle Z,Z \rangle_h$ is indeed the only meaningful way to introduce Fubini-Study metric?
Yes, this metric is well-known. If $G$ acts isometrically and freely on a Riemannian manifold $M$, then the quotient space $M/G$ has the so-called Riemannian submersion metric. Points of $M/G$ are in a bijection with $G$-orbits. A tangent plane at a point of $p\in M/G$ is represented by a normal subspace to the corresponding $G$-orbit. The $G$-action isometrically identifies the normal spaces to a $G$-orbit, and the thus gives an inner product on $T_p(M/G)$.
Perhaps one can simply say that the Fubini-Study metric is the one induced on the quotient of $S^{2n+1}\subset\mathbb{C}^{n+1}$ (i.e., the unit vectors with respect to the Hermitian metric) by the $S^1$-isometric action of multiplication by ${\mathrm{e}^{it}\ |\ t\in\mathbb{R}\ }$.
If one wants the other scale factors, then one just takes, instead of $S^{2n+1}$, any other nonzero level set of the Hermitian length function. Since this $S^1$ commutes with $\mathrm{SU}(n{+}1)$ (which acts by isometries on $S^{2n+1}$), one has that $\mathrm{SU}(n{+}1)$ acts via isometries on $\mathbb{CP}^n$.
Could you give reference where the construction of this
Riemannian (Hermitian) submersion metric $\bar{h}$ with respect
$ M \to M/G$
and Riemannian (Hermitian) mertic $h$ on $M$ is explicitly
described (preferably as "cooking recipe"?
If this is exactly the constuction I'm looking for then this precisely
gives the Fubini Study metric $f_{FS}$ on $\mathbb{C}^{n+1} / \mathbb{C}^*$
which is (accoring to https://en.wikipedia.org/wiki/Fubini%E2%80%93Study_metric#In_local_affine_coordinates) on affine chart $U_0$ given by
$$g_{i{\bar {j}}}=h(\partial {i},{\bar {\partial }}{j})={\frac {(1+|\mathbf {z} |^{2})\delta {i{\bar {j}}}-{\bar {z}}{i}z_{j}}{(1+|\mathbf {z} |^{2})^{2}}}$$
($\partial _{i}, \partial _{j}$ are canonical basis vectors
of tangent space at $[Z_0:...:Z_n]$.
Not sure what you mean by a "cooking recipe". Petersen's "Riemannian geometry" book covers the material; just look up "Riemannian submersion" and "projective space" in the index.
The $\partial\bar\partial \log \langle Z,Z\rangle$ is not really just a clever guess and is in fact canonical. You want a Kahler potential that is defined as a function of the Hermitian product (and nothing else) but that defines a Kahler form that is homogeneous of degree $0$, so that it can be pushed down to $\mathbb{C}P^n$. The logarithm is the only function that will do this.
|
2025-03-21T14:48:31.523310
| 2020-07-15T23:20:35 |
365745
|
{
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"D.-C. Cisinski",
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}
|
Stack Exchange
|
"Universal coefficent theorem" for pro-étale cohomology
In algebraic topology, for any space with finite homology type, the universal coefficient theorem states that for any abelian group $G$, we have
$$H^n(X,G)\cong \left( H^n(X,\mathbb{Z})\otimes G\right)\oplus \text{Tor}_1(H^{n+1}(X,\mathbb{Z}),G).$$
My question is whether the analogous statement is true for the pro-étale cohomology, namely if $R$ is a $\mathbb{Z}_\ell$-algebra, do we have
$$H^n_{proét}(X,\underline{R})\cong \left(H^n_{proét}(X,\underline{\mathbb{Z}_\ell})\otimes R\right)\oplus \text{Tor}_1(H_{proét}^{n+1}(X,\underline{\mathbb{Z}_\ell}),R)$$
for a sufficiently nice scheme? I'm mostly interested in the case of a smooth, projective scheme over some algebraically closed field (possibly of positive characteristic).
Also, would this decomposition respect that Galois action on the cohomology?
Note: $\mathbf Z_\ell$ needs to be endowed with a topology for it to give the 'right' cohomology in the pro-étale site. So there should probably be a natural topology on $R$ as well (somehow compatible with the one on $\mathbf Z_\ell$) for this to have any chance of success.
Would it be enough to assume that the structure morphism $\mathbb{Z}_\ell \rightarrow R$ is continuous?
I do not think it is reasonable to consider any topology on $R$: the point of this formula is to compare the (derived) tensor product with $R$ within the pro-étale topos with the (derived) tensor product externally (i.e. within the ordinary derived category of $\mathbf{Z}_\ell$-modules), where everything is discrete.
@Denis-CharlesCisinski I don't understand what you mean. If you take $R = \mathbf Z_\ell$ with the discrete topology (i.e. the true constant sheaf), then you should get something very different than the usual $\mathbf Z_\ell$ cohomology.
@R.vanDobbendeBruyn I mean $R$ should be discrete, not $\mathbf{Z}\ell$. This is what happen when you go from $\mathbf{Z}\ell$-linear coefficients to $\mathbf{Q}\ell$-linear ones: $\mathbf{Q}\ell=\mathbf{Z}\ell\otimes\mathbf{Q}$ with $\mathbf{Q}$ discrete. My point is that, in the "universal coefficients theorem", there is a tensor/tor with cohomology groups considered as discrete objects, so that $R$ has to be discrete. My answer below explains how to proceed (I underline the continuous $\mathbf{Z}\ell$ to differientiate from the discrete one).
@Denis-CharlesCisinski Ah I see, you're answering a slightly different question where instead of taking $H^n(X,R)$ you consider $H^n(X,\mathbf Z_\ell \otimes R)$. That seems fair (I guess there is probably no universal coefficients for a more general class of coefficients).
The only condition you need on $X$ for such a formula to hold is that it is coherent (=quasi-compact and quasi-separated).
Let $R$ be a discrete ${\mathbf{Z}_\ell}$-module. We consider the sheaf $\underline{\mathbf{Z}}_\ell$ on the pro-étale site defined as the limit of the constant sheaves $\mathbf{Z}/\ell^i\mathbf{Z}$. This is an algebra on the constant sheaf associated to the discrete ring $\mathbf{Z}_\ell$. Hence we may define $\underline{R}$ as the tensor product of $R$ with $\underline{\mathbf{Z}}_\ell$ (exercise: it is in fact the derived tensor product, since $\underline{\mathbf{Z}}_\ell$ has no $\ell$-torsion stalkwise; this is where pecularities of the pro-étale site have a role to play, in a proof which is otherwise very formal). Let $R\Gamma(X,-)$ denote the derived global sections on the pro-étale topos of $X$. To prove that the canonical map
$$R\Gamma(X,\underline{R})\leftarrow R\Gamma(X,\underline{\mathbf{Z}}_\ell)\otimes^L_{\mathbf{Z}_\ell}R$$
is invertible in the derived category of $\mathbf{Z}_\ell$-modules, assuming that $X$ is coherent, we may assume that $R$ is of finite type because $R\Gamma(X,-)$ commutes with filtered colimits (this is precisely what coherence is good for). If $R=S\oplus T$, it is sufficient to prove it for $S$ and $T$ separately. Hence, without loss of generality, we may assume that $R=\mathbf{Z}_\ell$, in which case this is trivial, or that $R=\mathbf{Z}/\ell^i\mathbf{Z}$, in which case this is always true as well (taking the cone of the multiplication by $\ell^i$ commutes with $R\Gamma(X,-)$).
|
2025-03-21T14:48:31.523590
| 2020-07-16T04:23:49 |
365748
|
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"authors": [
"Gilles Mordant",
"Sandeep Silwal",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365748"
}
|
Stack Exchange
|
Projecting a vector onto a random subspace
Let $A\in\mathbb{R}^{k\times d}$ be matrix with i.i.d. $\mathcal{N}(0,1/k)$ entries with $k<d$, and let $B=A^{\top}A$. I would like to compute the distribution of $Bx$ where $x\in\mathbb{R}^{d}$ is a fixed vector.
The matrix $B$ essentially projects onto a $k$-dimensional subspace of $\mathbb{R}^{d}$ (it has rank $k$ almost surely). However, computing the distribution explicitly using the expression for entries of $Bx$ is tricky since the entries are dependent. Any hints on how to proceed?
Seems like it would depend a lot on what $x$ exactly is. Consider the case that $k=1$. Then if $x = e_1$ then $Bx$ will depend a lot on just one random gaussian versus the case where $x$ is the all ones vector.
@SandeepSilwal Yes, the distribution of $Bx$ will be parameterized by $x$. The mean $\mathbb{E}[Bx]$ is clearly $x$, but it is not clear what the distribution is.
Just for fun, I made a few simulations for different x's. The first indications are that the solution seems far from easy to obtain. Some marginals of the vector obtained look Gaussian, some other absolutely don't and I did not even start digging into dependence patterns. Is there any setting in which you naturally encounter this question ?
@GillesMordant Thanks for taking the effort to perform a simulation! I was studying random projections and this came to mind. Usually, one would map $x\in\mathbb{R}^d$ to a smaller vector $Ax\in\mathbb{R}^{k}$. But $A^{\top}Ax$ also ``effectively'' does the same thing: its range is a $k$-dimensional subspace of $\mathbb{R}^{d}$. I was curious to see what the points in the range would be distributed like.
It could also be possible that your question can be rewritten as a function of an order two non-decoupled Gaussian chaos, which has been studied in the literature. I however never came across distributional results. I hope this is not a misleading track...
Thanks for the suggestion, I found some results on moments of Gaussian chaos. Computing the distribution seems trickier than I thought..
Are there some other properties you might be interested in, rather than the precise distribution? It might be easier if you just need the moments of the norm, bounds on the tail or so on.
@ThomasDybdahlAhle Yes, tail bounds on $Bx$ would also be useful (for example of the form $\mathbb{P}(|Bx-\mathbb{E}[Bx]|_{2}\ge t)$).
|
2025-03-21T14:48:31.523778
| 2020-07-16T04:38:42 |
365749
|
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"Alex Gavrilov",
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|
Stack Exchange
|
Extending a holomorphic vector bundle: a reference request
Let $Y$ be a complex manifold, $X\subset Y$ a compact submanifold, and $E\to X$ a holomorphic vector bundle. Can $E$ be extended
to a bundle over an open neighborhood of $X$ in $Y$? (Four years ago I have asked this question on MO Extending the tangent bundle of a submanifold for the case $E=T_X$.)
After tinkering with this problem for a while I found a necessary condition, there is an invariant in $H^2(X, \mathcal{N}_{X/Y}^*\otimes End(E))$ which must be zero for an extension to be possible. So far, so good.
Now I think about writing it down and submitting somewhere (assuming it is a new result). But, in a decent paper there are supposed to be references to known results in the same direction, right? And this is what the real problem is: damned if I have a clue where to look! It is all miles away from areas I am familiar with (mostly differential geometry), and this far I could not find anything remotely relevant. So, it would be nice if someone helps me with this.
It seems to me that your result follows by Proposition 1.1 of the paper
P. A. Griffiths: The extension problem in complex analysis. II: Embeddings with positive normal bundle, Am. J. Math. 88, 366-446 (1966). ZBL0147.07502,
that can be freely downloaded here. The statement is the following:
Proposition 1.1 (Griffiths 1966). If $\alpha$ is a holomorphic vector bundle $\mathbf{E} \to X$, then $$\omega(\alpha_{\mu-1}) \in H^2(X,\,\Omega(\mathrm{Hom}(\mathbf{E}, \, \mathbf{E})(\mu))).$$
Here $\omega(\alpha_{\mu-1})$ is the obstruction to extending $\mathbf{E}$ to the $\mu$th infinitesimal neighborhood $X_{\mu}$ of $X$ in $Y$, provided that you already have an extension $\alpha_{\mu-1}$ to $X_{\mu-1}$, and $$H^2(X,\,\Omega(\mathrm{Hom}(\mathbf{E}, \, \mathbf{E})(\mu))= H^2(X, \mathrm{End}(\mathbf{E}) \otimes \mathrm{Sym}^{\mu}(N_{X/Y}^*)).$$
In order to have an extension of $\mathbf{E}$ to a genuine analytic neighborhood of $X$ in $Y$, all these obstruction classes must vanish. In fact, if I understand correctly, you only provided the obstruction class for the extension of $\mathbf{E}$ to the first infinitesimal neighborhood $X \subset X_1$ of $X$ in $Y$.
Thank you, this is exactly what I missed. You are right, the invariant I have found is the first obstruction. The construction I used is more explicit then Griffiths', but it is probably still unpublishable.
In case anyone is interested, I posted this https://arxiv.org/abs/2007.12461 .
|
2025-03-21T14:48:31.523965
| 2020-07-16T04:47:12 |
365750
|
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"url": "https://mathoverflow.net/questions/365750"
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|
Stack Exchange
|
Number of integers in relation to Pythagorean triples with some quadratic relations
Given an integer $m>0$ possibly composite we can find non-negative integers which are not equivalent to $0\bmod m$ with
$$ab+cd\equiv0\bmod m$$.
Is there any integer quadruples bounded in $[0,m-1]^4$ are there that also satisfy
$$bd-ac\equiv0\bmod m$$
$$a^2+d^2=e^2$$
$$b^2+c^2=f^2$$
for some integers $0<e,f\leq\sqrt m$?
If $(a,d,e)$ and $(b,c,f)$ are primitive Pythagorean triples then $a,b$ are of form $2m'$ are $d,c$ are of form $m'^2-1$ and $e,f$ are of form $m
^2+1$.
So there are only two integers $m_1$ and $m_2$ that parameterize $a,d$ and $b,c$ respectively with
$$bd=2m_1(m_2^2-1)$$
$$ac=2m_2(m_1^2-1)$$
$$ab=4m_1m_2$$
$$cd=(m_1^2-1)(m_2^2-1)$$
and we seek
$$4m_1m_2+(m_1^2-1)(m_2^2-1)\equiv0\bmod m$$
$$2m_1(m_2^2-1)-2m_2(m_1^2-1)\equiv0\bmod m.$$
We also have $$\begin{bmatrix}a&d\\d&-a\end{bmatrix}\begin{bmatrix}b\\c\end{bmatrix}=\begin{bmatrix}b&c\\-c&b\end{bmatrix}\begin{bmatrix}a\\d\end{bmatrix}=\begin{bmatrix}\ell m\\\ell'm\end{bmatrix}$$ at some integers with $\ell\ell'\neq0$.
|
2025-03-21T14:48:31.524052
| 2020-07-16T06:27:35 |
365751
|
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|
Stack Exchange
|
mean curvature for codimension $>1$?
The mean curvature of a hypersurface in a Riemannian manifold is defined to be the trace of the second fundamental form. I was curious, does the notion of mean curvature generalise to higher codimension objects, if so how is this defined? If not, how come this notion does not make much sense in higher codimensions? Any reference or explanation is appreciated, thank you.
It is defined as the trace of the second fundamental form, but the second fundamental form is normal bundle valued. And usually you scale by the inverse of the dimension of the submanifold.
Thank you. do you happen to have reference by any chance?
Chapter 5.C in the Riemannian geometry book of Gallot-Hulin-Lafontaine.
|
2025-03-21T14:48:31.524131
| 2020-07-16T07:12:33 |
365754
|
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"Krishnarjun",
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"Peter Humphries",
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"https://mathoverflow.net/users/480516",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365754"
}
|
Stack Exchange
|
On the Chowla and twin prime conjectures
I'm reading https://terrytao.wordpress.com/tag/chowlas-conjecture/ and at some point it is mentioned that, the twin prime conjecture is a variant of Chowla's conjecture that $\sum_{n\leq x} \lambda(n)\lambda(n+2) =o(x)$, where $\lambda$ denotes the Liouville function. Does anyone have a reference/proof of this result ? Also, what would be de Polignac's conjecture in terms of $\lambda$ ?
This is not known to be true. Tao claims this is a "variant" of the twin prime conjecture on the blog post you reference. It's possible that if one had a sufficiently strong quantitative version of the claim one could deduce the twin prime conjecture, but this certainly isn't known from a $o(x)$ bound.
@MarkLewko, okay, thanks. So, do you have any reference/proof that ''a sufficiently strong quantitative verson of the claim entails the twin prime conj'' ? Also, how strong should be the quantitative version ?
The variant is replacing $\lambda(n)$ with $\Lambda(n)$, where $\Lambda(n)$ denotes the von Mangoldt function, and extracting a main term.
You can find some details here https://terrytao.wordpress.com/2011/11/21/the-bourgain-sarnak-ziegler-orthogonality-criterion/
|
2025-03-21T14:48:31.524360
| 2020-07-16T07:48:59 |
365756
|
{
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"authors": [
"Haim",
"Mohammad Golshani",
"Timothy Chow",
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|
Stack Exchange
|
Applications of Set theory vs. model theory in mathematics
I have a question that has occupied my mind for some time.
Let's first consider applications of set theory and model theory in mathematics.
Major applications of set theory are in topology, Banach spaces, $C^*$-algebras, functional analysis, dynamical systems, algebra,....
On the other hand, major applications of model theory are in number theory, arithmetic and algebraic geometry, algebra,...
What is now interesting to me is that when set theory is applied at some part, model theory has few to say about that part, and vice versa, when model theory is applied heavily in some part of mathematics, then set theory appears rare there.
Even though both fields have applications in algebra, when we restrict ourselves to sub-branches of algebra, again the above obsession is true.
I am wondering if my observation is correct and if so is there any reason for it? Or should we just wait for more convergence in the future?
Remark 1. On the other hand surprisingly there is deep connections between set theory and model theory.
Remark 2. One answer to the question might be, set theory in the sense of model theory is wild (and not tame), but this is not quite convincing to me.
Related to your second remark: I would say model theory is mostly about first order statements, whereas non elementary set theory deals with higher order statements.
I'm not sure that your observation is correct. For example, in addition to the applications of set theory to $C^*$-algebras, there are also many applications of model theory to the field: https://arxiv.org/pdf/1602.08072.pdf
@Haim Yes I know that, even for example I can say Shelah's proof of independence of Ax-Kochen isomorphism theorem from CH vs. the model theoretic works on the subject. But if for example you look at the paper you sent, I think you agree that the subjects of the applications are different from what set theory has, also I think the applications stated are not as deep as those given by set theory.
I am essentially looking for a part of mathematics outside of logic (say not a theory), where both set theory and model theory have deep applications.
@MohammadGolshani What would qualify as a "deep application"?
Dynamics is another example of an area where both set theory and model theory have many applications (again, I don't know which of them satisfy your criterion for being deep), one recent example is this paper by Hrushovski, that makes an interesting connection to the Kechris-Pestov-Todorcevic correspondence: https://arxiv.org/pdf/1911.01129.pdf
At the risk of stating the obvious: Sets have no structure. Models have structure. It shouldn't be too surprising that they would have applications to different kinds of questions, should it?
@TimothyChow It seems to me like an almost empirical fact that in order for a non-set theoretic statement to be susceptible to set theoretic methods, it should either be a statement involving subsets of Polish spaces with a very simple (Borel/analytic) definition (in which case methods from DST can be used) or it shouldn't be absolute (so forcing, large cardinals and inner models might be useful). It seems that non-set theoretic statements that satisfy either of the above requirements are harder to come by, whereas such requirements don't exist in the case of model theoretic applications.
|
2025-03-21T14:48:31.524598
| 2020-07-16T08:03:46 |
365757
|
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"url": "https://mathoverflow.net/questions/365757"
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|
Stack Exchange
|
Let $R$ be a local ring where 2 is invertible. Must there exist a faithfully flat $R$-algebra where the squaring map is surjective?
Let $R$ be a local ring where 2 is invertible. Must there exist a faithfully flat $R$-algebra where the squaring map $x\mapsto x^2$ is surjective?
This is certainly true for fields. For DVR's, you can take the strict henselization, and then take the colimit over all extensions taking square roots of the uniformizer.
For a general local ring, I'm a bit lost. Presumably you would start with the strict henselization, then continuously take square roots of stuff in the maximal ideal, though it's unclear to me that the resulting colimit would even range over a set, let alone if the resulting ring is faithfully flat over $R$.
If $R$ is a local Noetherian ring of characteristic $2$, then adding all square roots amounts to passing to the perfection (direct limit over $x \mapsto x^2$) which is flat if and only if $R$ is regular.
Now let $R$ be a local excellent domain of positive characteristic. Then you could ask for the (presumably stronger) condition that $R \to R^{+}$ is flat, where $R^{+}$ is the absolute integral closure (the integral closure of $R$ in an algebraic closure of its field of fractions). Theorem 4.13.(2) of https://arxiv.org/pdf/1803.03229.pdf says that this happens if and only if $R$ is regular.
The following works over any ring $R$: Take a family $\underline{X}:=(X_a)_{a\in R}$ of indeterminates indexed by $R$, and put $R_1:=R[\underline{X}]/I$ where $I$ is generated by $(X_a^2-a)_{a\in R}$. Then $R_1$ is free as an $R$-module (you can view it as $\bigotimes_{a\in R}R[X_a]/(X_a^2-a)$) and every element of $R$ becomes a square in $R_1$. Now just iterate the process to get $R\subset R_1\subset R_2\subset\dots$ and put $R_\infty=\varinjlim_n R_n$. This $R_\infty$ solves the problem.
Okay, I can see that $R_1$ is faithfully flat, since $R_1$ can be viewed as a colimit of $R_S$, where $S\subset R$ is a finite subset, and $R_S := R[(X_s)_{s\in S}]/(X_s^2-s)$. This is a directed colimit of faithfully flat $R$-algebras, and hence is also faithfully flat by https://stacks.math.columbia.edu/tag/090N. Though, out of curiosity, how are you seeing that $R_1$ is free? How do you define the infinite tensor product? Is a directed colimit of free modules free?
An infinite tensor product (of algebras, not modules!) can be defined as a coproduct, and constructed as a colimit of finite tensor products. If $(A_i)$ is a family of $R$-algebras, then $A=\bigotimes_i A_i$ is generated as $R$-module by tensors $\otimes_i x_i$ with finite support ($x_i=1$ for almost all $i$). If $B_i$ is an $R$-basis of $A_i$ for each $i$, it is easy to check that taking $x_i$ in $B_i$ gives rise to a basis of $A$. In the above example, a basis of $R_1$ is given by the finite products $\prod_{a\in S}x_a$ where $S$ runs through finite subsets of $R$ and $x_a=$ class of $X_a$.
For the last question, the $\mathbb{Z}$-module $\mathbb{Q}$ is the colimit of the free modules $\frac{1}{n}\mathbb{Z}$ ($n\geq1$) but is not free. In fact, every flat modules is a directed colimit of free modules (Lazard's theorem).
|
2025-03-21T14:48:31.524824
| 2020-07-16T09:25:15 |
365760
|
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|
Stack Exchange
|
"Natural" embeddings of free groups in general linear groups
There are many embeddings of the free groups $\mathbb{F}_n$ into $GL_m(\mathbb{Z})$. An often encountered example for $\mathbb{F}_2$ is the Sanov subgroup generated by
$$\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix}
\;\text{ and }
\;
\begin{bmatrix}
1 & 0\\
2 & 1
\end{bmatrix}.$$
But an example that one could consider particularly "natural" is the commutator subgroup generated by
$$\begin{bmatrix}
1 & -1\\
-1 & 2
\end{bmatrix}
\;\text{ and }
\;
\begin{bmatrix}
1 & 1\\
1 & 2
\end{bmatrix},$$
since it follows from a group theoretic construction.
Hence my question: what would be a natural choice of embedding for $\mathbb{F}_n$ when $n>2$?
If $\langle x,y\rangle$ is a "natural" embedding of $F_2$, then $\langle x^iyx^{-i}: 0\le i<n\rangle$ is a "natural" embedding of $F_n$.
I agree, this seems reasonably natural.
Thank you.
|
2025-03-21T14:48:31.524923
| 2020-07-16T09:25:41 |
365761
|
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|
Stack Exchange
|
An abstract characterization of line integrals
Let $M$ be a smooth manifold (endowed with a Riemann structure, if useful). If $\omega \in \Omega^1 (M)$ is a smooth $1$-form and $c : [0,1] \to M$ is a smooth curve, one defines the line integral of $\omega$ along $c$ as
$$I(\omega, c) = \int _0 ^1 \omega_{c(t)} (\dot c (t)) \ \mathrm d t \ .$$
This is clear, but this is just a formula that does not give any insight into the innards of the concept.
Is is possible to define the concept of line integral by purely abstract properties?
To give two analogies, the algebraic tensor product is defined by some universal property, and then it is shown that it exists and is essentially unique. Similarly, the Haar measure on locally-compact groups is defined as a regular, positive measure, that is invariant under (left) translations, and then is shown to exist and be essentially unique. Do you know of any similar approach for line integrals?
To clarify: if $\mathcal C$ is the space of smooth curves in $M$, I am attempting to understand line integration as a map $I : \Omega ^1 (M) \times \mathcal C \to \mathbb R$ uniquely determined by some properties: what are these properties? For sure, linearity in the first argument is among them. What else is needed?
Are you asking for something like the following: Let $\Omega$ be the set of maps $I: C([0,1],M)$ satisfying an appropriate list of universal properties. Then, for each $I \in \Omega$, there exists a $1$-form $\omega$ such that, for each continuous $C: [0,1] \rightarrow M$, $I(C)$ is the line integral of $\omega$ along $C$.
I don't think that I want "there exists a $1$-form $\omega$"; I believe that I want "for all $1$-forms $\omega$" instead. Alexander Betts' answer below seems extremely close to what I had in mind, but I'll leave the question open for several more days, to give it a chance to collect as many useful answers as possible.
My point is that I believe you can define the universal properties so that you do not even need to assume that what you are integrating $1$-forms along each curve $C$. The fact that the integrand is a $1$-form is already implied by the universal properties, if properly chosen. I think what I am proposing is even more general than what you are asking for.
Since I do not fully get your ideas, I have edited my question to hopefully make clearer what I am looking for.
I'll suggest here another possible characterisation, expanding on a suggestion of the OP in one of the comments. Again, this is an assertion that certain known properties of line integration characterise it uniquely; this doesn't provide a "new" construction of line integration. Unlike my previous answer, here all the action takes place on the one manifold $M$.
To avoid various technicalities, I'm going to redefine $\mathcal C_M$ to be the set of immersed paths, i.e. smooth paths $c\colon[0,1]\to M$ such that $\dot c(t)\neq0$ for all $t\in[0,1]$. I think this restriction could probably be removed with enough effort.
Theorem:
For any manifold $M$, line integration is the unique function $I\colon\Omega^1(M)\times\mathcal C_M\to\mathbb R$ satisfying the following properties:
(additivity in the path) Suppose that $c_1$ and $c_2$ are two immersed paths that are composable, i.e. all the derivatives $c_1^{(i)}(1)=c_2^{(i)}(0)$. Then $I(\omega,c_1c_2)=I(\omega,c_1)+I(\omega,c_2)$ for all $\omega\in\Omega^1(M)$. Here $c_1c_2$ denotes the composite path, defined by$$c_1c_2(t)=\begin{cases}c_1(2t)&0\leq t\leq1/2\\c_2(2t-1)&1/2\leq t\leq1.\end{cases}$$
(additivity in the $1$-form) We have $I(\omega_1+\omega_2,c)=I(\omega_1,c)+I(\omega_2,c)$ for all $\omega_1,\omega_2\in\Omega^1(M)$ and all $c\in\mathcal C_M$.
(locality) If $\omega\in\Omega^1(M)$ satisfies $\omega_{c(t)}(\dot c(t))=0$ for all $t\in[0,1]$, then $I(\omega,c)=0$.
(exact forms) If $f\colon M\to\mathbb R$ is smooth, then $I(\mathrm df,c)=f(c(1))-f(c(0))$.
The proof uses two lemmas.
Lemma 1: Let $c$ be an immersed path. Then there is a non-negative integer $N$ such that for all $0\leq k<2^N$, the restriction $c|_{[2^{-N}k,2^{-N}(k+1)]}$ of $c$ to the interval $[2^{-N}k,2^{-N}(k+1)]$ is an embedding.
Proof (outline): This follows from the standard fact that an immersion is locally an embedding (see e.g. this MO question), and that $[0,1]$ is compact.
Lemma 2: Let $c$ be an embedded path in $M$ and $\omega\in\Omega^1(M)$. Then there exists a smooth function $f\colon M\to\mathbb R$ such that $\omega_{c(t)}(\dot c(t))=\mathrm df_{c(t)}(\dot c(t))$ for all $0<t<1$.
Proof: The pullback $c^*\omega$ is a smooth $1$-form on $[0,1]$, hence is $\mathrm df_0$ for some smooth $f_0\colon[0,1]\to\mathbb R$. We want to show that $f_0$ extends to a smooth map $f\colon M\to\mathbb R$ (i.e. $f_0=f\circ c$).
To do this, we first extend $c$ to a smooth map $c\colon(-\epsilon,1+\epsilon)\to M$ for some $\epsilon>0$. This is possible by Borel's Lemma, which says that we can choose smooth maps $(-\epsilon,0]\to M$ and $[1,1+\epsilon)\to M$ having the same higher-order derivatives at $0$ and $1$ as $c$, respectively.
Decreasing $\epsilon$ if necessary, we may even assume that $c\colon(-\epsilon,1+\epsilon)\hookrightarrow M$ is an embedding. The tubular neighbourhood theorem then implies that the embedding $c$ extends to an embedding $\tilde c\colon(-\epsilon,1+\epsilon)\times (-1,1)^{d-1}\hookrightarrow M$, where $d=\dim(M)$. In other words, we have $c(t)=\tilde c(t,0,\dots,0)$ for all $t$.
We now extend $f_0$ as follows. By Borel's Lemma again, we may extend $f_0$ to a smooth function $f_0\colon(-\epsilon,1+\epsilon)\to\mathbb R$, and then extend this again to a smooth function $f_0\colon(-\epsilon,1+\epsilon)\times(-1,1)^{d-1}\to\mathbb R$. Multiplying by an appropriate bump function if necessary, we may assume that $f_0$ vanishes outside $(-\frac12\epsilon,1+\frac12\epsilon)\times(-\frac12,\frac12)^{d-1}$.
We've now constructed an extension $f=f_0\circ\tilde c^{-1}$ of $f$ on the open neighbourhood $\mathrm{im}(\tilde c)$ of the image of $c$. Moreover, we've ensured that this extension has compact support (it vanishes outside a compact subspace), so we can extend $f$ to all of $M$ by specifying that it is $0$ outside $\mathrm{im}(\tilde c)$. This yields the desired $f$. This proves Lemma 2.
Proof of Theorem: We show unicity. Let $I$ and $I'$ be two functions $\Omega^1(M)\times\mathcal C_M\to\mathbb R$ which satisfy the given conditions. We need to show that $I(\omega,c)=I'(\omega,c)$ for all $\omega\in\Omega^1(M)$ and all immersed paths $c$.
To do this, suppose first that $c$ is embedded. By Lemma 2 we can choose a smooth map $f\colon M\to\mathbb R$ such that $\omega_{c(t)}(\dot c(t))=\mathrm df_{c(t)}(\dot c(t))$ for all $c\in[0,1]$. Using additivity in the $1$-form, locality and the condition about exact forms, we find that $I(\omega,c)=I(\mathrm df,c)=f(1)-f(0)$. Since the exact same argument applies to $I'$, we have $I(\omega,c)=I'(\omega,c)$.
Now let us deal with the general case. By Lemma 1 we can choose a non-negative integer $N$ such that $c|_{[2^{-N}k,2^{-N}(k+1)]}$ is an embedded path for all $0\leq k<2^N$. A repeated application of the additivity property implies that $I(\omega,c)=\sum_{k=0}^{2^N-1}I(\omega,c|_{[2^{-N}k,2^{-N}(k+1)]})$ and similarly for $I'$. Since we already know that $I$ and $I'$ agree on embedded paths, we obtain that $I(\omega,c)=I'(\omega,c)$, as desired. This concludes the proof.
Remark:
If one only cares about the integrals of closed 1-forms, then this whole setup can be significantly simplified: one doesn't need to restrict to immersed paths, and one can replace the locality condition above with the more natural condition:
(locality') If $\omega$ vanishes on an open neighbourhood of the image of $c$, then $I(\omega,c)=0$.
Give me some time, please, to digest it! It seems that you get linearity ($I(a \omega, c) = a I(\omega, c)$) for free, which I find strange.
I don't know if it's exactly what you're looking for, but line integration is the unique way to assign a real number $I(\omega,c)\in\mathbb{R}$ to every pair of a smooth $1$-form $\omega$ on a smooth manifold $M$ with boundary and smooth path $c\colon[0,1]\to M$ such that:
(adjunction) if $f\colon M\to N$ is a smooth map of smooth manifolds with boundary, $\omega$ is a smooth $1$-form on $N$, and $c\colon[0,1]\to M$ is a smooth path in $M$, then$$I(f^*\omega,c)=I(\omega,f\circ c)$$
(normalisation) if $M=[0,1]$, $\mathbf 1\colon[0,1]\rightarrow[0,1]$ is the identity path, and $\omega=g(x)\mathrm{d}x$ for a smooth function $g$, then $I(\omega,\mathbf 1)=\int_0^1g(x)\mathrm{d}x$, where the integral denotes the usual Riemann line integral.
(To show this characterises line integration uniquely, apply the adjunction formula to the smooth path $c\colon[0,1]\to M$ to show that $I(\omega,c)=I(c^*\omega,\mathbf 1)=\int_0^1c^*\omega$.)
Remark:
This is the approach that one takes when defining iterated integration of a sequence $\omega_1,\dots,\omega_n$ of $1$-forms along a path $c$. For instance, we know how to double-integrate over the interval $[0,1]$: the double-integral of $g(x)\mathrm dx$ and $h(x)\mathrm dx$ is $\int_0^1\left(\int_0^xg(y)\mathrm dy\right)h(x)\mathrm dx$, and by demanding the same adjunction relation you get a way to define a double line integral $I(\omega_1\omega_2,c)$ for all pairs of smooth $1$-forms $\omega_1,\omega_2$ on a manifold $M$ with boundary, and all smooth paths $c\colon[0,1]\to M$. For more details, see the works of Kuo-Tsai Chen, who was the first to develop this theory systematically
This seems very close (at least in spirit) to what I had in mind. What I do not understand, though, is the domain of definition of $I$. If $\mathcal M$ is the class of smooth manifolds with boundary, it seems that the domain of $I$ is the disjoint union of $\Omega^1(M) \times \mathcal C _M$, where $M \in \mathcal M$ and $\mathcal C _M$ is the set of smooth curves $: [0,1] \to M$. Weird, though...
You're right that the domain of $I$ is this funny disjoint union, but perhaps this isn't such a strange domain to get. We probably want to think of $I$ really as a collection of functions $I_M\colon\Omega^1(M)\times\mathcal C_M\to\mathbb R$, one for each $M\in\mathcal M$, and we're saying that this whole collection is uniquely determined by some compatibility condition with respect to smooth maps, plus a normalisation condition when $M=[0,1]$.
(ctd) These sorts of structures are actually quite common: if we view $I_M$ as a map $\mathcal C_M\to\mathrm{Hom}(\Omega^1(M),\mathbb R)$, then this "compatibility with smooth maps" condition is just saying that these $I_M$ are the components of a natural transformation (in the sense of category theory). And natural transformations crop up all over the place.
I think that I can come up with version better suited to my needs, that does away with the disjoint union: property (1) should hold for smooth maps $f : M \to M$ (yes, not $N$), and property (2) should be replaced by the condition that $I(\mathrm d \varphi, c) = \varphi(c(1)) - \varphi(c(0))$ for every smooth $\varphi : M \to \mathbb R$. This means that we may work with a single, fixed manifold $M$. Also, one may take it now to be boundaryless (which is what I need). (The boundary was needed by the case $M = [0,1]$ which becomes irrelevant in this approach.)
Glad to hear it! Though I would be a little surprised if those two conditions uniquely determine $I$ in general. Your second condition only pins down line integrals of exact $1$-forms; it's unclear how this helps calculate integrals of inexact forms (non-closed forms seem especially difficult).
Oh, I thought that "normalization" meant fixing the multiplicative constant that cannot be fixed by condition (1) alone. I shall have a look at Chen's approach and see how he conceives it. Can you please help me with a reference containing the proof of what is in your answer (by Chen or anybody else)? Thank you.
The pullback of a one form to an interval is always exact, and by property 1 it suffices to know how to integrate over the interval, so the fundamental theorem (condition 2 of Alex M.) should suffice to characterize it.
@AlexM. why do you not like condition 1 in the original form? A manifold without boundary is a manifold with empty boundary (the class of manifolds with boundary contains the manifolds without boundary)
Does this work for higher forms and higher-dimensional submanifolds $c$?
After my initial enthusiasm, I am not so sure that I get you anymore: it seems that all you are saying is that line integrals are uniquely defined by the requirement that $\int _M \omega = \int _0 ^1 (c^* \omega) (t) \ \mathrm d t$ (the latter being a Riemann integral) - a thing which I agree with, but which is just a trivial restatement of the defining formula in my question. In particular, functoriality seems to be nowhere used in full generality, since we always only pull forms from $M$ back to $[0,1]$, no other manifolds are involved.
I guess the point I was trying to make is that two conditions that one should expect line-integration to satisfy turn out to determine it completely (admittedly, via a completely trivial argument). I think if one cares about defining line-integration on all $M$ simultaneously (which perhaps isn't the setup you're interested in), one can't help but notice that any "sensible" theory of line-integration is determined by what happens when $M=[0,1]$.
|
2025-03-21T14:48:31.525710
| 2020-07-16T09:40:42 |
365763
|
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|
Stack Exchange
|
Removing integral from norm by inequality
My first question on Math Overflow.
For my Mathematics Bachelor thesis I am looking at a paper called "Deep Limits of Residual Neural networks" by Matthew Thorpe and Yves van Gennip. (arxiv.org link)
In Lemma 4.6, a preliminary result to show that something Gamma-converges, they first show that a discretized ODE uniformly converges to the ODE itself.
Some info you need:
$X: [0,1] \to \mathbb{R}^d,$
$\sigma$ is Lipschitz continuous with constant $L_{\sigma}$,
$\dot{X}(t) = \sigma(K(t)X(t)+b(t)), \quad t\in[0,1], K: [0,1]\to \mathbb{R}^{d\times d}, b: [0,1] \to \mathbb{R}^d$.
The inequality that I cannot confirm:
$$
\sup_{s\in[0,t]} || \int_0^s \dot{X}(r)dr + x|| \leq \sup_{s\in[0,t]} L_\sigma s ||X||_{L^{\infty}([0,s])}||K||_{L^{\infty}} + L_{\sigma}||b||_{L^{\infty}} + ||x||.
$$
Now I expected $||b||_{L^{\infty}}$ to also be scalar multiplied by $s$. Here are my steps
\begin{align*}
\sup_{s\in[0,t]} || \int_0^s \dot{X}(r)dr + x|| & \leq \sup_{s\in[0,t]} || \int_0^s \sigma(K(r)X(r)+b(r)) dr|| + ||x|| \\
& \leq \sup_{s\in[0,t]} L_{\sigma}|| \int_0^s K(r)X(r)dr + \int_0^s b(r) dr|| + ||x||\\
& \leq \sup_{s\in[0,t]} L_{\sigma}\Big( \int_0^s ||K(r)X(r)||dr + \int_0^s ||b(r)|| dr\big) + ||x||\\
& \leq \sup_{s\in[0,t]} L_{\sigma}\Big( ||KX||_{L^{\infty}} \int_0^s dr + ||b||_{L^{\infty}}\int_0^s dr\Big) + ||x||\\
& \leq \sup_{s\in[0,t]} L_{\sigma}\Big( ||K||_{L^\infty}||X||_{L^{\infty}}\cdot s + ||b||_{L^{\infty}}\cdot s\Big) + ||x||.
\end{align*}
So I think the $s$ in front of $||K|| ||X||$ comes from integrating over $\mathcal{1}$, and therefore the same thing should happen for $b$.
Also: In the inequality in the paper they use the $L^{\infty}([0,s])$ norm for $X$ and the $L^{\infty}$ norm for $K$ and $b$ on the right side, but I cannot explain this. Could this have something to do with the missing $s$?
If I understand correctly, $0 \le s \le1$. If you estimate $s$ by $1$ in your chain, then you get the bound you are looking for.
@GiorgioMetafune So you're saying $t\leq 1$?
@GiorgioMetafune Later on they choose $t=\frac{1}{2L_{\sigma}||K||{L^{\infty}}}$ which is not necessarily $\leq 1$ right? Even later they switch from $||X||{L^{\infty}([0,t])} \leq 2L_\sigma ||b||{L^\infty} + 2||x||$ to $||X||{L^\infty ([0,1])} \leq 2(2^N - 1)L_{\sigma} ||b||_{L^\infty} + 2^N||x||$ with $N=\left \lceil{\frac{1}{t}}\right \rceil $. Maybe if I understand how they switch there, I see why $t\leq 1$?
|
2025-03-21T14:48:31.525877
| 2020-07-16T09:43:25 |
365764
|
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|
Stack Exchange
|
Roots of determinant of matrix with polynomial entries
Let $p_1, p_2,\dots, p_n$ and $q_1,q_2,\dots,q_n$ be a collection of complex polynomials. Let $A$ be a $n \times n$ matrix satisfying
$$a_{ij} = \begin{cases} p_i(x) & \text{ if } i = j, \\ q_i(x) & \text{ otherwise} \end{cases} .$$
is there any connection between the roots of the polynomials $p_i$'s and $q_i$'s and the roots of the polynomial $\det A$? if not, is this true under at least under any special assumptions?
Kindly share some references.
Thank you.
If $r_i(x)=p_i(x)-q_i(x)$ then the determinant is $$\left(1+\sum_{i=1}^n \frac{q_i(x)}{r_i(x)}\right)\prod_{i=1}^n r_i(x).$$
@BrendanMcKay Thank you very much.
Let $r_i := p_i - q_i$.
$${\bf A} (x) := \begin{bmatrix} p_1 (x) & q_1 (x) & \ldots & q_1 (x)\\ q_2 (x) & p_2 (x) & \ldots & q_2 (x)\\ \vdots & \vdots & \ddots & \vdots\\ q_n (x) & q_n (x) & \ldots & p_n (x)\end{bmatrix} = \mbox{diag} \left( {\bf r} (x) \right) + {\bf q} (x) {\Bbb 1}_n^\top$$
Using the matrix determinant lemma,
$$\det \left ( {\bf A} (x) \right) = \det \left( \mbox{diag} \left( {\bf r} (x) \right) \right) \left( 1 + {\Bbb 1}_n^\top \mbox{diag}^{-1} \left( {\bf r} (x) \right) {\bf q} (x) \right) = \color{blue}{\left( 1 + \sum_{i=1}^n \frac{q_i(x)}{r_i(x)} \right)\displaystyle\prod_{i=1}^n r_i (x)}$$
as mentioned by Brendan McKay some 20 minutes ago.
Thank you. The reference is very helpful.
is it possible to do this, if I assume some entries if this matrix are zero other entries are as it is? Thank you.
@GA316 The matrix determinant lemma is used for rank-$1$ updates of invertible matrices. I assume it can be generalized to non-invertible matrices. What kind of matrices do you have in mind?
Consider your matrix $A(x)$. Now, I make some entries zero randomly such a way the if $a_{ij} =0$ then $a_{ji} = 0$. Remaining entries of $A(x)$ are unchanged. Thank you.
@GA316 Once you destroy the "invertible matrix + rank-1 matrix" structure, you have to use another approach.
hmm. ok. Thank you. Do you know any different approach? kindly share some references. Thank you.
@GA316 The less structured the matrix, the more general the tool. One can always use Laplace expansion, though I avoid it like the plague.
ok. Thanks a lot.
|
2025-03-21T14:48:31.526057
| 2020-07-16T09:49:53 |
365766
|
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|
Stack Exchange
|
Examples of amenable Banach algebras which have non-amenable subalgebra
I am looking for examples of amenable Banach algebras which have non-amenable subalgebra
I know
1: Each amenable Banach algebra has a bounded approximate identity
2: If $I$ be a closed ideal in an amenable Banach algebra, then
$I$ amenable if and only if $I$ has a bounded approximate identity
Does anybody have an example?
Thank you for any suggestions!
Mateusz's answer mentions lots of good mathematics but I feel obliged to point out that the fundamental example which answers your original question in the negative is $M_2({\bf C})$. (Banach algebras behave very differently from ${\rm C}^*$-algebras and $L^1$-group algebras.)
The point is that the algebra
$$
{\bf C}[x] / (x^2) \cong
\left\{
\begin{pmatrix} a & b \cr 0 & a \end{pmatrix}
\colon a,b \in {\bf C} \right\}
\subset M_2({\bf C})
$$
is not amenable, and should not be "generalized amenable" in any sensible version of "generalized amenability".
Non-amenability can be seen in many ways, but the most direct one -- if you use the definition of amenability in terms of derivations -- is to note that the mapping
$$
\begin{pmatrix} a & b \cr 0 & a \end{pmatrix} \to b
$$
is a non-zero derivation. In some sense this is the philosophical idea behind derivations on associative algebras, they are the 1st-order terms that arise when one perturbs a homomorphism.
This is a better example. I guess I'm too attached to $\ast$-stuff.
@MateuszWasilewski In general it does seem that $\ast$-stuff is the way to go :)
@Yemon Choi, thank you this is very helpful information
There are plenty of examples, even for $C^{\ast}$-algebras. By the results of Connes and Haagerup, for $C^{\ast}$-algebras amenability is equivalent to nuclearity, so I will work with nuclearity, which is more familiar for operator algebraists.
For group $C^{\ast}$-algebras of discrete groups nuclearity is equivalent to amenability of the group. So, for example, the group $C^{\ast}$-algebra of the free group $F_n$ is not nuclear. On the other hand, it can be embedded in a nuclear $C^{\ast}$-algebra, as I will show below.
Whenever we have an action of a group $\Gamma$ on a compact space $X$, we can construct the (reduced) crossed product $C(X) \rtimes \Gamma$, which contains the group $C^{\ast}$-algebra $C^{\ast}_{r}(\Gamma)$ as a subalgebra. We now need an example of an action of a free group such that the crossed product is nuclear. We can use, for example, the action of the free group on its boundary. You can visualise the free group as a tree, using its Cayley graph, and the boundary in this case will be the boundary of this tree, i.e. space of infinite paths up to natural equivalence. This action is amenable, which is exactly a property needed for proving nuclearity of the crossed product. To sum up, the inclusion $C_{r}^{\ast}(F_n) \subset C(\partial F_n) \rtimes F_n$ gives you an example of a nonamenable subalgebra of an amenable algebra.
Mateusz Wasilewski,Thank you. This is very helpful.
A uniform algebra is a closed subalgebra of $C(K)$ for some compact Hausdorff space $K$. Uniform algebras consisting of differentiable functions possess nontrivial derivations (namely, the derivative). Examples include the disk algebra $A(D)$, and the Hardy space $H^{\infty}(D)$.
|
2025-03-21T14:48:31.526398
| 2020-07-16T10:35:17 |
365771
|
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|
Stack Exchange
|
Projection from closure of locally closed subscheme is Etale
Let $S$ an arbitrary scheme and denote by $\Delta: S \to S \times S$ the diagonal immersion and $p_i: S \times S \to S$ the both projections to first resp second factor. (in following we will wlog talk only the projection on first factor, ie we set $p:=p_1$)
In general when $S$ isn't separated the image $\Delta(S) \subset S$ is only locally closed and not closed in $S \times S$. Since the restricted projection $p \vert _{\Delta(S)}: \Delta(S) \to S$ is an isomorphism that's not interesting to study this map. But the extended restriction
$$p \vert _{\overline{\Delta(U)}}:\overline{\Delta(U)} \subset U \times U \to U$$ looks more interessant. Which scheme theoretic map properties $\mathcal{Q}$ does $p \vert _{\overline{\Delta(U)}}$ inherit from $p \vert _{\Delta(S)}$?
(ie $\mathcal{Q}$ stands for example beeing etale, smooth, flat etc)
Indeed the main motivation behind this question is the question if $p \vert _{\overline{\Delta(U)}}$ is always etale? (see This closely related question & discussion). Alexl gave the example where $S$ is the affine line with doubled origin and in this case it's quite obvious that $p \vert _{\overline{\Delta(U)}}$ is etale as well. But does this example cover all phenomena can occure by passing from $\Delta(S)$ to $\overline{\Delta(S)}$ as subscheme of
$S \times S$?
More generally assume that $f: X \to Y$ a morphism of schemes and $S$ is locally closed subscheme of $X$ and $\overline{S}$ it's schematic closure in $X$. Assume the restriction $p \vert _S:S \to Y$ is etale (or has another fancy morphism property $\mathcal{Q}$ but I'm primary interested on etaleness as archetypical property), does the "extended" restriction $p \vert _{\overline{S}}:\overline{S} \to Y$ has also property $\mathcal{Q}$?
Is there any criterion / theory treating this problem and probably gives an answer for which morphism properties $\mathcal{Q}$ the above behaves well?
Take a degree $d \geq 3$ covering of smooth projective curves $f \colon X \to Y$, branched at a single point $y \in Y$ and such that $$f^{*}(y)=kx_0 + x_1 + \ldots + x_{d-k},$$ with $k <d$. Then the restriction $$f \colon X-{x_0 } \to Y$$ is surjective and unramified, hence étale, but it extension to $X$ clearly is not.
So the attempt to generalize the issue was too optimistic. But in more concrete situation when $f$ is the diagonal map $\Delta: S \to S \times S$. Why is the restriction $p_1 \vert _{\overline{\Delta(S)}}:\overline{\Delta(S)} \subset S \times S \to S$ of the projection $p_1: S \times S \to S$ nevertheless etale? The issue that I wanted to understand here https://mathoverflow.net/questions/365735/affine-scheme-as-algebraic-space that a usual scheme $S$ is an algebraic space. So we need data $(U,R)$ with $U:=S$ and $R \subset U \times U$ closed (!).
The natural choice for $R$ is $R:=\overline{\Delta(U)}$ since if $S$ is not separated, the image $\Delta(U)$ isn't closed. That is we need etaleness of projection maps $p_i: \overline{\Delta(U)} \to U$. Do you know an argument why that's true?
|
2025-03-21T14:48:31.526628
| 2020-07-16T11:02:00 |
365773
|
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|
Stack Exchange
|
Rigid analytic varieties vs rigid spaces
In rigid analytic geometry, some sources refer to "rigid spaces", where others refer to "rigid analytic varieties". Do these two terms stand for the same thing, or is there a difference between them?
More details:
Comparing the definition of a rigid analytic variety in [BGR, Definition 9.3.1/4] with the definition of a rigid space in [FvdP, Definition 4.3.1] (for the strong topology), there seem to be two differences in the definitions:
A rigid space is a type of G-ringed space, whereas a rigid analytic variety is a type of locally G-ringed space.
The G-topology on a rigid analytic variety $X$ is required to satisfy condition (G2), namely that if $U\subseteq X$ is admissible open and $\mathfrak U=(U_i)_{i\in I}$ is a covering of $U$ by admissible opens, then $\mathfrak U$ is an admissible covering if and only if it has an admissible refinement. There is no such requirement in the definition of a rigid space in [FvdP].
Of these differences, I don't think the first causes any problems: rigid spaces are automatically locally G-ringed since affinoids are.
The second difference seems like it could be more problematic. If we take $X$ an affinoid space, then I think we can find a G-subtopology $T$ of the strong topology for which $(X,T,\mathcal O_X|_T)$ satisfies all the conditions of a rigid analytic variety except (G2). I'd suggest something like taking an strong-admissible covering $(X_i)_{i\in I}$, declaring the $T$-admissible sets in $X$ to be the strong-admissible ones, and declaring the $T$-admissible coverings of a $T$-admissible $U\subseteq X$ to be the trivial covering $\{U\}$ together with all strong-admissible coverings which refine the covering $(U\cap X_i)_{i\in I}$. This seems to satisfy all the properties required of a rigid space (in particular $(X_i)_{i\in I}$ is a $T$-covering of $X$ by affinoids equipped with the strong topology), but does not satisfy (G2) in general.
I'm tempted to suggest based on this that the definitions of a rigid analytic variety in [BGR] and a rigid space in [FvdP] should be the same, but that the definition in [FvdP] accidentally omits the condition (G2). Is this a reasonable conclusion?
References
[BGR] Siegfried Bosch, Ulrich Güntzer and Reinhold Remmert, Non-Archimedean Analysis. A Series of Comprehensive Studies in Mathematics, volume 261. Springer 1984.
[FvdP] Jean Fresnel and Marius van der Put, Rigid Analytic Geometry and Its Applications. Progress in Mathematics, volume 218. Birkhäuser 2004.
I am pretty sure that you are right and that [FvdP] just forgot one condition. I remember that there was a first version of the book which had quite a few typos... Maybe you could ask van der Put? I think he is still active.
Also, you misspelled the first name of the first author in your first reference.
Thanks, corrected that now!
They are absolutely the same thing.
|
2025-03-21T14:48:31.526829
| 2020-07-16T11:25:06 |
365774
|
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|
Stack Exchange
|
Vertex cover algorithm
Given a graph $G(V, E)$, remove the vertex (or one of the vertices) with the highest cardinality from $G$ and put it in a list $L$. Repeat until in $G$ there are only vertices with cardinality $0$ (no edges in the graph).
Now, for each vertex $v$ of $L$, from the first to the last (first in first out), check if its edges, in the given graph, are covered by the vertices that are after $v$ in the list. If it is, remove $v$ from $L$, otherwise leave $v$ in $L$.
At the end the vertices of $L$ are a vertex cover of the given $G(V, E)$, but is it a minimum vertex cover? I can not find an example where the algorithm does not find a minimum vertex cover.
Why MathOverflow instead of Computer Science SE or CS Theory SE?
Given that finding a minimal vertex cover is NP-hard it's unlikely that your algorithm works. Depending on how you break ties in case there is more than one vertex of maximum degree, paths are counterexamples.
@FlorianLehner What do you mean with "paths are counterexamples"? Both in odd and even paths (i.e. paths with odd/even number of vertices) the algorithm finds a minimum vertex cover.
@MarioGiambarioli As I wrote, it depends on how you break ties: in $P_5$ there are three vertices of degree 2, perhaps the algorithm chooses the central vertex first. After that we are left with four vertices of degree 1, the algorithm could pick the endpoints of the original path. So we end up with a cover of 3 vertices, but the unique minimum cover consists of 2 vertices.
Just to clarify: by cardinality, you mean what's generally called degree? and when you remove a vertex, you also remove all the edges incident to that vertex?
@FlorianLehner You are right, in $P_{5}$ the algorithm does not work. Thank you for the answer.
@GerryMyerson Yes, with cardinality I mean degree (and this is the right word, I am sorry). When it removes a vertex, obviously all the edges incident to that vertex are removed because an edge is defined by a couple of vertices (a single vertex does not define an edge.
|
2025-03-21T14:48:31.527011
| 2020-07-16T11:36:19 |
365775
|
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|
Stack Exchange
|
Equivalent parallel pair of $j$-morphisms
I have a question about a definition used in nLab article on $n$-groupoids: https://ncatlab.org/nlab/show/n-groupoid
What does it mean that "every parallel pair of $j$-morphisms is equivalent for $j>n$? Surely, that's not a research question, but I nowhere found an answer. Does that mean that the corresponending equilizers on $j$-level are equivalences? (just a guess of mine)
It means that between every two $j$-cells with the same domain and codomain ($j > n$), there is a $(j + 1)$-cell between them (which is necessarily an equivalence by the first condition of the definition on that page) exhibiting them as equivalent.
(Meta note: math.stackexchange is best suited for non-research-level mathematics questions.)
Addition to the meta note: also the category theory zulip community might be a good place to ask.
Further to @varkor's answer, I would recommend playing around with a homotopy-oriented proof assistant, for example Globular: it is an excellent way to acquire a hands-on intuition of "being equivalent up to higher cells". You can find a tutorial video for Globular by Jamie Vicary.
|
2025-03-21T14:48:31.527113
| 2020-07-16T12:24:39 |
365777
|
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|
Stack Exchange
|
About $\pi$, $e$ and transcendence
This is mostly curiosity on my part. I assume experts would have some up-to-date info.
Are $1$, $\pi$ and $e$ linearly independent over $\mathbb{Q}$?
Does the set $\{ m\pi+ne;\;\;m,n\in\mathbb{Z}\}$ contain any algebraic numbers?
Thanks.
see https://mathoverflow.net/q/33817/11260
@CarloBeenakker I checked that question before I posed mine. I do not understand what Wilkie's work has to do with the specific question. I am quite familiar with 0-minimality. Then I asked about linear independence.
on a trivial note, the set ${ m\pi+ne;;;m,n\in\mathbb{Z}}, m^2+n^2 \ne 0$ can contain at most one algebraic number with $(m,n) =1$ of course and $mn \ne 0$ for any such
Almost certainly $1, \pi, e$ are linearly independent over $\mathbb Q$, and also over the algebraic numbers (so in particular the only algebraic number in your set $\{m\pi + n e: m,n \in \mathbb Z\}$ is $0$ which comes from $m=n=0$). This would follow from Schanuel's conjecture, I think. But it is only a conjecture: we don't even know that $e + \pi$ is irrational.
Yes, Schanuel's conjecture implies that ${ 1, i \pi, e, e^{i \pi} }$ has transcendence degree 2, so $e$ and $\pi$ are algebraically independent (which is stronger than linear independence over the algebraic numbers).
|
2025-03-21T14:48:31.527228
| 2020-07-16T14:05:24 |
365781
|
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|
Stack Exchange
|
A characterization for a commutative ring with a special intersection property for prime ideals
Let $R$ be a commutative ring with $1$ with the property that for any infinite family $\{P_i\}_{i\in I}$ of distinct prime ideals of $R$ we have $\cap_{i\not= j} P_i\subseteq P_j$ for all but fnitely many $j\in I$. Is there any characterization for such a ring?
I suggest revising the title to include some keywords such as "infinite prime avoidance."
@Steven Landsburg: Clearly the statement is true if $P_i's$ are not distinct and since I have stated ** for any infinite family=for every infinite family** it means that we are working on an arbitrary family of distinct prime ideals.
I have the following obvious/extreme examples and counter-examples in mind and I am wondering if you have discovered richer and subtler illustrations to work with. Examples: (1) Finite rings, valuation rings (more generally uniserial rings) (2) UFDs, Dedekind domains. Counter-examples: rings which surject onto a direct product of infinitely many rings. I am also wondering whether anything can be said about meet-irreducible rings (https://en.wikipedia.org/wiki/Irreducible_ring).
|
2025-03-21T14:48:31.527328
| 2020-07-16T14:21:38 |
365782
|
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|
Stack Exchange
|
Kummer theory if $\ell = p$
Background. Let $k$ be a field and let $\ell$ be an integer which is divisible in $k$. Then one has a short exact sequence of abelian étale sheaves
$$ 0 \to \mu_\ell \to \mathbb{G}_m \xrightarrow{(\,\cdot\,)^{\ell}} \mathbb{G}_m \to 0\,\text{,}$$
from which one can deduce interesting applications. It's related to classical Kummer theory, and the connecting map $k^{\times} / \ell \to H^1_{\mathrm{et}}(\operatorname{Spec} k;\mu_\ell)$ gives rise to a map $K_n^{\mathrm{M}}(k)/\ell \to H^n_{\mathrm{et}}(\operatorname{Spec} k ; \mu_{\ell}^{\otimes n})$ (where $K_n^{\mathrm{M}}(\,\cdot\,)$ denotes Milnor K-theory), which appears in the Bloch–Kato conjecture.
Observation. If $\ell$ is not divisible in $k$, i.e. $k$ is of characteristic $p$ and $\ell$ is a multiple of $p$, then the above short exact sequence fails to hold the way it's stated, but it does hold true if we work with fppf sheaves instead.
Question. Are there interesting consequences of the resulting sequence? For instance, is there a version of Kummer theory, and an analogue of the Bloch–Kato conjecture, involving flat cohomology rather than étale cohomology?
Does the tensor product of $\mu_\ell$ with itself in the fppf world make sense and not degenerate?
You write that ℓ is divisible in k — do you mean invertible ?
@MatthieuRomagny I don't remember at all but it sounds reasonable that that was my intention.
|
2025-03-21T14:48:31.527456
| 2020-07-16T15:29:13 |
365788
|
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|
Stack Exchange
|
Reference request: boundedness for semistable principal bundles on a family of curves
We work over an algebraically closed field $k$.
Let $G$ be a reductive group and $X$ be a smooth projective curve over $k$. It is proven in [1, Theorem 1.2] that the moduli of semi-stable principal $G$-bundles over $X$ of a given degree (or "topological type") is bounded.
It seems that the proof there could be readily generalized to principal $G$-bundles on a given family of curves. The precise statement should be:
Let $\pi: \mathcal X \to S$ be a family of smooth projective curves over a scheme $S$ of finite type over $k$. Fix a degree $d$ for principal $G$-bundles (c.f. [1, Definition 3.2]). Then there exists a scheme $T$ of finite type over $S$, together with a principal $G$-bundle
$$
\mathcal P \to \mathcal X \times_{S} T,
$$
such that for any closed point $s$ of $S$, the base-change of the above family to $s$
$$
\mathcal P_s \to \mathcal X_{s} \times T_{s},
$$
consists of all semi-stable $G$-bundles on $\mathcal X_s$ of degree $d$.
The proof in [1] reduces the boundedness of the moduli of semi-stable principal $G$-bundles to the boundedness of the moduli of line bundles of a fixed degree. Hence it seems that the proof should be generalizable without essential modification.
However, I could not find a direct reference in the literature. Does anyone know such a result in the literature?
[1] Holla, Yogish I.; Narasimhan, M. S., A generalisation of Nagata’s theorem on ruled surfaces., Compos. Math. 127, No. 3, 321-332 (2001). ZBL1047.14018.
|
2025-03-21T14:48:31.527574
| 2020-07-18T07:31:38 |
365916
|
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|
Stack Exchange
|
Reductions to the MAX-3-DCC Problem
I am currently working on the Max-3-DCC problem that asks for the heaviest vertex-disjoint cycle cover of weighted directed graphs.
The problem has been reduced to 3-SAT in 1979 by L. Valiant in his article on The Complexity of Calculating the Permanent.
Later, around 2005 B. Manthey proved the APX-hardness of the problem in On Approximating Restricted Cycle Covers, followed by a series of articles on related problems of calculating restricted cycle covers.
I recently had new ideas for calculating Max-3-DCCs and was looking for benchmark instances similar to those for TSP, however neither a related question on cs.stackexchange https://scicomp.stackexchange.com/questions/35522/benchmark-instances-for-directed-3-cycle-cover?noredirect=1#comment67646_35522 nor extensively searching the internet brought about anything directly in that direction.
I have the impression that the Max-3-DCC is the Cinderella of NP complete problems; it is almost never "invited" to lists of such problems and hardly the subject of interest, expressed as publications.
I therefore put my hopes in getting access to benchmark instances via "simple" reductions from other NP complete problems:
Question:
which NP complete optimization problems, for which collections of freely accessible benchmark instances are available, can be transformed to Max-3-DCC and back in time linear in the size of the instance, which should be possible because most of Karp's problems are class NLIN according to this article.
My most wanted are of course the problems in Karp's list of 21, most prominently TSP and vertex-cover.
for clarification:
instances qualify as benchmarks for me only if their optimal solution is known and if known heuristics will fail to report it. Instances with best *known* solutions are also acceptable.
Apart from that the data should be freely accessible.
|
2025-03-21T14:48:31.527722
| 2020-07-18T07:56:50 |
365917
|
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|
Stack Exchange
|
A subcontinuous function, which is not continuous
Let $E$ be a Banach space and $T: E\rightarrow E$ be a mapping. $T$ is said to be subcontinuous if for any sequences $(u_n)_{n\in\mathbb{N}}$ in $E$ that converge strongly to $u$ the sequence $(T(u_n))_{n\in\mathbb{N}}$ converges weakly to $T(u)$.
I am looking for a subcontinuous function which is not continuous. I thought I can find something in $l^2$, but I didn't.
I already posted it on math.stackexchange, but to no avail
Thank you
I delete it, since i didn't have any interaction on Math SE..
Let $(e_n)$ be the standard orthonormal basis of $\ell^2$: recall that, as a sequence, $(e_n)$ converges weakly to $0$. Now define a map $f\colon\mathbb{R} \to \ell^2$ by $f(\frac{1}{n})=e_n$ and $f(t)=0$ if $t\leq 0$, and interpolating linearly between $\frac{1}{n}$ and $\frac{1}{n+1}$: this is continuous at every point except at $0$ where it is subcontinuous (because any sequence of the form $t_k\, e_{n_k} + (1-t_k)\, e_{n_k+1}$ with $0\leq t_k\leq 1$ and $n_k\to+\infty$, converges weakly to $0$ when $k\to+\infty$). If you want a subcontinuous function $\ell^2\to\ell^2$, just right-compose $f$ with a nonzero continuous linear map $\ell^2\to\mathbb{R}$, e.g., $e_0^*$.
What the you mean exactly by : interpolating linearly between...
That means if $x\in[(n+1]^{-1},n^{-1}]$ then $$f(x) = f(\frac1n) +(f(\frac1{n+1}) -f(\frac1n))\frac{x-\frac1n}{\frac1{n+1} - \frac1n} $$
The sequence $u_k$ must be general (and converges to $0$), I don't know why you assume that $0\leq u_k\leq 1$?
What I called $u_k$ obviously wasn't what was called $u_k$ in the question: I changed this to $t_k$ to avoid confusion. Terms with $u_k<0$ are unproblematic, all terms will be $\leq 1$ eventually, and terms between $0$ and $1$ can be written $t/n+(1-t)/(n+1)$ with $0\leq t\leq 1$ so their image by $f is $te_n+(1-t)e_{n+1}$.
|
2025-03-21T14:48:31.527862
| 2020-07-18T08:19:58 |
365919
|
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|
Stack Exchange
|
Approximate Sobolev embedding
It is well-known in $H^2(\mathbb R^3)$ embeds into $L^{\infty}(\mathbb{R}^3).$ Now consider a function $u \in \ell^{\infty}(h\mathbb Z^3)$ and a grid of points $x \in h\mathbb{Z}^3.$
We then define the finite-difference Laplacian
$$(\Delta_hu)(x):=\frac{\left(\sum_{i=1}^3 f(x+he_i)+f(x-he_i)\right)-6 f(x)}{h^2}$$
I wonder, is it true that for some universal $C>0$
$$\Vert u \Vert_{\ell^{\infty}(h\mathbb Z^3)} \le C (\Vert \Delta u \Vert_{\ell^2(h\mathbb Z^3)}+ \Vert u \Vert_{\ell^2(h\mathbb Z^3)})?$$
Here,
$$\Vert u \Vert^2_{\ell^2(h\mathbb Z^3)} = \sum_{x \in h\mathbb Z^3} h^3 \vert u(x) \vert^2 $$
The intuition behind this estimate is that we discretely approximate the continuous setting with $L^2$ and $L^{\infty}$ norms.
Please let me know if you have any questions.
$|u|{\ell^\infty(I)}=\sup{|u(i)|: i\in I} \le |u|{\ell^2(I)}$.
@JochenWengenroth sorry, the norm here is not the standard $l^2$ norm but a weighted one, so that we 'approximate' the continuous world...
Yes, this is true, and there is a proof which closely tracks your intuition. As you know, this estimate can be proved in the continuum by applying the Sobolev embedding twice, first to get $\nabla u \in L^p$ for $p<\frac{2d}{d-2}=6$, and then once more to get $u\in L^\infty$. So for simplicity let me discuss how to get discrete versions of the Sobolev embedding for only one derivative. You can put these together in the same way to get your bound.
We will transfer the Sobolev inequalities in the continuum to the lattice by brute force. First, extend your function $u$ define on the lattice $h\mathbb{Z}^d$ by making it be constant on all cubes of side length $h$ which are centered on a point of $h\mathbb{Z}^d$. Next, mollify this piecewise constant function with the standard mollifier, on length scale $h/10$. Name the resulting smooth function $v$, which is now defined in $\mathbb{R}^d$. You have the following pointwise bounds:
\begin{equation}
\left\| v \right\|_{L^\infty(z+[-h/2,h/2]^d)}
\leq
\sup_{z' \sim z} | u(z')|
\end{equation}
and
\begin{equation}
|D_h(z)|:=\frac1h\sup_{z'\sim z}|u(z) - u(z')|
\leq
\frac{C}{h^d} \int_{z+[-h/2,h/2]^d)} |\nabla v|.
\end{equation}
Here $\sim$ means nearest-neighbor in the $h\mathbb{Z}^d$ lattice. The first bound is pretty easy, the second is true because near the boundary between nearest neighbor cubes, there will be a positive-measure set of points for which the mollifier picks up the difference between $u(z)$ and $u(z')$, and therefore on this set $|\nabla v|$ will be at least proportion to this difference. The proportion of this set in the cube is lower bounded by a constant (which does not depend on $h$).
Now, applying the (continuum) Sobolev inequality to $v$ and putting everything together gives a (discrete) Sobolev inequality for $u$.
There is one more minor point I will mention, which is that you have defined the $H^2$ norm with respect to the Laplacian only (and not the full set of mixed second derivatives). But you can perform a discrete integration by parts (twice) to bound the $\ell^2$ of $D^2_hu$, the full set of (possibly mixed) second-order differences, by the $\ell^2$ norm of $\Delta_h u$ (mimicking the usual proof in the continuum).
|
2025-03-21T14:48:31.528342
| 2020-07-18T08:48:15 |
365921
|
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|
Stack Exchange
|
Which field extensions do not affect Chow groups?
Let $X$ be a (say, smooth projective) variety over a field $k$. For which $K$ it is known that the ("ordinary", that is, not higher) Chow groups of $X$ map onto that of $X_K$ bijectively?
This statement appears to be easy if $K$ is stably rational over $k$. What about other extensions; any references on this matter?
I assume that you aware that you are asking which projective $k$-varieties have an integral decomposition-of-the-diagonal.
I believe this is in general open. When K is purely transcendental over k, one can have injectivity maybe? But surjectivity does not seem immediate to me in general.
Though I think there are some works in this direction, by e.g. Vishik, Karpenko, Merkurjev, Zainoulline, S. Gille, R. Fino, S. Baek, etc. and I think you may find some partial answers from some of their papers. Like A. Vishik, in Geometric methods in the algebraic theory of quadratic forms, 25–101, Lecture Notes in Math., 1835, Springer, Berlin, 2004? This long paper seems to consider the function field case.
Yes, this question may be attacked by means of decomposing the diagonal; yet I am not sure that looking at smooth projectives is optimal here. On the other case, the injectivity question appears to be much easier; it suffices to have a degree 1 zero-cycle on any (and thus, on all) smooth X such that $k(X)=K$.
|
2025-03-21T14:48:31.528585
| 2020-07-18T09:12:58 |
365922
|
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|
Stack Exchange
|
Laplacian coupled with another equation over a two-dimensional rectangular region
I have the two-dimensional Laplacian $(\nabla^2 T(x,y)=0)$ coupled with another equation which is:
$$\frac{\partial t}{\partial x}+\alpha(t-T)=0 \tag 1$$
where it is known that $t(x=0)=t_i$.
The Laplacian is defined over $x\in[0,L], y\in[0,l]$ subjected to the following boundary conditions
$$\frac{\partial T(0,y)}{\partial x}=\frac{\partial T(L,y)}{\partial x}=0 \tag 2$$
$$\frac{\partial T(x,0)}{\partial y}=\gamma \tag 3$$
$$\frac{\partial T(x,l)}{\partial y}=\zeta \Bigg[T(x,l)-t\Bigg] \tag 4$$
$\gamma, \alpha, \zeta, t_i$ are all constants $>0$.
Attempt
I solved $(1)$ using the method of integrating factor and substituted in $(4)$ to get the boundary condition in the following form:
$$\frac{\partial T(x,l)}{\partial y}=\zeta \Bigg[T(x,l)-\Bigg\{\alpha e^{-\alpha x}\Bigg(\int_0^x e^{\alpha s }T(s,y)\mathrm{d}s+\frac{t_{i}}{\alpha}\Bigg)\Bigg\}\Bigg] \tag 5$$
Can someone suggest a way to handle the Laplacian subjected to b.c.(s) $(2), (3), (5)$ ?
If there exists an alternate way to handle this coupled system, those suggestions are welcome too.
Physical meaning
The problem describes the flow of a fluid (with temperature $t$ and described by $(1)$) over a rectangular plate (at $y=l$) heated from the bottom (at $y=0$). The fluid is thermally coupled to the plate temperature $T$ through boundary condition $(4)$ which is the convection or Robin type condition.
What you can find out without solving anything is $t(L,\ell)=\frac{\alpha\gamma}{\ell}L$, by the divergence Theorem. But you probably want the full picture inside.
|
2025-03-21T14:48:31.528736
| 2020-07-18T10:16:29 |
365925
|
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|
Stack Exchange
|
Reference for topological graph theory (research / problem-oriented)
I would be interested in recommendations for topological graph theory texts. I think Gross and Yellen has a great chapter on topological graph theory, and I find Mohar and Thomassen's Graphs on surfaces from 2001 a great reference as well.
Could you recommend more current references, ideally with focus on open problems / research?
Though old, the classic "Topological Graph Theory" by Gross and Tucker is quite useful
Maybe this is another useful reference for you, now I found the link:
Ralucca Gera, Stephen Hedetniemi, Craig Larson, Teresa W. Haynes (editors) (2018): Graph Theory: Favorite Conjectures and Open Problems
It is actually two volumes, and obviously more recent than the other reference I mentioned. It covers graph theory as a whole and does not focus on topological graph theory only. It is a collection of conjectures and open problems. I would judge it to be clearly at graduate and reserach level, but written in a very "inviting" way and starting from examples.
The reason why I think it could be interesting for you: it is full of research ideas and references. I took an hour or so to leaf through it a few weeks ago and was quite fascinated: many short articles, often starting with a few personal remarks and how the author got interested in a particular field, and then moving very fast to conjectures and open questions in that field. Remarkably, the second volume includes a comprehensive list of 70 conjectures, and closes with 600+ references.
Just to add one comment to this reference, the first volume is actually from 2016. The second volume is from 2018.
My recommendation, try Lando and Zvonkin (2004): Graphs on Surfaces and Their Applications.
I think it is a great book which applies graphs embedded on surfaces to solving problems from other fields of mathematics. The style is very refreshing, vivid, and lively, I would say. The style reminded me of Hatcher's chapter 0 in his Algebraic Topology text, and of Matousek's book "Using the Borsuk-Ulam Theorem".
I would think the target audience of this book is graduate and research level, for some topics the pace is high. Exellent list of references, I think over 300.
Edit: I was just thinking, maybe the following quote from this book gives you a good indication. The authors are talking about a topological graph here:
"It is not merely a topological object, a graph embedded into (or drawn on) a two-dimensional surface. It is also a sequence of permutations (or, if you prefer, it "is encoded by" a sequence of permutations), which provides a relation to group theory. And it is at the same time a way of representing a ramified covering of the sphere by a compact two-dimensional manifold. Considering the sphere as the Riemann complex sphere we obtain, on the covering manifold, the structure of a Riemann surface. And Riemann surfaces rarely walk by themselves. Usually they keep company with Galois theory, with algebraic curves, moduli spaces and many other exciting subjects."
A great and current reference is "Algorithms for embedded graphs" from Éric C. de Verdière, it is a 66 page synthesis of his course notes from 2017 (find here: http://monge.univ-mlv.fr/~colinde/cours/all-algo-embedded-graphs.pdf). Covers topological graph theory plus related algorithms e.g. to minimize edge length of embedded graphs.
See this quote from the link about the contents of these course notes. Note especially chapters 4, 6, and 7 regarding your OP question. Quote:
The first chapter introduces planar graphs from the topological and combinatorial point of view.
The second chapter considers the problem of testing
whether a graph is planar, and, if so, of drawing it without crossings in
the plane.
Then we move on with some general graph problems, for which
we give efficient algorithms when the input graph is planar. Then, we consider graphs on surfaces (planar graphs being an important
special case).
In Chapter 4, we introduce surfaces from the topological
point of view; in Chapter 5, we present algorithms using the cut locus to
build short curves and decompositions of surfaces.
In Chapter 6, we introduce two important topological concepts, homotopy and the universal
cover.
All these techniques are combined in Chapter 7 to provide algorithms to shorten curves up to deformation.
this is a great reference
I like a lot the book from Beineke & Wilson (editors) "Topics in Topological Graph Theory" from 2009 for that purpose. Take a look at the article "Open Problems" from Archdeacon in this book. It is just like 5 pages or so, but inspired me a lot. I think you could find it very useful.
I can recommend Topics in Chromatic Graph Theory (Encyclopedia of Mathematics and its Applications) with editors Lowell W. Beineke and Robin J. Wilson. It is from 2015, and if you are interested in chromatic topological graph theory topics, there are three relevant chapters for you:
Chapter 1: Colouring graphs on surfaces, chapter 4: Hadwiger's conjecture, chapter 8: Geometric graphs.
My interest is not so much in topological graph theory research (more interested in applications), but I have read these chapters as well, they are very good. It is definitely gradudate level with current research topics.
Here's the reference page from a computational topology course I took a while back.
To complement the answers from last year, now that the question has become active again, I could add these two monographs that approach the topic from different perspectives.
Referring to the "research-oriented" viewpoint requested by the OP: it helps to study a perspective away from the main stream, if you are research oriented; you will have a better chance of finding an unsolved problem that is doable --- published lists of unsolved problems are typically not of that type....
• The Foundations of Topological Graph Theory by Bonnington and Little:
This is an attempt to place topological graph theory on a purely
combinatorial yet rigorous footing. The sole requirement for
understanding the logical development in this book is some elementary
knowledge of vector spaces over the field $\mathbb{Z}_2$.
• Graphs on Surfaces: Dualities, Polynomials, and Knots by Ellis-Monaghan and Moffatt:
We discuss the interdependency between duality, medial graphs and knots; how this interdependency is reflected in algebraic invariants of graphs and knots; and how it can be exploited to solve problems in graph and knot theory.
|
2025-03-21T14:48:31.529178
| 2020-07-18T10:22:18 |
365926
|
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|
Stack Exchange
|
Shortest path on Riemannian manifold with boundary
Let $(M^n,g)$ be a smooth Riemannian manifold with non-empty boundary $\partial M$. Let $x\in \partial M$. Let $v\in T_x(\partial M)$ be a unit vector tangent to the boundary. Assume
$$II_{\partial M}(v)<0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\, (1)$$
where $II_{\partial M}$ denotes the second fundamental form of the boundary.
Let $\gamma\colon [0,l]\to \partial M$ be a normalized geodesic on the boundary which starts at $x$ in the direction $v$.
Is it true that if $l$ is sufficeintly small then $\gamma$ is a shortest path in $M$ (rather than in $\partial M$)? A reference would be very helpful.
Here the condition (1) seems to be crutial.
Remark: I think I have a proof of the positive answer to the question when $n=2$.
|
2025-03-21T14:48:31.529260
| 2020-07-18T14:22:21 |
365938
|
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|
Stack Exchange
|
Is the fiber product of two connected complex varieties over a connected base connected?
If $X,Y,Z$ are connected varieties over $\mathbb{C}$, with morphisms $X\rightarrow Z, Y\rightarrow Z$, is it true that the fiber product $X\times _Z Y$ is connected?
The statement is true for $S=Spec(\mathbb{C})$, but I can't see it for general $S$.
Any help would be appreciated.
Try $X,Y,Z$ all equal to $\mathbb{C}^*$ and the morphisms $z\mapsto z^2$.
@Mohan : Thank you for your answer. If I understand correctly, the underlying set of $\mathbb{C}^\times_{\mathbb{C}^} \mathbb{C}^*$ under the squaring map is ${(x,x): x\neq 0 }\sqcup {(x,-x):x\neq 0}$, am I correct?
@Mohan : I am actually interested in projective varieties. Would the answer be positive if we had all the varieties projective?
Whenever $X \to Y$ is a finite étale Galois cover of degree $d > 1$, the fibre product $X \times_Y X$ splits up as a disjoint union of $d$ sheets. This can happen for smooth projective varieties as well, e.g. the multiplication by $n$ map $E \to E$ on an elliptic curve (say $n$ invertible in $k$).
In light of the above comments, it may be more interesting to ask: if $Z$ is connected and $X\to Z$ and $Y\to Z$ have connected fibers is the fiber product connected?
@cgodfrey : That's a great point, and in fact, the maps I would like to consider DO have connected fibers. Do you have ideas about this case? That would be of great help.
This is ok when $X \to Z$ (resp. $Y \to Z$) is proper with geometrically connected fibres. Indeed, apply Tag 0377 to the proper map $X \times_Z Y \to Y$ (resp. $X \times_Z Y \to X$), which is a topological quotient since it is proper. (A similar argument can be made if you replace 'proper' by 'universally open', e.g. flat of finite presentation.)
@R.vanDobbendeBruyn : Than you so much for the reference!
@yojusmath: It is better to ask this in a separate question.
|
2025-03-21T14:48:31.529411
| 2020-07-18T16:01:31 |
365943
|
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"url": "https://mathoverflow.net/questions/365943"
}
|
Stack Exchange
|
Resolution of pairs in characteristic p
Let $R$ be a complete DVR of characteristic $p$, say $R=\mathbb{F}_p[[t]]$, and $X$ be a reduced scheme of finite type over $R$. Let also $X_s$ denote the special fiber of $X$. If I understand correctly, resolution of singularities in characteristic p, says that there is a regular scheme $X'$ of finite type over $R$ and a blowup $f:X'\to X$ in a nowhere dense center $Z$.
However, it gives us no information on what the new special fiber $X'_s$ looks like.
Q: (a) Is there a (conjectured) version of resolution of singularities in characteristic $p$ that also ensures that $X'_s$ is of a "nice" form (normal crossings divisor)?
(b) If so, does this mean that $X'$ locally looks like
$$\operatorname{Spec}(R[x_1,\dotsc,x_d]/(f_1^{e_1}f_2^{e_2}\dotsb f_n^{e_n}-t))$$
where the $\mathbb{F}_p$-schemes cut out by the $\bar f_i$'s are as in the definition of normal crossings?
It seems that the corresponding statement in characteristic $0$ is true and well-known (Theorem 3.1.5 in Temkin - Absolute desingularization in characteristic zero). Also, any references that address this (or closely related) issues would be useful.
Some background: I am not an algebraic geometer, but I found it convenient to assume, at least temporarily, some form of RoS in characteristic $p$ for a problem I am trying to solve.
|
2025-03-21T14:48:31.529525
| 2020-07-18T17:18:27 |
365946
|
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"authors": [
"Carlo Beenakker",
"Fedor Petrov",
"Iosif Pinelis",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365946"
}
|
Stack Exchange
|
Positivity of $ \int_{-\infty}^{\infty} \left\{{2^{1/\beta-1/2} \over v}\right\}^{it} { \Gamma\{(it+1)/\beta\}\over \Gamma\{(it+1)/2\} }dt$
I have the following function
$$
\int_{-\infty}^{\infty} \left\{{2^{1/\beta-1/2} \over v}\right\}^{it}
{ \Gamma\{(it+1)/\beta\}\over \Gamma\{(it+1)/2\} }dt
$$
where $1<\beta<2$, $v>0$. Need to show it is positive.
The inverse Mellin transform of
$$
\left\{2^{1/\beta-1/2} \right\}^{it}
{ \Gamma\{(it+1)/\beta\}\over \Gamma\{(it+1)/2\} }
$$
is
$$
{C \over v}\int_{-\infty}^{\infty} \left\{{2^{1/\beta-1/2} \over v}\right\}^{it}
{ \Gamma\{(it+1)/\beta\}\over \Gamma\{(it+1)/2\} }dt
$$
It seems $v$ should be positive
@TanyaVladi, yes, thank you
the integral is a complex number, so what do you mean by "positive" ?
@CarloBeenakker, it is not complex. Do you want me to draft the proof?
I just evaluated it numerically for $\beta=3/2$ and $v=1$ and find $1.50029 + 1.65104, i$ --- see https://www.wolframalpha.com/input/?i=Integrate%5B+%282%5E%28%28I+t%29%2F6%29+Gamma%5B2%2F3+%281+%2B+I+t%29%5D%29%2FGamma%5B1%2F2+%281+%2B+I+t%29%5D+%2C%7Bt%2C0%2CInfinity%7D%5D
@CarloBeenakker, I think the integral should be from $-\infty$ to $+\infty$
@CarloBeenakker, sorry yes, the integral is from $-\infty$
Why such strange expression in power $it$? Is not it an arbitrary positive number?
inverse Mellin transform of ${ \Gamma{(it+1)/\beta}\over \Gamma{(it+1)/2} $
$\newcommand\Ga\Gamma
\newcommand{\R}{\mathbb{R}}
\newcommand{\de}{\delta}
\newcommand{\ga}{\gamma}
\newcommand{\Si}{\Sigma}$
We have to show that for $a:=-\ln(2^{1/b-1/2}/v)\in\R$ and $b:=\beta\in(1,2)$,
\begin{equation*}
I(a):=\int_{-\infty}^{\infty} e^{-iat}R(t)\,dt>0, \tag{1}
\end{equation*}
where
\begin{equation*}
R(t):=\frac{\Ga\big((1+it)/b\big)}{\Ga\big((1+it)/2\big)}. \tag{2}
\end{equation*}
The key is Euler's product formula
\begin{equation*}
\Ga(z)=\frac1z\,\prod_{j=1}^\infty\frac{(1+1/j)^z}{1+z/j}
\end{equation*}
for $z\in\mathbb C\setminus\{0,-1,-2,\dots\}$, which yields
\begin{equation*}
\frac{\Ga(s+it)}{\Ga(s)}=\prod_{j=1}^\infty(1+1/j)^{it}
\Big/\prod_{j=0}^\infty\Big(1+\frac{it}{j+s}\Big); \tag{3}
\end{equation*}
here and in what follows, $s$ is any positive real number and $t$ is any real number.
Based on (3), it is easy to obtain
Lemma 1: $\ln|\Ga(s+it)|\sim-\pi|t|/2$ as $|t|\to\infty$.
The proof of Lemma 1 will be given at the end of this answer.
It also follows from (3) that
\begin{equation*}
R(t)=c\prod_{j=1}^\infty(1+1/j)^{iht}f_j(t), \tag{4}
\end{equation*}
where $c:=\Ga(1/b)/\Ga(1/2)>0$,
\begin{equation}
h:=\frac1b-\frac12=\frac{2-b}{2b},
\end{equation}
and
\begin{equation}
f_j(t):=\frac{1+it/(1+2j)}{1+it/(1+bj)},
\end{equation}
so that $f_j$ is the characteristic function (c.f.) of a random variable (r.v.) $X_j\sim p_j\de_0+(1-p_j)Exp(-1/(1+bj))$, where in turn $p_j:=(1+bj)/(1+2j)\in(0,1)$, $\de_0$ is the Dirac distribution supported on the set $\{0\}$, and $Exp(-1/(1+bj))$ is the exponential distribution with mean $-1/(1+bj)$, supported on the interval $(-\infty,0]$. Here and in what follows, $j$ is any natural number. Note that $EX_j=-\frac{hj}{(j+1/b)(j+1/2)}$ and $Var\,X_j\le1/(bj)^2\le1/j^2$. So, the series
\begin{equation}
\sum_{j=1}^\infty(X_j-EX_j)=:S
\end{equation}
converges almost surely. Hence, by (4)
\begin{equation*}
R(t)=ce^{ihc_1t}f_S(t),
\end{equation*}
where $f_S$ is the c.f. of the r.v. $S$ and
\begin{equation}
c_1:=\sum_{j=1}^\infty\Big(\ln(1+1/j)+EX_j\Big) \\
=\sum_{j=1}^\infty\Big(\ln(1+1/j)-\frac{j}{(j+1/b)(j+1/2)}\Big)\in\R
\end{equation}
(in fact, $c_1=(\ga b-2 b+b \ln4+2 \psi\left(1+1/b\right))/(2-b)$, where $\ga=0.577\dots$ is the Euler constant and $\psi:=\Ga'/\Ga$; however, the actual value of $c_1$ does not matter here).
So, $R$ is the c.f. of the r.v. $T:=hc_1+S$. Also, by Lemma 1, $R$ is in $L^1$. It now follows that the function $I$, defined by (1), is $2\pi$ times the density of the r.v. $T$. Thus, $I(a)\ge0$ for all real $a$, as desired.
It remains to provide
Proof of Lemma 1: By (3),
\begin{equation*}
\frac{|\Ga(s+it)|}{\Ga(s)}=\prod_{j=0}^\infty\frac{j+s}{|j+s+it|}
=\exp\{-\Si_{s,t}/2\},
\end{equation*}
where
\begin{equation}
\Si_{s,t}:=\sum_{j=0}^\infty\ln\Big(1+\frac{t^2}{(j+s)^2}\Big).
\end{equation}
Since $\ln\big(1+\frac{t^2}{(j+s)^2}\big)$ is nonincreasing in $j$, we have
\begin{equation}
J_{s,t}\le\Si_{s,t}\le J_{s,t}+\ln\big(1+\frac{t^2}{s^2}\big),
\end{equation}
where
\begin{equation}
J_{s,t}:=\int_0^\infty\ln\big(1+\frac{t^2}{(x+s)^2}\big)\,dx\sim\pi|t|
\end{equation}
as $|t|\to\infty$, which completes the proof of Lemma 1 and the entire answer.
(In fact, integrating by parts, for $t\ne0$ we find
\begin{equation}
J_{s,t}=\pi|t|-s \ln \left(s^2+t^2\right)-2 t \arctan(s/t)+2 s \ln s\sim\pi|t|.)
\end{equation}
The proof of Lemma 1 and the entire answer are now complete.
looks great, I do not quite see Fourier in the original problem, I have $\left{{2^{1/\beta-1/2} \over v}\right}^{it}$ though, where did it go?
@Vova : this power equals $e^{iat} $ for some real $a$.
I still do not see how you got $(1+ui)/(1+cui)$ from ${1+(it)/(1+2j) \over 1+(it)/(1+\beta j) }$, if you do the change of variables, your $u$ depends on $j$
@Vova : I have rewritten the answer in a probabilistic and more detailed way.
Great one, I really like the original line of proof as well. I have a follow up question. My $b$ is actually $b_{kl}$ and by knowing how $b$ survives integration, in which functional form it appears on the other end - I can find such $A_{kl}$ such that $A_{kl}*f(b_{kl})_{v}$ is positive definite for any $v$ and any $n$, $k,l=1,\cdots n$
@Vova : I don't think I understand your most recent comment. In any case, I think it would be a better idea to post the further question separately and in detail.
yes, I just posted, for whatever reason cannot paste the link here
|
2025-03-21T14:48:31.529875
| 2020-07-18T17:32:04 |
365947
|
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"authors": [
"Asaf Karagila",
"David Roberts",
"Harry Gindi",
"Ivan Di Liberti",
"Kevin Carlson",
"Malice Vidrine",
"Maxime Ramzi",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365947"
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|
Stack Exchange
|
When size matters in category theory for the working mathematician
I think a related question might be this (Set-Theoretic Issues/Categories).
There are many ways in which you can avoid set theoretical paradoxes in dealing with category theory (see for instance Shulman - Set theory for category theory).
Some important results in category theory assume some kind of ‘smallness’ of your category in practice. A very much used result in homological algebra is the Freyd–Mitchell embedding theorem:
Every small abelian category admits an fully faithful exact embedding in a category $\text{$R$-mod}$ for a suitable ring $R$.
Now, in everyday usage of this result, the restriction that the category is small is not important: for instance, if you want to do diagram chasing in a diagram on any category, you can always restrict your attention to the abelian subcategory generated by the objects and maps on the diagram, and the category will be small.
I am wondering:
What are results of category theory, commonly used in mathematical practice, in which considerations of size are crucial?
Shulman in [op. cit.] gives what I think is an example, the Freyd Special Adjoint Functor Theorem: a functor from a complete, locally small, and well-powered category with a cogenerating set to a
locally small category has a left adjoint if and only if it preserves small limits.
I would find interesting to see some discussion on this topic.
Very often one has the feeling that set-theoretic issues are somewhat cheatable, and people feel like they have eluded foundations when they manage to cheat them. Even worse, some claim that foundations are irrelevant because each time they dare to be relevant, they can be cheated. What these people haven't understood is that the best foundation is the one that allows the most cheating (without falling apart).
In the relationship between foundation and practice, though, what matters the most is the phenomenology of every-day mathematics. In order to make this statement clear, let me state the uncheatable lemma. In the later discussion, we will see the repercussion of this lemma.
Lemma (The uncheatable).
A locally small, large-cocomplete category is a poset.
The lemma shows that no matter how fat are the sets where you enrich your category, there is no chance that the category is absolutely cocomplete.
Example. In the category of sets, the large coproduct of all sets is not a set. If you enlarge the universe in such a way that it is, then some other (even larger) coproduct will not exist. This is inescapable and always boils down to the Russel Paradox.
Remark. Notice that obvious analogs of this lemma are true also for categories based on Grothendieck Universes (as opposed to sets and classes). One can't escape the truth by changing its presentation.
Excursus. Very recently Thomas Forster, Adam Lewicki, Alice Vidrine have tried to reboot category theory in Stratified Set Theory in their paper Category Theory with Stratified Set Theory (arXiv: https://arxiv.org/abs/1911.04704). One could consider this as a kind of solution to the uncheatable lemma. But it's hard to tell whether it is a true solution or a more or less equivalent linguistic reformulation. This theory is at its early stages.
At this point one could say that I haven't shown any concrete problem, we all know that the class of all sets is not a set, and it appears as a piece of quite harmless news to us.
In the rest of the discussion, I will try to show that the uncheatable lemma has consequences in the daily use of category theory. Categories will be assumed to be locally small with respect to some category of sets. Let me recall a standard result from the theory of Kan extensions.
Lemma (Kan). Let $\mathsf{B} \stackrel{f}{\leftarrow} \mathsf{A} \stackrel{g}{\to} \mathsf{C}$ be a span where $\mathsf{A}$ is small and $\mathsf{C}$ is (small) cocomplete. The the left Kan extension $\mathsf{lan}_f g$ exists.
Kan extensions are a useful tool in everyday practice, with applications in many different topics of category theory. In this lemma (which is one of the most used in this topic) the set-theoretic issue is far from being hidden: $\mathsf{A}$ needs to be small (with respect to the size of $\mathsf{C})$! There is no chance that the lemma is true when $\mathsf{A}$ is a large category. Indeed since colimits can be computed via Kan extensions, the lemma would imply that every (small) cocomplete category is large cocomplete, which is not allowed by the uncheatable. Also, there is no chance to solve the problem by saying: well, let's just consider $\mathsf{C}$ to be large-cocomplete, again because of the the uncheatable.
This problem is hard to avoid because the size of the categories of our interest is as a fact always larger than the size of their inhabitants (this just means that most of the time Ob$\mathsf{C}$ is a proper class, as big as the size of the enrichment).
Notice that the Kan extension problem recovers the Adjoint functor theorem one, because adjoints are computed via Kan extensions of identities of large categories, $$\mathsf{R} = \mathsf{lan}_\mathsf{L}(1) \qquad \mathsf{L} = \mathsf{ran}_\mathsf{R}(1) .$$ Indeed, in that case, the solution set condition is precisely what is needed in order to cut down the size of some colimits that otherwise would be too large to compute, as can be synthesized by the sharp version of the Kan lemma.
Sharp Kan lemma. Let $\mathsf{B} \stackrel{f}{\leftarrow} \mathsf{A} \stackrel{g}{\to} \mathsf{C}$ be a span where $\mathsf{B}(f-,b)$ is a is small presheaf for every $b \in \mathsf{B}$ and $\mathsf{C}$ is (small) cocomplete. Then the left Kan extension $\mathsf{lan}_f g$ exists.
Indeed this lemma allows $\mathsf{A}$ to be large, but we must pay a tribute to its presheaf category: $f$ needs to be somehow locally small (with respect to the size of $\mathsf{C}$).
Kan lemma Fortissimo. Let $ \mathsf{A} \stackrel{f}{\to} \mathsf{B} $ be a functor. The following are equivalent:
for every $g :\mathsf{A} \to \mathsf{C}$ where $\mathsf{C}$ is a small-cocomplete category, $\mathsf{lan}_f g$ exists.
$\mathsf{lan}_f y$ exists, where $y$ is the Yoneda embedding in the category of small presheaves $y: \mathsf{A} \to \mathcal{P}(\mathsf{A})$.
$\mathsf{B}(f-,b)$ is a is small presheaf for every $b \in \mathsf{B}$.
Even unconsciously, the previous discussion is one of the reasons of the popularity of locally presentable categories. Indeed, having a dense generator is a good compromise between generality and tameness. As an evidence of this, in the context of accessible categories the sharp Kan lemma can be simplified.
Tame Kan lemma. Let $\mathsf{B} \stackrel{f}{\leftarrow} \mathsf{A} \stackrel{g}{\to} \mathsf{C}$ be a span of accessible categories, where $f$ is an accessible functor and $\mathsf{C}$ is (small) cocomplete. Then the left Kan extension $\mathsf{lan}_f g$ exists (and is accessible).
Warning. The proof of the previous lemma is based on the density (as opposed to codensity) of $\lambda$-presentable objects in an accessible category. Thus the lemma is not valid for the right Kan extension.
References for Sharp. I am not aware of a reference for this result. It can follow from a careful analysis of Prop. A.7 in my paper Codensity: Isbell duality, pro-objects, compactness and accessibility. The structure of the proof remains the same, presheaves must be replaced by small presheaves.
References for Tame. This is an exercise, it can follow directly from the sharp Kan lemma, but it's enough to properly combine the usual Kan lemma, Prop A.1&2 of the above-mentioned paper, and the fact that accessible functors have arity.
This answer is connected to this other.
This is such a good answer!
Thanks, @HarryGindi! :)
This is a good answer, but I feel it would be even better with a very concrete example. The one I had in mind was actually given by David Roberts in his answer, so that's a good complement
@MaximeRamzi thanks! :-)
Re: my paper with Forster and Lewicki, I think I speak for my co-authors as well when I say that as a cheat goes, an NF-like universe isn't a very good cheat (and was never really meant to be; it was more a route to get at what kind of beast NF is). All of your usual size problems end up turning into $T$-functor problems, which are usually even more opaque. As an aside, I'm surprised our paper made anyone's radar!
@MaliceVidrine thanks for your comment. I found your paper very interesting! :) Don't you remember my email?
Oh! Now that I take a moment to stare at your name, I do! Apologies for my slow memory.
What happened to \quad and \qquad? Why do you feel the need to self-abuse with these single-spaces?
@AsafKaragila I was just not aware of that. I'll change it later. Thanks.
Here's an example that links more to mathematical practice outside category theory proper. Recall that for a small site $(C,J)$, where I take $J$ to be a Grothendieck pretopology on the small category $C$, any presheaf $C^{op} \to \mathbf{Set}$ has a sheafification, and this extends to give us a functor $[C^{op},\mathbf{Set}] \to Sh(C,J)$ from presheaves to sheaves, left adjoint to the inclusion. Sheafification can be described as two applications of the Grothendieck plus construction, which is a colimit indexed by a set of covering sieves.
Now we can also talk about large sites, and at least talk about individual (pre)sheaves even when we cannot form the categories of them (say because, like the Stacks project, we do not wish to use universes, or whatever). There is then a real obstruction to forming the sheafification. Famously the category of schemes (over a base scheme, if desired) with the fpqc (pre)topology has presheaves on it that don't admit a sheafification (see tag 0BBK). What is happening here is that the condition WISC (Weakly Initial Set of Covers) is violated. This condition says that for any object, there is a set of covering families such that every covering family is refined by one in that set. This allows the construction of the colimit in the plus construction, and can be also seen as a kind of solution set condition for the construction of the left adjoint to the inclusion of sheaves into presheaves. So in a sense, this is a special case of Ivan's answer.
Large sites are not all that uncommon in practice, even if they are glossed over. Ignoring essentially small examples (like that of the category of finite-dimensional manifolds), then the category of all topological spaces (or CGWH spaces, even) with the open cover topology is large but satisfies WISC; same for the category of schemes (or the relative case) with pretty much any topology coarser than fpqc; same for any category of infinite-dimensional smooth manifolds (again with the open cover topology).
Thus this condition WISC is very natural from both a category-theoretic point of view, and also from a sheaf-theoretic or even geometric viewpoint, being satisfied very often, but not always. From a set theoretic point of view (considering forcing as being an instance of forming sheaf toposes), it's actually quite hard to make it fail, and one cannot to that without proper class forcing (or the analogous thing in a topos-theoretic approach).
I should add: I understand that the fppf satisfies WISC, but I don't know if the ph topology (which is finer than the fppf topology) does. Of the topologies on categories of schemes listed in the Stacks Project, the fpqc and v-topologies definitely don't satisfy WISC. Anything coarser than fppf (so: syntomic, smooth, étale and Zariski topologies) does satisfy WISC. So the ph topology is the only one there I don't know about.
Johan de Jong informs me the h and the ph topologies both satisfy WISC. I believe this implies that in the chart at https://pbelmans.ncag.info/topologies-comparison/ (slightly incomplete), everything except the fpqc topology satisfies WISC.
Interestingly, the h topology isn't subcanonical (!), which in my mind makes it rather fine/large, but it is incomparable with fpqc (which is subcanonical), so WISC and subcanonicity are independent.
The other answers are good, but I would like to point out that Ivan's "uncheatable" lemma can in fact be cheated. The proof of that lemma (due to Freyd) makes inescapable use of classical logic, and in constructive mathematics it is possible to have a non-poset that is complete for the size of its own set of objects (a complete small category). It is even possible to have a category "of sets" with this property (e.g. those called "modest sets" in realizability). Then all Kan extensions into such a category exist, all sheafifications of modest presheaves exist, and presumably all Bousfield localizations of modest spectra exist (although the latter may get you into HoTT water when you try to do it constructively).
About the only thing the category of modest sets lacks is a subobject classifier (it is locally cartesian closed). So these days I prefer the following argument as the "least cheatable" (calling something "uncheatable" sounds like a challenge) manifestation of size issues in category theory.
Lemma 1: Any endofunctor of a complete small category has a fixed point.
Proof: If $C$ is complete-small, so is the category of $F$-algebras for any endofunctor $F:C\to C$. But any complete small category has an initial object (by essentially the same argument that any complete meet-semilattice also has all joins), and an initial $F$-algebra is a fixed point of $F$ (by Lambek's lemma).
Lemma 2: If $C$ is an elementary topos, the double powerset functor $X \mapsto \Omega^{\Omega^X}$ has no fixed point.
Proof: By Cantor's diagonalization argument.
Thus, no elementary topos can have all limits of the size of its collection of objects.
HoTT water, hah!
I really appreciate that you challenged my answer. I'll bring a gauntlet at the next conference. Get ready to fight.
Let me share a basic theorem in algebraic topology that hides a subtle set-theoretic point that turns out to be the item at stake. I am talking about Bousfield localization. Let me put it this way, consider the category $\mathcal{T}: = \mathsf{Ho}\mathcal{Sp}$ the homotopy category of spectra. Let $E \in \mathcal{T}$ and consider the smallest triangulated category stable with coproducts that contains $E$, denoted $\langle E \rangle$. Bousfield's theorem asserts that the inclusion functor
$$\langle E \rangle \hookrightarrow \mathcal{T}$$ has a right adjoint.
The idea of the proof is clear, Given a spectrum $X$ build step by step spectra $N_\alpha \in \langle E \rangle$ (indexed by cardinals) and consider its cofiber sequence
$$
N_\alpha \to X \to B_\alpha
$$
If one takes the (homotopy) limit of all $B_\alpha$ one arrives to an object in $\langle E \rangle^\perp$ whose fiber $N$ (the colimit of $N_\alpha$) is automatically the value of the adjoint. Up to some checking and some precisions this would be the proof. The problem is that one cannot take a class-indexed limit, unless one accepts a form of the universe axiom in which case these limits live outside our initial universe!
So, what is the way out? Bousfield clever argument was that there is an cardinal $\gamma$ such that $B_\gamma \in \langle E \rangle^\perp$ by using regular cardinals and arguments related to the presentability of the model category of spectra. With this reasoning, both $N_\gamma$ and $B_\gamma$ live in our universe, or otherwise said, the proof makes sense with our favorite choice of foundations (von Neumann-Gödel-Bernays, say).
Needless to say, there are other versions of this result that use the same set-theoretic trick to achieve a bound on the index cardinal, for instance an analogous result for derived categories.
All of this is related to the so called "small object argument".
|
2025-03-21T14:48:31.531053
| 2020-07-18T17:37:45 |
365948
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365948"
}
|
Stack Exchange
|
Equivalence relation induced by Kolmogorov quotients
Recall: given a (possibly non-$T_0$) topological space $X$, its Kolmogorov quotient $KX$ is the $T_0$ topological space formed by $X/\sim$ where $x\sim y$ if they are topologically indistinguishable. Denote the mapping $X\to KX$ of $x$ to its equivalence class $\pi$.
I have two loosely related terminology questions:
Is it okay to use the word "section" to refer to a mapping (and/or the image of such a mapping) $\gamma: KX \to X$ such that $\pi\circ \gamma = id$? (This would be the word from category theory, just wondering if there is another established terminology that is used.)
Is there a word for the equivalence relationship where two topological spaces $X$ and $Y$ are said to be equivalent if their quotients $KX$ and $KY$ are homeomorphic?
(I am particularly interested in the case where the non-$T_0$ topology comes from a semi-norm or a pseudo-metric; so if there is an answer to 2 when restricted to semi-normed spaces or pseudometric spaces, I'd be happy too.)
|
2025-03-21T14:48:31.531159
| 2020-07-18T19:31:41 |
365953
|
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"Ben McKay",
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"sort": "votes",
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}
|
Stack Exchange
|
Reference for the divergence theorem for embedded $C^1$-submanifolds of $\mathbb R^d$ with boundary
I'm aware of Gauss's theorem (aka the divergence theorem) for compact subsets $K$ of $\mathbb R^d$ with "$C^1$-boundary"$^1$.
I know that there are several generalizations of this theorem, but since I'm not familiar with general differential geometry, I'd like to find a reference which particularly considers the case of the manifold boundary $\partial M$ of a "$d$-dimensional embedded $C^1$submanifold of $\mathbb R^d$ with boundary"$^2$.
Is there a nice reference which presents and proofs this case in a rigorous way?
Please note that my main background is abstract measure/probability theory.
$^1$ i.e. for each $p\in\partial K$, there is an open neighborhood $U$ of $p$ and a $\psi\in C^1(U)$ with $K\cap U=\{\psi\le0\}$ and $\psi'(x)\ne0$ for all $x\in U$. $\partial K$ is denoting the topological boundary of $K$ and it is a $(d-1)$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$.
$^2$ i.e. each point of $M$ is locally $C^1$-diffeomorphic to $\mathbb H^d:=\mathbb R^{d-1}\times[0,\infty)$ and the manifold boundary $\partial M$ is a $(d-1)$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ (without boundary).
What you ask is a particular case of Stokes theorem for differential forms on (Riemannian) manifolds. I think a good reference, given what you write about your main topics, is do Carmo book on differential forms
M. Do Carmo, Differential Forms and Applications, Springer Science & Business Media, 2012
But, honestly, I think that for the working mathematician there are no good sources on this topic. Unfortunately to manage these topics (80% consist of definitions and notation) one has to workout a lot of examples and these are what textbooks lack.
There are lots of books covering Stokes theorem. My favorite is M. Spivak's "Calculus on manifolds" https://en.wikipedia.org/wiki/Calculus_on_Manifolds_(book).. As for "having to work out lots of examples", does not this apply to every subject?
Yes, but you can find a lot of books with solved exercises on limits or derivatives. But no one with worked out exercises on these things. The truth is that the theory is elegant in the abstract setting, but as soon as one has to go down to exercises it suddenly becomes a mess. And people avoid to put examples because it requires too much effort to write down these things, compared to examples on how to compute derivatives or limits or integrals.
Unfortunately, Do Carmo's book is not a good reference for this result. The book uses the term "differentiable" to mean $C^{\infty}$, and assumes this of all manifolds and forms in Stokes's theorem. You can see this when he proves that $d^2=0$ for differentiable forms, which is clearly not true for $C^1$ forms. The problem is to find a reference for the $C^1$ case. I think we need to look for something else, because I think do Carmo's proof only works for $C^1$ differential forms on $C^2$ manifolds with boundary.
@BenMcKay: regarding smoothness of the manifold one can work in a $C^\infty$ atlas that is compatible with the given $C^1$ atlas. Smoothness of the form is also not a problem because one can approximate $C^1$ forms by $C^\infty$ forms, so the $C^\infty$ Stokes theorem implies the $C^1$ version.
@IgorBelegradek: I agree. But if the problem is to find a reference for the final result, we might prefer a reference that carries out the steps that you indicate. Those steps, when completed in full, are on their own longer than the proof that do Carmo gives. One needs to define convolution, arrive at the relevant local results and be more careful about partitions of unity than do Carmo is to patch local results.
Do you read German?
Several German textbooks have very thorough treatment of this; for example
O. Forster, Analysis 3; Springer
|
2025-03-21T14:48:31.531442
| 2020-07-18T20:06:12 |
365955
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365955"
}
|
Stack Exchange
|
What is the reason behind the name 3n-display?
In the paper "The display of a formal $p$-divisible group" Zink defines some objects and calls them $3n$-display. A $3n$-display over $R$ is a quadruple $P$, $Q$, $F$, $F^1$ such that $P$ is a projective $W(R)$ module, $Q$ is a submodule and $F:P\to P$, $F^1 :Q\to P$ are functions satisfying some conditions. I don't know why these things are called $3n$-display and I don't see any relation to $3$ or $n$ in the definition. Does someone know the reason behind this name?
I have not found a source where Thomas Zink explains the name, however arXiv:1906.00899 explains it as an abbreviation of “not-necessarily-nilpotent” (or $3n$-) displays. See also Travaux de Zink, page 343.
|
2025-03-21T14:48:31.531531
| 2020-07-18T20:51:09 |
365958
|
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"JGWang",
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|
Stack Exchange
|
An integral involving Levy process with no positive jumps
Let $L_t$ be a Levy process that has no positive jumps, but is not strictly decreasing, i.e
$$
L_t = \gamma t + \sigma B_t + J_t,
$$
where $B_t$ is a Brownian motion, $J_t$ is a pure jump process with only negative jumps.Suppose further, that either $\sigma > 0$ or the Levy measure of $J_t$ is absolutely continuous. In such a case, it is known that the 1-dim distribution of $L_t$ is absolutely continuous.
Let $p(t,x)$ be the probability density function of $L_t$:
$$P(L_t \in B) = \int_B p(t,s) ds.$$
Fix $a > 0$.
Now for $t > 0$ and $x < a$, let
$$
q(t,x) = \int_0^t \frac{a}{s} p(s,a)p(t-s,x-a)ds.
$$
I have reason to believe that $q(t,x) \to p(t,a)$ as $x\nearrow a$, but haven't been able to prove it.
If $\gamma=1>0, sigma=0$, and $J_t$ is a Poisson process, the 1-dim distribution of $L_t$ doesn't have a density.
Agreed. Thank you. I'll correct it.
$L_t$ I notice is probably not the best symbol here. It us often used for local times. Anyway, here the definition is clear.
|
2025-03-21T14:48:31.531642
| 2020-07-18T21:51:59 |
365960
|
{
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"authors": [
"Alec Rhea",
"Anderson Brasil",
"David Roberts",
"Hanul Jeon",
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"url": "https://mathoverflow.net/questions/365960"
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|
Stack Exchange
|
Looking for references for NBG theory meant for the working mathematician (not for someone interested at Foundations of Mathematics)
As most mathematicians, I've always used sets as the main tool for doing mathematics. My knowledge of the subject is limited to what I've learned from Halmos's "Naive Set Theory" and, for years, it was more than enough for me, as I don't have any interest at working at the field of Foundations of Mathematics. But, recently, the subject of surreal numbers caught my attention and for the first time in my life I had to deal directly with Classes. I picture them as a generalization for the concept of set, allowing the existence of "special sets" (Proper Classes) which may not be members of any "set" (Class), and so far this basic understanding has been enough to follow the proofs at the books. Still, when I am not tutored by the books, I feel often afraid of doing something illegal for not knowing exactly what I am allowed to do with classes.
Therefore, I am looking for references of NBG theory which are meant for the working mathematician (not for people who want to become experts at Fundations of Mathematics). Basic texts teaching me the ability to work with classes without risks.
So far, all the material I've found (Zaring, for example) is too technical for me. What I am looking for is something like Halmos' book, but with classes. Is there anything similar? Where can I get a solid knowledge about classes without being annoyed by irrelevant (for me) technical details? Any reference would be appreciated.
If you are interested only in the operations you can do with classes in the sense of category theory (eg products, quotients etc), then you can rest in the knowledge that NBG classes form a Boolean pretopos with subobject classifier. It additionally has exponentials for sets. See eg my question and answer at https://math.stackexchange.com/q/231087/3835
@David Roberts, despite a certain introductory book in category theory is currently one of my objects of desire, currently my knowledge in the field is nearly zero. And, right now, I am looking for something that I can quickly learn.
If you just want to look up the axioms of NBG and MK, you may refer to the first some pages of Kameryn Williams' doctoral dissertation. However, the dissertation is definitely inclined to foundational interests, so not adequate for your purpose.
@AndersonBrasil sorry, I think it would be reasonable to check out the other answers and their reference. The point is that you can do cartesian products of a pair of classes (and so finitely many), quotients by equivalence relations, carve out subclasses by formulas, form disjoint unions of a pair of classes (and by extension finitely many) and these behave just in the same way they do for sets. You can form the class of all functions from a set to a class. This is most of what you need to know, not the nitty-gritty of exactly which axioms are available.
@HanulJeon, ty for the reference.
@DavidRoberts, the problem is that right now I don't have ANY knowledge of category theory. So I would probably take much more time that way, as it would require me learning at least the the basic notions of the theory. Of course, this option has a big advantage: I would acquire some knowledge of the subject (which is probably something advisable to any mathematician). But right now I don't have the time for this, I need to quickly learn to deal with Classes. That is why I am ruling out this option. But I really do appreciate your replies. Thanks a lot.
@Anderson ok, fair enough. One to come back to, then. Sorry it wasn't as useful for you.
I appreciate your efforts, @DavidRoberts. And I am pretty sure that your tip, recorded at this thread, will be useful for someone else in here.
Do you have a reason to prefer NBG over MK (Morse-Kelley theory of sets and classes)? If not, you might look at the development of MK in the Appendix (if I remember correctly) of Kelley's book "General Topology". It's surely intended for the working mathematician, since (again, if I remember correctly) Kelley had originally intended this book to be titled "What every young analyst should know".
As a working mathematician, I am agnostic about foundation models =p.
As I've said, my knowledge of set theory is basically Halmos' book, which is essentially ZFC without the axiom of regularity. My preference for NBG was only because (as far as I know) NBG ressembles ZFC more than MK ressemble ZFC. So I felt that it would be easier for me to adapt to NBG. Still, it is nice to have options. I will try to find the reference you gave me. Thanks a lot.
@AndersonBrasil I’d actually argue that MK is syntactically more similar to ZFC then NBG, even though it is a bit stronger in terms of consistency strength.
I've finally got a copy of Kelley's book. The appendix is indeed very interesting and might be very useful. Thanks.
This might be what you're looking for: Elliot Mendelson, An Introduction to Mathematical Logic, fifth ed., CRC Press, Boca Raton (2010). I think there is a 6th edition as well.
Another possibility would be: Smullyan, R. and Fitting, M. (2010): Set theory and the Continuum Problem, Dover Publications, New York.
For Mendelson NBG includes the axiom of global choice and for Smullyan and Fitting it just includes the axiom of choice.
I took a quick look at the Smullyan-Fitting and the second chapter seems that is going to be very helpful. Thanks a lot.
|
2025-03-21T14:48:31.532040
| 2020-07-18T22:05:59 |
365962
|
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|
Stack Exchange
|
Is a direct sum of flabby sheaves flabby?
Consider a family of flabby (= flasque) sheaves $(\mathcal F_i)_{i\in I}$ of abelian groups on the topological space $X$.
My question : is their direct sum sheaf $\mathcal F=\oplus _{i\in I} \mathcal F_i$ also flabby?
Here is the difficulty:
Given an open subset $U\subset X$ a section $s\in \Gamma(U,\mathcal F)$ consists in a collection of sections $s_i\in \Gamma(U,\mathcal F_i)$ subject to the condition that for any $x\in U$ there exists a neighbourhood $x\in V\subset U$ on which almost all $s_i\vert V \in \Gamma(V,\mathcal F_i)$ are zero.
Now, every $s_i$ certainly extends to a section $S_i\in \Gamma(X,\mathcal F_i)$ by the flabbiness of $\mathcal F_i$.
The problem is that I see no reason why the collection $(S_i)_{i\in I}$ should be a section in $\Gamma(X,\oplus _{i\in I} \mathcal F_i)$, since I see no reason why every point in $X$ should have a neighbourhood $W$ on which almost all the restrictions $S_i\vert W$ are zero.
Of course any direct sum of flabby sheaves is flabby if the space $X$ is noetherian, since in that case we have $\Gamma(U,\mathcal F) =\oplus_{i\in I} \Gamma(U,\mathcal F_i)$ for all open subsets $U\subset X$.
I have only seen the fact that direct sums of flabby sheaves are flabby (correctly) used on noetherian spaces, actually schemes, so that my question originates just from idle curiosity...
An example can be found in Bredon's Sheaf Theory book: Example 5.11.
No, a direct sum of flabby sheaves need not be flabby.
Take $X=\{1,1/2,1/3,1/4,\dots\}\cup\{0\}$ with the subspace topology from $\mathbb R$, and let $\mathcal F$ be the sheaf whose sections over an open $U\subseteq X$ are the functions $U\to\mathbb F_2$ (not necessarily continuous). This is a flabby sheaf. I claim that the infinite direct sum $\mathcal F^{\oplus\mathbb N}$ of countably many copies of $\mathcal F$ is not flabby.
To see this, let $U=X\setminus\{0\}$, and for $i\in\mathbb N$ let $s_i\colon U\to\mathbb F_2$ denote the function sending $1/i$ to $1$ and all other elements of $U$ to $0$. Thus each $s_i$ is a section of $\mathcal F$ over $U$. Observe that $s=(s_i)_{i\in\mathbb N}\in\Gamma(U,\mathcal F^{\oplus\mathbb N})$, since locally on $U$ all but finitely many of the sections $s_i$ are equal to zero (the topology on $U$ is discrete).
I claim that this section $s$ doesn't extend to a section of $\mathcal F^{\oplus\mathbb N}$ over all of $X$. Indeed, if $s$ extended to a section $\tilde s=(\tilde s_i)_{i\in\mathbb N}$, then there would be a neighbourhood of $0$ in $X$ on which all but finitely many of the $\tilde s_i$ were equal to $0$. But this would imply that $\tilde s_i(1/i)=s_i(1/i)=0$ for all sufficiently large $i$, which is impossible. Thus $s$ does not extend.
What about finite direct sums? (Or is that a basic sheaf theory fact?)
@ChrisGerig Flabbyness is preserved for arbitrary direct products and a finite sum of abelian sheaves is the same as their finite product.
Thank you, Alexander: this is a perfect counterexample. I hope it will find its place in some basic book using sheaves, or maybe become an item in the Stacks Project.
@GeorgesElencwajg It will. :-) I am writing a sequel to ams.org/open-math-notes/omn-view-listing?listingId=110823 that will be about homological methods in commutative algebra. The first half or so is about homological algebra per se, and there sheaves appear as a fundamental recurring example. I am putting Alexander's counterexample as a guided exercise.
Dear@Andrea, this is wonderful news! When I wrote about my hope, I actually thought that its realization was not very probable: I'm happy you are showing me that I was unduly pessimistic! I am delighted that you are writing a follow-up to the book I downloaded a few months ago from the AMS site, and which I much appreciated. Thanks a lot for that great document and my best wishes for the future one.
|
2025-03-21T14:48:31.532342
| 2020-07-18T22:45:12 |
365965
|
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|
Stack Exchange
|
Determining the kernel of the localization map when defining the localization by generators and relations à la Serre
All rings considered will be commutative and unitary. Let $A$ be a ring, $S \subseteq A$ a multiplicatively closed subset. The localization $\lambda_S : A \longrightarrow A[S^{-1}]$ can be characterized as a ring homomorphism $\lambda : A \longrightarrow B$ with the following three properties:
(LC1) $\lambda$ localizes $S$, i.e. $\lambda(s)$ is invertible in $B$ for all $s \in S$;
(LC2) for every $b \in B$ there is $s \in S$ such that $s b \in \text{im $\lambda$}$;
(LC3) $\ker \lambda = \{a \in A \,| \,\exists s \in S: sa = 0\}$.
One way to achieve this is to define the localization by means of generators and relations: take an indeterminate $T_s$ for each $s \in S$, form the polynomial ring $A[T] = A[T_s|s \in S]$ over $A$ in these indeterminates and quotient out the ideal generated by the $sT_s - 1$ , $s \in S$, thus defining the localization $A[S^{-1}]$:
\begin{equation*}
A[S^{-1}] := A[T_s|s \in S]\,/\,(sT_s-1|s \in S).
\end{equation*}
The structure map $\lambda_S : A \longrightarrow A[S^{-1}]$ then comes along as the composite
\begin{equation*}
A \longrightarrow A[T_s|s \in S] \longrightarrow A[S^{-1}].
\end{equation*}
See [1], pp. I-7-8. The question is how to verify properties (LC1-3) for this construction. In fact, (LC1-2) are straightforward, but (LC3) seems hard. It is known to be true, since it holds in the other widespread model of localization, given by $\mu_S : A \longrightarrow S^{-1}A$ with
\begin{equation*}
S^{-1}A := A \times S / \sim,
\end{equation*}
where $\sim$ denotes the equivalence relation
\begin{equation*}
(a,s) \sim (b,t) :\iff \exists u \in S:\, u(ta-sb) = 0,
\end{equation*}
and
\begin{equation*}
\mu_S(a) := a/1,
\end{equation*}
where, for $(a,s) \in A \times S$, $a/s$ denotes its equivalence class in $S^{-1}A$. Here, $(LC3)$ is trivial for $\mu_S$, holding by construction. Since both $\lambda_S$ and $\mu_S$ are universal among the ring homomorphisms localizing $S$, it holds for $\lambda_S$, too. But to show this directly for $\lambda_S$ using its definition, is surprisingly difficult: if $\lambda_S(a) = 0$, this means there are $s_1, \dots s_n \in S$ and polynomials $p_1(T), \dots, p_n(T) \in A[T]$ such that ($T_i:=T_{s_i}$)
\begin{equation*}
a = \sum_{i=1}^n p_i(T)(s_iT_i - 1).
\end{equation*}
From this I can conclude
\begin{equation*}
a = -\sum_{i=1}^n a_i \quad,\quad a_i := p_i(0)
\end{equation*}
but this is, for the time being, the end of the flagpole. In the best
of all possible worlds, I would have $p_i(T) = a_i$; this would give
$a_is_i = 0$ for $i=1, \dots, n$, and so $sa = 0$ with $s := s_1 \cdots s_n$, but I see no reason for that.
So does somebody know what is needed to make progress towards (LC3)?
[1] Serre, J.-P.,
Algèbre locale - Multiplicités
(Lecture Notes in Mathematics 11). Springer 1965
Here is a proof that $$\ker \lambda = \{ a \in A \, \vert \, sa = 0 \text{ for some } s \in S\} \quad (LC_3)$$ holds true assuming that the following definition is in use: $$A[S^{-1}] = A[T_s \vert s \in S] /\left(sT_s -1 \vert s \in S\right).$$
If $ta = 0$ for some $a \in A$ and some $t \in S$, then we have $$a = -(tT_t - 1)a \in (sT_s -1 \vert s \in S).$$ Hence the inclusion $\{ a \in A \, \vert \, sa = 0 \text{ for some } s \in S\} \subseteq \ker \lambda$ comes (almost) for free.
To prove the reverse inclusion, consider $a = \sum_{i = 1}^n p_i(T_1, \dots, T_n)(s_i T_i - 1) \in A[T_1, \dots, T_n]$ and reason by induction on $n \ge 1$.
Let us suppose that $n = 1$, i.e., $a = p_1(T_1)(s_1 T_1 - 1)$. Replacing simultaneously $a$ by $s_1^m a$ and $p_1(T_1)$ by $s_1^m p_1(T_1)$ for some $m > 0$ if need be, we can assume that either $p_1(T_1) = 0 = a$, or $\deg(s_1 p_1(T_1)) = \deg(p_1(T_1))$. As the latter identity is clearly impossible, the induction base is settled.
Suppose now that $n > 1$ and let $\overline{a}$ be the image of $a$ in $\overline{A}[T_1, \dots, T_{n - 1}] \simeq A[T_1, \dots, T_n]/(s_nT_n - 1)$ where $\overline{A} = A[T_n]/\left(s_n T_n - 1\right)$ (see claim below).
Since $\overline{a} = \sum_{i = 1}^{n - 1} \overline{p_i}(T_1, \dots, T_{n - 1}) (s_i T_i -1)$ where $\overline{p_i} \in\overline{A}[T_1, \dots, T_{n - 1}]$ is obtained from $p_i$ by assigning the last indeterminate $T_n$ to its image in $\overline{A}$, the induction hypothesis yields $s \overline{a} = 0$ for some $s \in S$. This means that $sa \in (s_n T_n - 1) \subset A[T_n]$ so that we can conclude by resorting to the case $n = 1$.$\square$
Note that we have used the following:
Claim. Let $R$ be a commutative and unital ring. Let $R[T_1, \dots, T_n]$ be the ring of multivariate polynomials over $R$ with $n$ indeterminates $T_1, \dots, T_n$. Let $P_1, \dots, P_k \in R[T_n]$ with $k \ge 0$. Then the natural isomorphism $R[T_1, \dots, ,T_n] \rightarrow (R[T_n])[T_1, \dots, T_{n - 1}]$ induces a ring isomorphism $R[T_1, \dots, T_n]/(P_1, \dots, P_k) \rightarrow \overline{R}[T_1, \dots, T_{n - 1}]$ where $\overline{R} \Doteq R[T_n]/(P_1, \dots, P_k)$.
When you say $\deg(s_1 p_1(T_1)) = \deg(p_1(T_1))$, did you perhaps mean to say on the left hand side $\deg(s_1 T_1 p_1(T_1))$?
@ZachTeitler Dear Zach, thanks for your interest and careful reading. I actually meant what I wrote initially. I have now expanded the aforementioned sentence to make the idea clearer: we multiply both hands by $s_1$ until either $p_1(T_1)$ gets annihilated (we are then done) or its degree does not decrease after a further multiplication by $s_1$. This would mean that $\deg(s_1p_1(T_1)) = \deg(p_1(T_1))$ and hence $\deg(s_1T_1p_1(T_1)) = \deg(p_1(T_1)) + 1$. The latter is impossible as $\deg(a) = 0$. If it still looks odd, awkward or wrong, I'll be glad to follow your suggestion.
It makes sense! Thank you.
Dear Luc Guyot, many thanks for your suggestions of a proof. I have not
been able to fully retrace your chain of argument as it seems to me that
some points remain unsaid. But the indication that an inductive proof
might be within reach was helpful, and I will give an ouline of
proof as I have conveived it below as a separate answer.
Dear @MathCrawler, I didn't design my answer as a suggestion. Is it too concise?. I understand that several points remain unclear to you. Without a detailed feedback, it is difficult to improve. (I just made some edits though.)
The proof of (LC3), in the given setting, is surprisingly difficult, or, at least, elaborate. Let $a \in A$ with $\lambda_S(a) = [a] = 0$ in $A[S^{-1}]$,
i.e. one has
\begin{equation}
\tag{1}
a \in (sT_s-1\,|\,s \in S).
\end{equation}
To show is that
\begin{equation}
\tag{2}
sa = 0
\end{equation}
for some $s \in S$. Because of (1), there are elements $s_1, \dots s_n \in S$ and polynomials $p_1(T), \dots,p_m(T) \in A[T]$ such that
\begin{equation}
a = \sum_{i=1}^n p_i(T) (s_iT_i - 1) \quad
\text{in $A[T]$},\quad,\quad T_i := T_{s_i}.
\end{equation}
As a first reduction, we may assume $p_i(T) = p_i(T_1, \dots,T_n)$ for all $i$, so that
\begin{equation}
\tag{3}
a = \sum_{i=1}^n p_i(T_1, \dots, T_n) (s_iT_i - 1)
\quad \text{in $A[T]$}.
\end{equation}
Namely, let $T' \subseteq T$ be those indeterminates which either equal some $T_i$, or which appear in at least one $p_i(T)$, $i = 1, \dots, n$, so that we may write $T' = \{T_1, \dots, T_n, T_{n+1}, \dots, T_q\}$.
By eventually introducing dummy terms with coefficient 0, we may assume $p_i(T) = p_i(T') = p_i(T_1, \dots, T_q)$, so that $$a = \sum_{i=1}^n p_i(T_1, \dots, T_q) (s_iT_i - 1)$$. Putting $p_i(T):=0$ for $i=n+1, \dots, q$ then gives
\begin{equation*}
a = \sum_{i=1}^q p_i(T_1, \dots, T_q) (s_iT_i - 1)
\quad \text{in $A[T]$},
\end{equation*}
which upon renaming $q$ by $n$ gives (3).
To prove that $sa = 0$ for some $s \in S$ we proceed by induction on $n$. For $n = 1$ we start with
\begin{equation*}
a = p(T_s) (sT_s - 1) \quad \text{in $A[T]$}
\end{equation*}
for some indeterminate $T_s \in X$. We abbreviate notation by writing $u := T_s$, so that we have the equation
\begin{equation}
a = p(u) (su - 1) \quad \text{in $A[T]$}.
\end{equation}
Let $p(u) = \sum_{k=0}^d a_k u^k$; then
\begin{equation*}
\begin{split}
p(u) (su - 1)
&= \sum_{k=0}^d sa_k u^{k+1}-\sum_{k=0}^d a_ku^k\\
&= \sum_{k=1}^{d+1} sa_{k-1} u^k -
\sum_{k=0}^d a_k u^k\\
&= sa_du^d + \sum_{k=1}^d(sa_{k-1}-a_k) u^k-a_0\\
&= a,
\end{split}
\end{equation*}
so that
\begin{equation*}
a_0=-a \quad,\quad a_k=sa_{k-1}\,,\,k=1,\dots, d-1
\quad,\quad sa_d = 0,
\end{equation*}
hence
\begin{equation*}
a_k = -s^ka \,,\, k=0, \dots, d \quad,\quad
sa_d = 0 ,
\end{equation*}
so that
\begin{equation*}
s^{d+1}a = -sa_d = 0,
\end{equation*}
as was to be shown. This establishes the base clause of the induction.
We now assume that $n \ge 1$, and that, with $k < n$,
\begin{equation*}
a = \sum_{i=1}^k p_i(T_1,\dots,T_n)(s_iT_i-1)
\quad \text{in $A[T]$}
\end{equation*}
implies that $sa = 0$ for some $s \in S$, and we want to show that the same is true for $k = n$. So we assume, with a given ring $A$, that $a \in \ker \lambda_S$ and (2) holds. We put $A' := A[T_n]/(s_nT_n - 1)$. The projection $A \longrightarrow A'$ then realizes(!) the localization $$\lambda_{S'} : A \longrightarrow A[S'^{-1}]$$ with $S' := \{s_n\}$; in particular, $A'= A[S'^{-1}]$. The canonical map
\begin{equation*}
A[T_n] \longrightarrow A[T] \longrightarrow
A[S^{-1}]
\end{equation*}
induces, by passing to the quotient, $$A'= A[S'^{-1}] \longrightarrow A[S^{-1}] = (A[S'^{-1}])[S^{-1}]$$, which realizes the localization
\begin{equation*}
\lambda_S' : A[S'^{-1}] \longrightarrow (A[S'^{-1}])
[S^{-1}].
\end{equation*}
The localization map $\lambda_S : A \longrightarrow A[S^{-1}]$ then factors as the composite of localizations
\begin{equation*}
A \longrightarrow A' \longrightarrow A[S^{-1}] =
A \longrightarrow A[S'^{-1}] \longrightarrow
(A[S'^{-1}])[S^{-1}].
\end{equation*}
Let $\overline{a} \in A' = A[S'^{-1}]$ be the image of $a \in A$ under $A \longrightarrow A'$. Then $\lambda_S(a) = \lambda_S'(\overline{a}) = 0$. and so, by (3),
\begin{equation*}
\overline{a} = \sum_{i=1}^{n-1}
\overline{p_i}(T_1, \dots, T_{n-1}) (s_iT_i - 1)
\quad \text{in $A'[T]$}
\end{equation*}
with $\overline{p_i}(T_1, \dots, T_{n-1}) = p_i(T_1,\dots, T_{n-1},1/s_n)$, $i=1, \dots, n-1$, since $s_nT_n - 1 = 0$ in $A' = A[S'^{-1}]$. Therefore, by the induction hypothesis, $s\overline{a} = \overline{sa} = 0$ for some $s \in S$. Thus $sa \in \ker \lambda_{S'}$, and so, by the base clause $n=1$ applied to $\lambda_{S'}$,
\begin{equation*}
s_n^{d+1}(sa) = (s_n^{d+1}s)a = 0,
\end{equation*}
which finishes the proof. As a byproduct of the proof we obtain that $s$ in (2)
may be chosen as a product of the $s_i$'s (with repeated factors), i.e. as
an element of the multiplicative closure of $\{s_1, \dots, s_n\}$.
|
2025-03-21T14:48:31.533049
| 2020-07-19T01:29:52 |
365973
|
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"Alf",
"Boby",
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"https://mathoverflow.net/users/37014",
"https://mathoverflow.net/users/69661",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365973"
}
|
Stack Exchange
|
Maximizing variance of bounded random variable through convex optimization
I am interested in maximizing the variance of a random variable $X$ supported on $[0,1]$. Formally,
$$\max_{P_X: X \in [0,1]} {\rm Var}(X),$$
where $P_X$ is a distribution of $X$. This question is well-studied on this forum. For example, see this question. The solution to this question is given by $P_X(0) = P_X(1) = \frac 12$.
My question is about a specific approach to solving this. Concretely, I am interested in applying the convex optimization method to this problem.
Here is the outline of the proof/method:
Note that space of distribution over $[0,1]$ $$\mathcal{P} = \left\{ P_X : X \in [0,1] \right\}$$ is convex and compact in weak-topology.
$P_X \to {\rm Var}(X)$ is concave.
Combining 1) and 2) we conclude that this is a well-defined convex optimization problem.
Find the KKT conditions by using the directional derivative. These are given by the following: $P_X$ is an optimizer if and only if
\begin{align}
&(x-E_{P_X}[X])^2 < {\rm Var}_{P_X}(X), x\in [0,1] \setminus {\rm supp}(P_X)\\
&(x-E_{P_X}[X])^2 = {\rm Var}_{P_X}(X), x \in {\rm supp}(P_X)\\
\end{align}
My question: How to solve the above equations and produce an optimal $P_X$?
Of course, at this point, we can just plug-in our distribution and check it. However, this would be cheating. I would like to solve for the optimal distribution starting with the above equations with no extra help. For example, at this point, we don't even know that the distribution is discrete.
Why is this interesting: Since variance is one of the simplest concave operators over the space of distributions, solving it via convex optimization method serves as an interesting example.
@RodrigodeAzevedo It's in a bullet point 1). I will clarify this.
I think that duality may help here. The (pre-)dual should be a convex minimization problem on the space of continuous functions. But I didn't think any further than that... On a second thought: I think that optimality conditions should say a bit more, namely that you have equality only on the support (i.e. the inequality is strict outside the support).
@Dirk Yes, there should be an inequality outside of support. I will correct it. I would be very interested in looking at a dual approach, so if you have time please add some details or thoughts on how in.
The second KKT condition implies that there exists a nontrivial quadratic function $q$ such that $q(X) = 0$ almost surely. Therefore $P_X$ is supported on at most 2 points, and by the first equation these must be 0 and 1. The second equation then yields that indeed he made must be balanced. Or are you looking for something less hands on than this?
@J. Thanks. This is nice. Also, yes, less hands-on would be good. I want to use this optimization problem as an introductory example to convex optimization over p.d. spaces. Therefore, it would be good to use things that are less hands-on and easily generalize.
@J. Basically your approach counts the number of zeros of $q(X)$. Since there are at most two zeros, the result follows. My question then is: This is essentially all we can do with these equations just count zeros? Or is there some inversion that is possible?
|
2025-03-21T14:48:31.533284
| 2020-07-19T03:40:19 |
365980
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexander Betts",
"https://mathoverflow.net/users/126183"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365980"
}
|
Stack Exchange
|
Outer Galois representations and Tate modules of Jacobian varieties
Let $X$ be a proper smooth curve over a field $k$. Then we have an exact sequence of profinite groups
\begin{equation*}
1 \to \pi_1(X_{\overline k}) \to \pi_1(X) \to G_k \to 1,
\end{equation*}
which induces an outer Galois representation
\begin{equation*}
G_k \to \mathop{\mathrm{Out}}\left(\pi_1(X_{\overline k})\right).
\end{equation*}
Show that after abelianization, we have an isomorphism of $G_k$-modules
\begin{equation*}
\pi_1(X_{\overline k})^\mathrm{ab} \cong \prod_\ell T_\ell(J_X),
\end{equation*}
where $J_X$ denotes the Jacobian variety of $X$ and the $G_k$-module structure on $\pi_1(X_{\overline k})^\mathrm{ab}$ is induced by the outer Galois representation.
Have you tried looking in Szamuely's book "Galois Groups and Fundamental Groups"? That's usually a good place to look for things like this.
|
2025-03-21T14:48:31.533371
| 2020-07-19T03:53:44 |
365981
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Federico Poloni",
"Rajesh D",
"https://mathoverflow.net/users/14414",
"https://mathoverflow.net/users/1898"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365981"
}
|
Stack Exchange
|
Asymptotic expansion involving a matrix equation
Let $b = [b_1,b_2,b_3,...b_n]^T$
$A = [a_{ij}]_{n \times n}$ such that $a_{i,j} = 1\forall 1\le i,j \le n$
$C = [c_{ij}(\lambda)]_{n \times n}$ such that $c_{ij}(\lambda) = O(\frac{1}{\lambda})$
$A+C$ is known to be a symmetric positive semi definite matrix
$I_n$ is an $n\times n$ Identity matrix
$e = [e_1,e_2,e_3,...e_n]^T$
$$b = (A+C+\frac{1}{\lambda^2}I_n)^{-1}e$$
I am trying to find an asymptotic expansion for $b_i$
Show that $b_i = O(\lambda)$.
For example If I take matrix $C$ as all zeros, $A$ is given as all ones, and the case $n=2$ I get $$b_i = O(\lambda)$$
Is there any good argument to generalize for any $n$, that $b_i = O(\lambda)$?
PS: The elements of all matrices are non negative, except matrix $e$, which can
take negative values.
Are you interested in the case $\lambda \to 0$ or $\lambda \to \infty$ here?
@FedericoPoloni : $\lambda \to \infty$
Re your answer:
Statement 1 follows from Weyl's inequalities and $\rho_1(C) = O(\lambda^{-1})$.
$2 \implies 3$ seems problematic; what if $C = \frac{1}{\lambda^3}I$, which should be possible given the definition of big-O? Then $\rho_1(M^{-1}) \sim \lambda^2 \not\in O(\lambda)$.
Statement 4 seems just the matrix norm inequality $\|M^{-1}e\| \leq \|M^{-1}\|\|e\|$.
$M$ is positive definite, as $A+C$ is psd.
In question I had stated $A+C$ is known to be $psd$ so $M$ is a pd. Hope that solves the problem.
@RajeshDachiraju Updated the counterexample.
Your example still assumes $\rho(M) = 0$ which is not possible.
$M$ is positive definite.
EDIT: Ok, fixed counterexample, hopefully.
Your counter example seems valid to me. Luckily in my problem $C$ goes as $\frac{1}{\lambda}$...perhaps $O$ is not suitable here, I should have used $\omega$ or something. I have a closed form expression for $C$ which is $\frac{1}{\lambda}$. Thank you for the answer.
@RajeshDachiraju Still won't work. What if $C$ is entrywise $O(\lambda^{-1})$ but has an eigenvalue that goes like $O(\lambda^{-3})$? For instance, take $n>2$ and let $Q$ be an orthogonal matrix such that $Ef=f_1$, where $f$ is the vector of all ones and $f_1$ is the first column of $I$. Then $QAQ^{-1} = diag(n,0,0,...0)$. Take then $QCQ^{-1} = diag(0,\lambda^{-1},\lambda^{-3},\lambda^{-3},\dots,\lambda^{-3})$. Then if $Q$ is "complicated enough" all entries of $Q$ will scale like $\lambda^{-1}$, but $QMQ^{-1} = diag(n+\lambda^{-2},\lambda^{-1},\lambda^{-3},\lambda^{-3},\dots,\lambda^{-3})$.
@RajeshDachiraju That I don't know.
Thanks for your valuable comments and answer.
My proof (some statements are without proof, for which I need references)
Let $X$ be any $n\times n$ square matrix and Let $\rho_1(X), \rho_2(X), \rho_3(X),...\rho_i(X),...\rho_n(X)$ denote the eigenvalues in descending order, of the matrix $X$.
Statement : 1* $$\rho_1(A+C) = n+O(1/\lambda)$$ $$\rho_i(A+C) = 0+O(1/\lambda),i = 2,3...n$$
Let $M = A+C+\frac{1}{\lambda^2}I_n$
Statement : 2 $$\rho_1(M) = n+O(\frac{1}{\lambda})$$ $$\rho_i(M) = 0+O(\frac{1}{\lambda}),i = 2,3,...n$$
statement : 3 $$\rho_1(M^{-1}) = O(\lambda),i = 1,2,3,...n-1$$ $$\rho_{n-1}(M^{-1}) = O(1)$$
statement : 4* The highest Eigen value of $M^{-1}$ is $O(\lambda)$, hence $b = M^{-1}e = O(\lambda)$, which means $b_i = O(\lambda), i = 1,2,3...n$
Statements 1 and 4, I have no proofs or references, which I need some help.
Statements 2 and 3 are true and sources readily available.
|
2025-03-21T14:48:31.533606
| 2020-07-19T03:55:14 |
365982
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Emil Jeřábek",
"Eugene Zhang",
"Gabe Conant",
"https://mathoverflow.net/users/120374",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/38253"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365982"
}
|
Stack Exchange
|
Question on $\aleph_0$-categorical nonhomogeneous structures
Macpherson in a survey of homogeneous structures, states that there are many $\aleph_0$-categorical structures which are not homogeneous. Here homogeneity is the ultrahomogeneity that is defined as every isomorphism between two finite substructures of a structure $M$ can be extended to an automorphism of $M$. $\omega$-homogeneity means that any finite partial elementary mapping can be extended so that its domain includes any given element.
I am confused on this because it is well known that a $\aleph_0$-categorical structure is both atomic and countably saturated, and both atomic and countably saturated structures are $\omega$-homogeneous. This actually means that a $\aleph_0$-categorical structure is ultrahomogeneous. Where is wrong here?
By the way, $\aleph_0$-homogeneity and $\omega$-homogeneity mean the exact same thing. So, in addition to Emil's answer, you can refer to the discussion on your other question about why $\omega$-homogeneity is not the same as ultrahomogeneity.
You are confusing several notions of homogeneity. Saturated structures, and therefore also $\aleph_0$-categorical structures, are homogeneous, but not necessarily ultrahomogeneous. This means that every finite partial elementary mapping extends to an automorphism.
$\omega$-homogeneity is in fact an even weaker property: it says that any finite partial elementary mapping can be extended so that its domain includes any given element. However, this is equivalent to the property above for countable structures.
Ultrahomogeneity of $\omega$-saturated structures implies quantifier elimination, hence it is not implied by any standard model-theoretic properties that are invariant by expansion of the language with definable predicates.
In more detail, let me try to deconfuse Macpherson’s terminology by reviewing the relevant properties (using more standard terminology that does not drop the ultra- prefixes) and their connections. In what follows, $M$ is a structure, and $\kappa$ is an infinite cardinal.
$M$ is $\kappa$-homogeneous if for every partial elementary map $f\colon M\rightharpoonup M$ such that $|f|<\kappa$, and for every $a\in M$, there exists a partial elementary map $g\supseteq f$ such that $a\in\operatorname{dom}(g)$.
$M$ is strongly $\kappa$-homogeneous if every partial elementary map $f\colon M\rightharpoonup M$ such that $|f|<\kappa$ extends to an automorphism of $M$.
If $\kappa=|M|$, and $M$ is $\kappa$-homogeneous, it is in fact strongly $\kappa$-homogeneous. Such structures are simply called homogeneous.
$M$ is $\kappa$-ultrahomogeneous if for every partial isomorphism $f\colon M\rightharpoonup M$ such that $|f|<\kappa$, and for every $a\in M$, there exists a partial isomorphism $g\supseteq f$ such that $a\in\operatorname{dom}(g)$.
$M$ is strongly $\kappa$-ultrahomogeneous if every partial isomorphism $f$ such that $|f|<\kappa$ extends to an automorphism of $M$.
$M$ is ultrahomogeneous if it is $\kappa$-ultrahomogeneous (or equivalently, strongly $\kappa$-ultrahomogeneous) for $\kappa=|M|$.
The basic properties are:
If $M$ is $\kappa$-saturated, it is $\kappa$-homogeneous.
If $M$ is atomic, it is $\omega$-homogeneous.
The following are equivalent:
$M$ is $\kappa$-ultrahomogeneous;
$M$ is $\kappa$-homogeneous, and every partial isomorphism $M\rightharpoonup M$ is elementary.
Likewise for strong $\kappa$-ultrahomogeneity.
If $M$ is in a finite relational language, or if it is $\omega$-saturated, the following are equivalent:
Every partial isomorphism $M\rightharpoonup M$ is elementary.
$M$ has quantifier elimination.
Consequently, if $M$ is in a finite relational language, or if it is $\omega$-saturated, the following are equivalent:
$M$ is $\kappa$-ultrahomogeneous.
$M$ is $\kappa$-homogeneous, and $M$ has quantifier elimination.
What is the exact difference between strong ultrahomogeneous and ultrahomogeneous? The first part of each definition is the same, ie. "if every partial isomorphism f such that". So the difference lies in the second part of each difinition.
No, that is important. For example, by back and forth, this definition of κ-ultrahomogeneous means there is an automorphism. So it is the same as strongly κ
-homogeneous. If I am not wrong, strongly κ-homogeneous should be stronger than κ
-homogeneous as the name suggests.
This definition of $\kappa$-ultrahomogeneous means there is an automorphism: nonsense. Of course it is strictly stronger. For example, let $M$ be the lexicographic product $(\omega_1+1)\times\mathbb Q$ as a linearly ordered set. Then $M$ is $\omega$-ultrahomogeneous (every dense linear order without endpoints is), but it is not strongly $\omega$-ultrahomogeneous, as the partial isomorphism $(0,0)\mapsto(\omega_1,0)$ does not extend to an automorphism of $M$ (there are $\aleph_1$ elements above the former element, but only $\aleph_0$ above the latter).
These two definitions are the same for countable models in which automorphism can follow for sure by back and forth.
It is better noted that for countable structure, both definitions are equivalent
|
2025-03-21T14:48:31.534167
| 2020-07-19T04:16:42 |
365984
|
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"LSpice",
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}
|
Stack Exchange
|
If an estimate is false on $L^{1}$, then it is false for the $\delta$ distribution?
Let $u=\int e^{\dot{\imath}K(x,y)} f(y) dy$. My advisor told me that we can disprove an integrability estimate
$$\|u\|_{L^p}\lesssim \|f\|_{L^{1}}\label{1}\tag{1}$$
by disproving it when $f=\delta$, the Dirac Delta distribution.
When I asked him for the reasoning for this, he told me $\delta$ is the limit of a sequence of $L^{1}$ functions with norm 1, in the sense of distributions. Indeed, if $f\in L^{1}(\mathbb{R}^{d})$ with $\|f\|_{L^{1}(\mathbb{R}^{d})}=1$, we can define $f_{n}(x)=n^d\,f(nx)$, and then we can change variables and apply the dominated convergence theorem to show that
$$\int_{\mathbb{R}^{d}}f_{n}(x)\phi(x)dx=\int_{\mathbb{R}^{d}}f(x)\phi\left(\frac{x}{n}\right)dx\longrightarrow \phi(0)\int_{\mathbb{R}^{d}}f(x)dx=\delta(\phi)$$
for every test function $\phi$.
My question is: how this convergence in the sense of distributions justify/implies that if the estimate \eqref{1} is false when $f=\delta$ then \eqref{1} is also false for a general $f\in L^1$ ?
I mean if the convergence were in $L^{1}$ norm, then the claim is obvious. So I guess my question is, does there exist a sequence $f_n$ of (normalized) $L^{1}$ functions such that
$$\int_{\mathbb{R}^{d}}|f_{n}-\delta|\rightarrow 0\qquad\qquad ?$$
Obviously, by the argument above, we have
$$\int_{\mathbb{R}^{d}}f_{n}(x)\phi(x)dx\rightarrow
\int_{\mathbb{R}^{d}}\delta(x)\phi(x)dx\Longrightarrow
\int_{\mathbb{R}^{d}}[f_{n}(x)-\delta(x)]\phi(x)dx\rightarrow0$$
for every test function $\phi$. Where to go from here ?
Not a justification for the $\operatorname L^1$ versus distributional convergence, but a point of language: What I would understand from your advisor's claim is that an estimate that fails for the $\delta$ function must fail for some integrable $f$, not for a generic integrable $f$ (in whatever sense that's meant).
First of all, observe that your "subquestion" is ill-posed because $\int_{\mathbb{R}^{d}}|f_{n}-\delta|$ does not make any sense: the Dirac delta distributions is of course not an $L^1$ function, it is merely a distribution $\mathcal D'$.
The right answer goes as follows (well, more or less, you should actually give more information about the kernel $K$, but let me sketch out the idea nonetheless).
Just as you correctly pointed out that $\int f_n \phi\to <\delta,\phi>=\phi(0)$, it is easy to see that $u_n(x)=\int e^{i K(x,y)}f_n(y)\,dy$ converges pointwise a.e. to $u(x)=e^{iK(x,0)}=<e^{iK(x,\cdot),\delta}>$.
Regardless of the kernel $K$, this function of $x$ has modulus $\left|e^{iK(x,0)}\right|\equiv 1$ and is therefore not in $L^p$, so clearly this tends to violate $\|u\|_p\lesssim \|f\|_1$.
In order to conclude completely rigorously, one can argue as follows:
I first claim that pointwise a.e convergence can be improved to distributional convergence. Indeed, from the pointwise bound $|u_n(x)|=\left|\int e^{-iK(x,y)}f_n(y)\,dy\right|\leq \int\left| e^{-iK(x,y)}f_n(y)\,dy\right|=\int |f_n(y)|\,dy=1$ we see that $\|u_n\|_\infty\leq 1$.
Hence for any test function $\phi\in C_c$, and from the previous ponitwise a.e. convergence $u_n(x)\to u(x)$, an easy application of Lebesgue's dominated convergence theorem yields
$$
<u_n,\phi>=\int u_n(x)\phi(x)\to \int u(x)\phi(x)\,dx=<u,\phi>.
$$
assume now by contradiction that your estimate $\|u\|_p\lesssim\|f\|_1$ holds. Then taking $f=f_n$ with $\|f_n\|_1=1$ you see that $\|u_n\|_p\lesssim 1$ would be bounded. By the Banach-Alaoglu-Bourbaki theorem you conclude that, up to a subsequence, there would exist some $v\in L^p$ such that $u_n\rightharpoonup v$ weakly in $L^p$. But since weak $L^p$ convergence is stronger than distributional convergence, and by uniqueness of the limit in the sense of distributions, step 1 implies $u=v$. This is impossible since $u$ is not $L^p$ but $v$ would be (as a weak $L^p$ limit.)
Note: for $p=+\infty$ the Banach-Alaoglu-Bourbaki still applies, and $u_n\overset{*}{\rightharpoonup} v$ weakly star. The weak-* convergence is still better that distributional convergence, so we're good. The case $p=1$ might be more tricky, I've never thought of that (I'm not from harmonic analysis but I'm sure this must be a classical issue)
Thanks a lot. I appreciate the time and effort you put into making the answer so clear.
|
2025-03-21T14:48:31.534456
| 2020-07-19T06:08:57 |
365989
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"HenrikRüping",
"Kun Wang",
"https://mathoverflow.net/users/30881",
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}
|
Stack Exchange
|
Invariant neighborhood in a G-CW complex
Let $G$ be a discrete group and $X$ be a $G$-CW complex. For any $x\in X$ and open neighborhood $U$ of $x$, I am interested in the question that whether we can find a $G_x$-invariant open neighborhood $V$ of $x$ that is contained in $U$. This is not true in general, for example, if $X$ is the cone over $\mathbb R$ with the $G=\mathbb Z$ action by translation, then the cone point $x_0$ has a neighborhood which does not contain any $G_{x_0}$-invariant open neighborhood. But I am wondering whether the following is true:
Any $G$-CW complex $X$ is $G$-homotopy equivalent to a $G$-CW complex $Y$ so that any open neighborhood $U$ of $y\in Y$ contains a $G_y$-invariant open neighborhood $V$.
As a special case, I am wondering whether the following is true:
Let $\mathcal F$ be a family of subgroups of $G$ which is closed under conjugation and finite intersection. Can we always find a model $E_{\mathcal F}G$ for the universal $G$-CW complex relative to $\mathcal F$ so that any open neighborhood of $x\in E_{\mathcal F}G$ contains a $G_x$-invariant open neighborhood?
Thank you!
I will call the property of a $G$-CW-complex that inside every neighborhood of a point one can find a $G$-invariant neighborhood property $A$.
As in your example, a graph where an edge stabilizer has infinite index in one of the adjacent vertex stabilizers does not have property $A$.
Further $G$-subcomplexes of $G$-CW-complexes with property $A$ have property $A$; given an open set in the subcomplex, extend it to an open set of the whole complex, choose that invariant neighborhood there and intersect back.
Now look at the free group in $a,b$ with the family of subgroups $\mathcal{F}$ containing the trivial group and all conjugates of $\langle a \rangle$ and $\langle b \rangle$. One model for $E_FG$ is given by the Bass-Serre tree, which does not have property $A$. But why can't there be a better model?
Then its one-skeleton would also have property $A$. But by the defining property of $E_FG$ we can find a point $p_a$ with $\langle a \rangle$ is contained in the stabilizer of $p_a$. Since $\langle a \rangle$ is maximal in $F$, it actually follows that $\langle a \rangle$ is the stabilizer of $p_a$. Choose $p_b$ analogously.
Since the one-skeleton is connected, we can choose a finite path from $p_a$ to $p_b$. we want to show that on that path there is some edge, whose stabilizer group has infinite index in the stabilizer of one adjacent vertex. If this was not the case, then $\langle a \rangle$ and $\langle b \rangle$ would be comensurable in the free group, which they are not.
Thanks, Henrik. The reason that I asked this question is I was reading the proof of Proposition 7.4 of the paper https://arxiv.org/abs/math/0701434 (I guess you probably have read this paper). On page 28, it says "Let $\tilde{U}$ be the preimage of $U\subset E$ under the $G$-equivariant map $e\mapsto h_s(p(y), e)$". But I don't think this map is $G$-equivariant when $y$ is fixed (it is $G$-equivariant as a map $(e, y)\mapsto h_s(p(y), e)$). So $\tilde{U}$ won't be $G_{\tilde e}$-invariant. Do you have any thought about this or there is something missing in my understanding.
Without having a detailed look I would guess it could one of the two:
What is the $G$-action used on a product of $G$-spaces. One would guess its the diagonal action, but maybe they choose just to act on $G$.
2)Maybe its just a typo and $\tilde{U}$ should just be $G_y$-invariant ? would this suffice?
@KunWang Does your gmail Address still work? (We exchanged a couple of emails 7 years ago). I just found in my PDF folder the slides of a talk I gave in 2014 about this. Of course a talk should be less formal than the paper, so maybe these are exactly the opposite of what you are looking for, but I can send them to you anyway.
Yes, Henrik. That would be very helpful. Thanks a lot!
|
2025-03-21T14:48:31.534716
| 2020-07-19T07:15:21 |
365993
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Thomas Schucker",
"https://mathoverflow.net/users/142151",
"https://mathoverflow.net/users/40804",
"mme"
],
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365993"
}
|
Stack Exchange
|
A pseudo-Riemannian version of a theorem by Fubini
Guido Fubini, ``Sugli spazii che ammettono un gruppo continuo di movimenti,'' Annali di Mat., ser. 3, 8 (1903) 54.: Let $M$ be a Riemannian manifold of dimension $d\ge 3$. Its isometry group cannot be of dimension $d\,(d+1)/2-1$.
Does this theorem remain true in the pseudo-Riemannian setting?
I am not very familiar with Lie algebras but I think the following argument works: This would give a codimension 1 subgroup of $O(p,q)$. A theorem of Tits (see here) then would give a nonzero homomorphism $\mathfrak o(p,q) \to \mathfrak{sl}_2$. But $\mathfrak o(p,q)$ is simple, so this implies $d = p + q \leq 2$.
@MikeMiller For $d=4$ the maximal isometry groups are the anti de Sitter group $O(2,3)$, the Poincar'e group $O(1,3)\ltimes R^4$ and the de Sitter group $O(1,4)$.
Thank you for correcting me! I will leave my comment up in case it helps anyone else who is unfamiliar.
I just found the answer by G. S. Hall (2003) in Class. Quantum Grav. 20 3745.
Theorem 8. Let $M$ be a connected smooth paracompact manifold of dimension $n ≥ 3$ admitting a smooth metric $g$ of signature $(p,q)$ with $p ≤ q$ and $q ≥ 3$. Let $K(M)$ be the vector space of all global smooth Killing vector fields on $M$. Suppose dim$K(M)\not=1/2\,n(n+1)$. Then dim$K(M)<1/2\,n(n+1)−1$.
|
2025-03-21T14:48:31.534947
| 2020-07-19T07:32:44 |
365994
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Lucian",
"https://mathoverflow.net/users/39602"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365994"
}
|
Stack Exchange
|
Algebraicity of a ratio of values of the Gamma function
The following ratio:
$$\frac{\Gamma(2/5)^3}{\pi\Gamma(1/5)}$$
has kept appearing in my research, and the only thing I know about its value is that it is $\cong 0.7567213$, whence the following two questions:
Is the value of this ratio an algebraic number?
What is the exact value of this ratio?
This number is expected to be transcendental. This answer gives a conceptual framework for studying the algebraicity of such $\Gamma$ ratios, and in fact a completely explicit criterion (which is only conjectural, when it comes to establishing transcendence).
Your number is equal to
\begin{equation*}
\frac{\Gamma(2/5)^3}{\Gamma(1/5)^2 \Gamma(4/5)}
\end{equation*}
up to multiplication by an algebraic number. This ratio is described by the vector of exponents $(-2,3,0,-1)$ and the criterion there is not met for $u=2$. In fancy terms, the Bernoulli distribution $B_1$ does not vanish on this vector.
...which itself can be rewritten as $~\dfrac{B\bigg(\dfrac25,~\dfrac25\bigg)}{B\bigg(\dfrac15,~\dfrac15\bigg)}~.$
|
2025-03-21T14:48:31.535036
| 2020-07-19T08:55:44 |
366001
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Laurent Moret-Bailly",
"https://mathoverflow.net/users/7666"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366001"
}
|
Stack Exchange
|
conditions on a morphism $f:X\rightarrow Y$ to ensure $X$ is reduced, given $Y$ is reduced?
Let $X,Y$ be finite type projective schemes over $\mathbb{C}$, and $f:X\rightarrow Y$ be a surjective morphism (but not an isomorphism). Suppose it is known that $Y$ is reduced, and the fibers of $f$ are reduced as well.
Are there any known results about $f$, which ensures that $X$ is also reduced?
There are similar results for connectedness, but I can't find anything for reducedness.
With these assumptions, the answer is no: take $X=Y\amalg Z$ where $Z$ is an non-reduced closed subscheme of $Y$.
In Lemma 1.4. of this article, it is proven that $f$ flat, $X$ pure dimensional and $Y$ irreducible ensures that $X$ is reduced in your case.
Maybe some of these requirments can be relaxed, I haven't thought really thought about it.
|
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