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2025-03-21T14:48:29.994063
| 2020-03-03T20:21:09 |
354087
|
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|
Stack Exchange
|
When is validity definable in $L_\alpha$?
Below, $\alpha$ is a countable p.r.-closed ordinal $>\omega$.
Let $\mathcal{L}_\alpha=\mathcal{L}_{\infty,\omega}\cap L_\alpha$ (note that this is not the same as $\mathcal{L}_{\omega_1,\omega}\cap L_\alpha$ since elements of the latter are explicitly countable). By the Barwise completeness theorem if $\alpha$ is admissible or a limit of admissibles the set of $\mathcal{L}_\alpha$-validities is definable in $L_\alpha$ - indeed, it's $\Sigma_1$ over $L_\alpha$ - the point being that we have a good notion of proof.
However, if $\alpha$ is neither admissible nor a limit of admissible this argument breaks down. For some such $\alpha$ the result still holds, e.g. if $L_\theta$ is interpretable in $L_\alpha$ where $\theta$ is the next admissible above $\alpha$ (and in particular whenever the supremum of the $\alpha$-recursive ordinals is admissible), but that seems rather rare (e.g. club-many countable ordinals are non-Gandy).
My question is:
For which $\alpha$ is $\mathcal{L}_\alpha$-validity definable in $L_\alpha$?
(I'd also be interested in the finer-grained question of $\Sigma_1$-definability; for what it's worth, I'm not sure which is harder.) "Obviously" this should fail for most $\alpha$ ... but at present I don't even have an example of a single $\alpha$ for which it fails!
Closely related is the following question:
Supposing $\varphi\in\mathcal{L}_\alpha$ is a validity, how high up in the $L$-hierarchy do I have to look to find a proof of $\varphi$?
I suspect the answer to this question is not in general "unboundedly high below the next admissible."
Note that unlike Barwise completeness, Barwise compactness does not apply to inadmissible limits of admissibles. For example, work in $\theta=\sup_{n\in\omega}\omega_n^{CK}$, for $i\in\omega$ let $\varphi_i$ be the sentence saying that $L_{\omega_i^{CK}}$ is an initial segment of the universe, and let $\psi$ be the sentence "$\omega$ is standard and there are only finitely many admissible ordinals." Then ${\varphi_i:i\in\omega}\cup{\psi}$ is $\mathcal{L}_\theta$-finitely satisfiable but clearly unsatisfiable.
|
2025-03-21T14:48:29.994222
| 2020-03-03T20:23:46 |
354088
|
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|
Stack Exchange
|
Number of $O_K$-submodules of given index (or given elementary divisors)
Let $K / \mathbb{Q}$ be any number field, and $O_K$ its ring of integers. Consider the $O_K$- submodule $M := O_K^d = \{(x_1, \ldots, x_d):x_i \in O_K\}$, and choose integral ideals $I_1 \subseteq \ldots \subseteq I_d$. Now the question is: how many distinct $O_K$-submodules $N$ does $M$ have such that $M/N \cong O_K/I_1 \oplus \ldots \oplus O_K/I_d$?
This is probably elementary, but stripped of all the conveniences I have in case $K=\mathbb{Q}$, I am not sure how to proceed without invoking too much theory.
|
2025-03-21T14:48:29.994298
| 2020-03-03T20:37:30 |
354089
|
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|
Stack Exchange
|
Is the graph minicut with the node cardinality constraint NP-hard?
I wonder whether the following problem is a well-studied NP-hard problem?
Get a graph $G$ and a number $k$, we partition the graph $G$ into two components where each component should have at most $k$ vertices and the number of edges in the cut is minimal.
In other words, is the mini-cut problem with the vertex budget constraint NP-hard?
Thanks.
How many vertices does $G$ have and are there lower bounds on the vertices in each component?
No, there are not lower bounds on the vertices in each component. You can assume that the number of vertices of $G$ is $n$.
Since we're looking for a bipartition, upper bounds on the part sizes are exactly the same as lower bounds.
Min-cut with bounded part sizes is the same as max-cut with bounded part sizes on the complement. Given any graph $H$ for which we would like to know the max-cut, form a new graph $G$ by adding $|H|$ isolated vertices. Then the max-cut of $G$ with bounded part sizes is the max-cut of $H$, with the part sizes balanced out using the isolated vertices. So max-cut can be reduced to max-cut (or min-cut) with bounded part sizes, and your problem in NP-hard.
Could you elaborate your first claim? I imagine something like that to be true, but don't see the exact correspondence.
@smpapers, you're right, that only works if the parts are balanced rather than bounded -- if the partition is into parts of size as equal as possible then they're equivalent because the total number of possible crossing edges is fixed. Happily the reduction I sketch is to balanced max-cut, and so balanced min-cut, which is a special case of bounded min-cut (so the general case is still NP-hard).
Great, I agree.
|
2025-03-21T14:48:29.994450
| 2020-03-03T21:27:22 |
354092
|
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"Andreas Blass",
"André Henriques",
"Henno Brandsma",
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|
Stack Exchange
|
Which compact metrizable spaces have continuous choice functions for non-empty closed sets?
Let $X$ be a compact metrizable space and let $\mathcal{K}_{ne}(X)$ be the collection of non-empty closed subsets of $X$ with the Vietoris topology (i.e. the topology induced by the Hausdorff metric for any compatible metric on $X$).
Question: When does there exist a continuous function $f: \mathcal{K}_{ne}(X) \rightarrow X$ such that for every $G \in \mathcal{K}_{ne}(X)$, $f(G) \in G$?
This feels like it should have been studied before, but I am unable to find a reference.
Some easy observations:
If $X$ has a continuous choice function for non-empty closed sets and $Y$ is a closed subspace of $X$, then $Y$ has a continuous choice function for non-empty closed sets.
$\inf : \mathcal{K}_{ne}([0,1])\rightarrow [0,1]$ is a continuous choice function for non-empty closed subsets of $[0,1]$. So we also have this for any closed subspace of $[0,1]$, such as Cantor space and any countable compact metrizable space.
The circle and the tripod (three copies of $[0,1]$ glued together at $0$) both do not have continuous choice functions for non-empty closed sets (in both spaces given a set with two points there is a continuous path that makes the points switch places while keeping them separate). So no spaces in which these embed do either.
Any finite disjoint union of spaces with continuous choice functions for non-empty closed sets also has a continuous choice function for non-empty closed sets (having elements in a clopen subset is a clopen condition in $\mathcal{K}_{ne}(X)$, so we can patch together the choice functions by cases).
A reasonable conjecture is that any such space embeds into $[0,1]$, but I could also see something tricky like the pseudo-arc having a continuous choice function for non-empty closed sets.
Nice question! Let $N$ be the one-point compactification of $\mathbb N$. Then the same kind of argument (by considering two-point subsets) also proves that $[0,1]\times N$ does not admit a continuous choice functions for non-empty closed sets.
@AndréHenriques Very nice. A similar argument should work for solenoids.
Indeed, a solenoid contains $[0,1]\times N$ as a subspace.
See https://mathoverflow.net/questions/74614 for the special case of continuous choice from two-point sets, which already excludes lots of spaces.
It's an old (1981) theorem by Jan van Mill and Evert Wattel (see this paper) that a compact space has a continuous selection iff it is orderable. (So has a linear order whose order topology is the topology on $X$). $F \to \min(F)$ and $F \to \max(F)$ are then the two only continuous selection functions IIRC. Even a continuous selecting function for $[X]^2$, the subspace of doubletons, is enough to get orderabilility.
If a space $X$ admits three distinct points $x_1,x_2,x_3\in X$ such that $X\setminus\{x_i\}$ is connected for every $i=1,2,3$, then $X$ does not admit a continuous choice function from the set of two-point subsets of $X$ back to $X$.
Indeed, such a choice function could be used to define three continuous maps $f_i:X\setminus\{x_i\}\to \{0,1\}$, and at least one of $f_1,f_2,f_3$ would have to be non-constant. Contradiction.
This argument excludes the pseudo-arc.
Do you know if this implies that any continuum with more than one point and a continuous choice function for non-empty closed sets must be homeomorphic to $[0,1]$?
It seems that the answer is yes. A continuum cannot have precisely 1 non-cut point and a continuum with precisely 2 non-cut points is homeomorphic to $[0,1]$.
@JamesHanson a metrisable continuum, yes, let's not restrict ourselves...
|
2025-03-21T14:48:29.994692
| 2020-03-03T21:39:18 |
354095
|
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"Amir Sagiv",
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|
Stack Exchange
|
Lorenz ODEs with negative parameters
Consider the Lorenz system
$$\dot{x}(t) = \sigma(y-x) \, ,$$
$$\dot{y}(t) = x(\rho-z) - y \, ,$$
$$\dot{z}(t) = xy-\beta z \, .$$
Usually one considers the parameters $\sigma, \rho,$ and $\beta$ to be positive. Is there an inherent problem in the system if these parameters are negative?
(cross posted after 9 days on MSE)
Excellent question, I don't understand why people downvoted the question. I accidently put a negative in front of one of the parameters and got some "funky" figures. It would be interesting to know what the physical meaning of such a system is though...
@ThomasW agreed, and I search for an answer or a reference quite thoroughly...
There are many well-known ODE systems from various fields of science that can be transformed into the Lorenz system or generalized Lorenz system after a change of variables. After such a transformation the parameters of the original system may become negative in terms of the Lorenz system. The reason for looking for such a transformation is simple: to provide general methods for studying all these systems.
See, for example,
Leonov G. A., Kuznetsov N. V. On differences and similarities in the analysis of Lorenz, Chen, and Lu systems. Applied Mathematics and Computation, 256, 334-343 (2015).
and references therein and around the paper.
Thank you! In such cases, what is known about the negative-parameter regime? From a short glimpse at the Leonov-Kuznetsov paper, I don't see a direct discussion on that.
For example, the attractor in the Chen system corresponds to a repeller in the Lorenz system (for some parameters, which include negative). Lorenz-like systems are extensively studied in many papers and it is hard to compare explicitly which regions of parameters are affected because most of theorems include several long inequalities in terms of the parameters. I remembered a paper, which explicitly deals (numerically) with the region of negative paremeters in the Lorenz system, but now I cannot find it. As I remembered it, there is nothing essentially new.
|
2025-03-21T14:48:29.994883
| 2020-03-04T01:12:24 |
354100
|
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|
Stack Exchange
|
Formula for volume of $n$-ball for negative $n$
Does the expression $$\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n,$$ which gives the volume of an $n$-dimensional ball of radius $R$ when $n$ is a nonnegative integer, have any known significance when $n$ is an odd negative integer?
You should explain the context why you are asking this question to make it more relevant.
Pure curiosity.
It can occur in the dimensional regularization of Feynman diagram integrals in quantum field theory, but I don't think it has too much significance there.
Here is a funny formula: with $\nu !=\Gamma (\nu+1)$, we have $$
\sum_{\nu\in \mathbb N\cup (\mathbb N+\frac12)}\frac{π^\nu}{\nu !} R^{2\nu}=e^{π R^2}
+\sum_{k\in \mathbb N}\frac{π^{k+\frac12}}{(k+\frac12)!} R^{2k+1}
=\sum_{\nu\in \mathbb N\cup (\mathbb N+\frac12)}\vert \mathbb B^{2\nu}\vert R^{2\nu}
=\sum_{n\in \mathbb N}
\vert \mathbb B^{n}\vert R^{n}.
$$
Maybe https://en.wikipedia.org/wiki/Reflection_formula is relevant here?
According to wikipedia, $\Gamma(\frac{1}{2}-m) = \frac{(-4)^mm!}{(2m)!}\sqrt{\pi}$. Substituting this in to the above formula gives $|B^{-(2m+1)}| = \frac{(2m)!}{(-\frac{\pi}{4})^mm!}$.
I already asked this question here: https://math.stackexchange.com/questions/1176034/what-are-the-negative-dimentional-n-sphere-and-n-cube There are some answers.
I will answer regarding the dimension $-1$.
An example of such space is a set of periodic lattices on a real line.
Indeed, you can see that the Hausdorff dimension of a periodic lattice is $-1$: If we scale down the lattice twice, it would be able to include two original lattices. Since scaling the fractal down twice makes it twice as big, its dimension is $\frac{\ln 2}{\ln (1/2)}=-1$.
The formula for the volume of a ball gives $\frac1{\pi R}$ for $n=-1$. This means the unit ball ($R=1$) is the $\pi$-periodic one.
If we reduce the radius (the step of the lattice divided by $\pi$), its volume increases: a lattice with step $1/2$ can be represented as two lattices of step $1$. Thus the volume of a lattice of step $1/2$ consists of two $1$-periodic lattices, and so has volume twice the volume of $1$-periodic lattice.
So, the volume of the $-1$-ball is the density of the lattice.
The ball is also a solid cube with side equal to the step of the lattice. Thus a unit cube is a lattice with step 1, its volume is 1. Thus, in dimensions 0,1 and -1 a ball is also a cube.
A ball with radius $1$ is a cube with side $\pi$ and a cube with side $1$ is a ball with radius $\frac1\pi$.
Interesting observation: in positive-dimentional space the ball becomes a point with reduction of its radius to zero. In zero-dimensional Euclidean space the ball is always a point, disregarding the "radius" and in this $-1$-dimensional space the ball becomes a zero-volume point when we increase the radius infinitely.
You can find some further ideas here.
This proposed value of the “volume” of the (-1)-ball does not agree with the formula given in the original post.
@JamesPropp oh yes, i missed a mistake
I do not see why “the unit sphere is a pi-periodic lattice” (and I doubt that it is a “known fact”). Also, Anixx’s style of answer is unlikely to discriminate between even and odd (negative) dimensionalities, so I think it’s on the wrong track.
@JamesPropp This answer is specifically about the dimension -1, and the generalization to other dimensions may require efforts. Particularly, because in dimension -1 the periodic lattice is both a sphere and a cube.
@JamesPropp And, in general, when we are talking about negative dimensions, the bodies in them are littices, spectrums and other (infinite but countable) collections of dots, but their volumes are determined based on their overall density (or the infinite part of numerocity if you prefer), not on the number of dots they include.
@JamesPropp also, take a look at this: https://tglad.blogspot.com/2017/08/reframing-geometry-to-include-negative.html
@JamesPropp a periodic lattice in -1-dimension is both a ball and a a cube. That's why I confined my answer to the dimension -1 only. If we take the dimension -2, the square lattice would be a cube, but not a sphere, and I cannot tell how a ball in this dimension should look like
I want to make another answer to this question, although it is is not really an answer, and I do not know whether it is relevant or not, but it may be useful.
Let's take
$$V(n,R)=\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n$$
Also, using the reflection formula for the Zeta function we can see:
$$n\zeta(1-n)\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}=(1-n)\zeta(n)\frac{\pi^{\frac{1-n}{2}}}{\Gamma(\frac{1-n}{2}+1)}$$
Interestingly enough, in the algebra of divergent integrals described here, there is a rule $\operatorname{reg}\omega_+^n=-n\zeta(1-n)$, (where $\omega_+=\int_{-1/2}^\infty dx$ is an infinite divergent integral and $\operatorname{reg}$ denotes regularization), the overall formula becomes
$$\operatorname{reg} V(n, \omega_+)=\operatorname{reg} V(1-n, \omega_+)$$
This is a very nice-looking relation but its deep meaning is unclear. Particularly because it is not evident what meaning should have the balls with infinite radius, and especially the balls of infinite radius in negative dimension :-). Still formally this relation looks beautiful.
See also this question: What is the connection between the Riemann Xi-function and n-sphere?
|
2025-03-21T14:48:29.995260
| 2020-03-04T02:05:32 |
354103
|
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|
Stack Exchange
|
Articles about Mather's Geometric Groups?
I'm trying to find some information about "Mather's Geometric Groups". But the information on that subject is quite scarce, the only thing I found was the "Mather's Geometric Lemma" in the book "Local Features in Natural Images via Singularity Theory" by J. Damon.
I really appreciate your help, thanks.
|
2025-03-21T14:48:29.995324
| 2020-03-04T02:31:31 |
354105
|
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|
Stack Exchange
|
Sato-Tate for length zero intervals
Let $E$ be an elliptic curve (without CM) over a number field $K.$ Is it known that $a(\mathfrak{p})=N(\mathfrak{p})+1-|E(\mathbb{F}_{\mathfrak{p}})|$ is neither zero, not $2\sqrt{N(\mathfrak{p})}$ for almost all primes (in the sense of density) ? In literature I know, Sato-Tate holds for totally real number fields with some additional conditions on the elliptic curve. I am actually asking for this particular case of length zero, but for arbitrary number fields.
Of course $a(p)$ is never irrational, so your second sentence is equivalent to $a(p)$ is non-zero for almost all primes -- or I misunderstand something. For curves with complex multiplication that is very wrong and for the others it depends on what you mean by "almost all primes". Maybe you could improve the question.
Actually, yes assume E is without CM, and when we are looking over arbitary number fields, $a(\mathfrak{p})=N(\mathfrak{p})+1-|E(\mathbb{F}_{\mathfrak{p}})|$. I m asking for density of prime ideals $\mathfrak{p}$ in $\mathcal{O}K$ such that $a(\mathfrak{p})$ is neither zero, nor $N(\mathfrak{p})^{1/2}$, and $\mathbb{F}{\mathfrak{p}}$ is the residue field at $\mathfrak{p}$.
I think it would be good to modify the question accordingly.
I just did, does it look good now ?
Yes, by the way https://mathoverflow.net/questions/115555/for-how-many-primes-does-an-elliptic-curve-over-a-totally-imaginary-field-have-s is related
The link actually answers my question generically for degree $2$ prime ideals. How about the degree $1$ case ? Also this shows an effective version generically. I wanted a much weaker (I only need zero density) result for a given non CM curve.
|
2025-03-21T14:48:29.995568
| 2020-03-04T02:32:24 |
354106
|
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|
Stack Exchange
|
Volume computation using probabilistic approach
Let $\mathbb{S}^{d-1}=\{v\in\mathbb{R}^d:\|v\|_2=1\}$, namely $d-$dimensional sphere. It is well-known that if a random vector $X$ is distributed uniformly on $\mathbb{S}^{d-1}$, then there exists i.i.d. standard normal random variables $N_1,\dots,N_d$ such that
$$
X \stackrel{d}{=}(N_1/N,\cdots,N_d/N),
$$
where $N=\sqrt{N_1^2+\cdots+N_d^2}$.
Now, my question is the following. Suppose that I am interested in understanding the size of a certain subset $S$ of $\mathbb{S}^{d-1}$ (with respect to $(d-1)-$dimensional Lebesgue measure). One can then set
$$
\mathbb{P}(X\in S) = {\rm Size}(S)/{\rm Size}(\mathbb{S}^{d-1}).
$$
Here, the presence of Gaussianity often makes the computation simple, thereby suggesting a potential approach for computing the size of $S$. Is this type of an approach well-known? Namely, is there any work on calculating the sizes of complicated sets using this type of probabilistic reasoning?
This approach is of course well known. Clearly, it just says that
$$P(X\in A)=P((N_1,\dots,N_d)\in C_A),$$
where $A$ is a Borel subset of the unit sphere $S^{d-1}$ and $C_A:=\mathbb R_+A$ is the corresponding cone.
The hard part is to compute the Gaussian measure, $P((N_1,\dots,N_d)\in C_A)$, of the cone $C_A$.
This is hard even when $A$ is a spherical simplex. The work on this begins with Schläfli (1858), whose paper (I think) is hard to read. A more general result was obtained by Plackett (1954), who gave a recursive formula (see formulas (7) and (6) there) for the centered Gaussian measure of a simplicial cone in $\mathbb R^d$ (not necessarily with the vertex at the origin) in terms of a certain integral functional of certain partial derivatives of the Gaussian measure of a varying simplicial cone in $\mathbb R^{d-2}$. See also references in Plackett's paper, including the one to Schläfli. (Plackett is actually dealing with the obviously equivalent problem, where the simplicial cone is just $c+\mathbb R_+^d$ for some $c\in\mathbb R^d$, but the Gaussian distribution is not necessarily the standard one.)
By $N$ you mean $(N_1,\ldots,N_d)$, not what OP calls $N$, right?
@BrendanMcKay : Of course. Thank you for your comment. This is now fixed.
Actually, this was done for the simplex in 1936 by Hellmuth Kneser (unfortunately published in Deutsche Mathematik) using exactly the OP's Gaussian extension method.
Iosif, thank you very much for your answer; and many thanks for the comments guys!
|
2025-03-21T14:48:29.995751
| 2020-03-04T03:47:34 |
354110
|
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|
Stack Exchange
|
The interaction between differentials on a graded ring and chain-homotopy equivalences
I am wondering about the following question:
Given a differential graded algebra $A$, how many other differentials can we put on the underlying graded ring of $A$, which are also chain-homotopy equivalent to $A$ ?
I'd like some references, perhaps, that may address the above question. Or if anyone here knows about results related to this question, please cite them here.
Thanks!
|
2025-03-21T14:48:29.995820
| 2020-03-04T05:34:02 |
354115
|
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"url": "https://mathoverflow.net/questions/354115"
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|
Stack Exchange
|
Solution to differential equation $f^2(x) f''(x) = -x$ on [0,1]
I'd like to solve a differential equation $$ f^2(x) f''(x)=-x $$ where $f(x)$ is defined on $[0,1]$ and has a boundary condition $f(0)=f(1)=0$.
I somehow found out that the solution is fairly close to $f(x) = x^{1/3} \phi^{2/3}(\Phi^{-1}(1-x))$ where $\phi$ and $\Phi$ are pdf and cdf of a standard normal distribution, but it fails to solve the differential equation exactly.
Thank for all comments!
Based on the solution structure of Emden–Fowler Equation, I was able to identify the values of constants that satisfy the boundary conditions. The followings are the details:
Define
\begin{equation}
Z_R(\tau) \triangleq \sqrt{3} J_{1/3}(\tau) - Y_{1/3}(\tau)
, \quad
Z_L(\tau) \triangleq - \frac{2}{\pi} K_{1/3}(\tau)
\end{equation}
where $J, Y, K$ are Bessel functions.
Further define
\begin{equation}
\bar{\tau} \triangleq \inf\{ \tau > 0; Z_R(\tau) = 0 \} \approx 2.3834
, \quad
a \triangleq \frac{1}{ \bar{\tau}^{4/3} Z_R'(\bar{\tau})^2 } \approx 0.2910
, \quad
b \triangleq a \left( \frac{9}{2} \right)^{1/3} \approx 0.1763.
\end{equation}
Then, the solution curve $\{ (x, f(x)) \}_{x \in [0,1]}$ is characterized by
\begin{equation}
\left\{ \left( x_R(\tau), y_R(\tau) \right) \right\}_{\tau \in [0, \bar{\tau}]} \bigcup \left\{ \left( x_L(\tau), y_L(\tau) \right) \right\}_{\tau \in [0, \infty]}
\end{equation}
where
\begin{equation}
x_R(\tau) \triangleq a \tau^{-2/3}\left[ \left( \tau Z_R'(\tau) + \frac{1}{3} Z_R(\tau) \right)^2 + \tau^2 Z_R(\tau)^2 \right]
, \quad
y_R(\tau) \triangleq b \tau^{2/3} Z_R(\tau)^2.
\end{equation}
\begin{equation}
x_L(\tau) \triangleq a \tau^{-2/3}\left[ \left( \tau Z_L'(\tau) + \frac{1}{3} Z_L(\tau) \right)^2 - \tau^2 Z_L(\tau)^2 \right]
, \quad
y_L(\tau) \triangleq b \tau^{2/3} Z_L(\tau)^2.
\end{equation}
In addition to this analytic solution, I also obtained a numerical solution by repeatedly computing
$$ f_{k+1}(x) \gets \left[ \left( f_k(x-2h) + f_k(x+2h) \right) + 4 \left(f_k(x-h)+f_k(x+h)\right) + \frac{8 x h^2}{f_k^2(x)} \right] \big/ 10 $$
on the grid $x \in \{2h,3h,\ldots,1-3h,1-2h\}$ for small $h$ with an initialization $f_0(x) \triangleq 0.5(1-(1-2x)^2)$.
The following figure shows these solutions:
$f(1)=0\ $ would imply $\ 0 = -1.\ $
@WlodAA We will have $f''(1) = -\infty$.
I see. Thank you.
This is a case of the Emden-Fowler equation, see http://eqworld.ipmnet.ru/en/solutions/ode/ode0302.pdf, which arises in astrophysics and was subject of intensive study. E. Hille wrote that "it has fantastically complicated singularities near x=0".
Surprisingly, this case of the Emden-Fowler equation is explicitly solvable:
see formula (2.3.27) in A. Polyanin and V. Zaitsev, Handbook of exact solutions
of ordinary differential equations, Chapman & Hill, 2003.
I copy the formula, without verifying it. Let
$$Z=C_1J_{1/3}(\tau)+C_2Y_{1/3}(\tau),$$
or
$$Z=C_1I_{1/3}(\tau)+C_2K_{1/3}(\tau),$$
where $J,Y$ are Bessel and $I$, $K$ are modified Bessel functions.
Then
$$x=a\tau^{-2/3}[(\tau Z^\prime+(1/3)Z)^2\pm\tau^2Z^2],\quad y=b\tau^{2/3}Z^2$$
satisfy $d^2y/dx^2=Axy^{-2}$ with $A=-(9/2)(b/a)^3.$
For the $+$ sign in $\pm$ take the first formula for $Z$, and for the $-$ the second one.
Remark. Emden-Fowler equation appears for the first time in the famous book by R. Emden, Gaskugeln (1907) and since then frequently arises in the study of stars and black holes.
The first version with $Y$ seems to be correct, but I think it does not lead to a solution of the original problem. However, the version with $K$ instead of $Y$ does not seem to solve the differential equation (with either version of the $\pm$ sign). Perhaps there is some transcription error?
Maple solves this using the Airy functions Ai and Bi, which are related to Bessel function of order $1/3, 2/3$.
@Neil Strickland: Yes, I made a misprint while copying from the book. Now corrected (second definition of Z has I instead of J).
I tried the following approach. Put $y=f(x)$ and $t=2x-1$ so the differential equation becomes $y^2\ddot{y}=-(t+1)/8$ with boundary conditions $y=0$ at $t=\pm 1$. We can then write $y=\sum_ia_it^i$. The differential equation gives a recurrence relation expressing all the coefficients $a_i$ in terms of $a_0$ and $a_1$. We can then truncate the power series to a given order $d$ and solve numerically for the boundary conditions. This seems to work in a well-behaved way, with good convergence at the endpoints and a result that is stable when we increase $d$. It looks like $a_0=0.450$ and $a_1=0.120$ to $3$ decimal places. Maple code is as follows:
with(plots):
Digits := 50:
d := 50:
y := add(a[i] * t^i,i=0..d):
sol0 := solve([coeffs(rem(expand(y^2 * diff(y,t,t) + (t+1)/8),t^(d-1),t),t)]
{seq(a[i],i=2..d)}):
y0 := expand(subs(sol0,y)):
sol1 := fsolve({subs(t= 1,y0),subs(t=-1,y0)},{a[0]=0.45,a[1]=0.1}):
aa[d] := subs(sol1,[a[0],a[1]]);
y1 := subs(sol1,y0);
y1x := subs(t = 2*x-1,y1):
Phi := unapply((1 + erf(x))/2,x):
phi := unapply(diff(Phi(x),x),x):
display(
plot(y1x,x=0..1,colour=red),
plot(x^(1/3) * phi(RootOf(1-x-Phi(_Z)))^(2/3),x=0..1,colour=blue)
);
This generates the following picture. The power series solution is in red and the function $x^{1/3}\phi(\Phi^{-1}(1-x))^{2/3}$ is in blue.
The coefficients $a_{2i}$ lie on a nice smooth curve, and the coefficients $a_{2i+1}$ lie on a similar curve shifted down slightly. Logs of the absolute values can be displayed as follows:
display(
listplot([seq(log(-coeff(y1,t,2*i)),i=3..(d-1)/2)],style=point,colour=red),
listplot([seq(log(-coeff(y1,t,2*i+1)),i=3..(d-1)/2)],style=point,colour=blue)
);
One could probably get further by finding an exact or approximate formula for these curves.
|
2025-03-21T14:48:29.996152
| 2020-03-04T06:56:45 |
354118
|
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|
Stack Exchange
|
Von Neumann Inequality in Banach spaces
It is known that the only Banach space that satisfies the von-Neumann inequality is the Hilbert space:
Theorem (see e.g. Pisier, "Similarity Problems and Completely Bounded Maps", p 27) For a Banach space $F$ the following conditions are equivalent:
for every linear operator $T$ on $F$ with $\|T\|\le 1$ and any algebraic polynomial $p$ we have $\|p(T)\|\le\sup_{|z|<1}|p(z)|$;
$F$ is isometric to a Hilbert space.
If we allow $F$ to be merely isomorphic to a Hilbert space, the norm on $L(F)$ is replaced with an equivalent one and we get the following weaker von-Neumann inequality:
There is $C>0$ such that for every linear operator $T$ on $F$ with $\|T\|\le 1$ and any algebraic polynomial $p$ we have $\|p(T)\|\le C\sup_{|z|<1}|p(z)|$.
What are the Banach spaces for which the weaker von-Neumann inequality holds?
Clearly, if the inequality holds for $F$ it holds for any isomorphic space, and so it is not a geometric property. Could it be an isomorphic characterization of Hilbert spaces? Or perhaps, something like super reflexivity would imply this property? Or maybe it just holds for all Banach spaces?
I am almost certain that it doesn't hold for $F=\ell^1$, and I have a suspicion that it won't hold for $\ell^p$ for $p\neq 2$
Small remark: If the weaker von Neumann inequality holds on a Banach space $F$; then every linear contraction on $F$ admits a $A(D)$-functional calculus (where $A(D)$ denotes the disk algebra). This seems to be a rather strong property.
I have just taken a quick look at the proof in Pisier's book (the result is attributed to Foias) and I wonder if one could follow the argument, replacing $1$ by $C$ where necessary, to end up with a purely Banach-space condition on the two-dimensional real subspaces of $F$. Have you checked if this works? It would then reduce the problem to a question about Banach space geometry.
@YemonChoi yes, I thought about it, but I don't know what to do with the condition the condition that there is $C$ such that $\frac{|y-\lambda x|}{|x-\overline{\lambda} y|}\le C$, if $|x|=|y|=1$.
This paper seems to be closely related On polynomialy bounded operators acting on Banach spaces. At least it gives examples of a space on which the implication won't hold.
@an_ordinary_mathematician thank you! I was aware of this paper, but I guess the relevant part is contained in the second paragraph. This gives a counterexample in the class of general Banach spaces, but does not help with a characterization
Interesting. What if you take B(X) where X is a non-Hilbertian renorming of a Hilbert space?
|
2025-03-21T14:48:29.996354
| 2020-03-07T19:13:43 |
354397
|
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|
Stack Exchange
|
Reference for Borel $\sigma$-algebra of topology of convergence in probability
I'm pretty sure I can prove the "Theorem" given further below (without very much difficulty), but it seems way too basic not to have been noticed before.
So I'm wondering if there are any papers/textbooks which state part or all of this "Theorem" - or at the least, which define the $\sigma$-algebra being considered.
Two random variables over a probability space $(\Omega,\mathcal{F},\mathbb{P})$ are said to be $\mathbb{P}$-equivalent if they agree $\mathbb{P}$-almost everywhere. Given a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a measurable space $(S,\mathcal{S})$, we write $L^0(\mathbb{P},\mathcal{S})$ for the set of all $\mathbb{P}$-equivalence classes of $(\mathcal{F},\mathcal{S})$-measurable functions.
Now define on $L^0(\mathbb{P},\mathcal{S})$ the $\sigma$-algebra
$$ \mathfrak{L}(\mathbb{P},\mathcal{S}) := \sigma( \, [X] \mapsto \mathbb{P}(E \cap X^{-1}(A)) \, : \, E \in \mathcal{F}, A \in \mathcal{S} \, ) $$
where $[X]$ denotes the $\mathbb{P}$-equivalence class of $X$.
Remark 1. The map $\,[X] \mapsto \int_\Omega h(\omega) g(X(\omega)) \, d\mathbb{P}\,$ is measurable with respect to $\mathfrak{L}(\mathbb{P},\mathcal{S})$ for every $h \in L^1(\mathbb{P})$ and every bounded measurable $g \colon S \to \mathbb{R}$, and likewise for all measurable $h \colon \Omega \to [0,\infty]$ and $g \colon S \to [0,\infty]$.
Remark 2. Just as with $L^\infty(\mathbb{P})$, the set $L^0(\mathbb{P},\mathcal{S})$ does not really depend fully on $\mathbb{P}$, but only on the equivalence class of $\mathbb{P}$ (under the standard notion of equivalence of measures). Furthermore, due to Remark 1, the $\sigma$-algebra $\mathfrak{L}(\mathbb{P},\mathcal{S})$ also only depends on the equivalence class of $\mathbb{P}$.
Are there any papers or textbooks that define the above $\sigma$-algebra $\mathfrak{L}(\mathbb{P},\mathcal{S})$? And are there any that state (either as a proved result or as an exercise) any parts of the following?
Theorem. Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $S$ be a separable metrisable topological space.
A sequence of $S$-valued random variables $X_n$ converges in probability to an $S$-valued random variable $X$ if and only if $\int_E g(X_n) \, d\mathbb{P} \to \int_E g(X) \, d\mathbb{P}$ for every $E \in \mathcal{F}$ and every bounded continuous $g \colon X \to \mathbb{R}$.
As a consequence: Suppose $\mathcal{F}$ is countably generated. Then the Borel $\sigma$-algebra of the topology of convergence in probability, defined on $L^0(\mathbb{P},\mathcal{B}(S))$ with reference to the topology of $S$, is precisely $\mathfrak{L}(\mathbb{P},\mathcal{B}(S))$.
In the case that $S=\mathbb{R}$, if $\mathcal{F}$ is countably generated, then the Borel $\sigma$-algebra of $L^p(\mathbb{P})$ is precisely $\mathfrak{L}(\mathbb{P},\mathcal{B}(\mathbb{R}))$ for any $p \in [1,\infty]$.
If $(\Omega,\mathcal{F})$ and $(S,\mathcal{B}(S))$ are both standard Borel spaces, then there exists an $(\mathfrak{L}(\mathbb{P},\mathcal{B}(S))\otimes\mathcal{F},\mathcal{B}(S))$-measurable function $e \colon L^0(\mathbb{P},\mathcal{B}(S)) \times \Omega \to S$ such that for every $x\in L^0(\mathbb{P},\mathcal{B}(S))$, the function $\omega \mapsto e(x,\omega)$ is a representative of $x$.
[I have proved the first statement on Math.SE, here. For the final statement, taking $\Omega=\mathbb{R}$ and $S=[0,1]$ or a countable subset thereof, just define $e$ by the Lebesgue differentiation theorem (which holds for arbitrary probability measures on $\mathbb{R}$).]
Remark 3. The $\sigma$-algebra $\mathfrak{L}(\mathbb{P},\mathcal{S})$ seems to me a very natural $\sigma$-algebra on $L^0(\mathbb{P},\mathcal{S})$. In particular, I think it serves as a way of way of overcoming the problem of the lack of a meaningful measurable structure on the set of all $(\mathcal{F},\mathcal{S})$-measurable functions (see the accepted answer of this MO question), if you're working in a setting where there is a natural equivalence class of measures on $(\Omega,\mathcal{F})$. This is especially justified by the final statement of the above "Theorem", which stands nicely in contrast to the negative result cited in the accepted answer to the linked MO question.
PS. I think Hans Crauel's Random Probability Measures on Polish Spaces has a lot of results addressing these kinds of questions, but in view of what Google will let me access, I don't think it addresses the particular question of the Borel $\sigma$-algebra of the topology of convergence in probability. (However, I don't currently have access to the full book, and so I can't say this with complete confidence.)
I realise this MO post was quite badly constructed before. I think I've now considerably improved it (as well as adding more to it).
|
2025-03-21T14:48:29.996670
| 2020-03-07T20:00:32 |
354402
|
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Stack Exchange
|
Pros and cons of specializing in an esoteric research area
If a mathematician specializes in a popular research area, then there are many job positions available, but at the same time, many competitors who are willing to get such job positions. For an esoteric research area, there are few competitors and job positions. There are very often pros and cons of such research areas.
What are some pros and cons of specializing in esoteric research areas that many people may not know?
Maybe it is a little hard to answer this question in full generality since circumstances vary. Hence, I especially want to listen to examples, personal experiences, and maybe urban legends.
Now, if you are interested, let me tell my personal circumstance to give some context to this question. I am a student from outside of North America who just graduated from my undergraduate institution. Since I decided to study abroad, I applied to several North American universities last December and was admitted to some of them. Now I am wavering between two universities. Denote those universities X and Y.
Among the specific research areas available at X (resp. Y), I am interested in two of them, say, A and B (resp. C and D).
I did some searching in those research areas, and I found out that there are many people who are researching C and D and some of them are in my country. However, there are only a few people interested in A and B and none of them are in my country. Based on my search, I think (with a little exaggeration) there are about 3 universities in the world where a graduate student can specialize in A. B is not as esoteric as A, but still, it seems there are not many people working on B. However, I think C and D are quite major research areas in my field of study.
[In this question, I used 'field of areas' as something in First-level areas of Mathematics Subject Classification and 'specific research areas' as something in Second or Third-level areas of it. ]
At first glance, I prefer X over Y, because I was very interested in A. Also, this is partially (maybe totally) because X is considered more `prestigious’ than Y. However, I’m a little nervous about specializing in esoteric areas such as A and B, because of the number of job positions and this kind of problem.
Anyway, I think it is not a bad idea to ask a question at MO and to listen to the pros and cons of specializing in esoteric fields to make a better decision. Any personal stories or examples will be really helpful. Thanks in advance.
I think the best thing is to work in an obscure area which has applications to a more popular area. But I understand this is not so easy.
If you plan to stay in academia, obscure areas may certainly limit significantly the places that would even consider you. Also, you will most likely need grants to advance professionally (tenure may even be contingent on them, depending on where you land). It will be hard to convince panels to fund your research when resources are already scarce for more "fashionable" areas.
It's hard to write a thesis. It's even harder to write a thesis in an area you don't enjoy, that you've only picked because you think it might give you better chances of getting a job some years hence. Better to work in an area you enjoy, and let the chips fall where they may.
Why not mention what A, B, C and D are? Presumably it's not a secret. And it would certainly be on-topic here.
Thanks for all the helpful comments! And @FaheemMitha, I didn't mention them because I want to keep anonymous on the internet, even in MO. If I tell what A and B are, then it is not hard to find out where X is. If I attend X then it can be very easy to figure out who am I. Maybe this can be a little bit too much, but anyway I want to keep anonymous.
Naming the areas in question would invite discussion of whether or not they are “hot” topics, which would almost certainly lead to this question being closed.
Are areas A and B new? Many popular areas were considered esoteric at their beginning?
"If a mathematician majors a popular research area, then there are many job positions available" really? :p
@AndyPutman, thank you for your suggestion, but I still want to focus on advice for students who want to major an (general) esoteric research area. Maybe I can make a new question for discussing an area is hot or not later.
By the way, I appreciate your answer! I followed your answer and it gave me some new information. Thank you!
@Taladris, A started in the 1980s, so I think age is not the main reason why this area is esoteric. I guess it's because of technical difficulty and theoretical depth. B is something like the brand new direction of an old theory.
@gented, isn't it generally right? Do you know any specific counterexample?
@Absinthe Where did you get the impression that there is plenty of research positions in mathematics (because truth is, there aren't)?
@Absinthe: I think you misunderstood me -- as it stands right now your question is on-topic (in my view), but I think that discussions about the pros and cons of specific, named areas are inappropriate and would vote to close such a question.
It's interesting that from the answers given here, one mostly gets the impression that in modern mathematics, it is not novelty and perspectives that are valued in themselves, but clannishness. :)
I think this is an important question, and something that is not talked about often enough. Just like we don't explain enough to undergraduates that the choice of their major has profound consequences for their career opportunities, we don't tell young mathematics researchers enough about the consequences of their choice of research area. (What's the analogue of the well-known joke about the English major serving food in your local restaurant?)
Instead of answering your question, let me give a small analysis of research dynamics, drawn from my own (limited) experience. The below is not supposed to be exhaustive, but I'm hoping to show that there are underlying mechanisms that cause certain areas to be more popular than others.
I'm hoping someone else will post an answer that gets closer to the core of your question, but I hope that this is at least somewhat useful.
What makes interesting research?
This may seem a little random when you're young, but there are actually some underlying mechanisms that are somewhat understandable. (Although admittedly I still haven't figured out a full answer to this question.)
First of all, just like universities and journals, top mathematics research gains some prestige from age. This means that solutions to old conjectures are valuable, and a new theory is appreciated more when it says something about the mathematics that existed before.
But there is a caveat: some areas are considered 'easy' or 'belonging to the past'. There are almost certainly areas that legitimately produce interesting mathematics but are not fashionable because they are considered 'too old'.
Secondly, mathematicians like powerful ideas. If you have some new technique that seems applicable in many situations, this is valued more than an ad hoc argument. If you have a powerful machine that people can use with their eyes closed, that's going to be cited a lot.
Thirdly, we prefer clean theorems. If you have a theorem without too many technical assumptions, then it's much easier to explain and motivate, and much easier for other people to use.
Moreover, connections to other areas are important too: if you prove amazing theorems on a (mathematical) island, then that's not as interesting as when you prove something that relates to other work.
A lot of the above seems driven by the fact that the mathematicians who evaluate your work do so on the basis of their own interests. This seems reasonable given that everyone's own expertise is limited, and their judgement biased towards their own work.
What makes fashionable research?
Areas of mathematics fall in and out of fashion, in part related to the criteria above. For example, if there is a new idea that seems to have potential, this can drive a research group until they feel the idea has been fully exploited. Especially new connections between different areas can spur a lot of activity, because now you have two communities working out these ideas.
When this happens, you can suddenly see a lot of researchers thinking about very similar things, and this is both a blessing and a curse. Indeed, it makes it easier to explain your results and to find collaborators, but it also means you have to keep up with progress all the time and risk getting 'scooped'.
I'm not entirely sure what the mechanisms for topics falling out of fashion are. One thing that can happen is that the main problem gets solved; for example I have the impression that the classification of finite simple groups ended an era of high activity in the area. But this doesn't always happen, because there could be other questions; for example the techniques for proving Fermat's Last Theorem are still very much alive in the Langlands programme.
I imagine it's also possible for a research area to dry out without the main problem being solved, although I haven't been around long enough to know an example offhand.
Note that timing is key: what is fashionable now may not be fashionable in the future, and sometimes researchers forecast the demise of their field. While it's hard to predict the future, it's probably a good idea to listen (critically) when people tell you something like this.
Finally, there are also some political factor involved: if your area is well-represented in the editorial boards of top journals, then you're going to have an easier time publishing in those journals. As Henriksen points out in There are too many B.A.D. mathematicians, this does not always happen for the right reasons.
So what to do next?
Many research groups are driven by a handful of famous questions and a much larger collection of more technical questions. You could ask the research groups you're considering what the main goal in their work is, and then evaluate this by the criteria above to see how easy a time you would have selling your research.
Usually the advice is to talk to some graduate students in the group to hear about their experiences, but in your case it might also be worthwhile to try to approach some postdocs¹ or senior researchers in the area. Another useful metric to look at is job placement: if an advisor has supervised many graduate students who did well on the academic market (short AND long term), this is useful information.
Finally, I should remark that these things matter much more when you're young, because the mathematics job market is highly catered towards the researchers who are lucky enough to have early success. Once you get a tenure-track of tenured job, it becomes easier to switch areas again. This is especially true if you work in an area with connections to other areas, although there are certainly examples of famous researchers switching to completely unrelated fields.
¹Be aware that postdocs are in the most precarious stage of their career, so their answers will be a bit more cynical than those of (young and naive) grad students or (successful) tenure-track or tenured faculty. This can be useful for getting more 'real' advice, but be sure to recognise the context in which it is given.
-1 for "But there is a caveat: some areas are considered 'easy' or 'belonging to the past'; for example I would be hesitant to publish a paper whose main result is in finite-dimensional linear algebra". Are you really saying that about the thousands of mathematicians who publish in Linear Algebra and its Applications or in Linear and Multilinear Algebra, and who speak at the ILAS conferences? Plus all mathematicians and physicists working in Quantum Information? What an ignorant thing to say.
@MartinArgerami: ah, you're right. I have removed the example.
Thanks. Upvoted :)
I agree that it is harder to get a job in some subfields than others. However, aside from some very broad things that are clear to almost everyone (e.g. there are very few people working on point-set topology), I think it is very hard to judge which subfields are "hot" as an undergraduate. Aside from things that other people have pointed out (like the fact that this is very much a moving target, and that most people you talk to have a very limited and narrow view of the subject as a whole), what "counts" as working in a hot field can be subtle. For instance, within the subareas I follow, there are certainly hot topics, but there are also topics that are almost entirely dead, and the difference between these topics is only obvious to an insider. If you don't have appropriate specialized knowledge, the topics look basically the same!
The advice I give to people considering graduate school is to focus less on potential areas (aside from broad things like "algebra" or "geometry"), and focus more on potential advisors. Go and google the names of your potential advisor's former students (which can be looked up on math genealogy, or on their website if they keep their cv up to date). See if they're getting the kinds of jobs that you aspire to. If none of them are, you should take this as an important signal.
Of course, not every student will be equally successful (after all, they're the ones who have to write the thesis!). What is more, very young advisors might not yet have much of a track record (but it is risky being someone's first couple of students!).
Great pragmatic advice.
I guess that a topic can be "hot" because it looks very promising, but lose its "hotness" because either it turned out that the problems in this area could be solved completely rather quickly, or it became transparent that the main problems in this area are so much out of reach that one cannot expect any substantial progress even with an increased research activity.
To this advice I would add that the place where you go to graduate school matters a lot --- for the less obvious reason that you can learn a lot from the other students who are there with you. Moreover, if you build links with your co-students who do other research they will help you keep in touch with many areas of mathematics that you do not specifically need for your PhD. Mathematics is one subject and it is easy to lose sight of this during your PhD work.
Pros:
the research community is small, so it's easier for you to make connections with the top people in the field
consequently, it should be easier to get good letters from top people and invitations to collaborate
you are more likely to get invited to speak at a single given conference in your area
you won't be directly compared to 20 other people your age working on the same area, for things like jobs and grants
some departments/employers don't care much about you're particular area of work, and will primarily look at things like your publication record and reference letters
it's easier to stay at the cutting edge of research, and not get scooped, or waste your time proving something you later find out was known 10 years ago
Cons:
the more esoteric your field is, the less well known and influential the "top people" may be outside of your community, making reference letters etc less impactful (in extreme situations, the top people themselves may not be that strong)
for academic jobs, most research departments will care some about what your field is, and how it fits in with the department interests
it may be hard convincing editors and referees new results merit publication in top journals
there is less grant money for your area, and you will need to be really good at selling it to get external grants
there may not be many conferences in your field, and if you go to other conferences, you may feel "left out"
there are fewer people for you to talk to about research, and possibly collaborate with (which for me means less fun and excitement)
Thoughts:
There are different levels of how esoteric a field might be, ranging from only a handful of people close to retirement are doing in it to a number of brilliant young people are working in it (maybe it is just "esoteric" because of the high level of technical mastery required for entry), and from being of tangential interest to almost no one versus people waiting for a potential breakthrough that will get coverage in major newspapers around the world.
If you want decent academic options after a PhD, Andy Putman's advice is good: see where people are publishing, sending students, etc to establish a minimum level of having decent prospects. Certainly there are areas where prospects are just bad. (Conversely, there are areas where right now prospects seem very good--I'm looking at you data science--but maybe the market will be flooded with data science applicants in 3 years, who knows? Well, possibly some data scientist already does...)
However, given 2 areas which are not "graveyards", trying to optimize between the above pros and cons based on the relative competitiveness and popularity of them (at least for academic job prospects) seems to me like playing the stocks without serious knowledge of the markets. You might win by chance, you might lose by chance, but you're trying to make bets based on pretty incomplete knowledge about a situation that is constantly changing. Thus the best choice is the one that would make you happiest at least in the short-medium term, which is typically some combination of the how appealing the area is to you and finding a successful advisor you're happy with.
My brother-in-law, a chemist, majored in inorganic biochemistry. Yes, inorganic. Yes, biochemistry. Yes, this subject exists: a handful of chemicals in our body, i.e. hemoglobin, are inorganic. Nevertheless, to my ears "inorganic biochemistry" sounds like an esoteric research area, except that it's in chemistry rather than math.
Unfortunately, the choice of the specialty made his Ph.D. less attractive for potential employers. It's harder to get a grant in the subject NSF didn't hear of. It's also much harder to identify the industry that would need a specialist in such a field. He was searching for employment for a while, and ended up joining US Army, the only organization he found that was researching the biological effects of inorganic chemicals, and got shipped to overseas.
I agree with Gerry Myerson's comment, one should do the things one likes. But, from the narrow viewpoint of post-graduate opportunities after specializing in a common versus an esoteric subject, I would recommend a common one.
I don't think this is very relevant to math, especially pure math, because pure math is extremely balkanized, and EVERY research area looks like inorganic biochemistry.
I think it really depends on your goals. The OP says he's thinking about which grad school to attend. If you're goal is to get a higher degree and then a job in industry, you would be well-advised to do research in something considered "useful." That's not a pre-requisite, and I know a ton of abstract homotopy theorists who now work as data scientists, but it's also fair to say that the transition out of academia is easier if you have some skills you can showcase to companies.
If your goal is to be a professor, I think it's important to realize just how hard it is to get a job these days. The AMS publishes data about the number of PhDs and the number of people who get a tenure track academic position every year. Only about 25% of math phds get a tenure track job. Many of those who get a job in a given year are coming from a postdoc rather than directly out of grad school. When we hire, we get more than 400 applicants for a single job, so even if you do everything right, the competition might still prevent you from getting a tenure track job. Really think about this. A lot of people are jaded when they can't get a job. Don't assume it will be easy or even possible.
If you want to go to a research university, I think you'd be well advised to pick a less esoteric field, or to connect your research to new and sexy things (e.g., lots of people are getting jobs now in topological data analysis, whereas almost none are getting jobs in chromatic homotopy theory).
If you are not wedded to the idea of being a professor at a research university, then do not overlook liberal arts colleges. These jobs have a slightly higher teaching load but can still provide a lot of research support, sabbaticals, etc. The essential point for these schools is that teaching is valued more than research, so if you think you might be happy at such a place then it's essential to develop your capabilities as a teacher. Much more important than your area of research is whether or not you can teach statistics or computer science. You get that skill and you will get a job, 100%. You might consider picking up a masters in CS or stats during grad school, as that would help you get a liberal arts job, and would also help if you go to industry. I got a master's degree in CS at the same time that I was getting a PhD in math, and it really did not take a ton of effort. One other point is that liberal arts colleges increasingly want faculty to do research with undergraduate students. Their undergraduate students can be very strong (more like the level of a master's student), but it might be wise to pick a research area that's easier to get into, like graph theory or knot theory, rather than something insanely esoteric. Again, computer science, applied statistics, and data science are very fertile areas for research with undergraduates.
As others have said: I think it's better to focus on the place and the advisor more than the specific field. The place meaning the university, its prestige, its culture etc. as well as the fact you have to live there many years and do very hard work while living there for not huge salary. Some places will be a lot less appealing than others and this matters!
And the advisor in terms of their outlook and culture, the types of jobs their students get, the kind of papers they write (separately from their field I mean are they long? short? technical and computational? or abstract and high-level? etc.). It is commonly said: speak to the grad students to get the scoop on advisors.
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2025-03-21T14:48:29.998580
| 2020-03-07T21:25:21 |
354409
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|
Stack Exchange
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Reeb stability counterexample: foliation in $S^{n-2}\times S^1\times S^1$ with non-diffeomorphic leaves
Reeb's global stability theorem requires the foliation to be of codimension 1. As a counterexample, in "Geometric theory of foliations", Camacho and Lins Neto present the following.
Consider the manifold $S^{n-2}\times S^1\times S^1$ with coordinates $(x_1,...,x_{n-1},\varphi,\theta)$ such that $\sum_{i=1}^{n-1}x_i^2 = 1$ , and two differential one-forms
$$\eta_1 = d\theta \quad\text{ and }\quad\eta_2= ((1-\sin(\theta))^2 + x_1^2) d\varphi + \sin(\theta) dx_1 .$$
Frobenius' integrability condition is satisfied as can be easily checked, so there's an integral manifold of codimension two for these differential forms.
There are some compact leaves homeomorphic to $S^{n-2}$ given by fixing $\theta = 0$ and $\varphi = \mathit{constant}$.
On the other hand, there are leaves of a second kind given by $\theta = \pi/2$ and $\varphi = \mathit{constant} + \frac{1}{x_1}$ and these are said to be non-compact.
Question 1: Is there any simple argument for non-compactness of these leaves?
Intuition: Some sequence in one of these leaves for which the first coordinate converges to zero, cannot converge in the same leaf "because it is not given by the same equation ($\varphi = \frac{1}{x_1}$)". I'm not sure if this is correct, and even if it was, I couldn't find some neat way to explain it.
Question 2: We have that $d\eta_2 = 2x_1 dx_1\wedge d\varphi $ on the leaves of the second kind but is zero when $x_1$ is. Does this give some topological difference for leaves?
The precise description of the leaves which are not spheres is as follows.
The leaves passing through points with $x_1\neq 0, \theta=\pi/2, \theta=constant+1/x_1$ are homeomorphic to $\mathbb{R}^{n-2}$ (you can work out an explicit parametrization).
The leaves passing through points with $x_1=0, \theta=\pi/2$ are homeomorphic to $S^{n-3}\times S^1$.
When $x_1$ tends to $0$ the non-compact leaves spiral around $S^{n-3}\times S^1$ as in the classical Reeb foliation.
In fact, the example is originally due to Reeb himself [CRAS 226 (1948), 1337-1339], as you can see here (in French): https://gallica.bnf.fr/ark:/12148/bpt6k31787/f1338.image (but few details are given). For a more recent exposition (when $n=4$, but the general case is similar) I recommend to have a look at Section 1.6 "Reeb Transition" in the paper by Pablo Lessa [Asian J. Math. 19 (2015), 433-464].
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2025-03-21T14:48:29.998782
| 2020-03-07T21:30:43 |
354410
|
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|
Stack Exchange
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Is there a compact, connected, totally path-disconnected topological group?
There exist homogeneous spaces such as the pseudo-arc, which are compact, connected, and totally path-disconnected. Is there a nontrivial, Hausdorff topological group with the same properties, i.e. that is compact, connected, and totally path-disconnected? What about a metrizable example?
(a) Here's the proof that such $G\neq0$ abelian doesn't exist. Path-free implies $Hom(\mathbf{R},G)=0$. Connected means $Hom(G,$finite$)=0$. So, $A$ being the Pontryagin dual: $Hom(A,\mathbf{R})=0$ (i.e. $A$ is torsion) and $Hom($finite$,A)=0$. So $A$ torsion and torsion-free, hence $A=0$. (b) There's no point in elaborating about Hausdorff, since $G$ every path from $G/\overline{{1_G})}$ lifts to a path in $G$. (c) The singleton is both connected and totally disconnected, so the question should ask $G\neq 1$ Hausdorff (or that $G$ doesn't carry the indiscrete topology).
I delete my answer because it is incomplete, and there is a complete answer already...
(I'm assuming the groups to be Hausdorff to avoid the discussion degenerate into idle banter.)
The answer is yes: $\{1\}$ is such a group.
The answer to the intended question (which is probably whether there's a nontrivial such group) is no.
Andrew M. Gleason. Arcs in locally compact groups. Proc. Nat. Acad. Sci. U.S.A. 36 (1950), 663-667. Link
1st line of MR review: The author gives an outline of the proof of the following theorem: Every locally compact group which is not totally disconnected contains an arc.
Edit: the above is for arbitrary locally compact groups, but for compact groups it's significantly easier. Indeed, if $G$ is a compact connected group, it follows from the Peter-Weyl theorem and the basic structure of compact connected Lie groups that there exists a group $H=A\times\prod_{i\in I}S_i$, where $A$ is a compact connected abelian group, and each $S_i$ is a simple, simply connected compact Lie group, and a surjective homomorphism $H\to G$ with totally disconnected kernel (this is for instance in Bourbaki, Lie, Chap 9, appendix). If $G\neq 1$, then $H\neq 1$, and then $\mathrm{Hom}(\mathbf{R},H)\neq\{1\}$ (since either $I$ is non-empty, or $A\neq 1$, and the abelian case is settled by Pontryagin duality as I mentioned in a comment. The composition map $\mathrm{Hom}(\mathbf{R},H)\to \mathrm{Hom}(\mathbf{R},G)$ being injective (because $\mathrm{Ker}(H\to G)$ is totally disconnected), one deduces $\mathrm{Hom}(\mathbf{R},G)\neq\{1\}$.
(Note: Peter-Weyl was established around 1925, and Pontryagin duality in the early 1930's; the basic structure of compact Lie groups was known before these dates; I'm not sure of an early reference for the structural result on compact connected groups but it follows easily so I guess was known to people working on Hilbert's 5th problem in the late 1940's).
Edit 2: one of the results in Hilbert's 5th problem is that for every connected locally compact group $G$, every neighborhood of $1$ contains a compact normal subgroup $W$ such that $G/W$ is Lie. Also it was proved by Iwasawa around 1950 that every connected Lie group $G$ has a maximal compact subgroup $K$, and that such $K$ is connected.
One this is granted, one reduces from the compact case to the locally compact case as follows: let $G$ be a nontrivial connected locally compact group. Let $W$ be a compact normal subgroup such that $G/W$ is Lie. Let $K/W$ be a maximal compact subgroup of $G/W$. I claim that $K$ is connected. Granting the claim and the compact case, we're done if $K\neq 1$. Otherwise $K=1$ and hence $W=1$, so $G$ is Lie and this case is fine.
If $K$ were not connected, $K$ would have the nontrivial profinite quotient $K/K^\circ$, and hence a nontrivial finite quotient, say with kernel $K'$. Hence there exists a symmetric neighborhood $N$ of $1$ in $G$ such that $NK'\cap K=K$. Let $W'$ be a compact normal subgroup of $G$ contained in $N$, such that $G/W'$ is Lie Since $K$ is maximal compact, we have $W'\subset K$, and $K/W'$ is a maximal compact subgroup of $G/W'$, but it is not connected, contradicting Iwasawa's result.
|
2025-03-21T14:48:29.999091
| 2020-03-07T21:54:53 |
354414
|
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|
Stack Exchange
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Problem arising from martingale solutions to SPDE: $Law(u)=Law(v)$ on $C([0,T]; X)$, can $Law(u)=Law(v)$ on $C([0,t]; X)$ for $t<T$?
I ask this question because I found in some papers of martingale solutions to SPDE, to prove the approximate solutions $u_n$ is a convergent sequence, one can use "stochastic compact" method to find another sequence $v_n$ (on another probability
space), which has the same distribution as the original approximate solutions $u_n$, like (1) below. Moreover, $v_n$ is a convergent sequence. One can prove the new sequence $v_n$ also satisfies the approximate scheme and then take limit to get a martingale solution $v$ to the target SPDE. But how can we know the initial distribution of $v(0)$ is the same as the given initial distribution? For example, does (3) below hold true particularly for $t=0$???. Many papers on different models omit this, so it seems very easy. Maybe I am too stupid to know why. If necessary, I can give the references to focus on a specific example. I also want to know wether we can have (2).
Thanks in advance!!!
Let $(\Omega_i, \mathcal{F}^i,\mathbb{P}_i,\{\mathcal{F}^i_t\}_{t\geq0})$ $i=1,2$ be two probability space. Let $T>0$. Let $u:\Omega_1\times[0,T]\mapsto X$ and $v:\Omega_2\times[0,T]\mapsto X$ be two $X$-valued process such that $u\in L^2\left(\Omega_1;C([0,T];X)\right)$ and $v\in L^2\left(\Omega_2;C([0,T];X)\right)$.
Assume that $$\mathbb{P}_1(u\in A)=\mathbb{P}_2(v\in A)\ \ \ \forall A\in\mathcal{B}(C([0,T];X)),\ \ \ (1)$$ where $\mathcal{B}(C([0,T];X))$ denotes the Borel sets of $C([0,T];X)$.
My questions are:
(a) For any $t\in[0,T)$, can we have
$$\mathbb{P}_2(u\in A)=\mathbb{P}_2(v\in A)\ \ \ \forall A\in\mathcal{B}(C([0,t];X))\ \ ???\ \ \ (2)$$
(b) For any $t\in[0,T]$, can we have $$\mathbb{P}_1(u(t)\in A)=\mathbb{P}_2(v(t)\in A)\ \ \ \forall A\in\mathcal{B}(X)\ \ ???\ \ \ (3)$$
What are $\mathbb{P_1}$ and $\mathbb{P_2}$?
@LSpice, the probability measures on $\Omega_1$ and $\Omega_2$, respectively. Sorry I missed the subtitle before.
Yes, both statements are true. For (a), note that for every Borel set $A$ of $C([0,t],X)$, the set $\{u \in C([0,T], X) : u|_{[0,t]} \in A\}$ is a Borel set in $C([0,T],X)$. This follows immediately from the fact that the restriction map from $C([0,T],X)$ to $C([0,t], X)$ is continuous and therefore Borel.
Likewise for (b), use the fact that the evaluation map $u \mapsto u(t)$ is a continuous map from $C([0,T],X)$ to $X$.
Thanks a lot. Very clear.
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2025-03-21T14:48:29.999279
| 2020-03-07T22:11:03 |
354415
|
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|
Stack Exchange
|
Is there a model-independent characterization of the gaunt strict $n$-categories amongst the weak $(\infty,n)$-categories?
Recall that a strict $n$-category $C$ is called gaunt if every $k$-morphism in $C$ with a weak inverse is an identity, for all $k$; let $Gaunt_n$ denote the strict 1-category of gaunt $n$-categories. Another way of saying this is that $C \in Gaunt_n$ iff $C$ has the unique right lifting property with respect to the canonical $n$-functor $E_k \to C_{k-1}$, where $E_k$ is the free $k$-equivalence and $C_{k-1}$ is the free $(k-1)$-morphism. Thus we have a characterization of the essential image of the fully faithful functor $Gaunt_n \to Cat_n^{str}$ into the strict 1-category of strict $n$-categories.
Now consider the composite inclusion $Gaunt_n \to Cat_n^{str} \to Cat_{(\infty,n)}$ into the $(\infty,1)$-category of weak $(\infty,n)$-categories. I believe this inclusion is also fully faithful; can we characterize its essential image? We can't repeat the same characterization as before, because the image of the canonical map $E_k \to C_{k-1}$ is already an equivalence in $Cat_{(\infty,n)}$.
Question: What is a (model-independent) characterization of the essential image $Gaunt_n \to Cat_{(\infty,n)}$?
I have some hope that there is a nice answer, from the following model-dependent considerations. In every model of $Cat_{(\infty,n)}$ I've thought about, the objects of $Cat_{(\infty,n)}$ are the fibrant objects of a model structure on some 1-category $\mathcal K$, defined by a (non-unique) right lifting property against the acyclic cofibrations of $\mathcal K$. It seems to me that in all cases, the following facts hold:
The functor $Cat_n^{str} \to Cat_{(\infty,n)}$ lifts to a canonical functor $Cat_n^{str} \to \mathcal K$;
The composite functor $Gaunt_n \to Cat_n^{str} \to \mathcal K$ is fully faithful;
The essential image of $Gaunt_n \subseteq \mathcal K$ can be characterized as the objects which satisfy the stronger unique right lifting property against the acyclic cofibrations of $\mathcal K$.
To flesh out (3) a bit, the lifting properties for fibrant objects in these model structures are generally characterized by "horn fillers" and "univalence maps" (a.k.a. "completeness" or "Rezk" or "2-out-of-6" maps). Unique lifting against the horn fillers seems to generally pick out objects which can be thought of as strict $n$-categories presented via a sort of "naive nerve" which doesn't quite handle equivalences appropriately; if in addition an object lifts against the univalence maps, it is forced to be gaunt (and in this case the "naive nerve" coincides with the "genuine nerve"). For example, if the Duskin nerve of a 2-category is univalent, then the 2-category is gaunt.
I find it striking that the model-dependent description of the essential image $Gaunt_n \subseteq \mathcal K$ seems to always take the same form across models $\mathcal K$, and I'm wondering if these parallel characterizations are really model-dependent avatars of something which can be said model-independently.
I'm just shooting from the hip here, but are they the representably discrete objects in the $(\infty,1)$-category of $(\infty,n)$-categories?
@AlexanderCampbell That sounds promising. Now that you put it that way, it does seem that the characterization should be via mapping-in properties rather than mapping-out ones. This is a bit counterintuitive to me, since commonly one thinks of gaunt $n$-categories as ones whose mapping-out properties should be good. But it's staring us in the face from the types of lifting properties one gets!
I definitely agree with Alexander guess. Whatever model of $(\infty,n)$-category you are using, discreteness make them into strict infinity categories as every higher isomorphisms asserting coherence of operations is an equality, and "gauntness" is exactly the Rezk completness condition. (this is especially easy to see with the $\Theta_n$-Space model).
I think you can say (inductively) that an $(\infty,n)$-category is gaunt if its underlying $\infty$-groupoid is discrete and all its mapping $(\infty,n-1)$-categories are gaunt.
Alexander Campbell's guess is correct.
Here is a reference.
Lemma 10.2 of this paper
Clark Barwick, Christopher Schommer-Pries, On the Unicity of the Homotopy Theory of Higher Categories, arXiv:1112.0040
shows that $Gaunt_n \simeq \tau_{\leq 0} Cat_{(\infty,n)}$. That is to say they are precisely the $(\infty,n)$-categories $G$ with the property that the space $Map(C,G)$ is homotopically discrete for all $C$.
They can also be described as the localization of $Cat_{(\infty,n)}$ at the single morphism $S^1 \times C_n \to C_n$, where $C_n$ is the free-walking $n$-cell, $S^1$ is the circle, and the map is projection. This description is model independent (for example the $n$-cell can be characterized model independently as in the proof of Lemma 4.8 in the same paper above), however it is easiest to check that this description is correct in a particular model such as Rezk's $\Theta_n$-spaces.
|
2025-03-21T14:48:29.999631
| 2020-03-07T22:31:52 |
354418
|
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|
Stack Exchange
|
Result of continuum tensor product of Hilbert spaces
Let's suppose that with number $\mu_1 \in \mathbb{R}$ we associate a Hilbert space $\mathcal{H}_{\mu_1}$ with countable basis $|1\rangle _{\mu_1}$, $|2\rangle _{\mu_1}$, $|3\rangle _{\mu_1}$, $\ldots$ Analogically
$\mathcal{H}_{\mu_2}$: $\mu_2 \in \mathbb{R}$ with basis $|1\rangle _{\mu_2}$, $|2\rangle _{\mu_2}$, $|3\rangle _{\mu_2}$, $\ldots$
$\mathcal{H}_{\mu_3}$: $\mu_3 \in \mathbb{R}$ with basis $|1\rangle _{\mu_3}$, $|2\rangle _{\mu_3}$, $|3\rangle _{\mu_3}$, $\ldots$
$\vdots$
And so on.
Could you tell me please, if we formally create the following space (continuum tensor product of separable Hilbert spaces)
$$
\mathcal{H} = \bigotimes\limits_{k\in \mathbb{R}} \mathcal{H}_{\mu_k},
$$
would it be a Hilbert space (obviously, not separable)?
P.S. Yes, I know that we can't enumerate all real numbers and chosen notation for indexes of $\mu$ isn't really good, but I did not come up with something better.
What do you mean by "continuous basis"? What do you mean by "next one space"? How do you define $\bigotimes$? Usually we define tensor products if Hilbert spaces in a way that makes the result a Hilbert space.
@NikWeaver: Sorry for my mistakes in English. I mean "counting basis" and "the following space". As for the definition of this tensor product, I'm not sure about its correctness. However, I suggested that construction like this one may exist.
It still sounds like you're not using the right word. Do you perhaps mean "countable basis" instead of "counting basis"? In any case, there is a construction for a possibly infinite tensor product of any collection of Hilbert spaces, indexed by any set, which could be the real numbers or any other set of the power of the continuum in your case.
The place to look here is the oeuvre of A. Guichardet. You can find online his notes ”Tensor products of $C^\ast$-algebras II. Infinite tensor products” which contains a chapter on such products of Hilbert space.
There are several ways you can define a Hilbert space tensor product $\bigotimes_{t \in X} H_t$, if each $H_t$ is a Hilbert space. The "full" tensor product is generated by all functions $h: X \to \bigcup H_t$ with $h(t) \in H_t$ for each $t$ and such that $\prod \|h(t)\|$ converges. We write $h$ as $\bigotimes h(t)$ and define the inner product by $\langle \bigotimes h(t), \bigotimes k(t)\rangle = \prod \langle h(t),k(t)\rangle$. (The convergence condition on $h$ is chosen to ensure this product converges, modulo some technicality.) Then take the linear span, factor out null vectors, and complete to get a Hilbert space. If infinitely many $H_t$ have dimension at least $2$ then this won't be separable.
That was the definition originally given by von Neumann, but it's not much used any more. The standard way to do things now is to fix a unit vector $u_t \in H_t$ for each $t$ and consider only those functions $h$ with $h(t) = u_t$ for all but finitely many $t$. If you do this then a countable tensor product of separable Hilbert spaces will be separable.
If $H_t = l^2$ for all $t$ as in your case, the most appropriate definition of a tensor product over $\mathbb{R}$ may be to take $\bigotimes_{t \in \mathbb{R}} H_t$ to be the symmetric Fock space $$\mathcal{F}_s(L^2(\mathbb{R})) = \bigoplus_{n \in \mathbb{N}} L^2(\mathbb{R})^{\otimes_s n}.$$
Here $\otimes_s$ denotes the symmetric part of the tensor product. This is a separable Hilbert space.
How is this a tensor product, you ask? Let $(e_n)$ be the standard basis of $l^2$. Then taking $u_t = e_0$ for all $t$, and using the second definition of tensor product given above, we get a Hilbert space with an orthonormal basis indexed by all functions $m: \mathbb{R} \to \mathbb{N}$ which are zero except at finitely many points. The basis itself will consist of the vectors $\bigotimes e_{m(t)}$. Using this basis, the tensor product naturally decomposes into infinitely many summands, one for each value of $\sum m(t)$. Only $\bigotimes e_0$ gives the value zero, so that is a summand of $\mathbb{C}$. To get the sum $1$ we need one $m(t) = 1$ and the rest zero, so those basis vectors span a copy of $l^2(\mathbb{R})$. To get the sum $2$ we need either one $m(t) = 2$ and the rest zero, or $m(t) = 1$ for exactly two values of $t$ and the rest zero; those basis vectors span a copy of $l^2(\mathbb{R})\otimes_s l^2(\mathbb{R})$. And so on. You end up with the Fock space over $l^2(\mathbb{R})$. Thinking about tensor products in that way, it is natural to incorporate the measurable structure of $\mathbb{R}$ by putting $L^2(\mathbb{R})$ in place of $l^2(\mathbb{R})$.
Full details are given in Section 2.5 of my book Mathematical Quantization.
Thank you for the answer. This construction looks very interesting!
You are welcome! This is a basic idea in quantum field theory --- the idea is that you have a Hilbert space at each point in space (state of the field at that point) and you want to take their tensor products to get the "composite system".
Why is von Neumann original definition not used anymore?
Yes, let me second @ceillac's question: why has von Neumann's construction fallen out of fashion? The currently fashionable construction looks a lot like an inductive limit of tensor products over finite subsets of the index set. Why is this approach "the good one"? Is it only because von Neumann's construction provided a non-separable space? How do we settle on what "good" means? (I suppose that if the Hilbert spaces involved are also algebras, then the natural choice is to take $u_t = 1 \in H_t$?)
I wouldn't say it's disliked because it's nonseparable, but the fact that it's nonseparable is a strong hint that it's not the "right" definition. Ultimately people use the constructions that help them prove theorems. In the case of infinite tensor products, as far as I know the main application is physics, where the part I describe is the physically relevant part and the rest is irrelevant.
(Elaborating on my last point, the modern tensor product is the finite energy states. The rest of von Neumann's construction is states with infinite energy.)
thank you @NikWeaver , I guess factoring out null vectors is not elegant as well. Also Von Neumann's definition might be dependent on the basis of t chosen (with change of basis t'=Rt't as physicists do to go from x the position to p the momentum )
Hi Nik, I would have thought a motivation for infinite tensor products would be to allow for infinite-particle states (since one can already construct a Fock space for QFT with an infinite sum of finite tensor products, allowing for states with arbitrarily large but finite numbers of particles). But does your second definition allow for infinite-particle states? At first glance it seems not, due to the finiteness restriction on $h(t)$, but I may not be understanding this correctly.
@WillG I'm merely reinterpreting Fock space, so no, there are no infinite particles states. Are there realistic physical theories that involve infinite particle states?
I see, thanks for the reply. I don't know of any theories with infinite particle states, but shouldn't there be? It seems reasonable to want a theory that describes a spatially infinite universe with constant particle density—this is the standard model of cosmology, after all! Just wondering about it, since that's what led me to this post.
Yeah, this really gets outside of my area of expertise, but I suppose KMS states would count as infinite particle states. But my impression there is that one takes the C*-algebra point of view and gets a Hilbert space via the GNS construction, and I am not sure there is any relation with a tensor product in von Neumann's sense. Sorry, I can't be very helpful with this.
|
2025-03-21T14:48:30.000282
| 2020-03-08T00:33:13 |
354422
|
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|
Stack Exchange
|
Modularity of elliptic curves with only minimal lifting
I have been trying to understand a bit of the basics of deformations of Galois representations. One point which leaves me curious now is that proving modularity lifting with arbitrary ramification on the Tate module, compared to modulo $p$, seems quite alot to ask. As that is true, it is not so much after all. But is it not possible to find a prime $p$ for each, say semistable, elliptic curve $E$ where the representation on its Tate module $\rho_{E,p}:G_\mathbb Q\rightarrow End_{\mathbb Z_p}(T_p(E))\cong GL(2,\mathbb Z_p)$ has minimal ramification? That is, find a prime $p$ which does not divide the order of the minimal discriminant of $E$ at all $l|N_E$, i.e. all primes of bad reduction of $E$ by the Néron-Ogg-Shafarevich criterion. Any large enough prime satisfies this condition so then my first question is: can you prove easily that the reduction modulo some large enough prime (or even an infinity of primes) $p$ is modular, as can be done for $p=3$ thanks to the Langlands-Tunnell theorem? This would then imply that $\rho_{E,p}$, thus $E$, is modular without proving "$R_\Sigma=T_\Sigma$" for all $\Sigma$, thus "$R_\text{unrestricted}=T_\text{unrestricted}$".
Of course Wiles uses the primes 3 and 5 only for $\rho_p$ so this prevents choosing the prime at will, and one has to assume ramification can increase arbitrarily on the Tate module. But did he or someone else after consider trying to use minimal modularity lifting only at some point?
If this is not possible for all semistable elliptic curves, for which elliptic curves would we expect it to be true? Could it be for Frey curves?
Are there Galois representations for which the induction argument used by Wiles to pass to arbitrary ramification on the liftings fails?
I hope you'll forgive me if this is easily found in standard references, or worse if I made a serious mistake, I find the subject quite demanding. Thanks.
EDIT: My first question didn't make much sense as it was, which proves I am a real newbie or the subject is really disturbing -probably both.
Langlands-Tunnell aims at generalizing that if $L/\Bbb{Q}$ ($L=\Bbb{Q}(E[p])$) is Galois with dihedral Galois group and the quadratic subfield is imaginary then its Artin representations are modular from class field theory and the (weight $1$) modularity of Hecke L-series of $\Bbb{Q}(\sqrt{-d})$ (usually we start with $L=\Bbb{Q}(x),x^3\pm x+1=0$)
All the modularity-lifting theorems in the world aren't going to help you if you don't have something modular to lift!
The reason $p=3$ is so important is that $GL_2(\mathbf{F}_3)$ is solvable, which allows you to use Langlands--Tunnell to show that the mod 3 representation is modular (at least if the image is large enough). Then Wiles' modularity lifting results allow you to get from this to full modularity. In order to use any other prime instead of 3, you'd need a variant of Langlands--Tunnell that worked for non-solvable Artin representations; and that's a massively harder problem than handling non-minimal ramification in modularity lifting (one which remains unsolved to this day).
Can there be solvable linear groups images of irreducible Galois representations on $E[p]$ for large primes? I have seen that Jordan and Mal'cev characterized such groups but can't deduce that they cannot receive such representations. I know little about elliptic curves. Are there cases of the Serre modularity conjecture which are "relatively easy" to prove for some elliptic representations, other than by Langlands--Tunnell, perhaps for those on Frey curves? Perhaps by methods unavailable in the 1990's. Thank you.
If E is a non-CM elliptic curve, it is "very rare" for the image of the Galois rep $\bar\rho_{E, p}$ to be anything smaller than the full $GL_2(\mathbf{F}p)$. It is known that given $E$ the image is this for all but finitely many $p$. It is conjectured that for any number field $K$ there is a constant $C$ such that if $p > C$ then $\bar\rho{E, p}$ is surjective for every non-CM elliptic curve $E / K$. So the answer to your question is "conjecturally no". (There is no royal road to Serre's conjecture.)
Is there a good reason to believe that the "Serre conjecture" at some (interesting?) prime is the hardest part of modularity conjectures for Galois representations on motives -I mean hardest knowing current lifting techniques? Does modularity of motives follow from Serre-type conjectures plus some reasonable improvement of known lifting results? Also, who made the big image conjecture you mention, is there a good review of what to expect for images of geometric Galois representations, perhaps with remarks on the differences between $\mathbb Z_p$ and $\mathbb F_p$ coefficients? Thank you.
Earlier I wrote some nonsense as comment, which I deleted. I was confused about things, among which the Fontaine-Mazur conjecture. I may well have written more nonsense though...
This is turning into one of those hydra-like MO threads where every answer generates two follow-up questions. MO is a good place for answering sharply defined questions. If you just want to gain some feel for the shape of a subject, then MO isn't well suited for that -- read some survey articles, or go talk to a human being.
Ok, thank you for your answers.
|
2025-03-21T14:48:30.000663
| 2020-03-08T01:50:23 |
354426
|
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|
Stack Exchange
|
Diagonal action on external product of trivial principal bundles
(Title, tags and phrasing of this problem might not very good, so please feel free to edit it.)
In the course of writing a long and technical proof, I recently came across the following problem:
Let $P$ be a space with left and right actions by a group $G$ (all in the analytic or algebraic category over $\mathbb{C}$, say), such that $G \times P \rightarrow P$, $(g,p) \mapsto (g p g^{-1})$ admits a cross section $X \subset P$, i.e. restriction of the action to $X$ induces an isomorphism $G \times X \overset{\sim}{\rightarrow} P$.
Consider the quotient $P' := G \backslash (P \times P)$ where the action of $G$ is $(p_1,p_2) \mapsto (p_1 g^{-1},g p_2)$.
Under what conditions can one conclude that the composition $G \times (X \times X) \rightarrow P \times P \rightarrow P'$ is also an isomorphism, when the action of $G$ on $X \times X$ comes from the diagonal $G \rightarrow G \times G$, i.e. when the map from $G \times (X \times X)$ to $P'$ is given by $(g,(x_1,x_2)) \mapsto (g x_1 g^{-1},g x_2 g^{-1}) = (g x_1,x_2 g^{-1})$?
Simply reusing the original isomorphism appears insufficient here, and therefore I was wondering whether there are any useful ideas and techniques that I am missing, or whether I'm on the wrong track entirely - any suggestions would be very welcome.
suppose $X$ is a point, and $P$ is isomorphic to $G$
Then, your first map is $g \mapsto (g, g^{-1})$ and the second one is a quotient by the action $(x, y) \mapsto (x g^{-1}, g y)$. It is clear that the composition of these maps is always $g \mapsto e$, so it is true if and only if the group is trivial.
Do you have an example when it is expected to be an isomorphism (for non-trivial $X$)?
I'm also not sure why you suppose that $P$ has both left and right action (while you are free to do so it is not a conventional definition of the principle bundle).
I don't follow your "first map" - if $X$ is any point of $P=G$ which is acting on itself by left and right multiplication, then $X$ is not a cross section (consider the identity element). But I agree that the phrasing and title is probably bad.
I'm looking at generalisations of Steinberg's cross section (as in the 2012 He-Lusztig AMS paper), here $G$ and $P$ are both subsets of a larger group. (So in particular, I'm wondering whether this claim is true for the Steinberg slice.)
The isomorphism $P = G \times X$ defines a map $pr_G : P \to G$. It seems that for this new map to be an isomorphism, you need for each $p_1,p_2$, the set of $g \in G$ such that $pr_G ( p_1 g^{-1} ) = pr_G ( g p_2)$ to consist of a single point. So it might make sense to try to understand $pr_G ( p_1 g^{-1} ) $ and $ pr_G ( g p_2)$ as maps $G \to G$.
That's a very good suggestion but unfortunately those inverse maps appear rather intractable in the cases I'm looking at, so I was hoping for something else...
|
2025-03-21T14:48:30.000891
| 2020-03-08T03:01:59 |
354429
|
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|
Stack Exchange
|
Holomorphic union of sets
Let $\mathbb{D}$ be the unit disk on the plane and let $U,V\subset \mathbb{D}$ be open and such that $U\cup V=\mathbb{D}$.
Is there a holomorphic map $\varphi:\mathbb{D}\times \mathbb{D}\to \mathbb{D}$ such that $\varphi(U\times V)=\mathbb{D}$?
Of course one can ask the same question for "ambient domains" other than $\mathbb{D}$.
Edit2: everything works, updating the answer.
Yes. Consider two cases.
Case 1. There is a point on the boundary contained both in the closure of $U$ and $V$. Do a Mobius transform transforming our disk into upper half-plane $\mathbb{H}$ and putting this point to $0$.
Now take the map $(x, y) \mapsto x+y$.
It is open, and its image (from $U \times V$) contains arbitrarily small translations of $U$ and $V$, hence contains $U$ and $V$, hence is $\mathbb{H}$.
Case 2. One of the subsets (say $U$ without loss of generality) contains $\partial\mathbb{D}$ in it's closure. Then, it contains a small annulus $1-\varepsilon<|r|<1$. There is a map from disk to itself which is surjective from this annulus: one can do such an automorphism of a disk that $0$ is in the image of the annulus and then use $z \mapsto z^2$. Composing this map with the projection on the first component we obtain the desired result.
I am not sure what you mean in the edit: since $U\cup V=D$, the union of the closures contains $\partial D$, and so $\partial U\cup \partial V=\partial D$. Since the latter is connected, $\partial U\cap \partial V\ne\varnothing$. Or did you mean how to drop the condition $U\cup V=D$? I would be very much interested in merely having $U\cap V=\varnothing$..
Annulus and concentric circle.
Actually my intuition says this might be a counterexample, but I'm not sure yet.
answer updated to full generality
Certainly not. If you take U, V to be small neighborhoods of distinct points this will fail. Are you sure this is what you need?
I've made them bigger. Thank you
@erz better, but you probably want to choose $\phi$ first and allow $U, V$ to be arbitrary with $U\cup V = \mathbb{D}$? Also is $\mathbb{D}$ open or closed for you?
I am not sure I understand your first question. $\mathbb{D}$ is both open and closed in itself. The motivation for my question is https://mathoverflow.net/questions/354133/generating-h-inftyx I am allowed to take a lot of functions, and compose them with a function of two variables. This is a natural extension in my opinion
|
2025-03-21T14:48:30.001092
| 2020-03-08T06:05:54 |
354435
|
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|
Stack Exchange
|
Simplest diophantine equation with open solvability
What is the simplest diophantine equation for which we (collectively) don't know whether it has any solutions? I'm aware of many simple ones where we don't know (whether we know) all the solutions, but all of these that I know have some solution.
Yes, I know that "simplest" is subjective. I'd be satisfied if it could be typeset in one line in a LaTeX document. Also, it would be nice if it could be easily memorized (no seven-digit numbers with nonobvious patterns:) though that's secondary.
$(x+1)(y+1)=2^{2^{1000}}+1$ [wanting solutions in positive integers only].
$x^5+y^5=z^5+w^5$ [positive integers, ${,x,y,}\ne{,z,w,}$].
Determining which integers $n$ are a sum of three cubes is a very famous open problem:
$$a^3 + b^3 + c^3 = n, \quad a,b,c \in \mathbb{Z}.$$
Conjecturally, $n$ is a sum of three cubes iff $n \not \equiv 4,5 \bmod 9$.
Note that this is really a family of Diophantine equations, rather than a single Diophantine equation.
This one, together with @MattF.'s comment, fits most what I wanted. "Perfect cuboid" is nice over positive integers, but encoding positive integers themselves in Z really complicates the equation. Thank you very much!
It's more complicated than the other answers by MattF and DanielLoughran, but the Erdős–Straus conjecture states that for every integer $n \ge 2$, there exist positive integers $x, y, z$ such that
$$\frac4n = \frac1x + \frac1y + \frac1z$$
Again, this is a family of Diophantine equations, related to Egyptian fractions, and computer research has verified the conjecture holds for all $n$ up to a rather large number, but whether it holds for all $n$ ($\ge 2$) is an open problem.
I'm sure that the smallest $n$ for which no solution is known is more than a 7-digit number.
@GerryMyerson In fact, it more than a 16-digit number..
|
2025-03-21T14:48:30.001255
| 2020-03-08T07:04:55 |
354438
|
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Stack Exchange
|
Concerning $\mathbb{C}(s_1,s_2,s_3,y)=\mathbb{C}(x,y)$, where $s_1,s_2,s_3$ are symmetric
Perhaps the following question is not in the level of MO questions, but it has not received comments in MSE, so I ask it here also:
Let $\beta: \mathbb{C}[x,y] \to \mathbb{C}[x,y]$ be the involution on $\mathbb{C}[x,y]$
defined by $(x,y) \mapsto (x,-y)$.
Let $s_1,s_2,s_3 \in \mathbb{C}[x,y]$ be three symmetric elements with respect to $\beta$.
It is not difficult to see that a symmetric element w.r.t. $\beta$ is of the following form:
$a_{2n}y^{2n}+a_{2n-2}y^{2n-2}+\cdots+a_2y^2+a_0$, where $a_{2j} \in \mathbb{C}[x]$.
Assume that the following two conditions are satisfied:
(1) Each two of $\{s_1,s_2,s_3\}$ are algebraically independent over $\mathbb{C}$.
Notice that the three $s_1,s_2,s_3$ are algebraically dependent over $\mathbb{C}$, since the transcendence degree of $\mathbb{C}[x,y]$ over $\mathbb{C}$ is two.
(2) $\mathbb{C}(s_1,s_2,s_3,y)=\mathbb{C}(x,y)$; this notation means the fields of fractions of $\mathbb{C}[s_1,s_2,s_3,y]$ and $\mathbb{C}[x,y]$, respectively.
Example:
$s_1=x^2+x^5+A(y), s_2=x^5y^2+B(y), s_3=x^3y^2+C(y)$, where $A(y),B(y),C(y) \in \mathbb{C}[y^2]$.
Question 1: Is it possible to find a 'specific' form of at least one of $\{s_1,s_2,s_3\}$?
A plausible answer may be: One of $\{s_1,s_2,s_3\}$ is of the form
$\lambda x^ny^{2m}+D(y)$ for some $D(y) \in \mathbb{C}[y^2]$, $\lambda \in \mathbb{C}^{\times}$, $n \geq 1$, $m \geq 0$; is it possible to find a counterexample to my plausible answer?
Perhaps it is better to first consider two (easier) questions replacing conditions (1) and (2) by:
(1') $\{s_1,s_2\}$ are algebraically independent over $\mathbb{C}$ +
(2') $\mathbb{C}(s_1,s_2,y)=\mathbb{C}(x,y)$; call this Question 1'.
(1'') $s_1 \neq 0$ +
(2'') $\mathbb{C}(s_1,y)=\mathbb{C}(x,y)$; call this Question 1''.
I guess that the answer to question 1'' is: $s_1=\lambda xE(y) + F(y)$, where
$\lambda \in \mathbb{C}^{\times}$ and $E(y),F(y) \in \mathbb{C}[y^2]$.
Remarks:
(i) In the above example we already have
$\mathbb{C}(s_2,s_3,y)=\mathbb{C}(x,y)$ and $\mathbb{C}(s_1,s_2,y)=\mathbb{C}(x,y)$.
(ii) We can write $x=\frac{u(s_1,s_2,s_3,y)}{v(s_1,s_2,s_3,y)}$ for some $u,v \in \mathbb{C}[X,Y,Z,W]$. Then, if I am not wrong, taking $y=0$ (if possible?) we obtain that $x=\frac{u(s_1(x,0),s_2(x,0),s_3(x,0))}{v(s_1(x,0),s_2(x,0),s_3(x,0))}$, hence $\mathbb{C}(s_1(x,0),s_2(x,0),s_3(x,0))=\mathbb{C}(x)$.
Question 2: Is there an example where all $s_1,s_2,s_3$ are necessary to obtain $\mathbb{C}(s_1,s_2,s_3,y)=\mathbb{C}(x,y)$? Namely, it is not possible to omit one of $\{s_1,s_2,s_3\}$ and still get $\mathbb{C}(x,y)$.
I guess that the answer is positive.
Thank you very much!
It seems that in the last comment to the following question, Jeremy Blanc presented an answer to my above question 1'', namely: $s_1=\frac{ax+b}{cx+d}$, where $a,b,c,d \in \mathbb{C}(x)$ are such that $ad-bc \neq 0$. https://mathoverflow.net/questions/299842/two-bivariate-polynomials-or-rational-functions-that-generate-mathbbcx-y?rq=1
Take $s_{1}=(x+x^{2})y^{2},,s_{2}=(x^{2}+x^{3})y^{2},,s_{3}=s1\cdot s2$. Given $s_{1}$ and $s_{2}$, one finds $x=s_{2}/s_{1}$, hence $y^{2}=s_{1}/(x+x^{2})$. And given $s_{1}$ and $s_{3}$, one obtains $s_{2}$ as $s_{3}/s_{1}$, hence again $x$ and $y^{2}$. and similarly when given $s_{2}$ and $s_{3}$.
@MatthévanderLee, thank you very much for your example (counterexample for plausible answer to question 1). What if we assume that all three $s_i$ are necessary for $\mathbb{C}(s_1,s_2,s_3,y)=\mathbb{C}(x,y)$? Is it possible to find their forms?
@MatthévanderLee, please, what if I slightly change my plausible (wrong) answer to the following: One of ${s_1,s_2,s_3}$ is of the form
$E(x)y^{2m}+D(y)$, where $E(x) \in \mathbb{C}[x]$, $D(y) \in \mathbb{C}[y^2]$, $m \geq 0$.
|
2025-03-21T14:48:30.001476
| 2020-03-08T09:33:01 |
354442
|
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|
Stack Exchange
|
On a statistic for permutations
Given a permutation $\pi$ we can write $\pi=s_{i_1} ... s_{i_l}$ as a product of simple transpositions $s_i=(i,i+1)$ in a minimal way.
Question 1: Is there an "official" name for the permutation statistic given by the cardinality of the set $\{i_1,...,i_l \}$ ? (such as Coxeter length seems to be the official name for $l$)
See http://www.findstat.org/StatisticsDatabase/St000019 for this statistic.
Question 2: Is there a reference that the generating function for this permutation statistic restricted to 321-avoiding permutations is given by Catalan's triangle: http://oeis.org/A009766 ?
Q1: I have seen ${i_1,\dots,i_l}$ being called support of $\pi$ as it is independent of the chosen word the Matsumoto's property that all reduced expressions are connected by braid relations.
Just looking at St000019: what is wrong with cardinality of support ?
@ChristianStump I guess "Support cardinality" might be a good name for this statistic, but maybe there is already a name for this used in some textbook/article that might be even shorter.
The support of a permutation can also mean the set of all of its non-fixed points. That is a different set.
I think Bridget Tenner is someone worth asking about this (or at least looking into her papers).
For Q1, the sequence corresponding to Stat19 is given in A263484 and comes from a 2005 article by Richard Stanley (developing ideas from Comtet) on connectivity sets, connectedness, cut points, etc.
It may not have a standard name because it is the reflection of the decomposition/block number Stat56, i.e., for $\pi \in S_n$, $\text{Stat19}(\pi) = n - \text{Stat56}(\pi)$. See A059438. Another related statistic is global ascents Stat234.
For Q2, the connection between block numbers of 321-avoiding permutations and Catalan's triangle was established by Adin, Bagno, and Roichman, arXiv 1611.06979, later J. Algebraic Combinatorics. The related abstract for Bagno's talk at the Permutation Patterns 2017 conference is more expository.
For the first part of your question, support doesn't appear to be an appropriate word, as in the context of permutations it is normally used for the set of non-fixed points. In general, a mathematical support means as an assistance to the main act.
To try to define the term, we can first define the process that arrives at the statistic:
Let $T_n$ be the set of all non-trivial simple transpositions on a set of size $n$, say $S$.
Let $\pi$ be a permutation of $S$, and let $$\;_\pi T_n=\left\{\{t_1,\dots t_k\}\in T_n^k:\pi\prod_\limits{i=1}^k t_i=S\right\}$$
i.e. $\;_\pi T_n$ is the set of sets of simple transpositions that take $\pi$ to the identity $S$.
Define:
$$\;_\pi U_n(t\in \;_\pi T_n) =\{u:u\in t\}$$
as the set of unique elements of a set $t$, and
$$_\pi U_n=\bigcup_t \;_\pi U_n(t)$$
as the set of sets that contain the unique elements of simple transposition sets from $\;_\pi T_n$.
Define $\;_\pi V_n=\{|v|:v\in \;_\pi U_n\}$ and then the statistic we want is $\min(\;_\pi V_n)$.
So, minimum cardinality of the identifying simple transposition sets for a permutation $\pi$, or perhaps, the minimum number of simple transpositions needed to return $\pi$ to $S$.
The actual statistic can be easily calculated by creating an array of $n-1$, and ticking every cell that is crossed by returning an element in a straight path to it's original position, and counting the number of ticks, max 1 per cell.
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2025-03-21T14:48:30.001732
| 2020-03-08T10:51:07 |
354448
|
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|
Stack Exchange
|
Do we know absolute bounds for the norm of Satake parameters?
If we consider the set of all unramified Satake parameters $S$ of all automorphic representations of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ as $n$ varies, do we know absolute (that is, independent of $n$) lower and upper bounds for the norm of elements of $S$?
Motivation: denoting by $\mathcal{M}_{a,b}$ the set of such automorphic representations whose set $S_{a,b}$ of Satake parameters at unramified primes $p$ fulfill $a\leq\vert\alpha_{p,j}\vert\leq b$, a proof that all the relevant Satake parameters are bounded above by a quantity independent of $n$ (and $p$ as well) would imply that if $\mathcal{M}_{a,b}$ is closed under the Rankin-Selberg convolution, then $a=b=1$.
The archimedean Satake parameters of a classical Maass form of Laplace eigenvalue $1/4+t^2$ are $\pm it$. The eigenvalues are unbounded, so the Satake parameters are also unbounded. So it is not clear what you are after.
I'm interested in the Satake parameters whose norm equals $1$ assuming Ramanujan conjecture.
We don't know any upper bound for $|\alpha_{p,j}|$ that is independent of $p$.
Thank you for your comment. Feel free to turn it into an answer so that I can accept it and the question get closed.
I turned my comment into an answer.
If $\alpha_{p,j}:=p^{\mu_{p,j}}$ are bounded independent of $p$ then $\Re(\mu_{p,j})\le 0$ for all $j$. At least for $PGL(n)$ this implies that $\Re(\mu_{p,j})=0$ for all $j,p$, i.e. Ramanujan.
We don't know any upper bound for $|\alpha_{p,j}|$ that is independent of $p$. On the other hand, we do know that each $|\alpha_{p,j}|$ is bounded by $p^{1/2}$, hence if $\pi$ is an automorphic representation whose Rankin-Selberg powers $\pi\otimes\dots\otimes\pi$ are all automorphic, then the Satake parameters of $\pi$ at every unramified prime lie on the unit circle. This simple observation is one of the motivations for the Langlands program.
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2025-03-21T14:48:30.002003
| 2020-03-08T11:31:05 |
354450
|
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|
Stack Exchange
|
What happened to Stark's book on the analytic theory of algebraic numbers?
I just read the excellent chapter 6 Galois Theory, Algebraic Numbers and Zeta Functions(*) in Waldschmidt, Michel, et al., eds. From number theory to physics. Berlin etc.: Springer, 1992 by Harold Stark.
At the beginning of this chapter, Stark writes
These lectures have been abstracted from my long promised forthcoming book [2].
The reference [2] points to
Stark, H.M.: The Analytic Theory of Algebraic Numbers. Springer, 1993
Google yields no result on this, except for the book by Waldschmidt et al. above, and an old (1975) survey article of Stark in the Bulletin of the AMS.
Does anyone know if this book ever came to completion in any shape or form ? I would love to read Stark's take on the subject, given his very clear style in (*).
Google Scholar indexes an article by Stark in the Bulletin of the American Mathematical Society with the same name, from 1975.
I stated this in my post already.
Oh, I'm sorry. You are right.
The book was never written.
Would you mind elaborating Stopple ?
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2025-03-21T14:48:30.002143
| 2020-03-08T11:39:21 |
354451
|
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|
Stack Exchange
|
Is there an area-preserving diffeomorphism of the disk which is nowhere conformal?
This question is a cross-post. Let $D \subseteq \mathbb{R}^2$ be the closed unit disk.
Does there exist a smooth area-preserving diffeomorphism $f:D \to D$ that does not have conformal points?
I want the singular values $\sigma_i$ of $df$ to be everywhere distinct, and $\det(df)=1$.
It is proven here that for any map $f:D \to \mathbb R^2$, the condition of being with distinct singular values is 'generic' in the following sense: There exist $f_n \in C^{\infty}(D, \mathbb{R}^2)$ such that $f_n \to f$ in $W^{1,2}(D, \mathbb{R}^2)$ and $df_n$ has distinct singular values everywhere on $D$.
It seems to me that this approximation procedure, applied to a map $f \in \text{Diff}(D)$, does not guarantee that the $f_n$ will map $D$ into $D$, let alone be diffeomorphisms. (e.g. I think that the convergence $f_n \to f$ cannot be made uniform in general). However, perhaps this genericity phenomena can still be used somehow.
This answer provides the following example for a one-parameter family of such diffeomorphisms $D\setminus \{0\} \to D \setminus \{0\}$:
$$f_c: (r,\theta)\to (r,\theta+c\log r).$$ This description is given in terms of polar coordinates both in the domain and the range. For each non-zero $c ֿ\in \mathbb R$ we get a diffeomorphism, with fixed distinct singular values whose product is $1$.
Edit:
Can we answer the infinitesimal version of the question? That is, let $f_t$ be a smooth family of area-preserving diffeomorphisms. Does each $f_t$ has a conformal point? This answer treats the "formally infinitesimal" case.
Maybe a fixed point theorem could be used.
I don't know the concepts you refer to, but is a nonconformal map the same as a map with no conformal points?
Let me try to prove an infinitesimal version where an area preserving diffeomorphism is replaced by a Hamiltonian vector field.
Let H be a function in the unit disc, constant on the boundary. Consider the Hamiltonian vector field $(H_y,-H_x)$ and the respective infinitesimal diffeomorphism
$(x,y) \mapsto (x+\epsilon H_y, y-\epsilon H_x).$
When is it conformal at a point? When the Jacobian is a dilation. That is, if the matrix is
$\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}$,
then $a=d, b+c=0$. In our case, the Jacobian is
$\begin{pmatrix}
1+\epsilon H_{xy}& \epsilon H_{yy}\\
-\epsilon H_{xx}& 1-\epsilon H_{xy}.
\end{pmatrix}$.
Hence we want to find solutions of
$H_{xx}=H_{yy}, H_{xy}=0.$
Consider the vector field $V:=(H_{xx}-H_{yy}, 2H_{xy})$. If it has zero on the boundary, we are done. If not, let's calculate it in polar coordinates $(\alpha,r)$ on the boundary circle $r=1$. Chain rule calculations simplify by the fact that, on the boundary, $H_\alpha=0$. My calculation yields:
$V = H_{rr} (\cos 2\alpha, \sin 2\alpha).$
The index of this field equals 2, so there are two zeroes, multiplicities counted, inside the disc.
Thanks, that is an interesting approach. However, there is something that I don't understand: $
(x,y) \mapsto (x+\epsilon H_y, y-\epsilon H_x)$ is not the exact $\epsilon$ time flow of the vector field $(H_y,-H_x)$; it is only a formal approximation of it, right? So it is not clear to me that this actually answers the continuous version of the question regarding the flow of a divergence-free vector field.
In other words, one could ask the following question: Let $\phi_{\epsilon}$ be a smooth family of area-preserving diffeomorphisms. Does each $\phi_{\epsilon}$ have a conformal point? Does your argument answers this version of the question? I am not sure of it, since the map you describe is not really the flow of the vector field.
I's similar to the divergence of a vector field. Given a diffeomorphism of a manifold with a volume form, one can ask how it distorts the volume at every point; the answer is that it's the determinant of the Jacobian. For an infinitesimal diffeomorphism, that is, a vector field, one considers its $\epsilon$-flow, calculates the respective Jacobian, and takes the first non-trivial term in $\epsilon$ (of degree 1), which is the divergence of the field.
OK, thanks. But I still don't understand something. We want to prove that the $\epsilon$-flow $\phi_{\epsilon}$ has conformal points, right? that is we want to prove that $d\phi_{\epsilon}(x)$ is conformal for some point $x$ in the disk. It seems to me that you have proved that $d\tilde \phi_{\epsilon}$ has a conformal point, where $\tilde \phi_{\epsilon}(x,y) := (x+\epsilon H_y, y-\epsilon H_x)$ is a first-order approximation of $\phi_{\epsilon}$. So, I am asking: can we deduce from that the original "true" $d \phi_{\epsilon}$ has conformal points?
That is: Is it true that if $\tilde \phi_{\epsilon}$ has conformal points, then so does $ \phi_{\epsilon}$?
No, the infinitesimal version doesn't imply the truth of the original "finite" conjecture. Another analogy that might be helpful: there are various results in symplectic topology claiming that certain symplectomorphisms of closed symplectic manifolds have fixed points. An infinitesimal version is that a Hamiltonian vector field has zeros, and this is easy: these zeros are the critical points of the Hamiltonian function. Arnold's conjectured in the 1960s that the "finite" versions also held, and this stimulated the development of symplectic topology in the 1980s and later.
Thanks. Just to be sure: I of course understand that even the "true"/"full" infinitesimal version - which is a question about a family $\phi_{\epsilon}$ of area-preserving maps, does not settle the question of existence of some diffeo' which does not need to be part of a flow. Furthermore, what you have solved was something weaker than the true flow version of the question, i.e. something which I might call 'formally infinitesimal', right? I do understand that this is a good step in the right direction. I just wanted to know whether this can be immediately pushed to the (true) flow-version.
It seems that this is not immediate. I will now take some time to understand the rest of the details of your answer. Thanks again.
|
2025-03-21T14:48:30.002553
| 2020-03-08T12:29:13 |
354453
|
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|
Stack Exchange
|
pullback of Frobenius morphism and pull back of a projection
Consider a smooth projective curve $X$ over an algebraically closed field $k$, and denote by $\omega_{X}$ its canonical line bundle.
Let $\mathcal X = X × S$ and let $q$ be the first projection with $X,S$ schemes.
Is it true that $f^*\Omega_{\mathcal X|S} = q^* \omega^p_{X}$ with $f$ the absolute frobenius of $\mathcal X$? how can we see it?
These are the exact notations taken from this paper , I guess I have to read: $f^*\Omega_{\mathcal X|S} = q^* \omega^{\otimes p}_{X}$?
Thank you.
From hartshorne p $180$ the canonical sheaf of a $k$-scheme $X$ is $\omega_X=\bigwedge^n \Omega_{X|k}$
So you mean that $X$ is of finite type over a field $k$? Please state your assumptions (in the original question). Also the tag Frobenius algebras is inappropriate.
I just removed the tag, thank you. I updated the question with the assumption from the paper. I think of the caconical line bundle as a locally free $\mathcal O_X-$module of rank $1$ (i.e. an invertible sheaf). Since $X$ is a smooth curve (of dim $1$), I guess $\omega_X=\Omega_{X|X}$ which is invertible.
|
2025-03-21T14:48:30.002664
| 2020-03-08T14:10:26 |
354458
|
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Stack Exchange
|
A problem about the connectivity of vertices that must have the same color for any proper minimal coloring of a graph
I've also posted this problem in Math Stack Exchange (here), and it now has an answer in there.
I'm trying to solve a problem about connectivity of entangled vertices in a graph.
Two vertices $u, v$ of a finite graph $G(V, E)$ are said to be entangled if for any proper coloring $c:V(G)\rightarrow\mathbb{N}$ with $\chi(G)$ colors we have $c(u) = c(v)$, that is, they must have the same color.
What I'm trying to prove is that, given two entangled vertices $u, v\in V(G)$, there is $w\in V(G)$ (possibly equal to $v$) also entangled with $u$ so that there is a set of size $\chi(G)-1$ of disjoint paths from $u$ to $w$.
EDIT: The proof cited below was incorrect, as shown by the accepted answer.
I was able to prove, using the vertex-connectivity version of Menger's theorem and induction, that the previous statement is true if $v$ is the only vertex in $G$ entangled with $u$, so I've been trying to show that if there is not a set of size $\chi(G)-1$ of disjoint paths from $u$ to $v$ (considering $u$ and $v$ entangled), there is still a vertex in $G-v$ entangled with $u$, but without success.
Another idea I had was showing that the minimal (in the number of edges) subgraph of $G$ for which there is still a vertex entangled with $u$, has exactly one vertex entangled with $u$.
I would appreciate some help with this subject.
This statement is true when $\chi(G) \leq 4$, and false when $\chi(G) \geq 5$. The proof of the first statement is long and sprawling, for which I apologise.
Notation: Throughout the post, for non-adjacent vertices $x$ and $y$, we use $\kappa(x,y)$ to denote the minimum cardinality of an $x-y$ separator. By Menger's Theorem, our graph under consideration will have a family of $\kappa(x,y)$ internally disjoint $x-y$ paths. Given a vertex $u$, we use $X_u$ to denote the set of all vertices entangled with $u$.
False for chromatic number 5 or more. Given an integer $n\geq 5$, we construct a graph $H_n = (V,E)$ with $\chi(H_n) = n$, that contains vertices $u$ and $v$ such that $X_u = \{u,v\}$ and $\kappa(u,v) = 3$. Let $V = \{u,v,x\}\sqcup A \sqcup B \sqcup C$, where $A$ and $C$ are sets of $n-2$ vertices, and $B$ is a set of $2$ vertices.
Make $A$, $B$ and $C$ cliques. Let $u$ be adjacent to every vertex of $A$, every vertex of $A$ adjacent to every vertex of $B$, every vertex of $B$ adjacent to every vertex of $C$, and every vertex of $C$ adjacent to $v$. Finally, let $x$ be adjacent to $u$, $v$ and every vertex of $A$.
By inspection, $\chi(H_n) = n$. To see that $H_n$ is as promised, we say without loss of generality that the vertices of $B$ get colours $1$ and $2$. It is then clear that either $c(x) = 1$ and $c(u)=c(v)=2$, or $c(x) = 2$ and $c(u)=c(v)=1$, so $u$ and $v$ are entangled with eachother, but not with any vertex of $B$, so $X_u = \{u,v\}$. Further, $B\cup \{x\}$ is a $u-v$ separator, so $\kappa(u,v) = 3$. This is inspired by the construction given in this answer.
True for chromatic number 4 or less. We need some new machinery. We say that a vertex $u$ and a set $S$ of vertices in a graph $G$ are weakly entangled if in every optimal colouring $c$ of $G$, there exists a vertex $w_c$ in $S$ such that $c(u)=c(w_c)$.
Lemma 1. Let $u$ and $v$ be entangled vertices in an $n$-chromatic graph $G$. If $S$ is a $u-v$ separator with $|S|\leq n-2$, then $u$ and $S$ are weakly entangled.
Proof. Assume to the contrary that $G$ has an optimal colouring $c$ such that $c(u)\neq c(w)$ for all $w$ in $S$, and let $G_v$ be the component of $G-S$ containing $v$. Since $|S|\leq n-2$, we may assume without loss of generality that $\{c(w) : w\in S\}\subseteq \{1,2,\dots,n-2\}$ and that $c(u)=c(v)=n-1$. Since no vertex of $S$ receives colour $n-1$ or $n$, we may swap these two colours in $G_v$ to obtain a new optimal colouring of $G$. In this new colouring, $c(u)=n-1$ and $c(v) = n$, contradicting their entanglement. QED.
Lemma 2. If $u$ and $v$ are entangled vertices in a graph $G$ with $\chi(G)\geq 3$, then $\kappa(u,v) > 1$.
You can prove Lemma 2 using the same "assume not, and swap colours on one side of the separator" as in the proof of Lemma 1.
Proposition 3. Your statement is true for any graph $G$ with $\chi(G) \leq 3$.
Proof. This is trivial for $\chi(G)\leq 2$. For $\chi(G) = 3$, your statement follows from Lemma 2 and Menger's Theorem.
This leaves us with the hard case in which $\chi(G)=4$. Lemmas are needed.
Lemma 4. Let $G$ be a graph with $\chi(G)\geq 4$. If $u$ and $v$ are entangled vertices, and $S$ is a $u-v$ separator with $|S|=2$, then $S\cap X_u \neq \emptyset$ (i.e., the separator contains a vertex entangled with $u$).
Proof. Assume to the contrary that $S=\{x,y\}$ is a $u-v$ separator and that neither $x$ nor $y$ are entangled with $u$. We assume without loss of generality that in every optimal colouring of $G$, the vertices $u$ and $v$ get colour 1. By Lemma 1, at least one of $x$ and $y$ get colour 1 in every optimal colouring of $G$. Since we can permute colours, we may also assume that in every optimal colouring $c$, we have that $\{c(x), c(y)\}\subseteq \{1,2\}$.
By our assumption to the contrary, there must exist optimal colourings $c_1$ and $c_2$ such that $c_1(x)=c_2(y) = 1$ and $c_1(y)=c_2(x) = 2$ (if no such colourings existed, then $u$ would be entangled with at least one of $x$ and $y$). Let $c_2'$ be the colouring obtained from $c_2$ by swapping colours 1 and 2. Let $G_u$ and $G_v$ be the components of $G-S$ containing $u$ and $v$ respectively. Observe that $c_1$ and $c_2'$ agree on $S$. By colouring $G_u\cup S$ with $c_1$ and colouring $G_v\cup S$ with $c_2'$, we obtain an optimal colouring of $G$ in which $u$ and $v$ have different colours, contradicting their entanglement. QED.
Theorem 5. Let $G$ is a graph with $\chi(G)=4$, and let $u$ be a vertex of $G$. If $|X_u| \geq 2$, then there exists some vertex $w$ in $X_u$ such that $\kappa(u,w)\geq 3$.
Proof. We may assume that $G$ is 2-connected (we can discard everything apart from some block of $G$ that contains $u$ and some other vertex of $X_u$). Let $v$ be a vertex of $X_u-\{u\}$ that minimises $d(u,v)$, and assume contrary to the Theorem statement that for all $w\in X_u$, we have $\kappa(u,w) = 2$. Among all the $u-v$ separators with exactly two vertices in $G$, let $S=\{x,y\}$ be a separator that minimises the distance $d(u,S)$, and assume without loss of generality that $d(u,x) = d(u,S)$. We further choose $S$ such that among all two-vertex $u-v$ separators containing $x$, the distance between $u$ and the vertex $y$ in $S-\{x\}$ is as small as possible. (All in all, we have minimised $d(u,v)$, then minimised $d(u,x)$ given the choice of $v$, then minimised $d(u,y)$ given the choices of $v$ and $x$).
Since $S$ separates $u$ and $v$, any geodesic from $u$ to $v$ must contain a vertex of $S$. By Lemma 4 and the minimality of $v$, we deduce that the vertex $x$ lies on all $u-v$ geodesics and is not in $X_u$, and that the vertex $y$ is in $X_u$ (so $u$ and $y$ are entangled).
We now construct an auxillary graph to show that $\kappa(u,y)\geq 3$. First, let $C_u$ and $C_v$ be the components of $G-S$ containing $u$ and $v$ respectively, and let $G_u = G[C_u\cup S]$ be the subgraph of $G$ induced by the vertices of $C_u$ and $S$. Similarly let $G_v = G[C_v\cup S]$. Form a new graph $G^*$ from $G_u$ by adding a vertex $y'$ with $N(y') = N(y)$, and adding a vertex $z$ such that $N(z) = \{x,y,y'\}$ (see the diagram, we are "cloning" $y$).
Note that if $T$ is a $u-z$ separator in $G^*$, then $T-\{y'\}$ is a $u-v$ separator in $G$. By the minimality of distances from $u$ to $v$, $x$ and $y$, we thus deduce that $\kappa(u,z)\geq 3$ in $G^*$. So there exist internally disjoint $u-z$ paths $P_1$, $P_2$ and $P_3$ in $G^*$. Without loss of generality, $y$ and $y'$ are in $P_1: u, \dots, y, z$ and $P_2: u,\dots, y', z$, and $x$ is in $P_3$. By deleting the end vertex $z$ from $P_1$ and $P_2$, and replacing $y'$ with $y$, we get two internally disjoint $u-y$ paths from $P_1$ and $P_2$. By following $P_3$ from $u$ to $x$, and then following an $x-y$ path in $G_v$, we obtain a third internally disjoint path. QED.
Commentary: Lemma 4 is the linchpin here. In a 3-separator between entangled vertices, you can change which vertex gets the entangled colour in different optimal colourings (at least one vertex must get it by Lemma 1). In a 2-separator, you cannot.
Here is a proof to a related claim that hopefully will give you some ideas.
Claim.
Let $X$ be an equivalence class of the entanglement relation on $V(G)$. Then for all distinct $u,v \in X$, there exist
$\chi(G)-1$ edge-disjoint paths in $G$ between $u$ and $v$.
Proof. Let $k=\chi(G)$ and $V_1, \dots, V_k$ be a partition of $V(G)$ into stable sets. By relabelling, we may assume that $X \subseteq V_1$. Observe that all vertices of $X$ must be contained in some component of $G[V_1 \cup V_2]$. Otherwise, we may recolour to obtain a $k$-colouring of $G$ where two vertices of $X$ are coloured differently. In particular, for all distinct $u,v \in X$, there is a $u$--$v$ path in $G[V_1, \cup V_2]$. Repeating the argument for $i=2, \dots, k$, gives the $k-1$ edge-disjoint $u$--$v$ paths in $G$. $\square$
Note that the claim proves something stronger and weaker than what was asked in the original question. It is weaker because the paths are edge-disjoint not vertex-disjoint. But it is stronger since it holds for all distinct pairs $u,v \in X$. Moreover, the paths constructed in the proof are almost vertex-disjoint. The only vertices they have in common are in $V_1$.
|
2025-03-21T14:48:30.003232
| 2020-03-08T14:57:18 |
354464
|
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"Libli",
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|
Stack Exchange
|
schemes having same reduced underlying space and same cotangent sheaf are isomorphic?
Let $X,Y$ be two closed subschemes of $\mathbb{A}^n_{\mathbb{C}}$ for some fixed $n$. Assume that I have an isomorphism of the underlying reduced spaces:
$$f : X_\text{red} \stackrel{\sim}\longrightarrow Y_\text{red}$$ which induces an isomorphism of $\mathcal{O}_{X_\text{red}}$-module:
$$ f^{*} \left(\Omega_{Y/\mathbb{C}} \otimes \mathcal{O}_{Y_\text{red}} \right) \stackrel{\sim}\longrightarrow \Omega_{X/\mathbb{C}} \otimes \mathcal{O}_{X_\text{red}}.$$
Can I lift $f$ to an actual isomorphism $X \stackrel{\sim}\longrightarrow Y$? Note that in the situation I am interested $Y$ is very singular. If $Y$ were smooth, I guess the theory of infinitesimal liftings would give the answer.
This is related to this question. Indeed, the existence of such a lift would guarantee the isomorphism between the rings introduced by the OP in that other question.
Consider the simplest example:
$$
X = \mathrm{Spec}(\mathbb{C}[\epsilon]/\epsilon^2), \qquad
Y = \mathrm{Spec}(\mathbb{C}[\epsilon]/\epsilon^3).
$$
Definitely, $X_{\mathrm{red}} \cong Y_{\mathrm{red}}$. Also, a simple computation shows that
$$
\Omega_{X/\mathbb{C}} \cong \mathbb{C},
\qquad
\Omega_{Y/\mathbb{C}} \cong \mathbb{C}[\epsilon]/\epsilon^2,
$$
hence $\Omega_{X/\mathbb{C}} \otimes \mathcal{O}_{X_{\mathrm{red}}} \cong \Omega_{Y/\mathbb{C}} \otimes \mathcal{O}_{Y_{\mathrm{red}}}$, but of course $X \not\cong Y$.
Ok, good point!
|
2025-03-21T14:48:30.003342
| 2020-03-08T17:41:20 |
354473
|
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"Dmitri Pavlov",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354473"
}
|
Stack Exchange
|
Embeddings into $\mathbf{Cat}$ preserving cohomology
Let $\mathbf{C}$ be a category in which we have some notion of (co)homology (say, groups with the usual cohomology, posets with the homology of the geometric realisation). We say that an embedding $F: \mathbf{C} \to \mathbf{Cat}$ of $\mathbf{C}$ preserves (co)homology if, for each object $X$ of $\mathbf{C}$, the (co)homology of $X$ agrees with the (co)homology of $F(X)$ (defined as the higher (co)limits of a suitable (that is: "naturally" derived from what we're taking our coefficients in over on $\mathbf{C}$) sheaf on $F(X)$).
For example, the usual functor $F_1$ embedding the category $\mathbf{Grp}$ of all groups into $\mathbf{Cat}$ by treating a group $G$ as a category with one object $\ast$ and $(\mathrm{Hom}(\ast,\ast),\circ) \cong G$ preserves cohomology: for a group $G$, the $n$th cohomology of $G$ with coefficients in a $G$-module $M$ is exactly the $n$-th higher limit of the presheaf on $F_1(G)$ taking the value the underlying abelian group of $M$ at $\ast$ and sending the morphism corresponding to each $g \in G$ to the endomorphism given by the action of $g$ on $M$.
Similarly, the embedding $F_2$ of the category of posets into $\mathbf{Cat}$ sending a poset $P$ to the category whose objects are the elements of $P$ with exactly one morphism $x \to y$ if $x \leq y$ in $P$, and no morphisms otherwise preserves cohomology, for essentially trivial reasons.
So, that brings us to my question: which other cohomology-preserving functors from "interesting" categories exist? In particular, can we do this for CW-complexes (or some other reasonable subcategory of $\mathbf{Top}$)?
For topological spaces, one can take the category of simplices of the singular simplicial set.
|
2025-03-21T14:48:30.003495
| 2020-03-08T17:42:56 |
354474
|
{
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"Federico Poloni",
"Johann Cigler",
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"url": "https://mathoverflow.net/questions/354474"
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|
Stack Exchange
|
Matrix-valued periodic Fibonacci polynomials
Consider the Fibonacci polynomials $f_n(x)$, defined by the recursion $f_n(x)=xf_{n-1}(x)-f_{n-2}(x)$ with initial values $f_0(x)=0$ and $f_1(x)=1$. It is well known that the values of these polynomials
are periodic with period $6$ for $x=1$ and $x=-1.$
There are also some matrices $x$ with integer coefficients for which the sequence $f_n(x)$ is periodic.
For example let $$x=g_k=\left(g(i,j)\right)_{i,j=0}^{k-1}$$ be the matrix with entries $g(i,j)=1$ if $|i-j|=1$ or $i=j=0$ and $g(i,j)=0$ else.
Then the sequence $f_n(x)$ is periodic with period $2(2k+1).$
Let for example $k=2$. Here we get a sequence with period $10:$
$\left ( \begin{matrix} 0 & 0 \\ 0 & 0
\end{matrix} \right )$, $\left ( \begin{matrix} 1 & 0 \\ 0 & 1
\end{matrix} \right )$, $\left ( \begin{matrix} 1 & 1 \\ 1 & 0
\end{matrix} \right )$, $\left ( \begin{matrix} 1 & 1 \\ 1 & 0
\end{matrix} \right )$, $\left ( \begin{matrix} 1 & 0 \\ 0 & 1
\end{matrix} \right )$, $\left ( \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right )$, $\left ( \begin{matrix} -1 & 0 \\ 0 & -1
\end{matrix} \right )$, $\left ( \begin{matrix} -1 & -1 \\ -1 & 0
\end{matrix} \right )$, $\left ( \begin{matrix} -1 & -1 \\ -1 & 0
\end{matrix} \right )$, $\left ( \begin{matrix} -1 & 0 \\ 0 &-1
\end{matrix} \right )$, $\left ( \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right )$
,$\dots$.
There are also some other matrices $x$ with integer coefficients such that the sequence $f_n(x)$ is periodic. I would be interested to find all such matrices with integer coefficients. Is there anything in the literature?
We seem to have $f_n(x)=\sum_{k=0}^{\frac{n-1}{2}} \binom{n-1-k}{k} (-1)^k x^{n-2k-1}$
Up to a change of basis, we can assume the matrix is in Jordan form $M = \bigoplus J_i$. It is known that for every polynomial $p$ one has
$$
p(J_i) =
p(\begin{bmatrix}
\lambda & 1\\
& \lambda & 1\\
& & \ddots & \ddots\\
& & & \lambda & 1\\
& & & & \lambda\\
\end{bmatrix})
$$
$$
\begin{bmatrix}
p(\lambda) & p'(\lambda) & \frac{p''(\lambda)}{2} & \dots & \frac{p^{(n-1)}(\lambda)}{n!}\\
& p(\lambda) & p'(\lambda) & \dots & \frac{p^{(n-2)}(\lambda)}{(n-1)!}\\
& & \ddots & \ddots &\vdots\\
& & & p(\lambda) & p'(\lambda)\\
&&&&p(\lambda)
\end{bmatrix}.
$$
In particular, periodicity depends on periodicity of the sequence on the eigenvalues of the matrix, and on the sequence of derivatives in case there are nontrivial Jordan blocks.
So if the matrix is diagonalizable, its period is the lcm of all periods of the eigenvalues. If it has Jordan blocks, you have to work out periods of derivatives of the sequence as well.
So you can reduce your problem to a scalar one. Have you solved that completely? Do you know if there are other periodic values other than $\pm 1$ ?
@ Federico Poloni: If I already have an integer valued matrix as in the special case $k=2$ where the matrix has two different eigenvalues which satisfy $a^2=a+1$, then it reduces to the sequence $f_n(a).$ This is a one-dimensional case, but with non-integer $a$ . In this case it is easy to verify that it is periodic with period $10$. But the problem is to find such matrices with integer coefficients.
OK, I see. I read the first revision and it did not mention the fact that the entries must be integers. I am deleting this answer soon as it does not answer the revised question anymore.
@FedericoPoloni I think you could maybe update the answer instead. See, suppose we are given some matrix $A$ with a cyclic vector (1 Jordan block is a special case). Then, consider our vector space as $\mathbb{C}[x]$ module (multiplication by $x$ acts by $A$), it is identified with $\mathbb{C}[x]/P(x)$ where $P(x)$ is a minimal polynomial of $A$.
Now, if we take $P(x) = gcd(f_n(x), f_{n+1}(x)-1)$, then multiplication by $x$ satisfies the desired periodicity (and universal in a sense that any other matrix of period $n$ has the same or lesser Jordan blocks with the same eigenvalues).
Obviously, it doesn't exactly give the full classification (because what is this gcd and when it is nontrivial is not clear), but we could probably work it out quite easily.
@LevSoukhanov The details are not fully clear to me but it looks promising -- you should write it up as an answer yourself!
Also, the scalar problem should be easy, right? You have $S = \begin{bmatrix}
0 & 1\
-1 & x\
\end{bmatrix}$ and want to know when it is periodic. It is more or less clear: it has determinant one and so only requirement is that its trace equals $2\cos(\frac{2\pi p}{q})$. It closes the scalar problem. I believe there shouldn't be any non-trivial Jordan blocks in the matrix version of the problem (but didn't do it yet).
|
2025-03-21T14:48:30.003785
| 2020-03-08T18:49:55 |
354475
|
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"authors": [
"Iosif Pinelis",
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|
Stack Exchange
|
Chernoff-type bound for sum of Bernoulli random variables, with outcome-dependent success probabilities
Let $X = (X_1, X_2, \ldots, X_n)$ be a sequence of (not necessarily independent) Bernoulli random variables where for each $i$, the success probability $\Pr[X_i = 1]$ itself is a random variable depending on the sequence $(X_1, \ldots, X_{i-1})$. For any assignment $X'=(X'_1, \ldots, X'_n)$ define $$\mu^\star(X') = \sum_{i=1}^n \Pr[X_i = 1 \mid X_1=X'_1, \ldots, X_{i-1}=X'_{i-1}].$$
Is it possible to show that with probability $1-\epsilon$,
$$
\sum_{i=1}^n X_i = \mu^\star(X) \pm \Theta\left(\sqrt{\mu^\star(X) \log \epsilon^{-1}}\right)?
$$
Note that if the success probabilities were fixed a priori, this would be implied by Chernoff bound. On the other hand, using Azuma's inequality on an appropriate martingale, a bound of $\sum_{i=1}^n X_i = \mu^\star(X) \pm \Theta\left(\sqrt{n \log \epsilon^{-1}}\right)$ could be proved (see this relevant question) which unfortunately depends on the sequence's length. Any ideas about how to get the dimension-free variant?
Won't be dimension independent due to the log under the square root in your aimed result
That's because of the success probability. I'll change it to avoid confusion.
Perhaps answers and comments to this question contain useful references? For example, version 4 of Bernstein's inequality on Wikipedia might give the desired bound, but unfortunately I have no time to check this now.
I think the bound as stated will likely fail to hold in general, even though I don't have a counterexample at this point. However, if under the square root $\mu^(X)$ is replaced by a real number $m>\mu^(X)$ such that $\ln\epsilon^{-1}=O(m)$, then the resulting bound can be proved, I think.
@MateuszKwaśnicki Thanks for the references. I did look at the question you linked before posting this question. I don't think the methods mentioned in the answers there or the martingale variant of Bernstein's inequality can be used to prove this since in our case we don't really have a useful bound on the variance of the entries as the probabilities are revealed along the way and worst-case estimates seem far off.
@IosifPinelis Are you suggesting that $\sum_{i=1}^n X_i = \mu^\star(X) \pm \Theta(\sqrt{\mu^\star(X)\log \epsilon^{-1}} + \log \epsilon^{-1})$ is provable? If yes, that would also be equally good and works for the applications I had in mind. Do you mind to explain how you think it can be proved?
@Mathman : No, this is not quite what I am suggesting: In what I am suggesting, $m>0$ must be a real number, not a random variable.
@IosifPinelis Oh I see, so it must be a real number that is larger than all possible outcomes of $\sum_i X_i$? How would the proof for that go?
@Mathman : No, the condition is $m>\mu^(X)$, not $\mu>\sum_i X_i$. If the condition $m>\mu^(X)$ quite suits you, I'll write the answer.
@IosifPinelis Please go ahead, it will be very useful and suits the application I had in mind.
$\newcommand\ep{\delta}$$\newcommand\de{\epsilon}$For $j=0,\dots,n$, let $S_j:=\sum_1^j d_i$, where $d_i:=X_i-E_{i-1}X_i$ and $E_{i-1}$ is the conditional expectation given $X_1,\dots,X_{i-1}$, with $E_0:=E$ and $S_0:=0$. Clearly, $(S_j)$ is a martingale.
By Theorem 8.7, if
$|d_i|\le a$ and $\sum_1^n E_{i-1}d_i^2\le b^2$ for some real $a,b>0$ and all $i$, then
\begin{equation}
P(|S_n|\ge r)\le2\exp\Big\{-\frac{b^2}{a^2}\,\psi\Big(\frac{ra}{b^2}\Big)\Big\} \tag{1}
\end{equation}
for $r\ge0$, where $\psi(u):=(1+u)\ln(1+u)-u$.
In our case, recalling that the $X_i$'s take values in the set $\{0,1\}$, we have $|d_i|\le1$, so that we can take $a=1$, and also
$$E_{i-1}d_i^2=E_{i-1}(X_i-E_{i-1}X_i)^2\le E_{i-1}X_i^2=E_{i-1}X_i,$$
whence
$$\sum_1^n E_{i-1}d_i^2\le\sum_1^n E_{i-1}X_i=\mu^*(X).$$
We now assume that for some real $m>0$ and $\ep\in(0,1)$ we have
$$\mu^*(X)\le m\quad\text{and}\quad\ln\frac1\ep\ll m;\tag{2}$$
as usual, we write $u\ll v$ or $v\gg u$ to mean $|u|=O(v)$, where the constant in $O(\cdot)$ is universal.
Then we may take $b^2=m$. Also, in (1), take $r=\sqrt{m\ln\frac1\ep}$. Then $\frac{ra}{b^2}=\frac rm=\sqrt{\frac1m\,\ln\frac1\ep}\ll1$ by (2), and hence $\frac{b^2}{a^2}\,\psi\big(\frac{ra}{b^2}\big)\gg\frac{b^2}{a^2}\,\big(\frac{ra}{b^2}\big)^2=\frac{r^2}{b^2}=\ln\frac1\ep$. Now (1) yields
$$P\Big(\Big|\sum_1^n X_i-m^*(X)\Big|\ge\sqrt{m\ln\frac1\ep}\Big)
=P(|S_n|\ge r)\le2\ep^c $$
for some universal real constant $c>0$. Finally, letting $\de:=2\ep^c$, we get the result that, as you said in your comment, suits the application you had in mind.
Thank you so much far taking the time to write this. It turns out that @MateuszKwaśnicki was right.
|
2025-03-21T14:48:30.004221
| 2020-03-08T19:56:02 |
354480
|
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|
Stack Exchange
|
When will a monad satisfy Moggi’s “equalizer” property?
I’m interested in finding when ($V$)-monads will satisfy the universal property that $\eta$ equalizes $\eta_T, T\eta$. I’m particularly interested in the case for monads on presheaf categories or that arise from morphism of limit theories.
You mean "$\eta$ is the equalizer of ... " ? because $\eta$ always equalize $\eta T$ and $T \eta$, that's part of the definition of a monad... Also the question seem too vague.
Thanks, I think I need to add a bit more context. I am fairly sure there is a class of monads where this property does not hold, for example the powerset monad or the monad arising from a degenerate Lawvere theory.
It's not really the monad that has this property but the category: it's not difficult to construct a new category that has it. See for example my www.paultaylor.eu/ASD/sobsc Also, this condition was observed long before Moggi.
Thank you for directing me to your work on Abstract Stone Duality, that’s just the sort of material I was hoping to find!
|
2025-03-21T14:48:30.004599
| 2020-03-08T20:46:30 |
354483
|
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|
Stack Exchange
|
Grayson filtration and weight filtration
I had a question that might be well-known but I'm not sure where to find it. Grayson defined a filtration on the algebraic $K$-theory of affine regular rings via commuting automorphisms which you can find here. You can define the same thing for general regular schemes. My question is that is it known that for a regular scheme (or smooth varieties over a field) the filtration coincides with the filtration coming from the Adams operation?
Walker here has proved it with the assumption of resolution of singularities. Suslin later on proved that the motivic spectral sequence corresponding to the Grayson filtration has the same $E^2$ page as the Friedlander-Suslin spectral sequence. But it obviously does not imply whether the Adams and Grayson filtrations coincide rationally or not.
My impression is that Adams operations are "well known" to act coherently on all levels of the weight spectral sequence for K-theory (of smooth varieties); probably, this fact was established by Gillet and Soul´e. It easily follows that the filtration induced by this spectral sequence on rational K-theory is the Adams one. You may find an argument of this sort (that yields even more information) in the Yagunov's paper "Motivic cohomology spectral sequence and Steenrod operations", Compos. Math. 152 (2016), no. 10, 2113–2133.
Now I should add a few words on versions of weight (or motivic) spectral sequences. The Friedlander-Suslin version is not popular now, since it depends on certain doubtful arguments of Bloch and Lichtenbaum. However, there is the Grayson version, and the Voevodsky-Levine version, and they are known to be isomorphic (see the paper I cited for more detail).
I had a slight confusion about these motivic filtrations. In order to get them there is a sheafification process. So let's say since we know weight zero part vanishes it implies that the map from weight>=1 part of grayson filtration to K-theory space, induces isomorphism on higher homotopy groups locally. Or does it also induce the isomorphism globally (at least rationally)? For example take a non-affine variety and consider the grayson filtration for it (Without any sheafification).
In the Walker's paper that I have referred, he has proved that the naive grayson filtration without any sheafificaiton process (he works with $\mathbb{P}^1$ instead of $\mathbb{G}_m$) on any smooth variety over a field of char 0 is rationally the same as Adams filtration.
I am not an expert, sorry. I think that rationally everything "should" commute with sheafifications, since the filtration splits and all morphisms between K-theory of various schemes "should" be compatible with these splittings.
|
2025-03-21T14:48:30.004806
| 2020-03-08T21:28:52 |
354488
|
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Stack Exchange
|
Weight 3 modular form associated to singular abelian surfaces?
Given an extremal K3 surface $S$ over $\mathbb{Q}$ (i.e. a K3 surface with maximal Picard rank) there is a 2-dimensional Galois representation on the transcendental lattice $T(S)$, and an associated cusp modular form of weight three. This is of course the modularity of extremal K3 surfaces, generalizing the famous result on elliptic curves. See, for example, Theorem 2.2 of (https://arxiv.org/pdf/1212.4308.pdf).
So are there weight 3 modular forms similarly associated to singular abelian surfaces over $\mathbb{Q}$? Can anyone point me to any literature where these are computed?
A singular abelian surface is an abelian surface with maximal Picard rank 4, and any such is isogenous to a product of elliptic curves with CM. The transcendental lattice (middle cohomology modulo the algebraic cycles) is then 2-dimensional and should conceivably carry a Galois representation. Which should have an associated modular form. Is this out there in the literature somewhere?
This paper discusses modularity of products of elliptic cuvers. It refers to some examples by Howe for products of elliptic curves with CM. Maybe check there?
May I ask why it is of weight 3 ?
@reuns Someone can probably give a better comment than me, but if a $d$-dimensional variety is modular, I think it is expected that the weight should be $d+1$. For example, elliptic curves correspond to weight 2 forms, and rigid Calabi-Yau threefolds give weight 4 forms. So K3 and abelian surfaces should be weight 3.
For the intuition: for $f\in M_k(\Gamma)$ the period polynomials make $\Bbb{C}[X]{\deg \le k-2}$ into a $\Gamma$ module, quotienting $\Bbb{C}[X]{\deg \le k-2}$ by the action of $\Gamma$ gives a dimension $k-1$ complex torus which in general is not algebraic. If it is then it is a dimension dimension $k-1$ abelian variety.
This more of comment. If $A$ is an abelian surface with Picard rank $4$, then the associated Kummer surface $X$ has Picard rank $20$. So the answer to your question is presumably yes. To be a bit more explicit, the etale cohomology $$H^2(\bar X_{et}, \mathbb{Q}_\ell)= H^2(\bar A_{et}, \mathbb{Q}_\ell)\oplus \mathbb{Q}_\ell(-1)^{16}$$
So you'll see the same Galois representation $S^2H^1(E)$ in both surfaces, where $E$ is the underlying CM elliptic curve. This presumably is what gives rise to your modular form.
Interesting, I didn't know about this nice connection to extremal K3s. Thanks! So you're expecting that the modular form for $A$ and it's Kummer K3 should be the same? Also, do you mind if I clarify what you mean by $S^{2}$? I was guessing symmetric product, but I thought we wanted specifically 2-dimensional Galois representations.
There is $\epsilon$ more to this, because if $E$ is CM, then $S^2 H^1(E)$ is reducible. $H^1(E)$ is the induction of a 1-diml representation $\psi$ of $G_K$, where $K$ is the CM field, and $S^2 H^1(E)$ splits as (cyclo char) + (induction of $\psi^2$) and the induction of $\psi^2$ corresponds to a wt 3 mod form.
Yes, sorry I was a bit hasty, I meant a subquotient of $S^2H^1(E)$
@DavidLoeffler Is there something in the literature which covers what I'm asking about? Ideally, I'd like to be able to look these weight 3 forms up, if it's that easy.
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2025-03-21T14:48:30.005037
| 2020-03-08T23:59:54 |
354497
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Stack Exchange
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Is there an symplectic field theory compactness theorem applicable in the context of Floer cohomology of a symplectomorphism?
Is there any reference in the literature about results regarding symplectic field theory (SFT) compactness for a neck-stretch in the context of Floer homology of a symplectomorphism $\phi \colon (M,\omega) \to (M,\omega)$, where $M$ is a closed symplectic manifold?
I would be interested in stretching $M$ along a tubular neighbourhood of a contact type codimension $1$ closed submanifold $S$.
I can see two problems:
if we consider the classical interpretation of $HF^*(\phi)$ as in [DS94] then we are considering $J$-holomorphic strips $u \colon \Bbb R \times [0,1] \to M$ with boundary condition $u(s,1)=\phi u(s,0)$. The target space would be closed, the submanifold $S$ would be closed but the domain wound be biholomorphic to a closed disk with two punctures on the boundary. Since we can assume that $$\text{lim}_{s \to \pm \infty} u(s,t) = x_{\pm} \in \text{Fix}(\phi)$$ then $u$ would extend to a map from the disk to $M$. But the only version of SFT for curves with boundary I’m aware relies on some control on where these boundaries are mapped (some lagrangian boundary conditions say)
Therefore I focused on the interpretation of $HF^*(\phi)$ as the count of horizontal sections of the symplectic mapping torus $\Bbb R \times M_{\phi}\to \Bbb R \times S^1$ (See the PhD thesis of Seidel for example). In this scenario, we are interested in $J$-hol sections of the symplectic mapping torus (call it for brevity $T_{\phi}$) whose domain is $\Bbb R \times S^1$. The upside now is that the domain doesn’t have boundary but the drawback is that $T_{\phi}$ is not compact and (assuming that $\phi \colon S\to S$) the neighbourhood on which I want to stretch would be a tubular neighbourhood of $\Bbb R \times S_{\phi}$ which again, is not compact.
On [CM05] the authors state the compactness theorem for curves into a closed symplectic manifold stretched along a closed stable hypersurface $M$, but then claims that the proof goes through in the case $X$ is a symplectic cobordism $X$ (which is not compact since we attach suitable ends to its boundary components) and $M$ is a closed stable hypersurface.
Can someone point me out where compactness of $M$ is used in that paper?
My hopes are that since we are dealing with finite energy sections maybe we can circumvent the fact that $\Bbb R \times S_{\phi}$ is not compact and rely instead on the closeness of $S_{\phi}$
And more importantly,
Are there instances of the result I’m interested in, already available in the literature?
ADDENDUM I’ve been thinking about it for a couple of days, and I realized that another possible problem is that we need to assume some non-degeneracy of the Reeb orbits (Morse-Bott condition) and I don’t see how to impose it in this setting without adding it in the hypothesis. In the paper of [BEHWZ03] it is said that these trajectories can be assumed to be non-degenerate (paragraph 2.3) and that it's a property of $J$, the almost complex structure that it's assumed to be symmetrical, cylindrical and adjusted to the symplectic form $\omega$. So is this non-degeneracy of the period orbits a generic property of $J$?
REFERENCES
[CM05] Cieliebak, K.; Mohnke, K. Compactness for punctured holomorphic curves. Conference on Symplectic Topology. J. Symplectic Geom. 3 (2005), no. 4, 589--654. MR2235856
[DS94] Dostoglou S., Salamon D. Self-dual instantons and holomorphic curves. Annals of Mathematics 139.3 (1994): 581-640.
[BEHWZ03] Bourgeois, F.; Eliashberg, Y.; Hofer, H.; Wysocki, K.; Zehnder, E. Compactness results in symplectic field theory. Geom. Topol. 7 (2003), 799--888.
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2025-03-21T14:48:30.005289
| 2020-03-09T04:00:39 |
354503
|
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"Lucia",
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Stack Exchange
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The multiplicative constant in the estimate for $S_a(x)=\sum_{n\leq x} d(n)^a$
Let $a$ be a positive real constant and let $d(n)$ denote the number of divisors of $n.$ Define
$$
S_a(x)=\sum_{n\leq x} d(n)^a.
$$
For $a=1,$ the following is well known
$$
S_1(x)=\sum_{n\leq x} d(n)=x \log x + (2 \gamma -1) x +{\cal O}(\sqrt{x})
$$
while for more general $a$, one has
$S_a(x) \sim C(a) x (\log x)^{2^a -1}$ where
$$
C(a) = \Gamma(2^a)^{-1} \prod_p \left( 1 - \frac{1}{p} \right)^{2^a} \left( \sum_{k \geq 0} \frac{(k+1)^a}{p^k}\right).
$$
I am wondering about accurate estimates of the constant $C(a),$ especially for $a\in (0,1).$
Edit: I took @Lucia's suggestion and obtained $C(1/2)\approx 1.14778.$
Here is a plot of the quantity $C(a)$ as a function of $a\in (0,1)$ where three curves are plotted for 100,150 and 200 terms with no appreciable change in the computational result.
*I am still wondering, however, if analytical lower bounds are known for $C(a)$ where $a\in(0,1),$ or for other larger values of $a.$ In particular, the fact that the maximum is achieved a bit to the right of $a=1/2$ looks intriguing.
Why don't you just compute $C(a)$ using the product? 100 terms should give you the first couple of digits.
@Lucia thanks, how do I compute/estimate the sum involving inverse prime powers?
@AlessandroLanguasco, thanks. I will try your suggestion.
It converges rapidly (like a geometric progression). So what's the problem?
@Lucia, I took your suggestion and the computation rapidly converged. $C(1/2) \approx 1.14778$ was the result. Also tried values of $a$ from $0.05$ to $0.95$ at intervals of $0.05$ and got a unimodal sequence for $C(0.05k)_{k=1}^{19},$ with minimum value $1.02173,$ and maximum value $1.1498,$ with the sampled maximum achieved at $a=0.55$.
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2025-03-21T14:48:30.005443
| 2020-03-09T08:43:29 |
354511
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Stack Exchange
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Conjugate points depends on choice of geodesic?
Is there a compact Riemannian manifold $M^n$ and $p, q\in M^n$, such that $q$ is conjugate to $p$ along some geodesic $\gamma$, but $q$ is not conjugate to $p$ along another geodesic $\tilde{\gamma}$?
Yes, take an oblate ellipsois and $p$ the south pole. The conjugate locus will be a sort of astroid with center at the north pole. See for example figure 7 here: https://projecteuclid.org/download/pdf_1/euclid.em/1087568023
Take now $\gamma$ to be a short meridian, in such a way that $\gamma(0)=p$ and $\gamma(\pi)$ is the north pole, and $\gamma$ remains a minimizing geodesic until there. After reaching the north pole the geodesic will cease to be optimal of course, and at some point $\gamma(\pi + \tau)$ will be conjugate to $p$ along $\gamma$ (this is precisely the point at which $\gamma$ intersects for the second time the astroid).
But then there exists another geodesic joining $p$ with $\gamma(\pi+\tau)$, which will be of course free of conjugate points.
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2025-03-21T14:48:30.005550
| 2020-03-09T09:23:31 |
354514
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Stack Exchange
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How to prove a result related to prime number theorem in research paper of Rivoal and Zudilin
Question is ->I am studying research paper: A note on odd zeta values and I am unable to think how to deduce a result which the authors don't prove. This result has to be proved assuming the prime number theorem and it's on Page 12 of the paper :
Prove that $\lim_ {n\to\infty} \frac{\log(\Phi_n) } {n} =\int_0^{1}
\rho_0 (t) d(\psi(t) + 1/t) $, where $\psi(t) $ = $\frac {\Gamma'(t) }
{\Gamma(t) } $.
where $\Phi(n)$ and $\rho(n)$ are described in this image:
Can someone please tell how to prove this result ?
I shall be really thankful.
This question already had an accepted answer. Why have you bumped it with an edit?
If we start by partitioning the range of summation into intervals on which $\rho_0(\frac np)$ is constant, we obtain
\begin{align*}
\log \Phi_n &= \sum_{2\sqrt n<p\le n} \rho_0\big( \tfrac np \big) \log p \\
&= \sum_{k=6}^{3\sqrt n-1} \sum_{\frac{6n}{k+1} < p \le \frac{6n}k} \rho_0\big( \tfrac np \big) \log p \\
&= \sum_{k=6}^{3\sqrt n-1} \rho_0\big( \tfrac k6 \big) \sum_{\frac{6n}{k+1} p \le \frac{6n}k} \log p = \sum_{k=6}^{3\sqrt n-1} \rho_0\big( \tfrac k6 \big) \big( \theta\big( \tfrac{6n}k \big) - \theta \big( \tfrac{6n}{k+1} \big) \big),
\end{align*}
where $\theta(x) = \sum_{p\le x} \log p \sim x$ by the prime number theorem. Thus
\begin{align*}
\log \Phi_n &\sim \sum_{k=6}^{3\sqrt n-1} \rho_0\big( \tfrac k6 \big) \big( \tfrac{6n}k - \tfrac{6n}{k+1} \big) \sim 6n \sum_{k=6}^\infty \frac{\rho_0(k/6)}{k(k+1)}.
\end{align*}
A similar partitioning calculation on the integral will hopefully lead to the same result.
in the fifth line of your answer how did you change $\sum_{2√n<p$\leq$n$ = $\sum_{k=6}^{3√n-1}$ $\sum_{6n/(k+1) p$\leq$6n/k) . Can you please elaborate why it is right?
There was a typo (now fixed). We're just splitting up the interval of summation from the first line into $3\sqrt n-1$ consecutive intervals of summation.
how did you deduced that p lies between 6n/k+1 and 6n/k ? I am not able to think about it. Can you please tell.
On the first line, every prime between $2\sqrt n$ and $n$ appears exactly once, and smaller or larger primes don't appear at all. On the second line, do any primes smaller than $2\sqrt n$ or larger than $n$ appear? Given a prime $p$ between $2\sqrt n$ and $n$, for how many values of $k$ will $p$ appear in the inner sum?
We're just splitting up the range of summation into intervals on which the function is simpler. A strong analogy would be the computation $$\int_0^N \lfloor x\rfloor^2,dx = \sum_{k=0}^{N-1} \int_k^{k+1} \lfloor x\rfloor^2,dx = \sum_{k=0}^{N-1} k^2 \int_k^{k+1} 1,dx.$$
thanks!! Now that question is clear. But I have one question still if you don't mind, in 3 rd = how in RHS of = you Put $\rho_0(n/p) $ out of summation by putting$\rho_0(k/6) $ . I can get only up to k+1/6 > n/p $\geq$ k/6 . I am sorry but I can't get it.
Check out the definition of $\rho_0$ ... on an interval of the form $[\frac k6,\frac{k+1}6)$, how many different values does $\rho_0$ take? (This, in fact, is the motivation for that specific splitting-up choice.)
@DxdxDxdx: Please restrict to one question per post. BTW if your professor has abandoned you, then you might as well switch to a different topic and work on it on your own. Or find another professor.
@DxdxDxdx: I take back "Please restrict to one question per post". Perhaps you can also discuss this topic with friends/classmates. You can also contact the authors of the paper, they are mathematicians just like those who respond here. They will be delighted to hear you write a thesis based on their work.
@DxdxDxdx: I don't have time for this right now. Also, it is not too much fun to help students understand a particular line from a paper that is not close to my research. I think your best shot is to contact the authors. As I said, they are mathematicians just like me, but they have the advantage that they wrote the paper, and you write a thesis about their work. So don't be shy, just drop them an email!
@YemonChoi Your comment is unkind. I request you to delete it. Can you please do that?
@Arnold are you referring to my comment asking why you have bumped the question to the top page through editing? I don't think that is an unkind comment; I think it is a reasonable question, given that you already accepted an answer to this question more than a year ago.
@YemonChoi No. I am not talking about that comment. The comment which I asked you to delete has been deleted by you or the moderators since I also flagged that comment. It's nothing personal.
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2025-03-21T14:48:30.005858
| 2020-03-09T10:24:30 |
354515
|
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|
Stack Exchange
|
On the hereditary Lindelof topological spaces
I received the following interesting point in (1). I could not find any reference or clear proof. Any suggestion?
Theorem. A topological space $X$ is hereditary Lindelof if and only if for any subspace $Y\subset X$, the $\sigma$-algebra generated by any base of the topology of $Y$ coincides with the Borel $\sigma$-algebra of $Y$.
If $X$ is hereditarily Lindelof then any open subset of $Y$ is Lindelof and therefore it is the union of countably many basic (for a given predetermined base for $Y$) open sets. Hence The $\sigma$-algebra generated by the base contains (so it is equal to) the Borel $\sigma$-algebra.
The other direction is not true. Let $X=\omega_1$ with the topology of initial segments. A basis for this topology coincides with the topology itself (except perhaps for the empty and the whole sets). So trivially the Borel $\sigma$-algebra and the $\sigma$-algebra generated by any base coincide. The same is true for any $Y \subseteq X$. However $X$ is not Lindelof.
Thanks a lot. It will be great if you hint me concerning a weaker topological condition that would equivalent to coincidness the $\sigma$-algebras
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2025-03-21T14:48:30.005972
| 2020-03-09T11:19:58 |
354516
|
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|
Stack Exchange
|
Law of a step function and its generalization to two dimensions on an appropriate spaces
Let's consider two discontinuous functions defined on $D$ and $D \times [0,T]$, respectively:
A step function: $u_1(x)=\begin{cases}
u_{L}, x<c_1, \\[2ex]
u_{R}, x>c_1,
\end{cases}$
A "generalization to two dimensions": $u_2(x,t)=\begin{cases}
u_{L}, x<c_2\cdot t, \\[2ex]
u_{R}, x>c_2\cdot t.
\end{cases}$
Here $x \in D \subseteq \mathbb{R}$, $t \in [0,T]$ and $c_1,c_2,u_{L},u_{R}$ are constants. Additionaly in the book "Stochastic equations in infinite dimensions, Da Prato G., Zabczyk J., 1992", we could find the definition of the law:
If $X$ is a random variable from $(\Omega,\mathcal{F})$ to $(E,\mathcal{S})$ and $P$ a probability measure on $\Omega$, then by $\mathcal{L}(X)$ we will denote the image of $P$ under the mapping $X$: $\mathcal{L}(X)(A)=P\{\omega \in \Omega:x(\omega)\in A\},\forall A\in \mathcal{S}.$ The measure $ \mathcal{L} (X)$ is called the distribution or the law of $X$.
As far as I know the probability law of $u_1$ and $u_2$ is the Dirac mass at $u_1$ and $u_2$, respectively, seen as a measure on an appropriate function space. Although I can't remember where I have read this in the literature. If my recollection is wrong, please correct me.
If we assume that the last paragraph above is correct, my question is: What would be those appropriate spaces in the cases of the functions $u_1$ and $u_2$?
Example I got last week for the $u_1$: the law of the function $u_1$ is the Dirac measure concentrated at $u_1$ on the space of cadlag function from [0,1] to $\mathbb{R}$. But for the problem I have in my mind cadlag functions probably won't work.
For the function $u_2$ I don't have any examples. I think that the space of $BV$ functions from $(D\times [0,T])$ to $\mathbb{R}$ should be one of the appropriate spaces. But my ideal appropriate spaces should look as $C([0,T];\mathcal{M}(D))$ or similar - they should be Banach space-valued.
I work on one problem for a few weeks that concerns these two functions and in order to apply the technique that was recommended me (in order to solve it), I need to consider the laws of this two functions. I need help with this. Any appropriate space you recommend me is welcome. Thanks everyone in advance.
Your functions $u_1$ and $u_2$ are not completely defined. For instance, $u_1(c_1)$ is undefined.
If you do not need to distinguish functions differing only on a set of Lebesgue measure $0$, then you may consider $u_1$ a point in the Banach space $B_1:=L^\infty(D)$, and $u_2$ a point in the Banach space $B_2:=L^\infty(D\times[0,T])$, where $D$ is any nonempty Lebesgue-measurable subset of $\mathbb R$.
For $(\Omega,\mathcal F,P)$, you can take any probability space. For each $i\in\{1,2\}$, you can take any $\sigma$-algebra $\mathcal S_i$ over $B_i$, and then the probability law of the (nonrandom) random element $\Omega\ni\omega\mapsto u_i\in B_i$ will be the Dirac measure on $B_i$ at $u_i$.
Added in response to a comment by the OP: Of course, you can identify any function $w\colon X\times Y\to Z$ with the function $\tilde w\colon Y\to Z^X$ by the formula $\tilde w(y)(x):=w(x,y)$ for all $(x,y)\in X\times Y$.
In the case of your function $u_2\colon D\times[0,T]\to\mathbb R$, the corresponding function $\tilde u_2$ may be considered as follows:
$$[0,T]\ni t\mapsto\tilde u_2\in \tilde B_2:=L^\infty\big([0,T],L^\infty(\mathbb R)\big),$$
where $D\ni x\mapsto\tilde u_2(t):=u_2(x,t)\in\mathbb R$. Then the probability law of the (nonrandom) random element $\Omega\ni\omega\mapsto\tilde u_2\in\tilde B_2$ will be the Dirac measure on the Banach space $\tilde B_2$ at the point $\tilde u_2$.
Thank you for the answer. For my problem it doesn't important what $u_1(c_1)$, $u_2(c_2 \cdot t, t)$ is. You could take $u_R$ for both of them. My guesses for $u_1$ would be $L^{\infty}(D)$ too, but also $BV(D)$ - space of functions with bounded variation, $\mathcal{M}(D)$ - space of Radon measures and $\mathcal{D}^{'}(D)$ - space of distributions. I am not sure if I am right. For the function $u_2$ the space $L^{\infty}(D \times [0,T])$ sounds good. I guess also that the spaces $BV(D \times[0,T])$, $\mathcal{M}(D \times [0,T])$, $\mathcal{D}^{'}(D \times [0,T])$ could work too?
@Mark : Yes, you can use many other spaces instead of $L^\infty$. However, for $BV$ and $\mathcal D'$, you need to assume that $D$ is open and also replace $[0,T]$ by $(0,T)$.
Thanks for the correction on $BV$ and $\mathcal{D}^{'}$. For $u_1$ spaces such as $\mathcal{M}$ and $L^p, p\in[0,\infty]$ sounds good for application. However I am not sure for the $u_2$. For it I need some Banach space-valued spaces. Last year I asked question on MSE for the function $u_2$ (edited today) (https://math.stackexchange.com/questions/3254479/a-function-of-time-and-space-as-a-collection-of-functions-of-space-that-is-param). There is a big list there at the end of it. I think that at least some of those Banach space-valued spaces could work. Am I wrong? And thanks again.
@Mark : I have added a response to your latest comment.
@Mark, $\mathcal D^{'}$ $\mathcal D^{'}$ (as in 1 2) sets the prime too high. You want $\mathcal D'$ $\mathcal D'$ (as in 3).
@LSpice, thanks for the correction. It is one of those things I didn't learn correctly from the start, and now I make that mistake from time to time.
@IosifPinelis, thank you for your detailed answer. I don't know if I will end up using your space $\tilde B_2$. But that is great example. It was recommended me to use space such as $\tilde B_2:=C([0,T];G)$ with appropriate $G$ and with appropriate convergence. On bounded domains $L^{\infty} \subset L^1 \subset \mathcal{M} \subset \mathcal D '$. So I thinkI will use space $\tilde{B}2:=C([0,T];\mathcal{M})$ i.e. $G=\mathcal{M}$ and of course with weak convergence of measures (i.e. $C{w}([0,T];\mathcal{M})$ on that big list in my MSE question). Hope that I am not mistaken. Thanks again.
|
2025-03-21T14:48:30.006469
| 2020-03-09T11:53:42 |
354519
|
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|
Stack Exchange
|
Explicit branching rules from $G(n+m)$ to $G(n) \times G(m)$ (where $G = \operatorname{SL}$, $\operatorname{SO}$ or $\operatorname{Sp}$)
Is there in the literature any explicit combinatorial description of the branching rules from $\operatorname{SL}(n+m)$ to $\operatorname{SL}(n) \times \operatorname{SL}(m)$, from $\operatorname{SO}(n+m)$ to $\operatorname{SO}(n) \times \operatorname{SO}(m)$, and/or from $\operatorname{Sp}(2n+2m)$ to $\operatorname{Sp}(2n) \times \operatorname{Sp}(2m)$?
To give an idea of what I mean by "explicit", for example, explicit descriptions of the branching rules from $\operatorname{SL}(n+1)$ to $\operatorname{SL}(n)$, from $\operatorname{SO}(n+1)$ to $\operatorname{SO}(n)$ and from $\operatorname{Sp}(2n+2)$ to $\operatorname{Sp}(2n)$ are well-known, due to Weyl, Murnaghan and possibly Zhelobenko. They are given for example on the Wikipedia page for "Restricted representation", or in Knapp's "Lie groups beyond an introduction" (2d edition) in section 9.3. So ideally, I would like each of the rules from $G(n+m)$ to $G(n) \times G(m)$ to be sufficiently explicit that the rule from $G(n+1)$ to $G(n)$ basically appears as the special case $m = 1$.
For the rules I am asking about, the only thing I have found so far is this paper by Howe, Tan and Willenbring. However they give the branching rules only for a subset of the possible highest weights (what they call the "stable range"). Is there a description of these branching rules that also works outside the stable range?
Also, in $\operatorname{SL}(n+m)$, the subgroup $\operatorname{SL}(n) \times \operatorname{SL}(m)$ is Levi, so we can use Littelmann's branching rules in terms of the path model (see e.g. on Wikipedia). However making this explicit is still a nontrivial problem of combinatorics. Even if it turns out to be an easy one, I would still rather cite its solution than have to write it up on my own.
More generally, is there anywhere a survey with a list of the branching rules for which combinatorial descriptions are known? I would expect such a list to be found, if not in the "official" published literature, then possibly on some website - e.g. the Atlas project (although it seems to me that they only care about infinite-dimensionnal representations - correct me if I am wrong) or somewhere in the Sage documentation (like in the vicinity of here).
I'm not sure there is a "clean" answer. For example, this paper discusses the branching for $G(n)$ to $G(n - 2) \times G(2)$ for various groups $G$, and the answer is not particularly elegant: https://doi.org/10.1006/jabr.1994.1064
For "rectangular shaped" representations, there are some beautiful multiplicity-free formulas which appear for instance in this paper of Okada: https://www.sciencedirect.com/science/article/pii/S0021869397974081
(Hi, Ilia!) This doesn't really answer your question and I'm sure you already know this, but both LiE and Sage are able to compute the branchings you describe (given a specific representation).
|
2025-03-21T14:48:30.006684
| 2020-03-09T12:38:57 |
354520
|
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"Johannes Trost",
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}
|
Stack Exchange
|
Inequality involving Gaussian integral
I'm looking to prove the following inequality:
$$
\left| \int_0^1 e^{-x^2} \sin(x) \, dx \right| \leq \frac{1}{2} \left(1- \frac{1}{e}\right)
$$
So far I have no idea on how to prove it. Anybody?
The right hand side is $\int_{0}^{1} e^{-x^2} x dx$. And the claim follows from $0\le \sin x \le x$ for $0\le x \le 1$.
Writing $\sin x=\frac1{2i}(e^{ix}-e^{-ix})$ and then $e^{-x^2}e^{\pm ix}=e^{-1/4}e^{-(x\mp i/2)^2}$, we get
$$\int_0^1 e^{-x^2}\sin x\,dx=\frac{\sqrt{\pi } \left(i\, \text{erf}\left(1+\frac{i}{2}\right)+2
\text{erfi}\left(\frac{1}{2}\right)-\text{erfi}\left(\frac{1}{2}+i\right)\right)}{4
e^{1/4}},$$
which is $0.294\ldots$; here $\text{erf}$ and $\text{erfi}$ denote, as usual, the error function and the imaginary error function, respectively.
On the other hand, $\frac12(1-\frac1e)=0.316\ldots>0.294\ldots$. This establishes your inequality.
|
2025-03-21T14:48:30.006784
| 2020-03-09T13:20:00 |
354526
|
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|
Stack Exchange
|
Impredictable subsets of $\mathbb{N}$
(I previously asked a similar question on cstheory.SE; I have simplified the notion, which presumably changes it but does not change the key properties I'm interested in.)
This is about a strange recursion-theoretic notion I encountered, I am unable to make much sense out of it. Some concrete questions below, but I'm also interested in just connections to existing recursion theory notions, I do not recognize these things but I'm no expert on the topic.
Let $P$ be the partial computable functions. Let me be lazy and identify Gödel numbers of partial computable functions with their indices, so actually $P = \mathbb{N}$.
I also think of them as Turing machines. Write $T \subset P$ for the total computable functions. If $\psi \in P$, I write $\psi(p){\downarrow}$ for the fact that computation of $\psi$ converges on input $p$,
and for a subset $A \subset \mathbb{N}$, I write $A \upharpoonright n = A \cap [0, n]$.
For a total (not necessarily computable) function $\phi : \mathbb{N} \to \mathbb{N}$, a subset $A \subset \mathbb{N}$ is $\phi$-impredictable if
$$ \exists \psi \in T: \forall \chi \in P: \exists^\infty p: \psi(p) \in A \iff \chi(p, A \upharpoonright \phi(p)){\downarrow} $$
and impredictable if it is $\phi$-impredictable for all $\phi \in T$. We say $A \subset \mathbb{N}$ is strongly $\phi$-impredictable if
$$ \exists \psi \in T: \psi > \phi \wedge \forall \chi \in P: \exists^\infty p: \psi(p) \in A \iff \chi(p, A \upharpoonright (\psi(p)-1))\!\downarrow $$
and strongly impredictable if it is strongly $\phi$-impredictable for all $\phi \in T$.
In words, $\phi$-impredictable means that there is a function $\psi$ that outputs positions $\psi(p)$ on the discrete number line, and these positions have the magical property that if you pick any Turing machine $\chi$, then infinitely many times it happens that $\chi$ guesses correctly whether $\psi(p)$ is in $A$ (in the $\Sigma^0_1$ sense) given access to only $p$ and some initial segment of $A$. The variants of impredictability above are the different ways we may pick this initial segment.
The word "impredictable" of course means roughly the same as "unpredictable", and indeed an impredictable subset must somehow be rather unpredictable (because no machine can guess the values incorrectly). I use it also as a mnemonic for "I'm predictable"; all Turing machines accidentally predict a term infinitely many times, so in some sense these subsets are very predictable.
Some observations that I believe are easily seen to be true:
If $A$ is strongly impredictable, then it is impredictable.
If $A$ is $\Pi^0_1$, then $A$ is not impredictable, indeed not $\phi$-impredictable for any $\phi \in T$.
If you replace $\psi(p)-1$ by $\psi(p)$ in the formula for strong impredictability, then you can just read off whether $\psi(p) \in A$ from the oracle, and thus $\psi(p)-1$ is the maximal number that makes sense in the formula.
A slightly less trivial observation is:
For every $\phi \in T$, there exists a $\phi$-impredictable $\Sigma^0_1$ subset.
I wrote a proof, but basically you just do it, so I'm not sure it's worth including.
Here are my questions:
Is the halting problem (strongly) impredictable? For some $\phi$? (You may choose your favorite definition of the halting problem.)
and if not...
Is there a (strongly) impredictable recursively enumerable subset of $\mathbb{N}$?
If no machine can guess the values incorrectly, then doesn't that make this a notion of predictability rather than of unpredictability?
It can feel that way, hence the "I'm predictable" pun. However, you could just as well say "guesses incorrectly" and just think that halting represents not thinking $n \in A$. But I'm thinking mainly of recursively enumerable languages so in this context I prefer to think "halting equals one"; I've considered many notions like this and if I keep flipping the conventions I can't keep track of them. Also, as explained in the question, if every machine is forced to guess correctly (or incorrectly), then the language is rather unpredictable, for example necessarily undecidable (and much more).
No, the halting set is not impredictable. By the recursion theorem, there is an infinite computable sequence of values such that I control their entry into $K$ (the halting problem). Number these values as $(x_{e, p})_{e < p \in \omega}$. Let $\phi$ be such that $\phi(p) > x_{e, p}$ for all $e < p$. The point is that I can change $K\upharpoonright \phi(p)$ by enumerating various $x_{e,p}$, and I have enough that I can do this $n$ times. Then I claim $K$ is not $\phi$-impredictable.
For each $\psi_e$, we'll build a $\chi_e$ defeating it. We only consider $p > e$. As long as we see $\psi_e(p)\!\uparrow$ or $\psi_e(p) \not \in K_s$, we define $\chi_e(p, K_s\upharpoonright \phi(p))\!\downarrow$. As soon as we see $\psi_e(p)\!\downarrow \in K_s$, we enumerate $x_{e,p}$ into $K$ and make $\chi_e(p,K\upharpoonright \phi(p))\!\uparrow$.
Edit: Adding detail about the use of the recursion theorem. First, $K$ is uniformly 1-complete, so there is a total computable $f$ such that for all $i$ and $n$, $n \in W_i \iff f(i,n) \in K$, where $(W_i)_{i \in \omega}$ is the usual listing of r.e. sets, and also $n \mapsto f(i,n)$ is injective for each $i$.
Now, run infinitely many copies of the above construction, indexed by $i$. Each copy builds its own $\phi$ and $\psi_e$. Each also builds an r.e. set $D_i$. The $i$th copy assumes that $\{f(i,n) : n \in \omega\}$ is the sequence of $x$ that it controls. The instruction "enumerate $f(i,n)$ into $K$" actually means: if $f(i,n)$ has already entered $K$, halt the construction; otherwise, enumerate $n$ into $D_i$ and pause the construction (possibly forever) until $f(i,n)$ enters $K$.
For each $i$, there is a $g(i)$ such that $D_i = W_{g(i)}$. Further, $g$ is computable. So by the recursion theorem, there is a $j$ with $W_j = W_{g(j)}$. So $D_j = W_j$, and $n \in D_j \iff n \in W_j \iff f(j, n) \in K$. So the $j$th copy of the construction is correct about its assumptions: it can cause $f(j, n)$ to enter $K$ by enumerating $n$ into $D_j$ and then waiting; also, no $f(j, n)$ it cares about will enter $K$ before the construction enumerates $n$, because if that happened the construction would halt, and in particular we would have $n \not \in D_j$, contradicting $n \in D_j \iff f(j, n) \in K$.
So the $j$th copy of the construction succeeds.
$\square\square\square$
Yes, there's a strongly impredictable r.e. set. For each $\phi_e$, you have a mother strategy that waits for $\phi_e(p)$ to converge, then chooses a large $x > \phi_e(p)$ and defines $\psi_e(p) = x$. This pair $(p,x)$ is then handed to one of its daughter strategies.
There's a daughter for each $\chi_i$. The basic module is to wait for $\chi_i(p,A_s\upharpoonright (x-1))\!\downarrow$, then enumerate $x$ into $A$, if it's not restrained from doing so. Assuming $x$ was enumerated, also restrain $A\upharpoonright (x-1)$. Give this action priority $e+i+n$, where $n$ is the number of prior restraints imposed by this daughter which are still being respected. The enumeration is permitted provided no restraint created by an action with smaller priority applies.
Then one argues by induction that every daughter succeeds infinitely often.
Sounds amazing. I'll need some time to process this though.
On the face of it this answer is so far beyond my level that I'm afraid to ask anything. I will ask about where I get stuck first, just in case it magically clicks... The first proof seems to use some interpretation of the (Kleene?) recursion theorem I've not heard of, naively it seems to me that machines either halt or not, and I don't get to choose anything. I wonder where I can see it elaborated on? (But maybe I should read a book and get back to this later.)
@VilleSalo Yes, Kleene's recursion theorem. There must be an example of using it in this manner somewhere in Soare's text, but I'm having trouble finding one at the moment. I've edited more detail into my answer. Hope this helps.
I think I understood the first part now, more or less, that's already worth accepting over, so I'll do that and start working on the second part.
This is not an answer, this is just an attempt at explaining and elaborating @DT's amazing use of the recursion theorem in my own words (though in the end it's pretty close to what they said). This is orthogonal to the impredictability issue, and perhaps everyone knows this stuff and it's a standard way to think of KRT, but still it was eye-opening for me so I figured I'd share. (And also I'll know if I misunderstood.)
So, suppose we write a program parametrized by $i$, which (in addition to whatever its main duty is) on the side enumerates some set $D_i$, in some arbitrary way. We think of this as having output access to the $i$th recursively enumerable set $W_i$, i.e.\ our program pretends that $D_i = W_i$. Of course, $W_i$ is a fixed object, chosen by the great Turing in the sky, and we do not get to modify it.
Or do we?
Let us do a trick. We modify our program program enumerating $D_i$ as follows: whenever we notice $n \in W_i$ and have not yet output $n \in D_i$, we throw a fit, and stop enumerating numbers altogether. Whenever we enumerate $n \in D_i$, right after that we step into a loop and wait for $n \in W_i$, possible forever. (This modification is really just checking that the equality $D_i = W_i$ that we want seems to hold so far.)
Let $E_i$ be the new set that we enumerate with this modified program. Observe that if $D_i \neq E_i$, then this can only happen because either we throw a fit, implying $n \in W_i \setminus E_i$, or we wait forever for some $n \in W_i$ right after enumerating $n$ into our set, implying $n \in E_i \setminus W_i$. In other words, $D_i \neq E_i$ implies $E_i \neq W_i$. Equivalently, $E_i = W_i \implies D_i = E_i$.
Now, observe that we have described just another program that enumerates a set $E_i$ given a parameter $i$, and so we can easily find a computable function $g$ such that $E_i = W_{g(i)}$. By the recursion theorem, $W_j = W_{g(j)} = E_j$ for some $j$. By the previous paragraph, we must have $D_j = E_j = W_j$. This means that at least on one parameter $j$, we indeed paradoxically get to choose the sequence $W_j$.
(Of course, we don't get to choose $j$, and it's not so paradoxical once you realize that for free we have "free output access" to e.g. some $W_{h(j)}$ for computable $h$, by just making $W_{h(j)}$ follow our construction. Still this is pretty magical.)
|
2025-03-21T14:48:30.007460
| 2020-03-09T13:27:47 |
354530
|
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"RayyyyySun",
"darij grinberg",
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|
Stack Exchange
|
Linear algebra - For symmetric matrix X $\in S^n$, prove the $a^T X a$ = $\det X \det(X_{n-1})$ , where $a_i$ = $(-1)^i M_{in} $
Suppose we have a symmetric matrix X$\in S^n$, and $X_k$ denotes the submatrix consists of first $k$ rows and columns of X. If $\det X < 0$, but $\det X_1, ..., \det X_{n-1} > 0$. Let $a_i=(-1)^i M_{in}$, where $M_{in}$ is the determinant of the $(i,n)$-th minor of X, that is, the matrix by removing the $i$-th row and the$n$-th column of X. Prove $a^T X a = \det X \times \det X_{n-1}$.\
I know that $\det X$ = $\sum_{i=1}^n |X_{in}|a_i$, and $\det X_{n-1} (-1)^n = a_n$, but have no idea how to show $a^T X a = \det X \times \det X_{n-1}$ from these conditions. By the way, I also noticed that $X_{n-1}$ is positive definite and tried to use Schur complement, but also couldn't figure it out. Can someone help me with this? Thanks a lot!
Recognise $a^T $ as the last row of the adjugate matrix of $X$, and recall that the product of this adjugate matrix with $X $ is $\det X$ times the identity matrix.
Aha, thanks for the hint :) I will try to work it out!
|
2025-03-21T14:48:30.007589
| 2020-03-09T13:32:58 |
354531
|
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|
Stack Exchange
|
Constant curvature difference surfaces
Surfaces of Constant Mean Curvature CMC and constant Gauss Curvature Product $K$ are well-known in differential geometry of surfaces of revolution. Denote half sum and half difference curvatures as
$$ k_1+ k_2 = 2 H_s; \, k_1- k_2 = 2 \, H_d. $$
Significance of $H_d$
Now I assume that $H_d$ may as well be of equal interest with fundamental importance. The assumption is motivated by its representation in Mohr's Circle radius of curvature for stress, shell curvatures, moment of inertia (in Figure) among other such tensors. As is known, material stress failure theories in structural mechanics operate on stress/strain and other tensoral differences as shear entities. It occurs as an important curvature invariant in the equation of Mohr's Circle :
$$\boxed{ (k_n-H_s)^2 +\tau_g^2= H_d^2} $$
I made a search in some textbooks that could be accessed. I was not lucky in finding references for the principal curvature difference profiles. Also the surface cannot be viewed as a particular case of CMC surfaces due to its sign change of curvature and so it is distinctly different. The calculation has a different expression and plots differently.
If $\phi$ is slope of tangent to meridian, primes on meridian arc
Const $H=H_s$ Mean curvature Delaunay meridians
$$\phi^{'}+\frac{\cos \phi}{r} =2 H_s$$
With initial radius $r=r_1$ at $\phi=0$
$$ \cos \phi = \frac{H_s (r^2-r_1^2)+r_1}{r}$$
$$(-k_1,k_2)= (H_s(r^2+r_1^2)-r_1,\, H_s(r^2-r_1^2)+r_1)\,$$
Const $H_d$ Difference curvature meridians
First Order ODE:
$$-\phi^{'}+\frac{\cos \phi}{r} = 2 H_d \tag1$$
First order ODE has the disadvantage of numerically computing indefinitely below points of singularity on $r=0$ axis so there appear multiple spindle meridian profiles below this symmetry axis.
Second Order ODE:
$$\phi^{''}= 2 H_d \tan\phi\, (\phi^{'} +2 H_d) \tag2 $$
$$ \cos \phi = \frac{r}{r_1}+2\, H_d\, r\, log \,\frac{r}{r_1}\tag3 $$
$$(k_1,k_2)=(\frac{1}{r_1}+2H_d(1+log \,\frac{r}{r_1}),\frac{1}{r_1}+2H_d \,log \,\frac{r}{r_1} ) \tag4 $$
$$ @\,r=0,\phi \rightarrow \pi/2$$
Second order ODE has the advantage of stopping computation at point of singularity so there appears no profile below symmetry line $r= 0.$
Shown below are profiles of constant difference $H_d$ of principal curvatures.
Three distinct shapes occur.
Progressive loops $ H_d <-0.5$ above $r=1$ ;
Two types below $r=1$
Ovaloids between cylinder and sphere $ 0>H_d>-0.5;$ and,
Profiles with Inflection Point for $ H_d >0 $ occurring at $ r=r_1e^{-(1+1/(2 r_1H_d))}. $
All profiles meet the axis of symmetry normally, however these are not umbilical points.
$ H_d$ Unduloids" />
Thanks in advance for your comments and for any references available on the topic.
The difference between the principal curvatures is not a geometrically well-defined notion. First, how do you choose the two curvatures, you would need a choice. Second, you would excluded umbilical points, or your two principal curvatures would coincide everywhere which directly implies that you have a (piece of a) plane or sphere.
@Sebastian: The square of the difference of the principal curvatures is well-defined, it's $4(H^2-K)$; setting it equal to a constant is reasonable. There are many such surfaces (locally), most of which are not surfaces of revolution. It's true that such surfaces cannot have umbilic points, and any compact surface with $H^2-K>0$ being constant must be either a torus or a Klein bottle (if such exist, which is not obvious). The PDE that describes such surfaces is hyperbolic, though, so elliptic methods won't generally be of much help, and the few examples with symmetry won't tell you much.
Thanks! Just to distinguish I used symbol $H_d^2= H^2-K = H_s^2-K>0$
Dear Robert Bryant, of course, you are right, and $H^2-K$ is well-defined. As you wrote, this implies that the constant difference curvature condition is well-defined, even though the difference is in general not.
Dear Prof. Robert Bryant: Trying to focus only on $\sqrt{H^2-K}$ profile images. Found that there are three types: progressive loops,alternating spindles and regressive loops that may illustrate your answer in part. .Intend to edit out DeLaunay unduloids as they are well known. Hope it should be in order.
Went in for second order ODE to obtain profiles upto $r=0$
I see a somewhat different description of the surfaces of revolution with $H^2-K$ equal to a positive constant, say $H^2-K=\delta^2$, where $\delta>0$ is constant. Note that scaling a surface by a constant $\lambda\not=0$ produces a new surface with $H^2-K$ divided by $\lambda^2$, so the actual magnitute of $\delta>0$ is not important for the possible shapes of such surfaces. Thus, your pictures strike me as somewhat misleading.
Instead, the key factor determining the shape of the corresponding surface of revolution is another constant, which can be described as follows: First, let $\bigl(f(s),g(s)\bigr)$ be a unit speed curve in the plane, and consider the surface of revolution parametrized by
$$
X(s,\theta) = \bigl(f(s)\,\cos\theta,\ f(s)\,\sin\theta,\ g(s)\bigr).
$$
One easily computes that the principal curvatures of this surface are
$$
\kappa_1 = g'(s)f''(s)-f'(s)g''(s)\quad\text{and}\quad \kappa_2 = -g'(s)/f(s),
$$
where, of course, $f'(s)^2+g'(s)^2 = 1$. Then the equation we want to study is
$$
g'f''-f'g'' + g'/f = 2\delta.
$$
Using the relation $f'f''+g'g'' = 0$, this becomes the relations
$$
f'' = g'(2\delta-g'/f)\quad\text{and}\quad g'' = -f'(2\delta-g'/f).
$$
Scaling the surface by $\lambda$ induces the scaling
$$
(f,g,s,\delta) \mapsto (\lambda f,\lambda g, \lambda s, \delta/\lambda),
$$
so we can reduce to the case $\delta=\tfrac12$ without loss of generality. Set $h=g'$
and we have the equation $h' = -f'(1-h/f)$ or $fh'-hf'+ff'=0$, which can be made exact by dividing by $f^2$, yielding the first integral
$$
\frac{h}{f} + \log|f| = C,
$$
so $h = f(C-\log|f|)$, and, since $(f')^2 = 1-h^2$, we must have $|f|(C-\log|f|) \le 1$. In particular, we have
$$
\frac{(\mathrm{d}f)^2}{\bigl(1-f^2(C-\log|f|)^2\bigr)} = (\mathrm{d}s)^2,
$$
and, since $(\mathrm{d}g)^2 = h^2\,(\mathrm{d}s)^2$, we arrive at the relation
$$
\bigl(1-f^2(C-\log|f|)^2\bigr)\,(\mathrm{d}g)^2 - f^2(C-\log|f|)^2\,(\mathrm{d}f)^2 = 0
$$
It is the constant $C$ that determines the shape of the resulting profile curve and hence the surface. Obviously, the null curves of this quadratic form have to lie inside the region where $f^2(C-\log|f|)^2 \le 1$, which are bounded by lines defined by
$|f|(C-\log|f|) = \pm1$.
For example, when $C>1$, there are three functions $\rho_i$ on $(1,\infty)$ such that
$$
0 < \rho_1(C) < \mathrm{e}^{C-1} < \rho_2(C) < \mathrm{e}^C < \rho_3(C)
$$
such that the function $1-f^2(C-\log|f|)^2$ is nonnegative on the three intervals,
$(-\rho_3(C),-\rho_2(C))$, $(-\rho_1(C),\rho_1(C))$, and $(\rho_2(C),\rho_3(C))$. Consequently, there wil be two distinct solutions $(f,g)$: One where $f$ and $g$ are periodic and $|f|$ is bounded by $\rho_1(C)$, and another where $f$ is periodic, oscillating between a minimum of $\rho_2(C)$ and a maximum of $\rho_3(C)$.
Meanwhile, when $C<1$, there is a single function of $C$, say $\rho_4(C)>\mathrm{e}^C$ such that the function $1-f^2(C-\log|f|)^2$ is nonnegative on the interval $(-\rho_4(C),\rho_4(C))$, and there will be a single solution (up to translation and reflection in $g$) for which $f$ lies in this interval.
There is a similar story for $C=1$, except that there is an exact solution, $(f,g)=(1,s)$ (the cylinder of radius $1$) and then two solutions, one with $0<f<1$ that asymptotically approaches the cylinder from the interior but pinches to a point on the axis of rotation in finite distance, and one with $1<f<M$ (where $M\approx3.591121477$ satisfies $M(1-\log M) = -1$) that is asymptotic to the cylinder of radius $1$ in both directions, but bulges out to a maximum diameter of $M$ at its 'equator'. The resulting surface of revolution has a circle of self-intersections.
It is probably worth noting that, near $f=0$, i.e., the axis of rotation, the function $g$ is not a $C^2$ function of $f$ (obvious, since, otherwise, the surface would have an umbilic there), but has an expansion of the form
$$
g = g_0 \pm \tfrac14 f^2\bigl(2C+1-\log|f|\bigr) + O(f^3)
$$
Thus, the surface is only $C^1$ when it intersects the axis of rotation.
Note that all of these solutions have $\delta=\tfrac12$, but they have quite different behaviors.
After the last ( second order ode formulation) do all the profiles now look alright?@Robert Bryant
@Narasimham: Yes, now they do look better. Your earlier ones were seriously off. You still don't have the solution that, in my notation is the case where $C=1$ and the solution is external to the cylinder and asymptotic to it at both ends with only one self-intersection. I confess, though, that I find your notation a bit strange because all my solutions have $H^2-K=\tfrac14$, which is $H_d^2=\tfrac14$ in your notation. I think that what is happening is that your solutions, other than the cylinder of radius 1 (where yours agrees with mine), are differently scaled versions of my solutions.
With direct integration I got $\dfrac{h}{f}-2H_d,log|f| =C,$ which verifies on direct differentiation. Am afraid with particular $ H_d= \frac12$ we lose all (non-cylindrical) general solutions. And also the sign.
@Narasimham: Probably, you chose a different orientation of the surface from the one I chose, which explains why you got $-2H_d$ instead of $+2H_d$. I get all of the solutions that you got, plus one more.
Alright, I took $\phi$ rotations per sign convention counterclockwise positive.
..a small addition to Professor Bryant's answer
For this class of surfaces, where $k1-k2 = constant$, we have introduced the name "Costant Skew Curvature Surfaces" (CSkC-surfaces) and we have studied an aspect concerning the Bonnet-surfaces.
It is well know that the famous question that Bonnet asked was: "When does there exist an isometric embedding $x:M^2 \rightarrow R^3$ such that the mean curvature function of the immersion is $H$?"
Our work was born from the question: Can a surface be CSkC and Bonnet at the same time, and, if that is the case, what does it represent?
We showed that the CSkC-surfaces with principal curvatures ($k_1$ and $k_2$) nonconstant cannot contain any Bonnet-surfaces, so if and only if $k_1$ and $k_2$ are both constant the class of CSkC-surfaces can admit Bonnet-surfaces.
This means that the only CSkC-surfaces for which exists a nontrivial isometric deformation preserving the mean curvature $H$ are (patch of) a circular cylinders.
see (Alexander Pigazzini, Magdalena Toda): https://projecteuclid.org/euclid.jgsp/1518577293
Another aspect of the CSkC-surfaces is that if we setting $k_1-k_2=constant$ renders the shape equation for an elastic membrane equivalent to the Schrodinger equation for a particle on the same surface.. then the same equations have the same solutions...
see (Victor Atanasov, Rossen Dandoloff): https://iopscience.iop.org/article/10.1088/0143-0807/38/1/015405/pdf
By relating the two works we can for example say that:
"if you want to look for an isometric deformation that preserve this relation (shape equation equivalent to the Schrodinger equation) then you can get the result only if the surface is (patch of) circular cylinder, and then search through its isometric deformations".
Can we see some profiles of CSkC to compare with results obtained above ? 2) How do K and Christoffel symbols vary in Bonnet surfaces (where H is conformally conserved) ? Can we plot K of Bonnet surface in some CAS (Like we can plot H in K− preserving isometric mappings) ? @Alexander Pigazzini
@Narasimham- Your questions are very interesting but I have not investigated, even the observation in V.Atanasov-R.Dandoloff paper, was pointed out to me by prof. Atanasov, but we have not gone further
Thank you, I had long wondered why the triplet had not been equally in centre stage in surface theory as a category...so investigated their physical shapes for $H_d$ surfaces of revolution at least. The triplet $ H_d, \sqrt K, H_s $ as a natural combination should be of interest.
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2025-03-21T14:48:30.008479
| 2020-03-09T13:42:17 |
354532
|
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Stack Exchange
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Distance between two polyhedra that takes incidence structure into account
Suppose that we have two polyhedra $P_1$ and $P_2$ in $\mathbb{R}^3$. I would like to define such a metric $\rho(P_1, P_2)$ that depends on several factors, but currently I don't know how to do it better.
The distance is intended to be smaller if:
"big facets" of $P_1$ have one-to-one correspondence with "big facets" of $P_2$,
"big facets" of $P_1$ and $P_2$ that correspond to each other are close to each other within Hausdorff distance,
"big facets" $f_{1,1}$ and $f_{1,2}$ of $P_1$ that share some edge $e_1$ have corresponding "big facets" $f_{2,1}$ and $f_{2,2}$ that also share some edge $e_2$, and $f_{1,1}$ is close to $f_{2,1}$, $f_{1,2}$ is close to $f_{2,2}$, and $e_1$ is close to $e_2$ within Hausdorff distance,
the same as (3), but "share some edge $e_1$" is replaced with "have close edges $e_{1,1}$ and $e_{1, 2}$" and so on.
Actually, the facet is considered to be "big" if its area $area(f)$ is "big". We may introduce some parameter $\varepsilon$ which will show whether the facet is big: $area(f) > \varepsilon$. But it seems that it would be better if the metric is parameter-free.
So my question consists of the following parts:
Is the concept described above already studied in the literature? Or at least anything related to this?
What is the most convenient way to make the metric definition? It seems that this is a hybrid of Hausdorff distance in Euclidean space and graph distance.
Actually, I'm trying to make the reconstruction in this metric, and to prove the properties of any algorithm I need to make a strict definition of this metric. So it would be great to make it as simple as possible.
Any ideas are welcome!
Two suggestions of literature that might lead you somewhere useful (but neither of these references in themselves seem to address your several criteria).
The first below uses curvature distributions. The second uses Minkowski's
mixed volumes. Both introductions survey "shape similarity" measures.
Shum, Heung-Yeung, Martial Hebert, and Katsushi Ikeuchi. "On 3D shape similarity." In Proceedings CVPR IEEE Computer Society Conference on Computer Vision and Pattern Recognition, pp. 526-531. IEEE, 1996.
PDF download.
Roerdink, Jos BTM, and Henk Bekker. "Similarity measure computation of convex polyhedra revisited." In Digital and Image Geometry, pp. 389-399. Springer, Berlin, Heidelberg, 2001. Book link.
Thanks for reference! After looking first one, I can say that this measure reflects the incidence structure somehow. But in the second paper, the mixed volume is $\sum{h_A(u_i)S(f_i)}$ where $f_i$ are facets of B and $u_i$ are their outer normals. What I can say is that if we divide one of facets onto two facets, the value of formula doesn't change. So it doesn't reflect the incidence structure. But the formula looks very attractive. Maybe we can modify it in a way that to make it reflect the incidence structure: $$\sum{h_A(u_i)c(S(f_i))}$$ where $c()$ is concave function. What do you think?
|
2025-03-21T14:48:30.008714
| 2020-03-09T14:10:26 |
354536
|
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Stack Exchange
|
Common basis for permutation matrices
How can I check whether there exists a common basis with respect to which two matrices and are permutation matrices?
More explicitly, let $A$ and $B$ be two unitary matrices whose eigenvalues form complete sets of roots of unity (meaning there is a unitary transformation to a permutation matrix for each). How can I check whether their exists a single unitary $U$ such that
$$
P_A = U A U^\dagger, \quad P_B = U B U^\dagger,
$$
such that $P_A$ and $P_B$ are permutation matrices?
Another restatement: is there a subset of $\mathbf{C}^n$ that forms an orthonormal basis, and which is invariant by both $A$ and $B$? (the existence of such a subset for each of $A$ and $B$ being the assumption)
When you say "how can I check", are you imagining that you have been given $A$ and $B$ explicitly, and you want to do a computation... Or are you asking for theoretical conditions that are equivalent? One obvious necessary condition is that $\langle A, B\rangle$ is isomorphic to a subgroup of $S_n$... I'm not sure how close it is to being sufficient.
Yes, what I am imagining is being given two matrices and trying to determine whether there exists such a U. This could either be using the properties of A and B, or it could be determined through some algorithm designed to find U. Could you also clarify what you mean by the angled bracket notation please?
@MarkSapir and all the subgroup they generate. This is why I added the group tag. For instance one can wonder, if the product of, say all $2^n$ $n$-fold products ($n$ being the size) of $A,B$ are conjugate to permutation matrices, whether it's enough to ensure that they can be simultaneously conjugate into the subgroup of permutation matrices. (Already one can wonder about an algorithm to check if two matrices generate a finite group.)
I don't know if it helps (or if I should add to the question), but in the case I'm interested in $A$ and $B$ are representations of the generators of the modular group (90deg rotation and Dehn twist). They are also symmetric in the original basis.
Clearly if $AB$ is not unitarily similar to a permutation matrix then there is no such $U$. If $AB$ is similar to a permutation as well as $A$ and $B$ separately, can I say anything? Can I say something in the particular case that $A$ and $B$ are representations of the generators of the modular group?
The angled backed notation means "the group generated by $A$ and $B$" -- i.e. all matrices obtained by taking finite length products of $A$, $B$ and their inverses.
Here's a possible method:
We might as well assume that $A$ is already a permutation matrix -- just conjugate $A$ and $B$ by the same unitary matrix $C$ and relabel.
Now if we are looking to conjugate $B$ to a permutation matrix AND retain $A$ as a permutation matrix, then we'll need to conjugate by something that centralizes $A$ -- so our conjugating matrix, call it $X$, must lie in $C_G(A)$, where $G=U(n)$.
(Note: you might wonder about the possibility that $A$ could be conjugated to a different permutation matrix instead, therefore allowing a larger choice for $X$; however this question assures us that if we were to do this, we could then conjugate by a permutation matrix to return $A$ to its original form, without affecting whether or not $B$ is a permutation matrix.)
$C_G(A)$ is easy enough to calculate -- especially if $A$ is made up of cycles of different lengths. Now let $\mathcal{B}$ be the natural basis. We use the reformulation of @YCor to consider what happens when we apply $X$ to $\mathcal{B}$. So, for instance, if
$$A=\begin{pmatrix} & 1 & \\ & & 1 \\ 1 & & \end{pmatrix}$$
then the elements of $C_G(A)$ are unitary matrices of the form
$$
X=\begin{pmatrix} a & b & c \\ c & a & b \\ b & c & a \end{pmatrix}.$$
If we apply the matrix $X$ to the natural basis, then we obtain a basis of the form
$$\left\{\begin{pmatrix}a \\ c \\ b \end{pmatrix}, \begin{pmatrix} b \\ a \\ c\end{pmatrix}, \begin{pmatrix} c \\ b \\ a \end{pmatrix}\right\}$$
where $a,b,c$ are restricted to ensure the basis is orthonormal.
These are all of the orthonormal bases for which $A$ is the permutation matrix written above. Now $B$ will be a permutation matrix with respect to one of these bases, say $\mathcal{B}_1$, if and only if, for all $v\in \mathcal{B}_1$, $Bv\in\mathcal{B}_1$. One simply has to test whether any of the listed bases have this property.
Remarks:
I'd like to find a nice, neat way to do that last bit -- check whether the matrix $B$ fixes any of the bases thrown up by considering the centralizer of $A$... But it's not obvious to me that one can do this in any efficient sort of a way...
The centralizer of $A$ will be a lot larger if there are cycles of the same length -- and this will yield a lot more possible bases to test at the end. One should clearly label $A$ and $B$ so that $A$ has as few cycles as possible.
So, algorithmically, it might be most efficient to use the observations in the comments about products of $A$ and $B$: take a few products of $A$ and $B$ of some bounded length -- if any of these can't be conjugated to permutation matrices, then, obviously, you stop. If they ALL can be conjugated to a permutation matrix, then choose whichever product resulted in a permutation matrix with fewest repeated cycles.
Added later: In fact, we can make that final step (where we study the action of the matrix $B$ on a bunch of bases) much more efficient if we check the isomorphism type of the group $\langle A, B\rangle$; this must be isomorphic to a subgroup of $S_n$, in which $A$ must correspond to the given permutation matrix (so, for instance, if $A$ was the matrix in the example above, then its image in $S_n$ would be the 3-cycle $(1,2,3)$). This allows us to explicitly list the possible permutations corresponding to $B$. For each of these possibilities, one has an explicit action on the basis $\mathcal{B}_1$.
So, for instance, in the example above, if $B$ were to correspond to the element $(1,2)$, then its action on the basis $\mathcal{B}_1$ would need to swap the first two basis elements. Checking this specific property is much more straightforward than checking if $B$ preserves any of the possible bases obtained using the centralizer of $A$.
One can conjugate a permutation matrix to a permutation matrix without centralising it. EDIT: Oops, I see you address this in a later paragraph.
I have not added it to my question (yet) but can we do any better if we know that $A$ and $B$ are representations of the generators of the modular group?
Can you give an example where A, B and AB are all separately unitarily equivalent to permutations but that A and B can't be transformed to permutations simultaneously?
One partial answer, for which you don't need the knowledge that the spectra are complete sets of roots of unity.
First step : verify that the group $G$ generated by $A$ and $B$ is finite (it should be contained a permutation group of $n$ elements). For this, explore the free group in two letters $a$ and $b$ (together with $a^{-1}$ and $b^{-1}$). Consider the disk $D_k$ of words $w$ of length $\le k$, and let $N_k$ be the number of distinct elements $w(A,B)$ for $w\in D_k$. If $N_k$ comes to exceed $n!$, the basis does not exist. If not, you have reached a $k$ so that $N_{k+1}=N_k$ ; the group is finite and you still have a chance, provided this $N=N_k$ is $\le n!$.
Next step : In the latter case, you must verify that the trace of $w(A,B)$ is a non-negative integer, for every $w\in D_k$.
I suspect that the converse is true, and that it is a result about linear representations of finite groups. Does anyone have a clue ?
Not only does $w(A,B)$ have integer trace, but also non-negative integer trace. This is actually enough to show that if $G$ is finite, then its order does indeed divide $n!$, by a result of Blichfeldt, which has appeared on MO in relation to several questions.
Specifically, if $n_{1},n_{2}, \ldots, n_{t}$ are all the distinct values taken by ${\rm trace}(w(A,B))$ as $w(A,B)$ runs over all non-identity elements of $G$, then , assuming $G$ is finite, its order divides $(n-n_{1})(n-n_{2}) \ldots (n-n_{t})$.
@GeoffRobinson, do you have a reference for that result?
@LSpice : See this question |: https://mathoverflow.net/questions/240317/why-is-this-character-expression-an-integer/240328#240328
|
2025-03-21T14:48:30.009300
| 2020-03-05T17:56:08 |
354250
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Stack Exchange
|
Mathematical physics without partial derivatives
Remark: All the answers so far have been very insightful and on point but after receiving public and private feedback from other mathematicians on the MathOverflow I decided to clarify a few notions and add contextual information. 08/03/2020.
Motivation:
I recently had an interesting exchange with several computational neuroscientists on whether organisms with spatiotemporal sensory input can simulate physics without computing partial derivatives. As far as I know, partial derivatives offer the most quantitatively precise description of spatiotemporal variations. Regarding feasibility, it is worth noting that a number of computational neuroscientists are seriously considering the question that human brains might do reverse-mode automatic differentiation, or what some call backpropagation [7].
Having said this, a large number of computational neuroscientists (even those that have math PhDs) believe that complex systems such as brains may simulate classical mechanical phenomena without computing approximations to partial derivatives. Hence my decision to share this question.
Problem definition:
Might there be an alternative formulation for mathematical physics which doesn't employ the use of partial derivatives? I think that this may be a problem in reverse mathematics [6]. But, in order to define equivalence a couple definitions are required:
Partial Derivative as a linear map:
If the derivative of a differentiable function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ at $x_o \in \mathbb{R}^n$ is given by the Jacobian $\frac{\partial f}{\partial x} \Bigr\rvert_{x=x_o} \in \mathbb{R}^{m \times n}$, the partial derivative with respect to $i \in [n]$ is the $i$th column of $\frac{\partial f}{\partial x} \Bigr\rvert_{x=x_o}$ and may be computed using the $i$th standard basis vector $e_i$:
\begin{equation}
\frac{\partial{f}}{\partial{x_i}} \Bigr\rvert_{x=x_o} = \lim_{n \to \infty} n \cdot \big(f(x+\frac{1}{n}\cdot e_i)-f(x)\big) \Bigr\rvert_{x=x_o}. \tag{1}
\end{equation}
This is the general setting of numerical differentiation [3].
Partial Derivative as an operator:
Within the setting of automatic differentiation [4], computer scientists construct algorithms $\nabla$ for computing the dual program $\nabla f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ which corresponds to an operator definition for the partial derivative with respect to the $i$th coordinate:
\begin{equation}
\nabla_i = e_i \frac{\partial}{\partial x_i} \tag{2}
\end{equation}
\begin{equation}
\nabla = \sum_{i=1}^n \nabla_i = \sum_{i=1}^n e_i \frac{\partial}{\partial x_i}. \tag{3}
\end{equation}
Given these definitions, a constructive test would involve creating an open-source library for simulating classical and quantum systems that doesn’t contain a method for numerical or automatic differentiation.
The special case of classical mechanics:
For concreteness, we may consider classical mechanics as this is the general setting of animal locomotion, and the vector, Hamiltonian, and Lagrangian formulations of classical mechanics have concise descriptions. In all of these formulations the partial derivative plays a central role. But, at the present moment I don't have a proof that rules out alternative formulations. Has this particular question already been addressed by a mathematical physicist?
Perhaps a reasonable option might be to use a probabilistic framework such as Gaussian Processes that are provably universal function approximators [5]?
Koopman Von Neumann Classical Mechanics as a candidate solution:
After reflecting upon the answers of Ben Crowell and gmvh, it appears that we require a formulation of classical mechanics where:
Everything is formulated in terms of linear operators.
All problems can then be recast in an algebraic language.
After doing a literature search it appears that Koopman Von Neumann Classical Mechanics might be a suitable candidate as we have an operator theory in Hilbert space similar to Quantum Mechanics [8,9,10]. That said, I just recently came across this formulation so there may be important subtleties I ignore.
Related problems:
Furthermore, I think it may be worth considering the following related questions:
What would be left of mathematical physics if we could not compute partial derivatives?
Is it possible to accurately simulate any non-trivial physics without computing partial derivatives?
Are the operations of multivariable calculus necessary and sufficient for modelling classical mechanical phenomena?
A historical note:
It is worth noting that more than 1000 years ago as a result of his profound studies on optics the mathematician and physicist Ibn al-Haytham(aka Alhazen) reached the following insight:
Nothing of what is visible, apart from light and color, can be
perceived by pure sensation, but only by discernment, inference, and
recognition, in addition to sensation.-Alhazen
Today it is known that even color is a construction of the mind as photons are the only physical objects that reach the retina. However, broadly speaking neuroscience is just beginning to catch up with Alhazen’s understanding that the physics of everyday experience is simulated by our minds. In particular, most motor-control scientists agree that to a first-order approximation the key purpose of animal brains is to generate movements and consider their implications. This implicitly specifies a large class of continuous control problems which includes animal locomotion.
Evidence accumulated from several decades of neuroimaging studies implicates the role of the cerebellum in such internal modelling. This isolates a rather uniform brain region whose processes at the circuit-level may be identified with efficient and reliable methods for simulating classical mechanical phenomena [11, 12].
As for the question of whether the mind/brain may actually be modelled by Turing machines, I believe this was precisely Alan Turing’s motivation in conceiving the Turing machine [13]. For a concrete example of neural computation, it may be worth looking at recent research that a single dendritic compartment may compute the xor function: [14], Reddit discussion.
References:
William W. Symes. Partial Differential Equations of Mathematical Physics. 2012.
L.D. Landau & E.M. Lifshitz. Mechanics (Volume 1 of A Course of Theoretical Physics). Pergamon Press 1969.
Lyness, J. N.; Moler, C. B. (1967). "Numerical differentiation of analytic functions". SIAM J. Numer. Anal. 4: 202–210. doi:10.1137/0704019.
Naumann, Uwe (2012). The Art of Differentiating Computer Programs. Software-Environments-tools. SIAM. ISBN 978-1-611972-06-1.
Michael Osborne. Gaussian Processes for Prediction. Robotics Research Group
Department of Engineering Science
University of Oxford. 2007.
Connie Fan. REVERSE MATHEMATICS. University of Chicago. 2010.
Richards, B.A., Lillicrap, T.P., Beaudoin, P. et al. A deep learning framework for neuroscience. Nat Neurosci 22, 1761–1770 (2019). doi:10.1038/s41593-019-0520-2.
Wikipedia contributors. "Koopman–von Neumann classical mechanics." Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 19 Feb. 2020. Web. 7 Mar. 2020.
Koopman, B. O. (1931). "Hamiltonian Systems and Transformations in Hilbert Space". Proceedings of the National Academy of Sciences. 17 (5): 315–318. Bibcode:1931PNAS...17..315K. doi:10.1073/pnas.17.5.315. PMC 1076052. PMID 16577368.
Frank Wilczek. Notes on Koopman von Neumann Mechanics, and a
Step Beyond. 2015.
Daniel McNamee and Daniel M. Wolpert. Internal Models in Biological Control. Annual Review of Control, Robotics, and Autonomous Systems. 2019.
Jörn Diedrichsen, Maedbh King, Carlos Hernandez-Castillo, Marty Sereno, and Richard B. Ivry. Universal Transform or Multiple Functionality? Understanding the Contribution of the Human Cerebellum across Task Domains. Neuron review. 2019.
Turing, A.M. (1936). "On Computable Numbers, with an Application to the Entscheidungsproblem". Proceedings of the London Mathematical Society. 2 (published 1937). 42: 230–265. doi:10.1112/plms/s2-42.1.230. (and Turing, A.M. (1938). "On Computable Numbers, with an Application to the Entscheidungsproblem: A correction". Proceedings of the London Mathematical Society.
Albert Gidon, Timothy Adam Zolnik, Pawel Fidzinski, Felix Bolduan, Athanasia Papoutsi, Panayiota Poirazi, Martin Holtkamp, Imre Vida, Matthew Evan Larkum. Dendritic action potentials and computation in human layer 2/3 cortical neurons. Science. 2020.
How about the discrete equivalents to derivatives ,i.e., difference equations
There are important discrete physical systems, for example quantum spin systems, which one can formulate without partial derivatives. In general, quantization probably helps - you base your description on an operator algebra, and nobody forces you to cast, e.g., $[x,p]=i$ in the position or momentum representations. Cf. the ladder operator formulation of the harmonic oscillator.
Maxwell's equations, and others like them, have integral formulations that eliminate most if not all derivatives (and are more "geometrical"). Do you exclude that just as you exclude Cauchy's formula?
@PiyushGrover I understand that discrete/continuous approximations are methods for implementing partial differentiation computations. As such they wouldn't count as alternative formulations.
@FrancoisZiegler That is an interesting counter-argument. However, within a classical framework if we are to simulate the work done by any force wouldn't we need to somehow approximate second-order derivatives of position?
Do you consider the acceleration in $F=ma$ a partial derivative? Arguably, it is a (total) derivative with respect to time $\frac{d^2x}{dt^2}$, not a partial one $\frac{\partial^2x}{\partial t^2}$. If the answer is no, then it seems to me that you can do all Newtonian physics with it.
@FedericoPoloni Doesn't the total derivative generally require the computation of partial derivatives? Ref: https://mathworld.wolfram.com/TotalDerivative.html Otherwise, the force term emerges in the Euler-Lagrange equations as a partial derivative which is worth noting since the Lagrangian formulation is computationally efficient.
Have I understood you correctly?
I think the way you set up the question, the answer can only be No. (1) We know that mechanics can be formulated using derivatives. (2) Such models correctly reproduce the behavior of natural phenomena (excluding extreme regimes where quantum effects become important). (3) Any alternative formulation must reproduce the same behavior of natural phenomena. From your comments, it seems that (3) is all you require for your notion of equivalence. By that logic, (1) and (2) imply that any alternative formulation of mechanics will be equivalent to the one with derivatives.
@IgorKhavkine To be precise, (3) isn’t sufficient for equivalence or my position would be tautological. If an alien civilisation is able to do (3) without (1) i.e. simulate classical mechanical phenomena without computing derivatives then their formulation doesn’t depend upon continuous/discrete approximations of derivatives.
@MattF. My understanding of integral formulations is that once you start simulating the work done by that field(gravity/electromagnetic) on an object you would need an approximation of second-order derivatives. In this case, I think you are referring to Gauss' law? Ref: https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity
@AidanRocke I am not suggesting to use Euler-Lagrange equations, just $F=ma$, which should be sufficient for Newtonian mechanics. In that formulation, only functions of one variable, the time $t$, appear. So "total derivative" is a bit improper as a term, and I don't think that partial derivatives are required. Of course, one could argue that also derivatives of a single-variate function are partial derivatives with respect to that lone variable, but then your question should probably have said "without derivatives" in the first place.
@AidanRocke Well given the motivation, it seems like if the organisms can do addition and subtraction, then discrete (difference) equations should do the job of replacing derivatives. If organisms cannot do addition or subtractions, I do not know what physics (if any) can be simulated.
@PiyushGrover If I understand you correctly, finite-difference methods would then be used for solving differential equations? Ref: https://en.wikipedia.org/wiki/Finite_difference_method#Explicit_method
@AidanRocke This wasn’t so much a counter-argument as a putative example of what you may want, or not. Basically integration by parts, or Stokes’ theorem, allows one to recast laws in an integral form avoiding derivatives although the setting is still “differential” geometry, sorry. For another example, recasting the nominally very “second-order” law that particles move on geodesics, see §§13–15 of Sternberg’s Einstein lecture General Covariance and the Passive Equations of Physics.
I think your edit is very confusing. Why do you introduce automatic differentiation, and then not use it later? Under which conditions do you claim that it produces exact derivatives? (Spoiler: in the most common setting, IEEE arithmetic on a computer, it doesn't.) Why pseudocode in an unusual language rather than a formula? Why does the comment call it "automatic" when you claim it's numerical? Are you interested in formulating physics without partial derivatives, or in making actual computations? In which model: on a silicon computer, or in "organisms with spatiotemporal sensory input"?
@FedericoPoloni Those are good questions. I think the nature of interdisciplinary problems is that a useful formulation of the problem for a broad audience is half the challenge. Regarding automatic differentiation, the derivatives are exact up to round-off error which is the notion of equivalence I am using in (2). As for formulating physics vs. making actual computations, I think those are fundamentally related questions. Leibniz and Newton did both simultaneously.
@FedericoPoloni Digital computers are used by most computational neuroscientists and computational biologists as models for biological computation.
But if all this stuff about computing derivatives is tangential to the actual question, why did you include it in the first place?
@FedericoPoloni How is it tangential? I think the challenge of formulating physics using a set of mathematical operations and methods for performing actual derivative computations are related. The latter allows you to define a notion of equivalence and the former requires you to clarify the alternative formulation you are using. Naturally, this alternative formulation must be computable.
@FedericoPoloni In fact, all the answers so far are spot on but from the comments it appears that my problem statement could have been improved so this is what I tried to do. In particular, a number of people felt that my notion of equivalence could have been made more explicit.
Would it be fair to interpret your question as: can one describe physics without differential calculus?
@MichaelBächtold Without partial derivative computations you wouldn’t have a multivariable differential calculus so I think that’s a very reasonable interpretation.
I added a link to gmvh's answer, but couldn't find Ben Crowell's. Is it this one?
As to question 2, there are certainly plenty of non-trivial discrete models in statistical physics, such as the Ising or Potts models, or lattice gauge theories with discrete gauge groups, that require no partial derivatives (or indeed any operations of differential calculus) at all to formulate and simulate.
Similarly, quantum mechanics can be formulated entirely in the operator formalism, and an entity incapable of considering derivatives could still contemplate the time-independent Schrödinger equation and solve it algebraically for the harmonic oscillator (using the number operator) or the hydrogen atom (using the Laplace-Runge-Lentz-Pauli vector operator).
So an answer to question 1 might be "at least anything that can be written as a discrete-time Markov chain with a discrete state space, as well as anything that can be recast as an eigenvalue problem", and other problems that can be recast in purely probabilistic or algebraic language should also be safe (although it might be hard to come up with their formulations without using derivatives at some intermediate step).
As to question 3, I personally don't believe that an approach to classical mechanics or field theory can be correct if it isn't equivalent (at least at a sufficiently high level of abstraction) to formulating and solving differential equations. But the level of abstraction could conceivably be quite high -- for an attempt to formulate classical mechanics without explicitly referring to numbers (!) cf. Hartry Field's philosophical treatise "Science without Numbers".
I believe Hartry Field avoids explicitly referring to numbers by assuming that physical space satisfies Hilbert's axioms for geometry, including the Archimedean and completeness axioms. From this one can derive a structure isomorphic to $\mathbb{R}$, so he actually does assume $\mathbb{R}$, implicitly.
As I said, eventually you have to be able to describe differential equations and all of that (which of course includes having $\mathbb{R}$ at your disposal). And I agree that Hartry Field implicitly assumes (the consistency of) $\mathbb{R}$; as far as I can tell, his nominalism is ultimately more a matter of presentation.
See my comment to another answer: linear algebra alone is, in some sense, also equivalent to derivatives.
After reflecting upon your answer, I wonder whether Koopman–von Neumann classical mechanics might be a candidate solution? Ref: https://en.wikipedia.org/wiki/Koopman%E2%80%93von_Neumann_classical_mechanics
I'm not familiar with KvN mechanics, but from the Wikipedia entry it doesn't really seem to meet your criteria -- note that the Liouville operator contains partial derivatives of the Hamiltonian function, and that simply putting those in as arbitrary operators won't work, since they would have to be related by the integrability condition on the gradient of the Hamiltonian.
Well if you take out partial derivatives, at least quantum field theory and in particular conformal field theory will survive the massacre. The reason is explained in my MO answer:
$p$-adic numbers in physics
One can use random/quantum fields $\phi:\mathbb{Q}_{p}^{d}\rightarrow \mathbb{R}$ as toy models of fields $\phi:\mathbb{R}^d\rightarrow\mathbb{R}$. In this $p$-adic or hierarchical setting, Laplacians and all that are nonlocal and not given by partial derivatives.
Most equations in physics are local and therefore need partial derivatives in order to be formulated. What should remain, in the very hypothetical scenario proposed in the question, is everything pertaining to nonlocal phenomena.
I'd query the contention that organisms or even inorganic matter compute in the sense described.
For example, if I drop a stone on the surface of the earth, it falls in a straight line. To call this as 'computing' a straight line seems rather a stretch of the word computation; to my thinking, to compute, means that one ought to be conscious that one is carrying out a computation. That is the person who dropped it is computing the straight line - and not the stone itself. It merely moves in a straight line. We know it moves in a straight line, and hence by dropping it, are describing a straight line.
This is an excellent answer, because it finally brings into focus the question of what we mean by computation. One way to think of it is that we, as humans, arrange two physical systems to behave in ways that can be mapped into each other: Say we are trying to predict what system A will do. If we can arrange system B to do the "same" thing, then by observing system B, we can predict A. B could be a traditional general purpose computer, but doesn't have to be. Now, there is no
reason for us to hobble ourselves in performing the mapping by, say, outlawing
derivatives ...
... it may well be that our understanding of both systems A and B, and therefore the construction of the mapping necessary for computation, hinges on using derivatives, even if system B does not "perform derivatives" in the traditional general purpose computer sense.
You may be interested in the historical note that I added to the question as well as this paper on brain computation: https://igi-web.tugraz.at/PDF/LNCS-10000-Theories_006_v1.pdf
One way for reformulating all of (classical) mechanics is Peridynamics, which does away with derivatives. It is essentially a non-local reformulation.
Javili, Ali, et al. "Peridynamics review." Mathematics and Mechanics of Solids 24.11 (2019): 3714-3739.
For one example of non-trivial physics without partial derivatives, one can look into Volume 1 of the Feynman lectures. In Chapter 28, Feynman starts to develop electrodynamics without partial derivatives — they only appear in Volume 2.
Instead of the Maxwell equation, Feynman uses a somewhat complex formula for the field that is generated by a single moving charge. The formula only has ordinary derivatives — the first and second time derivative of the speed of the charged particle — but becomes a bit unusual in that uses retarded time: The field at a distant point is determined by the movement of the particle some time ago, to correct for the finite speed of light.
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2025-03-21T14:48:30.011035
| 2020-03-05T19:39:13 |
354257
|
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If $u_n \to u$ in $H^1_0(\Omega)$, does $\chi_{\{u_n = 0\}} \to g$ for some $g$ in some space, for a subsequence?
Let $\Omega$ be a bounded and smooth domain.
Suppose we have $u_n \to u$ in $H^1_0(\Omega)$. We know that for a subsequence, $\chi_{\{u_n = 0\}} \rightharpoonup f$ to some $f$, weak-* in $L^\infty(\Omega)$. In general we cannot say that $f=\chi_{\{u=0\}}$.
But my question is, is it possible to have a strongly convergent subsequence of $\chi_{\{u_n = 0\}}$? I.e. can we obtain
$$\chi_{\{u_n = 0\}} \to g$$ for some $g$ in some (probably $L^p$) space, for a subsequence?
Motivation: If I cannot identify the weak limit as something nice (the desired indicator function), then the next best thing is to show that a strong limit exists. You might expect a strong limit because the $u_n$ are nicely convergent.
You at least need some conditions on $u$. If $u=0$ then the zero sets of the $u_n$ can be whatever you like, if you just rescale them small enough to make their $H^1$ norms small.
I think you'll run into the same problem whenever $m(u=0)>0$. If $u$ is a.e. nonzero then since in particular $u_n \to u$ in measure, I think you must have $m(u_n = 0) \to 0$ and therefore $\chi{u_n = 0} \to 0$ in whatever $L^p$ you like.
So to summarize: I claim that if $m(u=0)=0$ then $\chi_{{u_n = 0}} \to 0$ in every $L^p$, $p < \infty$, without needing to pass to a subsequence. For every other $u$, there exists a sequence $u_n \to u$ in $H^1_0$ such that $\chi_{{u_n = 0}}$ has no convergent subsequence in $L^1$ (and hence not in any other $L^p$ either). Would that resolve your question, and shall I post an answer to that effect?
@NateEldredge thanks for your replies. Feel free to answer it, I'm not sure if there is a weaker space under which there is a convergence so maybe I'll put a bounty on it when I can
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2025-03-21T14:48:30.011204
| 2020-03-05T20:25:36 |
354258
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Stack Exchange
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What is the maximal number of increasing length-$k$ subsequences?
Let $\rho \in \{1,\dots,r\}^n$ be a string of length $n$ in the alphabet $\{1,\dots,r\}$. A subsequence of $\rho$ of length $k$ is the precomposition $\rho\circ \iota$, where $\iota: \{1,\dots,k\} \to \{1,\dots,n\}$ is an injection. Let $O_k(\rho)$ be the number of subsequences of $\rho$ which are strictly increasing. Let $O_k(r,n)$ be the maximum value of $O_k(\rho)$ as $\rho$ ranges over all strings of length $n$ in the alphabet $\{1,\dots,r\}$.
Questions:
"What is $O_k(r,n)$?" (I think there is no "closed-form" formula, but anything to put it into context would be helpful.)
In particular, how close is $O_k(r,n)$ to the trivial upper bound $O_k(r,n) \leq {n \choose k}$?
Is it easy or hard to construct examples of $\rho$'s attaining the maximum value $O_k(\rho) = O_k(r,n)$?
I think I could probably find out a lot if I could locate a relevant OEIS sequence, but I'm hindered by the fact that there is more than one parameter here.
Are you assuming k is much less than r? A good test case is when n=qr, and rho is the concatenation of r constant sequences each of length q. Gerhard "Trying To Size Up Problem" Paseman, 2020.03.05.
@GerhardPaseman I'm interested in the full range of values of $k$, from $k=1$ to $k=r$. Perhaps it is a good idea to look at different regimes separately, though...
Surely the answer to (3) is that it's hard, because if it were easy then it would be easy to answer (1)?
When $n\gg r\gg k$, Gerhard Paseman's "test case" yields $O_k(r,n)\sim\binom{n}{k}$.
Generalising Gerhard Paseman's "test case", let $n=qr+s$, where $0\leq s\lt r$, and let
$$\rho=1^{q+1}2^{q+1} \;\ldots\; s^{q+1}(s+1)^q \;\ldots\; r^q$$
be the equitable string, in which each of $1,\ldots,s$ occur $q+1$ times and each of $s+1,\ldots,r$ occur $q$ times.
Then,
$$
O_k(\rho)
=\sum_{j=0}^k\binom{s}{j}\binom{r-s}{k-j}(q+1)^j q^{k-j}
=q^k \binom{r-s}{k}{}_2F_1(-k, -s, r-s-k+1; 1 + 1/q).
$$
I'd be inclined to conjecture that this is in fact the value of $O_k(r,n)$.
If the remainder $s$ is zero, then we simply have
$$\binom{r}{k}\frac{n^k}{r^k}$$
which is easy to analyze.
Thanks! It's interesting that your conjectured result depends so mildly on $k$ (and uses the same maximizer independent of $k$). Perhaps I shouldn't be surprised, though, since having lots of ordered length-$k$ subsequences is pretty strongly correlated with having lots of ordered length-$l$ subsequences for $k,l >1$.
The more I consider this, the more I'm convinced that a rigorous proof of the conjecture should be possible.
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2025-03-21T14:48:30.011398
| 2020-03-05T22:38:55 |
354266
|
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Given $\,m=\prod_k {p_k}^{\alpha_k}\,$ and the function $\,g(m)=\sum_k \alpha_k(p_k-1)^2$, find all solutions of the equation $\,g(2n)=n$
Let's consider the unique decomposition of a natural number $\,m\,$ into its prime factors:
$$\prod_k {p_k}^{\alpha_k}$$
Then, let's define the following arithmetic function (completely additive) $\,g:N\rightarrow N$:
$$g(m)=\sum_k \alpha_k(p_k-1)^2$$
The question is: $\,$ find all solutions of the equation $\;g(2n)=n$.
My first approach to the problem has been the following.
If $\,n\,$ is a solution and $\,n+1\,$ is prime, then also $\,(n+1)\cdot n\,$ is a solution.
In fact, if $\;g(2n)=n$, then $\;g(n)=n-1$. Furthermore
$$g(2\cdot(n+1)\cdot n)=1+n^2+g(n)=1+n^2+n-1=n^2+n$$
and
$$(n+1)\cdot n=n^2+n$$
So
$$g(2\cdot(n+1)\cdot n)=(n+1)\cdot n$$
Now, certainly $\,n=1\,$ is a solution. Therefore, the previous reasoning allows us to find further solutions immediately:
$$n=1\cdot 2=2$$
$$n=2\cdot 3=6$$
$$n=6\cdot 7=42$$
$$n=42\cdot 43=1806$$
But, unfortunately, $\,1807=13\cdot 139$.
Then, using brute force, I could ascertain that up to $\,10^8\,$ there are only the following solutions:
$n=1$
$n=2$
$n=6=2\cdot 3$
$n=9=3\cdot 3$
$n=42=2\cdot 3\cdot 7$
$n=182=2\cdot 7\cdot 13$
$n=1806=2\cdot 3\cdot 7\cdot 43$
$n=2666=2\cdot 31\cdot 43$
$n=6873=3\cdot 29\cdot 79$
$n=101702=2\cdot 211\cdot 241$
$n=361802=2\cdot 7\cdot 43\cdot 601$
$n=389846=2\cdot 421\cdot 463$
$n=74042282=2\cdot 6007\cdot 6163$
Many thanks for all your comments.
(Addendum)
Analyzing the list of solutions found, we note that those of the form $\,2pq$, with $\,p\,$ and $\,q\,$ prime ($p>q$), seem to enjoy the following property:
$$p-q=j(j+1)$$
for suitable values of $\,j$.
By imposing this constraint on our equation we obtain:
$$g(2\cdot 2pq)=1+1+(p-1)^2+(q-1)^2=2+(j(j+1)+q-1)^2+(q-1)^2$$
and
$$2pq=2(j(j+1)+q)q$$
By imposing the equality $\,g(2\cdot 2pq)=2pq\,$ and by isolating $\,q$, we obtain:
$$q=\frac{(j(j+1)-1)^2+3}4$$
In this way, with a minimum calculation time, the following solutions can be found (for $\,j<100$):
$n=2\cdot7\cdot13=182\;\;(j=2,\,solution\,already\,found)$
$n=2\cdot31\cdot43=2666\;\;(j=3,\,solution\,already\,found)$
$n=2\cdot211\cdot241=101702\;\;(j=5,\,solution\,already\,found)$
$n=2\cdot421\cdot463=389846\;\;(j=6,\,solution\,already\,found)$
$n=2\cdot6007\cdot6163=74042282\;\;(j=12,\,solution\,already\,found)$
$n=2\cdot188791\cdot189661=71612579702\;\;(j=29)$
$n=2\cdot1624351\cdot1626901=5285316532502\;\;(j=50)$
$n=2\cdot4062241\cdot4066273=33036361795586\;\;(j=63)$
$n=2\cdot6530581\cdot6535693=85363745055266\;\;(j=71)$
How did you arrive at such a $g$ function? It seems that pure curiosity wouldn't suffice (at least to me).
Also posted to m.se, https://math.stackexchange.com/questions/3572359/given-m-prod-k-p-k-alpha-k-and-the-function-gm-sum-k-alpha-k without notice to either site.
You have two more solutions up to $10^9$; these are $131236202 = 2 \cdot 8011 \cdot 8191$ and $303638726 = 2 \cdot 12211 \cdot 12433$.
Though I've shut off my program now, at least you might want to know this: up to $2\cdot 10^9$ there are three more solutions; these are $1269025562 = 2 \cdot 7 \cdot 2551 \cdot 35533$ and $1377547562 = 2 \cdot 26803 \cdot 26407$ and $1593244802 = 2 \cdot 28057 \cdot 28393$.
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2025-03-21T14:48:30.011602
| 2020-03-05T22:53:30 |
354267
|
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Stack Exchange
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"Expected Value" of a solution to a differential equation
I'm going to write this question in a very informal way as I'm looking for guidance, rather than a specific answer to a specific problem. So I took a course on stochastic processes and Martingales recently. My current study interest is to try and understand the (possibly many) relations between stochastic processes and IVPs. I am investigating the following question:
How can stochastic processes provide information on behavior of solutions to differential equations?
What Motivated the Question:
So I am given a function $u:\Omega\times[0,T]\to\mathbb{R}$ satisfying the IVP: $$ \begin{cases}\dfrac{\partial u}{\partial t}=F(u,D^\alpha u),\\ u(x,0)=u_0(x).\end{cases}$$
As we all know, a lot of the times these equations are very difficult to solve explicitly and even more difficult to understand their behavior (say in the $L^\infty$ sense for example). Of course, a lucky case is when the solution satisfies some known modulus principle. But this is far from a general case. So I want to ask myself a supposedly simpler questions:
What is the expected behavior of a solution, given the initial data $u_0$?
So after a bit of thought I asked myself: What if one can construct a stochastic process $\{X_t\}$ such that $X_0$ "holds" the initial data $u_0$ and then $X_t$ "predicts" the behavior of $u(t)$ using the above equation? By predict, I mean where do I expect my $u(x,t)$ to be given that $u_0(x)$ for some $t$ "close enough" to 0. Such a process would of course be constructed using the function $F$ and imposed properties on the solution. In the language of stochastic processes, I want to make explicit the following:
$$'E[u(t)\mid u_0]': \text{The expected value of $u$ at time $t$ given $u_0$}.$$
Very informally, I'm using the symbol $E[u(t)\mid u_0]\sim E[u(t)\mid\sigma(u_0)]$ where $\sigma(u_0)$ is the "information" I get from $u_0$ (not necessarily the $\sigma$-algebra of $u_0$).
Also, is it possible to find a stochastic process $\{X_t\}$ and an outcome $\omega$ such that $X_t(\omega)\sim u(t)$?
I hope I am making sense here. Do you think it is a good question? If there is any remark let me know in the comments so that I can edit. Some guidance would be of great help!
Thank you in advance.
So in the comments, I found out that something like Feynman-Kac is one answer to my question. But Feynman-Kac gives an explicit solution. What I am looking for is more of "behaviour analysis" of solutions to PDEs using stochastic processes.
In some sense, that is what the field of dynamical systems is all about. Especially Ergodic theory. See: https://en.wikipedia.org/wiki/Ergodic_theory
sounds like feynman-kac I think
Is something like this paper the type of thing you are thinking of? A probabilistic method of computing solutions for PDEs, which gets the right answer on average?
This is not my field of expertise, therefore I decided to post a comment instead of an answer. My advice is to have a look at the book of Mark I. Freidlin, Functional integration and partial differential equations, Annals of Mathematics Studies, No. 109, Princeton University Press. In this monograph the Feynman-Kac formula is used jointly with a perturbation method to study the properties (time asymptotic and other) of several kinds of PDEs, including quasi linear equations of KPP type.
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2025-03-21T14:48:30.011846
| 2020-03-05T23:38:36 |
354269
|
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Stack Exchange
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Digit summation of squared numbers
In olympiad teaching period, we have a session that students must try to design a good problem for others. Many times we arrive to good questions but sometimes there are some challenges. In one of our exam we gave this problem to our students:
Find the 5-digit numbers that the summation of the digits of their square is maximum?
In the problem designing session, we got this problem (the notations are corrected and redefined. Also, I give here the generalized form of question):
Let $D_n^2(k)$ denotes the total number of $n$ digits number such that the summation of the digits of its square is less than or equal to $k$. What is the behaviour of $D_n^2(k)$?
I think this general form is so hard. Is the question famous or previously studied somewhere?
Is there any approximation for the values of $D_n^2(k)$ based on $n$ and $k$?
You might start by looking at the links at OEIS sequence A004159.
This is loosely related to an early proposed (by Von Neumann) pseudo-random number generator the Middle square method, which would square an $n$ digit number (producing a $2n$ digit number), extract the middle $n$ digits, then iterate. Some cryptanalysis of this could theoretically be based on understanding the digits of the square of a number, but I'm unfamiliar with it.
Dear Mark, I am familiar with that method which you mentioned. But, I can not see how we can use it for this question!
Dear Israel, I did not find something there. Am I did wrong search?
Dear Mark Sapir, you are right. Is there any chances that for large $n$ we can assume identical distribution for digits in the squared numbers? Would you give any references for your claim and also for your conjecture?
Wow, I was not born 45 years ago. Anyway, thank you for your comment. I found some references for this conjecture.
@ShahroozJanbaz You didn't find the links or you didn't find anything relevant in them?
@Robert Israel I did not find relevant thing!
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2025-03-21T14:48:30.012027
| 2020-03-05T23:41:00 |
354270
|
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"Moishe Kohan",
"Richard Stanley",
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|
Stack Exchange
|
Are triangulations of n-dimensional manifolds determined by lower-dimensional skeleta?
Suppose that $M$ is an $n$-dimensional manifold equipped with a triangulation $T$. Given $n\ge 1$, in order to recover $T$ (up to an isomorphism of simplicial complexes) one needs to know at least the 1-dimensional skeleton of $T$. How large $k=k(n)$ does one need to recover the triangulation $T$ from the $k$-skeleton of $T$?
More formally. Let $X^k$ denote the $k$-skeleton of a simplicial complex $X$. Given $n\ge 1$, define $k(n)$ as the least $k$ such that for any two $n$-manifolds (without boundary) equipped with triangulations $X, Y$, if $X^k$ is isomorphic to $Y^k$ then $X$ is isomorphic to $Y$ (as a simplicial complex).
Question. How much is known about the function $k(n)$?
For instance, is it true that $k(n)\ge 2$ for all $n\ge 3$? (In other words, is it true that for each $n\ge 3$ there exist triangulated $n$-manifolds with isomorphic 1-skeleta of triangulations but non-isomorphic triangulations?) Is it true that $k(n)\ge n/2$ if $n$ is sufficiently large? Etc.
One can ask various other versions of this question. For instance, instead of requiring triangulations $X, Y$ to be isomorphic, only require the corresponding triangulated manifolds to be PL isomorphic. One can also restrict to pairs of homeomorphic $n$-manifolds.
Edit. I just found
Jerome Dancis, Triangulated $n$-manifolds are determined by their $[n/2] + 1$-skeletons. Topology Appl., 18(1):17–26, 1984.
He proves that compact triangulated $n$-manifolds are determined by their $\lceil (n+1)/2 \rceil$-dimensional skeleta. Together with Richard Stanley's reference, this yields that in the setting of compact manifolds
$$
k(n)= \lceil (n+1)/2 \rceil,
$$
except in the case of surfaces.
There are superexponentially many (in both $n$ and $d$) triangulations of a $(d-1)$-dimensional sphere (or even a simplicial $d$-polytope) with $n$ vertices such that every $\lfloor d/2 \rfloor$-element set of vertices form a face. On the other hand, if every ($\lfloor d/2\rfloor +1$)-element set of vertices form a face, then $n=d+1$ and the triangulation has just one maximal face. See https://en.wikipedia.org/wiki/Neighborly_polytope. Some results are also known for other manifolds.
@RichardStanley: Interesting. This means that $k(d-1)\ge [d/2]$, at least if $d$ is high enough. I will have to read more about this counting result.
|
2025-03-21T14:48:30.012215
| 2020-03-05T23:54:50 |
354271
|
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"Marcus M",
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|
Stack Exchange
|
Sum of sequences of random variables, with variable success probabilities
Consider two sequences of (not necessarily independent) Bernoulli random variables $X_1, X_2, \ldots, X_n$ and $Y_1, Y_2, \ldots, Y_n$. Suppose that for any $i$, we have $\Pr[X_i = 1] = \Pr[Y_i = 1] = p_i$, but the actual value of $p_i$ is determined only after observing $\{X_1, \ldots, X_{i-1}, Y_1, \ldots, Y_{i-1} \}$. That is, each $p_i$ is also a random variable.
Intuitively it seems that the two sums $\sum_i X_i$ and $\sum_i Y_i$ should typically be very close to each other. Are any such bounds known? Is there e.g. a simple argument to show that with high probability $\sum_i X_i = \sum_i Y_i \pm O(\sqrt{n \log n})$?
Are you assuming that given $p_i$, $X_i$ and $Y_i$ are independent of each other?
@MarcusM Intuitively, I don't think that should matter much for the sums whether $X_i$ and $Y_i$ are independent since they have the same success probabilities. Would it help if we assume they are independent?
You're right it shouldn't matter, actually. I'll add an answer.
Define the martingale $M_n = \sum_{i = 1}^n (X_i - Y_i)$ with the filtration $\mathcal{F}_n = \sigma( \{X_j,Y_j\}_{j=1}^n )$. Then $|M_{n + 1} - M_n| = |X_{n+1}-Y_{n+1}| \leq 1$ and it is indeed a martingale since $$\mathbb{E}[M_{n+1}\,|\,\mathcal{F}_n]= \mathbb{E}[X_{n+1} - Y_{n+1}\,|\,\mathcal{F}_n] + M_{n} = M_n\,. $$
By Azuma's inequality, we then have $$\mathbb{P}(|M_n| \geq \lambda) \leq 2 \exp\left( -\frac{\lambda^2}{2n}\right)\,.$$
Taking $\lambda \gg \sqrt{n}$ shows your desired behavior.
Very cute. Thanks for the super fast response.
|
2025-03-21T14:48:30.012340
| 2020-03-06T00:28:42 |
354274
|
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|
Stack Exchange
|
Distribution of the $k$-th largest eigenvalue of in the sample covariance matrix?
Let us assume we've a rectangular data matrix $X=[x_1 \dots x_n] \in \mathbb{R}^{p \times n}$, where the $x_i \in \mathbb{R}^{p \times 1}$ are iid column vectors. I'm not assuming here that the entries of the matrix $X$ are iid, but if you so need to answer the question, you can assume that first, and then perhaps we can see what happens when we put a covariance structure on $X$.
I'd like to know if there're some known results on the distribution of the $k$-th largest eigenvalue of the sample covariance matrix $C:= \frac{1}{p}X^TX$. References to such works would be greatly apprecited, as will be your answer as well!
You're looking for results on the distribution of the $k$th singular value of $X$. The high dimensional probability book has some results. See chapters 4 and 6.5 in particular where assuming i.i.d. sub-Gaussian they get strong concentration for all singular values, and with weaker distributional assumptions they get sharp expectation bounds (on only the largest singular value).
The Marchenko-Pastur distribution is somewhat relevant, but assumes i.i.d. entries.
@Mark Thank you for the book - it's a good on eit seems. I took a quick look at both chapter 4 and chapter 6.5, but I failed to notice any asymtotics which I originally meant to ask. Really my fault that I didn't make this clear - going to ask a separate question. What I'm looking for is the the distribution of the $k$-th largest eigenvalue when both the sample size $n$ and dimension $p$ foes to infinity.
@Mark Please see the modified question here where I ask for the asymtotic distribution of the $k$-th singular value - https://mathoverflow.net/questions/354288/tracy-widom-type-results-for-asymptotic-distribution-of-k-th-largest-eigenvalu
|
2025-03-21T14:48:30.012487
| 2020-03-06T01:52:07 |
354276
|
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|
Stack Exchange
|
No intermediate denominators growth for holonomic functions?
My question concerns holonomic sequences of rational numbers, meaning assignments $a(n) : \mathbb{N} \to \mathbb{Q}$ fulfilling a linear recurrent relation of the form
$$
a(n+k) = \sum_{i=0}^{k-1} p_i(n)a(n+i), \quad n = 1, 2, \ldots,
$$
where $p_i(x) \in \mathbb{Q}(x)$ are rational functions. In other words, the generating function $f(X) := \sum_{n} a(n)X^n \in \mathbb{Q}[[X]]$ and fulfills a linear ODE with polynomial coefficients.
Define $D_n$ to be the least common denominator of the rational numbers $a(1), a(2), \ldots, a(n)$. That is the minimum positive integer such that all $a(i) \in \frac{1}{D_n}\mathbb{Z}$ for $i \leq n$. A generalization (which I have no clue how to establish) of the prime number theorem would be that $\lim_{n \to \infty} \frac{1}{n}\log{D_n}$ converges (possibly to $+\infty$).
Question. If $D_n$ is unbounded, does it grow at least exponentially in $n$?
@MarkSapir: For $k=1$ it shouldn't be too hard to prove that this dichotomy is indeed true. It is also easy when $\sum_{i=0}^{k-1} \deg{p_i} \in {0,1}$.
@MarkSapir: But the $p_i(n)$ need not be constant. What you write applies to the $p_i \equiv \mathrm{const}$ case, i.e. when $f(X) \in \mathbb{Q}(X)$ is rational. Example: if $a(n) = 1/n$, or $f(X) = -\log(1-X)$, we have $D_n = [1,\ldots,n] = e^{n + o(n)}$, by the prime number theorem. I have no idea how to prove the $k=2$ case; rather trying to get a sense if the statement looks plausible in general.
|
2025-03-21T14:48:30.012630
| 2020-03-06T04:48:07 |
354277
|
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"HCH",
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|
Stack Exchange
|
Nontrivial Poisson relations for affine Poisson algebras
Let $A$ be a polynomial algebra over a field of characteristic $0$ in the variables $x_1,\dots,x_n$. Consider polynomials $f_1,\dots,f_m\in A$ and let $I$ be the ideal they generate in $A$. Moreover, assume that $\{\:,\:\}$ is a Poisson bracket on $A$, i.e., that it is a Lie bracket on $A$ and a derivation in each argument. We say that $I$ is a Poisson ideal in A if $\{I,A\}\subset I$. This is equivalent to saying that there exist polynomials $Z_{ij}^k\in A$ such that
\begin{align}
\{x_i,f_j\}=\sum_{k=1}^m Z_{ij}^k f_k,
\end{align}
where $j\in\{1,\dots m\}$ and $i\in\{1,\dots n\}$. In this case the bracket descends to $A/I$ and we say that $A/I$ is an affine Poisson algebra.
So far my attempts to construct examples with nonzero $Z_{ij}^k$ have failed. My question is: are there examples with nonzero $Z_{ij}^k$? In the case when the Poisson structure is constant or linear there might be conceptual reasons for the vanishing of the $Z_{ij}^k$. Does anybody know of results in this direction?
I correct myself: I am seeking for examples with $Z_{ij}^k$ not in $I$. In other words: the class of $Z_{ij}^k$ in $A/I$ should be nonzero.
Lie ideals yield Poisson ideals of the Lie-Poisson structure; look for example in the upper triangular matrices.
Let $L$ be a finite-dimensional non-abelian Lie algebra over a field $\mathbb{F}$ and consider the symmetric algebra $S(L)$ of $L$, which you can identify with the polynomial ring $\mathbb{F}[x_1,x_2,\ldots]$ where $x_1,x_2,\ldots,x_n$ is an $\mathbb{F}$-basis of $L$ over $\mathbb{F}$. Then the Lie bracket of $L$ can be uniquely extended to a Poisson bracket of $S(L)$ so that this commutative algebra becomes a Poisson algebra. Now, take an ideal $I$ of $L$ and note that $J=I\cdot S(L)$ is a Poisson ideal of $S(L)$. By using the structure constants of $L$, you can now easily find a lot of examples of the kind you are looking for.
Lie ideals in Lie algebras also define Poisson ideals of the associated Lie-Poisson structure.
Consider the Lie-Poisson structure associated to the Lie algebra of upper-triangular $3\times 3$ matrices. The Poisson structure matrix with respect to the generators $E_{11}, E_{12}, E_{13}, E_{22}, E_{23}, E_{33}$ is given by$$\left(\begin{array}{rrrrrr}
0 & E_{12} & E_{13} & 0 & 0 & 0 \\
-E_{12} & 0 & 0 & E_{12} & E_{13} & 0 \\
-E_{13} & 0 & 0 & 0 & 0 & E_{13} \\
0 & -E_{12} & 0 & 0 & E_{23} & 0 \\
0 & -E_{13} & 0 & -E_{23} & 0 & E_{23} \\
0 & 0 & -E_{13} & 0 & -E_{23} & 0
\end{array}\right)$$
We have for example $\{E_{13},E_{11}\} = -E_{13}$ and $\{E_{13},E_{33}\} = E_{13}$ and the other brackets with $E_{13}$ are zero, so $I = \langle E_{13} \rangle$ is a Poisson ideal, and the coefficients in the nontrivial relations are $\pm 1 \not\in I$.
In a Poisson algebra with a Lie-Poisson structure we can also form the ideal generated by all monomials of degree $2$. Continuing example above we have e.g. $$\{E_{11}E_{33},E_{13}\}=E_{11}\{E_{33},E_{13}\}+E_{33}\{E_{11},E_{13}\} = -E_{11}E_{13}+E_{33}E_{13},$$ so again there are nontrivial relations with constant coefficients which do not belong to the ideal.
I apologize for the slow processing. The case of Poisson ideals coming from linear Poisson structures have been systematically studied in the literature. Similarly there are the symplectic reductions. For details and examples one may consult the recent https://arxiv.org/pdf/2107.04204.pdf
For polynomial Poisson structures of higher degree I cannot find any systematic study in the literature.
|
2025-03-21T14:48:30.012996
| 2020-03-06T06:15:15 |
354279
|
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"Chris Wuthrich",
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|
Stack Exchange
|
class group size of cyclotomic field subextension
In the following, let $\mathbb{Q_1}$ denote the subfield of degree $p$ over $\mathbb{Q}$ in the $p^2$- cyclotomic extension.
What is the best known upper bound for the size of its class group, $\text{Cl}(\mathbb{Q_1})$?
It is known that the bounds of the order of $p^{(p-1)/2}$ are known for the class number of the $p$-cyclotomic fields. I was wondering if something similar is known for $Q_1$. For instance, if some bound in the order of $O(p^{p/2})$ or at least $O(p^p)$ can be proved.
It sounds like you want to bound the $p$-primary part of the class group.
@ChrisWuthrich yes.
The $p$-primary part of the class group of $\mathbb{Q}_1$, and in fact all $\mathbb{Q}_n$ in the cyclotomic tower of $\mathbb{Q}$, is trivial for all $p$. This is contained in Proposition 13.22 of Washington's "Introduction to cyclotomic fields".
|
2025-03-21T14:48:30.013094
| 2020-03-06T06:27:07 |
354280
|
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|
Stack Exchange
|
Kashiwara's definition of normal crystal
Let $\mathfrak{g}$ be a symmetrisable Kac-Moody algebra, and $U_q(\mathfrak{g})$ its associated quantum group. Each integrable module of $U_q(\mathfrak{g})$ admits a crystal basis, as was first shown by Kashiwara. In particular, the integrable irreducible module $V(\lambda)$ associated to each dominant weight has a crystal base $\mathcal{B}(\lambda)$.
On the other hand, we can define the notion of a $\mathfrak{g}$-crystal abstractly, as a set together with some partial operators satisfying some conditions (as in Section 1.5 of this paper of Kashiwara's). There are many more $\mathfrak{g}$-crystals than those arising from integrable representations, so we can ask for necessary and sufficient conditions for a $\mathfrak{g}$-crystal to be isomorphic to a crystal arising from an integrable $U_q(\mathfrak{g})$-module.
Question 1: Is there some characterisation of this kind of $\mathfrak{g}$-crystal, beyond that definition?
In that same paper of Kashiwara's, he defines a $\mathfrak{g}$-crystal $B$ to be normal if for every subset $J$ of the Dynkin diagram for which $\mathfrak{g}_J$ is finite-dimensional, the restriction $B_J$ is isomorphic to the crystal of an integrable $U_q(\mathfrak{g}_J)$-module. I cannot tell if this condition implies that $B$ is isomorphic to the crystal of an integrable $U_q(\mathfrak{g})$-module. I also cannot find this definition of normal appearing almost anywhere else in the literature, aside from this other paper of Kashiwara's.
Question 2: What kind of subset of $\mathfrak{g}$-crystals is Kashiwara's normal condition cutting out?
"We provide a simple list of axioms that characterize the crystal graphs of integrable highest weight modules for simply-laced quantum KacMoody algebras." Stembridge, A local characterization of simply-laced crystals. https://www.ams.org/journals/tran/2003-355-12/S0002-9947-03-03042-3/
@SamHopkins For integrable highest-weight modules, there are the Stembridge axioms (for simply-laced type), and also the notion of a "closed family" in arbitrary type (due to Joseph, I think). I am asking about arbitrary integrable modules though, not just highest/lowest weight.
Ah, apologies, I missed that aspect.
|
2025-03-21T14:48:30.013272
| 2020-03-06T08:26:23 |
354285
|
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"Sylvain JULIEN",
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|
Stack Exchange
|
About a possible extension of Siegel-Walfisz theorem
The Siegel-Walfisz theorem is stated in https://en.m.wikipedia.org/wiki/Siegel–Walfisz_theorem.
I want to know if it can be extended unconditionally to a modulus $q$ such that any factorization $q=\prod_{i}q_{i}$ fulfills $(q_{i},q_{j})=(q_{i}q_{j})^{\delta_{ij}/2}$ (hence for $q$ being the product of distinct primes) and $p_{i}\mid q$ implies $p_{i}\ll \log^{A} x$ or if some strong conjecture like GRH is required.
Should it be $(q_iq_j)^{\delta_{ij}/2}$ by a chance?
Right, thank you for pinpointing the typo. I'll edit.
For large $q$ such a result does not hold. This follows from the results in the paper "Limitations to the Equi-Distribution of Primes I" by Friedlander and Granville. Specifically, their Proposition 1 implies that for any constant $B>1$ and arbitrarily large $Q$ (assuming GRH, all sufficiently large $Q$ work), if $q$ is a number in the range $\left(\frac{Q}{(\log Q)^{1/8}},Q\right)$ with no prime factors below $\log q$, then for some $x\in(Q(\log Q)^B,3Q(\log Q)^B)$ and $a$ relatively prime to $q$ we have $\psi(x;q,a)-\frac{x}{\varphi(q)}\neq o(x)$.
It follows that for any $A\geq 1$, the uniform estimate $\psi(x;q,a)=\frac{x}{\varphi(q)}+o(x)$ cannot hold for all $q$ satisfying your condition. In case you are curious about $A<1$, it is impossible for a number $q$ to be a product of distinct primes smaller than $\log q$. Indeed, if $q$ is a product of $N$ primes below $\log q$, we have $N\leq\log q\leq\log 2^N<N$.
Your answer is instructive but there must be a typo cause the interval to which $x$ belongs is empty.
Also, the hypothesis is "a number $q$ with no prime factors below $\log q$" but does the theorem apply if some prime factors of $q$ are below $\log q$ while others are bigger?
@SylvainJULIEN Thanks, the typo was a missing factor. I don't know to what extent the result holds if $q$ has small factors. I suspect that if it has few factors below $\log q$ (something like, their product is $(\log q)^{O(1)}$), then the result remains valid.
|
2025-03-21T14:48:30.013424
| 2020-03-06T09:00:51 |
354288
|
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|
Stack Exchange
|
Tracy Widom type results for asymptotic distribution of the $k$-th largest eigenvalue of the sample covariance when $n, p \to \infty$?
Earlier I asked a question: Distribution of the $k$-th largest eigenvalue of in the sample covariance matrix?, but I forgot to mention that I'd like results for asymtotic regime. So, I'm posting here a modified question.
I'm new to random matrix theory, but per my understanding Tracy-Widom law describes the the asymptotic distribution of the largest eigenvalue of any square real symmetrix matrix with iid entries on the diagonal and above it, of dimension $n \times n$ as $n \to \infty$. (Please correct me if I'm wrong!).
What I'm trying to do is to connect, or at least find a resource that connect Tracy-Widom with Marcenko-Pastur law in a somewhat detailed way, as follows.
Let us assume we've a rectangular data matrix $X=[x_1 \dots x_n] \in \mathbb{R}^{p \times n}$, where the $x_i \in \mathbb{R}^{p \times 1}$ are iid column vectors. I'm not assuming here that the entries of the matrix $X$ are iid, but if you so need to answer the question, you can assume that first, and then perhaps we can see what happens when we put a covariance structure on $X$. Let, as often in the random matrix domain, let $n, p \to \infty, p/n \to c \in (0, \infty)$.I'm interested in the limiting distribution of $k$-th largest eigenvalue, when $k$ is fixed and $k$ - is varying.
Precisely, my questions are:
(1) What's the limiting distruition of the $k$-th largest eigenvalue of $\frac{1}{p}XX^{T}$, as $p, n \to \infty, p/n \to c?$
(2) Also what's the limiting distribution of the $k$-th largest eigenvalue of $\frac{1}{p}XX^{T}$, as $k, p, n \to \infty, p/n \to c, k/p \to c', c\in (0, \infty), c' \in (0,1) ?$
To show you an idea what I'm after, I'll mentiong what I found from my search:
(1)I found this paper that seems to be relevant: https://projecteuclid.org/download/pdfview_1/euclid.aoap/1481792600, but they deal with the the limiting distribution of the largest eigenvalue.
(2) This paper by Tracy and Widom: https://arxiv.org/pdf/hep-th/9211141.pdf, describes in Section E the probability density for the $k$-th largest eigenvalue. But I think there the underlying matrix is a real (or complex) symmetric (or Hermitian) matrix with iid entries on the diagonal and above the diagonal, and not sample covariance matrix.
Any help will be sincerely appreciated, as I'm super new to RMT!
|
2025-03-21T14:48:30.013561
| 2020-03-06T10:01:06 |
354291
|
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|
Stack Exchange
|
Elliptic curves and archimedean place
here is my question :
Let $K$ be any field, $ E \to spec(K)$. Let $ v \in M_K $ an archimedean place.
We know that $ \overline{K_v} \simeq \mathbb{C}$ and there exists $\tau_v \in \mathbb{H}$ such that :
$$ E(\overline{K_v}) \simeq \mathbb{C} / \mathbb{Z} + \tau_v \mathbb{Z} $$
My question is :
1) can we have an upper bound for $Im(\tau_v)$ by a function depending on $v$ ?
2) if we fix an ample invertible sheaf $L$ which give an embedding $E \to P^n(K)$ can we compute $\tau$ in function of the data of $L$ ? Or $L^{\otimes n}$ ?
$\tau$ can be computed by integrating a $1$-form over two different loops in $E(\mathbb C)$ (i.e. elliptic integrals).
I know that if I define this norm (Falting Norm) on the pull back of $\Omega^1_{\mathcal{E}}$ of the Néron model $\mathcal{E}$ over valuation ring of E : $$ |s|^2 = \dfrac{i}{2} \int_{E(\overline{K_v})} s \wedge \overline{s} $$ And if we take $\mathrm{d} z$, we have : $$ |dz|^2 = Im(\tau_v) $$ Do you have an other example ? Or maybe an ohter choice of section $s$ which give more informations ?
|
2025-03-21T14:48:30.013666
| 2020-03-06T10:11:26 |
354292
|
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|
Stack Exchange
|
Copies of $\mathbb{Z}\oplus \mathbb{F}_2$ in non-affine, irreducible Coxeter groups
Let $\left(W,S\right)$ be a non-affine, irreducible Coxeter system and assume that $W$ contains a copy of $\mathbb{Z}\oplus\mathbb{Z}$ (this is equivalent to $W$ being not word hyperbolic). Does this imply that $W$ contains a copy of $\mathbb{Z}\oplus\mathbb{F}_{2}$ as well? Here $\mathbb{F}_{2}$ denotes the free group in two generators. If this is not true: Is it at least true for $S$ large enough?
If a group $G$ contains a copy of $\mathbb{Z}^2$ then it is not word-hyperbolic. Did you mean the negation of your first line?
Yes, of course. Thank you!
What do you mean by that? Does this extra condition imply that $W$ contains $\mathbb{Z} \oplus \mathbb{F}_2$?
Consider the free product $W = \tilde A_2 * A_1 * A_1 * \ldots * A_1$. This Coxeter group is not affine, and it has a copy of $\mathbb Z^2$ within the $\tilde A_2$ part, so it satisfies your condition.
Now assume $W$ contains a copy of $\mathbb F_2 \oplus \mathbb Z$, such that $u,v$ generate $\mathbb F_2$ and $w$ generates $\mathbb Z$. This means that $A := \langle u,w\rangle$ and $B := \langle v,w\rangle$ are two disjoint copies of $\mathbb Z^2$ whose intersection is $\langle w\rangle = \mathbb Z$.
Now I claim (proof in comments) that all copies of $\mathbb Z^2$ in $W$ are finite-index subgroups of some conjugate of the $\tilde A_2$ part. So if $A$ and $B$ lie within the same conjugate, they intersect in another finite-index subgroup of that conjugate (which is $\mathbb Z^2$ again), but if $A$ and $B$ lie within different conjugates, they have trivial intersection. In neither case the intersection is $\mathbb Z$.
This is a contradiction to $A \cap B = \mathbb Z$, hence our assumption is false and $W$ does not contain a copy of $\mathbb F_2 \oplus \mathbb Z$.
One directly sees that if a free product $A\ast B$ contains a 1-ended group $C$, then $C$ is conjugate to a subgroup of $A$ or $B$. This applies to $C=F_2\times Z$, or more generally to any direct product of two infinite f.g. groups.
@YCor How does one see this directly?
If $C$ is a 1-ended f.g. group, every free inversion-free action on a tree fixes a vertex. Apply this to the action of $A\ast B$ on its Bass-Serre tree: $C$ fixes a vertex, and the vertex stabilizers are the conjugates of $A$ and $B$ precisely.
Ah, that's nice.
|
2025-03-21T14:48:30.013845
| 2020-03-06T10:39:40 |
354293
|
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|
Stack Exchange
|
Does every measurable subset of $\mathbb R$ of non zero Lebesgue measure contain arbitrarily long arithmetic progressions?
A subset $E$ of $\mathbb R$ is said to contain arbitrarily long arithmetic progressions, if for every natural $n$, there exists $a, d \in R, d$ nonzero, such that $a + kd$ is in $E$ for all natural $k \leq n$.
Does every measurable subset of $\mathbb R$ of non zero Lebesgue measure contain arbitrarily long arithmetic progressions?
Yes. For fixed $n$, we approximate our set $E$ from above by an open set $U=\sqcup \Delta_i$ ($\Delta_i$ are disjoint intervals) with such accuracy that one of intervals $\Delta_i$ satisfies $|E\cap \Delta_i|>(1-\frac1{n+1})|\Delta_i|$, where $|\cdot|$ denotes Lebesgue measure. Now if $\Delta_i=(a,a+(n+1)t)$, we consider $n+1$ sets $E_i:=(E-it)\cap (a,a+t), i=0,1,\ldots,n$. The sum of there measures equals $|E\cap \Delta_i|>nt$, thus there exists a point covered by them all. It corresponds to an arithmetic progression inside $E$ with difference $t$ and $n+1$ terms.
Lemma (pigeonhole principle): if $n+1$ subsets of a set $X$ have total measure greater than $n$ times measure of $X$, then there exists a point covered by them all. For example, you may assume the contrary and look at the sum of integrals of characteristic functions of these sets (which equals to the integral of the sum of these functions).
|
2025-03-21T14:48:30.013959
| 2020-03-06T10:57:30 |
354295
|
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|
Stack Exchange
|
On the birational equivalent class of algebraic surfaces with Picard number $1$
An open subset $U$ of a projective surface $Z$ is big if $\mathrm{codim}_Z(Z\setminus U)\geq2$.
Let $X$ and $Y$ be smooth complex projective surface. If there exists a birational map $f:X\dashrightarrow Y$ which is an isomorphism between big open subsets of $X$ and $Y$, I say that $X$ and $Y$ are birational equivalent.
Does exist a smooth complex projective surface $Y$ which is not birational equivalent to another one $X$ with Picard number $\rho(X)=1$?
Does exist a classification of such surfaces $Y$?
Let $X$ and $Y$ be smooth complex projective surface. If there exists a birational map $f: X \dashrightarrow Y$ which is an isomorphism between big open subsets of $X$ and $Y$, I say that $X$ and $X$ are birational equivalent. It is generally not such a good idea to give new definitions to terms that are already in common use.
In any case, a map of the kind you describe between smooth (or normal) surfaces is necessarily an isomorphism.
Edit: After rereading your question, it seems that you just want one surface $Y$ which is not birationally equivelant to a surface with picard rank $1$. This is a bit easier. It is not clear if you want $Y$ to have Picard rank $1$, but in both cases there are examples. For an example with Picard rank $1$ take $\mathbb{CP}^{2}$, for an example without picard rank $1$ take $\mathbb{P}^{1} \times E$ where $E$ is an elliptic curve. Below I show that there can be surfaces with picard rank $1$ which are not birational.
Yes, for example fake projective planes https://en.wikipedia.org/wiki/Fake_projective_plane. A fake projective is a smooth projective surface with the same Betti numbers as $\mathbb{CP}^{2}$.
They necessarily have Kodaira dimension $2$, which is different from the Kodaira dimension $\mathbb{CP}^{2}$ i.e. $-\infty$. This is not so hard to see, from some classical classification results on algebraic surfaces. A projective variety with $b_{2}=1$ is either Fano, Calabi-Yau or general type (just consider whether $K_{X} \in H^{2}(X,\mathbb{R})$ is positive, 0, or negative). Fano surfaces (also called del Pezzo surfaces) are classified into 10 topological types and have no fake projective planes. Calabi-Yau surfaces with $b_{1}=0$ have just two topological types, K3 and Enqriques surfaces, having $b_{2}=22$ and $b_{2}=10$ respectively.
A fake projective plane $X$ has Picard rank $1$, since Picard rank is bounded above by $b_{2}(X)=1=b_{2}(\mathbb{CP}^{2})$ by definition and is positive because there is an ample line bundle.
Thank you for your answer. In poor words: I'm studying a property of Higgs bundles over smooth complex projective surfaces $X$, which is invariant under previous "birational equivalence". I'm proving a theorem on such Higgs bundles which works (I guess) when $\rho(X)=1$; so I asked my self: "is $\rho(X)=1$ useless?" By your very clear examples, the answer is "It is not!" Thank you.
This question is not to clear to me: a very general K3 surface and $\mathbb{P}^2$ are amongst an infinite amount of counterexamples that you can find...
I don't understand the edit: what is $\mathbf{CP}^2$ an example of?
Something with picard rank 1 which is not birational to something with picard rank one (other than itself)… Yeah, I am also confused what the OP is asking for. In any case, what is clear is that Picard rank=1 is not a strong enough assumption to put any restriction on the birational type...
|
2025-03-21T14:48:30.014169
| 2020-03-06T11:04:44 |
354296
|
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|
Stack Exchange
|
Spectral sequence associated with a Postnikov tower (Solved by myself)
Suppose $E$ is an $S^1$-spectra of simplicial Nisnevich sheaves. For any $r\in\mathbb{Z}$, we have a distinguished triangle
$$E_{\geq r+1}\longrightarrow E_{\geq r}\longrightarrow F_r\longrightarrow E_{\geq r+1}[1]$$
in $SH_s^{S^1}(k)$, where $_{\geq r}$ denotes the truncation functor (homological index, which is $_{\leq -r}$ in term of cohomological index). We have $F_r\cong H(\pi_r(E))[r]$.
Let $U\in Sm/k$,
$$D_{p,q}^1=[\Sigma^{\infty}U_+[p+q],E_{\geq -p}], E_{p,q}^1=[\Sigma^{\infty}U_+[p+q],F_{-p}].$$ We know that
$$E_{p,q}^1=H^{-q-2p}(U,\pi_{-p}(E)).$$
What does this spectral sequence converge to and what are the $E_{p,q}^2$ terms?
You should add your answer as an answer, not as part of your question. I can do this for you, if you don't object. But you should add details if you do, certainly more than what you already have.
@Praphulla Koushik I'm sorry. The question was asked serveral days ago and I had been working on this.
Sorry for misunderstanding.. I did not see the history of the question.. :)
The convergence of this spectral sequence is because the Nisnevich topology of schemes has a cohomological dimension. The corresponding filtration is given by
$$F^pH_n=Im([\Sigma^{\infty}U_+[n],E_{\geq -p}]\longrightarrow [\Sigma^{\infty}U_+[n],E])$$. cf. https://pdfs.semanticscholar.org/4702/f5e28ad71b82c67e0bb9405cd3e2a1931e5f.pdf
To identify the $E_{p,q}^1$, one could prove that
$$[\Sigma^{\infty}U_+,F[n]]\cong H^n(U,\pi_0(F))$$
for $F\in SH^{S^1}_s(k)^{\heartsuit}$ by using the $\delta$-functor theories. The case for $n=0$ follows from the $t$-structure. For the effaceability, one could show that $H(A)$ is stably fibrant when $A$ is an injective Nisnevich abelian sheaf. For this, one shows that $K(A,n)=\Gamma(A[-n])$ is Brown Gersten fibrant and then it's easy to prove $H(A)$ is an $\Omega$-spectrum. cf. J. F. Jardine, 'Generalized Etale Cohomology Theories'.
|
2025-03-21T14:48:30.014305
| 2020-03-06T11:24:49 |
354299
|
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|
Stack Exchange
|
Weak maximum principle for a perturbation of the Laplacian
This question is motivated by similar considerations for the Kohn-Laplacian in the Heisenberg group, but it seems that I cannot even give an answer in the Euclidean case, so here we go.
Suppose that I consider the differential operator (in polar coordinates)
\begin{equation}
D:=\frac{\partial^2}{\partial r^2}{\color{red} -}\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}.
\end{equation}
Writing the Laplacian, $\Delta $ in polar coordinates we see in fact that
\begin{equation}
D+\frac{2}{r}\frac{\partial}{\partial r }=\Delta.
\end{equation}
For the Laplacian it is well known that the WMP (Weak Maximum Principle) holds (even the strong one), and since $D$ is a perturbation of the Laplacian by a differential operator of first order, we expect that the WMP to hold for $D$ as well, the problem being that the coefficient of $\frac{\partial}{\partial r}$ has a singularity.
So, is there something we can say about the WMP for $D$ ?
EDIT:
Some clarification about the solutions of the PDE $Du=0$. One can verify that a radial solution is $u_0(r,\theta)=r^2$. Suppose now that we have a compactly supported positive Borel measure $\mu, supp(\mu)=K$. Is it true that $$ u:=u_0*\mu $$
satisfies the WMP in $\mathbb{R}^2\setminus K$ ?
(I assume there is a typo, and you meant $\partial/\partial r$ rather than $\partial r / \partial r^2$.) If your domain does not touch the origin, you're good to go: even the strong maximum principle is satisfied, see Theorem 3.5 in Gibarg–Trudinger. On the other hand, if the domain contains $0$, then it is not immediately clear what one means by a solution.
You need a "boundary condition" at 0 since the 1d operator D^2-1/r D (the generator of a Bessel process) has 0 as an exit boundary (see for example Chapter Vi, Section 4 of the book Engel-Nagel "One parameter semigroups..."). This explains what happens on radial functions; in the non radial case expansion in spherical harmonics shows that the radial case is the worst. In case you need I can send some references where these operators have been studied in detail, in Nd.
@MateuszKwaśnicki, Typo corrected, thanks.
@GiorgioMetafune Thanks a lot for the response. If you could send me some references for such operators would be great.
Please, write to me<EMAIL_ADDRESS>
|
2025-03-21T14:48:30.014484
| 2020-03-06T12:21:55 |
354302
|
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|
Stack Exchange
|
Does the cardinality of coordinate projections of the rational points of affine varieties over finite fields also tend to $\infty$?
We know (basically by Lang-Weil) that for an absolutely irreducible n-dimensional affine variety $V$ the cardinality $\#V(F_{l})$ tends to $\infty$ for $l$ large enough. We could now look at the set of rational points $V(\mathbb{F}_{l}) \subset \mathbb{F_{l}}^{n}$. I want to know whether it is true that the cardinality of the projection to any or at least one coordinate $\#\pi_{k}(V_{\mathbb{F}_{l}})$ also tends to $\infty$ or if it could be possible that there could be a strictly increasing sequence of primes $(p_{j})_{j \in \mathbb{N}}$ such that there exists some $n \in \mathbb{N}$ such that $\#\pi_{k}(V({\mathbb{F}}_{p_{j}})) < n$ for all $j \in \mathbb{N}$. Maybe one can formulate the question in a more modern algebraic geometric way to make the answer easier?
In this generality, no. Your $V$ could be contained in a hyperplane $x=0$ and projection on the $x$-axis will have only one point over any field.
Good point! Maybe I should ask if there will always be one coordinate for which the cardinality tends to $\infty$? Or is there also an obvious counter example?
As amended, is true. It's trivial for curves and the general case can be reduced to this case.
Sorry, but I don't see why it is trivial for curves. Could you elaborate?
|
2025-03-21T14:48:30.014596
| 2020-03-06T12:33:05 |
354304
|
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"Pietro Majer",
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|
Stack Exchange
|
Density of continuous functions to interior in set of all continuous functions
Let $M$ be an $m$-dimensional manifold and $N$ be an $n$-dimensional manifold with boundary. Suppose also that the topology on $N$ can be described by a metric. Thus, the set $C(M,N)$ can be endowed with the topology of uniform convergence on compacta.
Intuitively it seems that every continuous function $f:M\rightarrow N$ can be approximated by continuous functions of the form $g:M\rightarrow\operatorname{int}(N)$; where $\operatorname{int}(N)=N- \partial N$, $\partial N$ denoting the boundary of $N$. But is this formally true? I.e., is it true that
$$
\overline{C(M,\operatorname{int}(N))} = C(M,N)?
$$
Prototype construction: Let $N=[0,b)$ then any function $f$ can be approximated by:
$$
f_n= \min\left(\frac1{n},f\right)
,
$$
of course, these are continuous but not smooth (since we don't need smoothness this is not an issue). I expect this type of construction can be generalized.
A boundary of a paracompact manifold has a collar neighborhood, i.e. $U\subset N$ that includes $\partial N$ and is homeomorphic to $\partial N\times [0,1)$ via a map $\psi$ that maps $\partial N$ onto $\partial N\times \{0\}$. Therefore, I will be talking about the points in $U$ as if they were in $\partial N\times [0,1)$.
For $n>1$ define a sequence of continuous maps $\varphi_{n}:N\to int(N)$ which is the identity on $N\backslash (\partial N\times [0,\frac{1}{n}))$, and such that $\varphi_{n}(x,t)=(x,\frac{1}{n})$, if $(x,t)\in \partial N \times [0,\frac{1}{n})$. Clearly, $\varphi_{n}$ converges to the identity map in the compact-open topology.
Now, if $f:M\to N$, the sequence $\varphi_n\circ f$ converges to $f$ in the compact open topology, since composition of map is a continuous operation with respect to the compact-open topology.
Note that the fact that $M$ is a manifold is not used, while the fact that $N$ is a manifold is used rather lightly. I wonder under which condition on a connected metric spaces $M$, $N$ and $F\subset N$ the set $C(M,N\backslash F)$ is dense in $C(M,N)$?
Originally (without putting much thought into it) I suggested $F$ to be merely closed nowhere dense and not separating $N$, but @Pietro Majer swiftly refuted that "conjecture".
As to the last question, consider $N=M=D$, the closed unit disk of $\mathbb{R}^2$, and $F={0}$. Then any map $f:D\to D$ close to the identity, say $|f-\text{id}|_{\infty,D} <1$ has degree $1$ wrto $0$, so it can't be in $C(D,D\setminus{0})$.
|
2025-03-21T14:48:30.014887
| 2020-03-06T12:58:16 |
354306
|
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|
Stack Exchange
|
Tensor-indecomposable modules
Let $A$ be a finite dimensional algebra.
Call an $A$-bimodule $M$ tensor-indecomposable in case $M$ is not isomorphic to $X \otimes_K Y$ for a left $A$-module $X$ and a right $A$-module $Y$.
Question 1: In case $A$ is a connected (acyclic) quiver algebra, is it true that the bimodules $A$ and $D(A):=Hom_K(A,K)$ are tensor-indecomposable?
Edit: I guess in case they were not tensor-indecomposable $X$ or $Y$ must be 1-dimensional, else we would get a contradiction that $A$ (or $D(A)$) is not indecomposable. But then the right or left action would not be faithful in case $X$ or $Y$ is simple. Is this a correct proof that $A$ and $D(A)$ are tensor-indecomposable? (The proof should probably work for any bimodule $M$ that is indecomposable faithful as a left and right $A$-module.
Question 2: Is there a homological criterion when a bimodule $M$ is tensor-indecomposable?
What is $D(A)$?
@BugsBunny I added the definition.
I think Q1 is clear as soon as $A\neq K$. BTW, $A=K$ is a silly counterexample.
Your quiver has a sink $a$ and a source $b$. There are no arrows from $a$ to $b$. Then $A$ has no element $x\neq 0$ such that $ax=x=xb$. But any tensor product, supported on all vertices, must have such an element.
A proof for $D(A)$ is similar and left to an interested reader :-))
No clue about Q2.
|
2025-03-21T14:48:30.015014
| 2020-03-06T14:56:07 |
354313
|
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|
Stack Exchange
|
Numerical evidence that $\pi$ is not normal in base two
Confusion is possible, but we got numerical evidence against
popular belief about the normality of $\pi$ in base two.
According to wikipedia
a real number is said to be simply normal in an integer base b if its infinite sequence of digits is distributed uniformly in the sense that each of the b digit values has the same natural density 1/b. A number is said to be normal in base b if, for every positive integer n, all possible strings n digits long have density b^-n.
Working with precision ten thousands binary digits and n=2,
the counts of the strings in $\pi$ are:
$(11: 1661, 10: 2505, 01: 2505, 00: 1659)$
$10$ occurs about 1.5 times more than $11$.
$\pi$ appears to be simply normal in base four.
The same discrepancy happens for $\sqrt{2}$, $\log{3}$ and
large random integers.
Is $\pi$ not normal in base two and $n=2$?
Computations were done with sagemath and pari/gp.
Added The shorter of the two programs, are there obvious bugs in it?
sage: pre=10^4
sage: gp.default('realprecision',pre)
0
sage: sp=gp.binary(gp.Pi())
sage: sp2=eval(str(sp[2]));sp3="".join(str(_) for _ in sp2)
sage: sp3.count('11'),sp3.count('10'),sp3.count('01'),sp3.count('00')
(5586, 8289, 8290, 5529)
How could this happen for random integers? Also, shouldn't your numbers add up to 10000?
They should still add up to 10000 (give or take a few): the first 10000 bits contain 9999 consecutive bit pairs
You are probably counting substrings without overlaps?
The numbers I get are $2510,2505,2505,2480$ (give or take 1 on each). Could you perhaps share the code you have used?
@მამუკაჯიბლაძე This has to be it - if you every time you see $11$ you delete it and move on to the next one, you're essentially counting occurrences of $011$, $01111$, $0111111, \dots$, which gives $1/2^3 + 1/2^5 + 1/2^7 + \dots = 1/6$.
@Wojowu I edited with the program, are there bugs in my program?
@WillSawin I edited with the program, are there bugs in my program?
From the data, we can tell with high confidence, without looking in the spec, that count does not allow overlaps.
e.g. see line 230 of https://github.com/python/cpython/blob/master/Objects/stringlib/fastsearch.h
For example, "00000000".count('00') gives 4
@მამუკაჯიბლაძე Many thanks :)
I'm voting to close this question as off-topic because the comment strongly suggests the numerical data is just wrong.
A place to test sage code: https://sagecell.sagemath.org/
I also just want to point out that when I said the numbers should add up to 10000, I obviously meant plus or minus 1. (or even plus or minus 2 depending on how you count the digits before the "decimal" point) so the 110 is not a counterexample.
@WillSawin Many thanks for the debugging :) I wrote a lot of wrong stuff trying to "rationalize" the bug...
This is more of a comment. I wanted to share the code I have used to provide values in my comment above.
s00 = 0
s01 = 0
s10 = 0
s11 = 0
a = 0
b = 0
P = pi
for i in range(10000):
a = b
b = floor(P)%2
P = 2*P
if a==0 and b==0:
s00 += 1
if a==0 and b==1:
s01 += 1
if a==1 and b==0:
s10 += 1
if a==1 and b==1:
s11 += 1
print s00,s01,s10,s11
This code returns 2510 2505 2505 2480. I'm afraid I cannot comment on what is wrong with your code.
Thanks, your code seems reasonable, will check
The bug in my code was found, I incorrectly didn't count overlaps.
|
2025-03-21T14:48:30.015261
| 2020-03-06T15:08:46 |
354314
|
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}
|
Stack Exchange
|
A question about the Buchsbaum-Eisenbud-Horrocks Conjecture
It's known that Mark E. Walker proved the "weaker" version of Buchsbaum-Eisenbud-Horrocks' Conjecture (BEH). Although the claim was stated to hold in arbitrary field $k$, Walker's proof does not seem to include the case when $k$ is of characteristic $2$.
QUESTIONS. Has weak BEH been proved or disproved when $k$ is of $char=2$? Why does this special characteristic resist, philosophically or technically?
The issue is that the cyclic Adams operation $\psi^2_{cy}$ is not defined when the characteristic of the base field is 2. There are some notes here by Mark Walker himself that do a good job of explaining Adams operations for commutative algebraists.
|
2025-03-21T14:48:30.015342
| 2020-03-06T15:18:45 |
354316
|
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"Igor Khavkine",
"Yaddle",
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|
Stack Exchange
|
Spectral properties of operators mapped to zero by some polynomial
Let $T$ be a bounded operator on a Banach space $X$ and suppose that there is a non-constant polynomial $p$ such that $p(T) = 0$. It seems to be well known that the spectrum of such an operator coincides with the point spectrum and consists of finite order poles of the resolvent, only. But I can not find any citable reference for this fact.
However, it is not hard to see for me that the spectrum coincides with the point spectrum and that is contained in the set of zeroes of $p$ and hence isolated. But I have no clue, why each of this spectral point is a pole of finite order of the resolvent of $T$.
Take the annihilating polynomial and shift its argument by $\lambda I$. That is, define the polynomials $q$ and $q_0$ by the identity $$(T-\lambda I) q(\lambda,T-\lambda I) - q_0(\lambda) I = p((T-\lambda I) + \lambda I).$$ Then your hypothesis $p(T)=0$ implies the following formula for the resolvent: $$(T-\lambda I)^{-1} = \frac{q(\lambda, T-\lambda I)}{q_0(\lambda)}.$$ It is now obvious that the poles of the resolvent only come from the vanishing of the scalar polynomial denominator $q_0(\lambda)$, which I think should be sufficient for your purposes.
This argument is a variant of the classic one for writing the inverse of a matrix using its characteristic polynomial and applying the Cayley-Hamilton theorem. I don't know the literature on spectral theory well enough to guess where someone might have written down this specific argument.
Why does $q$ have two arguments? Is that a typo?
@Yaddle, $q$ separately polynomially depends on $\lambda$ and $T-\lambda I$. More generally $p(x+y) = x q(y,x) - q_0(y)$, the way I defined things.
|
2025-03-21T14:48:30.015471
| 2020-03-06T16:58:52 |
354319
|
{
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"Bart Jansen",
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"url": "https://mathoverflow.net/questions/354319"
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|
Stack Exchange
|
Minimizing weight values while preserving the collection of maximum-weight independent sets
Consider an undirected graph $G = (V,E)$ with a weight function $w \colon V \to \mathbb{N}$ on its vertices. Let $\alpha_w(G)$ denote the maximum weight of an independent set in $G$, i.e., the maximum over all sets $S \subseteq V(G)$ consisting of pairwise nonadjacent vertices, of the value $\sum _{v \in S} w(v)$. Then the collection $\mathcal{I}_w(G)$ of maximum-weight independent sets of $G$ is defined as:
$\mathcal{I}_w(G) := \{ S \subseteq V(G) \mid S \text{ is an independent set and } \sum_{v \in S} w(v) = \alpha_w(G) \}$.
The same collection of maximum-weight independent sets can be generated by multiple different weight functions on $G$. For example, multiplying all weight values by the same constant leads to the same collection of maximum-weight independent sets. Let's say a weight function on an $n$-vertex graph is $f(n)$ bounded if each vertex weight lies in the range $\{1, \ldots, f(n)\}$. I am interested in the following question:
Is it true that for any weight function $w$ on an $n$-vertex graph $G$, there exists an equivalent $2^{o(n)}$-bounded weight function? That is, does there always exist a weight function $w'$ which assigns integer weights in the range from $1$ to $2^{o(n)}$ such that $\mathcal{I}_w(G) = \mathcal{I}_{w'}(G)$?
When we restrict $G$ to be a forest, there is an elementary argument to show that an equivalent weight function exists that is $O(n)$-bounded. In general, theorems from integer linear programming theory can be used to prove that there is always an equivalent weight function which is $2^{O(n \log n)}$-bounded, even in much more general settings. Can better bounds be obtained in this restricted setting of maximum weight independent sets?
The $\lceil n / 2 \rceil$ upper bound for forests doesn't look right. For a star tree with $n$ vertices and the only maximum weight set consisting of only the center we need weights up to $n$.
I think weight zero is also allowed in the def.
Actually, this doesn't help much if you ask for two maximum weight sets: the center or all the leaves.
You are right; my claimed bound of $\lceil n/2 \rceil$ for forests was in a slightly more relaxed model where I allow vertices to be merged into one, so that it occurring in a max independent set means all vertices that were merged belong to that maximum independent set. In the model as stated, $Theta(n)$ is the right bound for forests.
|
2025-03-21T14:48:30.015992
| 2020-03-06T17:06:05 |
354320
|
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|
Stack Exchange
|
Large scale analysis of matrix multiplications
Let $\mathbf{A}_{m\times n}$ and $\mathbf{B}_{m\times n}$ be two random i.i.d matrices with zero mean and unit variance. Then, are the following large-scale analysis true (m,n go to infinity with fixed ratio)?
\begin{align}
\mathbf{A}^\mathrm{H}(\mathbf{B}\mathbf{B}^\mathrm{H})^{-1}\mathbf{A}=\frac{1}{m}\mathbf{A}^\mathrm{H}(\frac{1}{m}\mathbf{B}\mathbf{B}^\mathrm{H})^{-1}\mathbf{A},
\end{align}
as $\frac{1}{m}\mathbf{B}\mathbf{B}^\mathrm{H}\rightarrow\mathbf{I}_m$, where $\mathbf{I}_m$ is an identity matrix, we have
\begin{align}
\frac{1}{m}\mathbf{A}^\mathrm{H}(\frac{1}{m}\mathbf{B}\mathbf{B}^\mathrm{H})^{-1}\mathbf{A}&\rightarrow\frac{1}{m}\mathbf{A}^\mathrm{H}\mathbf{A}\\
&=\frac{n}{m}\frac{1}{n}\mathbf{A}^\mathrm{H}\mathbf{A}\\
&\rightarrow \frac{n}{m}\mathbf{I}_n.
\end{align}
If $m,n \to \infty$ with $\frac{m}n \to \lambda \in (0,\infty)$ (let's suppose that $n$ depends on $m \in\mathbb{N}$), and if you are saying that each $\mathbf{B} = \mathbf{B}_m$ is an $m\times n$ random matrix whose entries are i.i.d. drawn from a fixed (across $n$) probability distribution $\mu$ with zero mean and unit variance, then the behavior of $$ \frac1m \mathbf{B}\mathbf{B}^\dagger $$ is known to be potentially very far from $\mathbf{I}_m$, and even much worse its inverse. This is the content of the Marcenko-Pastur law, https://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution
On the other hand, products of matrices of the type you described appear in multivariate Fisher statistics, which is covered in Chapter 2.5 in a great book by Yao et. al., Large sample covariance matrices and high-dimensional data analysis. A revised edition can be found here: https://www.researchgate.net/publication/272093238_Large_Sample_Covariance_Matrices_and_High-Dimensional_Data_Analysis
|
2025-03-21T14:48:30.016116
| 2020-03-06T17:19:08 |
354321
|
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|
Stack Exchange
|
Convergence of Bayesian posterior
Let $\Delta [0,1]$ denote the set of all probability distributions on the unit interval.
Let $\mu \in \Delta [0,1]$ denote an arbitrary prior. Importantly, $\mu$ does not necessarily admit a density nor necessarily has full support.
The decision-maker observes a sequence $X^T=(X_1,...,X_T)$, where $X_t \in \{0,1\}$ and assumes that all realizations $X_t$ are such that $X_t \sim \mathrm{Bernoulli}(\theta_0)$, where $\theta_0 \in [0,1]$ is unknown.
(It is irrelevant for the question whether the data is indeed iid or not.)
Question: Is there $A \subset [0,1]$ such that $A$ has at most two points such that for all open neighborhood $U$ of $A$ $\lim_{T\to\infty}\mu|X^T(U)=1$?
My conjecture: The posterior accumulates on the set $A$ that minimizes the KL divergence with respect to the empirical mean. That is, because the data is assumed to be iid by the decision-maker, then some version of Berk (1966) would apply and $A=\arg \min_{\theta \in \text{supp} \mu} D(\theta||\theta_0)$, where $D$ denotes the Kullback-Leibler divergence.
Is this line of reasoning correct?
Are there any references available with respect to this case (limit behavior of Bayesian posteriors with potentially misspecified prior in a discrete setting)?
Edit:
The argument follows trivially from the definition of the Bayes posterior.
Let $p_t$ denote the empirical mean of the sequence $X^t$ and $H(p,q)=-\sum_i p_i \ln q_i$. Then,
$\mu|X^t(A)=\frac{\int_A \theta^{t p_t}(1-\theta)^{t(1-p_t)}\mu(d\theta)}{\int_\Theta \theta^{t p_t}(1-\theta)^{t(1-p_t)}\mu(d\theta)}=\frac{\int_A e^{-t H(p_t,\theta)}\mu(d\theta)}{\int_\Theta e^{-t H(p_t,\theta)}\mu(d\theta)}$.
It follows that the posterior will accumulate only on the points in the support that minimizes the KL divergence with respect to the empirical mean. There are at most two such points.
New Question:
Fix a $(X_t)_{t}$ such that $\frac{||X_t||_1}{t}=:\theta_t \to \theta_0$ and suppose that $\arg \min_{\theta \in \text{supp}\mu}D(\theta||\theta_0)=\{\theta_H,\theta_L\}$, with $\theta_H>\theta_0>\theta_L$.
(1)
We know that $\mu|X^t(\{\theta_H,\theta_L\})\to 1$.
Is it the case that the posterior converges, i.e. $\mu|X^t \to \mu_\infty$ where $\mu_\infty(\{\theta_H\})=1-\mu_\infty(\{\theta_L\})=\alpha$, for some $\alpha \in [0,1]$?
(2) If so, can we say something about $\alpha$? E.g. $\alpha =\mu\left(\{\theta \in [0,1]:\theta>\theta_0\}\right)$?
(3) Let $\Delta_t(\epsilon):=\ln\frac{\mu|X^t(B(\epsilon,\theta_H)))}{\mu|X^t(B(\epsilon,\theta_L)))}-\ln\frac{\mu|X^{t-1}(B(\epsilon,\theta_H)))}{\mu|X^{t-1}(B(\epsilon,\theta_L)))}$, where $B(\epsilon,\theta)$ denotes the open ball of radius $\epsilon$ around $\theta$.
If instead we have that $X_t \sim Bernoulli(\theta_0)$, is it the case that for $\epsilon>0$ small, $\Delta_t(\epsilon)\to Y$, where $Y=\theta_H/\theta_L$ with probability $\theta_0$ and $Y=(1-\theta_H)/(1-\theta_L)$ with probability $1-\theta_0$?
In other words, will the change in log odds of $B(\epsilon,\theta_H)$ against $B(\epsilon,\theta_L)$ be approximately distributed as that of the log-odds of the prior with only $\theta_H$ and $\theta_L$ in the support?
|
2025-03-21T14:48:30.016319
| 2020-03-06T17:22:58 |
354322
|
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"LSpice",
"Mikhail Borovoi",
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|
Stack Exchange
|
Topology of the projective symplectic group
Consider the projective symplectic group $\mathrm{PSp}(n+1)$ over $\mathbb{C}$. This parametrizes $(n+1)\times (n+1)$ symplectic matrices modulo scalar multiplication.
Is $\mathrm{PSp}(n+1)$ irreducible?
Consider $4\times 4$ symplectic matrices. A matrix $A$ has a symplectic representative (modulo scalar) if and only if $A^{t}\Omega A = \lambda\Omega$ for some $\lambda\in\mathbb{C}^{*}$, where $\Omega$ is the standard symplectic form. Set $N = A^{t}\Omega A$. The condition $N = \lambda\Omega$ translates into the following equations in the entries of $A$:
$$N_{12} = a_{00}a_{21}-a_{01}a_{20}+a_{10}a_{31}-a_{11}a_{30}=0$$
$$N_{14} = a_{00}a_{23}-a_{03}a_{20}+a_{10}a_{33}-a_{13}a_{30}=0$$
$$N_{23} = a_{01}a_{22}-a_{02}a_{21}+a_{11}a_{32}-a_{12}a_{31}=0$$
$$N_{34} = a_{02}a_{23}-a_{03}a_{22}+a_{12}a_{33}-a_{13}a_{32}=0$$
$$N_{13}-N_{24} = a_{00}a_{22}-a_{01}a_{23}-a_{02}a_{20}+a_{03}a_{21}+a_{10}a_{32}-a_{11}a_{33}-a_{12}a_{30}+a_{13}a_{31}=0$$
Consider the variety $X$ defined by these equations in the $\mathbb{P}^{15}$ of $4\times 4$ matrices modulo scalar. MacAulay2 tells me that $X = X_1\cup X_2$ has two irreducible components both of dimension $10$ and of degree $12$ and $20$ respectively.
This is where the confusion comes from. What am I missing here?
Irreducible in the Zariski topology, I guess?
Exactly. That's it.
What is the point of using $n+1$ instead of $n$ (or $2n$, if you want to emphasize parity).
Like any semisimple group over $\mathbb C$, your group (which I, concerned with rationality issues, would prefer to call PGSp) is generated by irreducible unipotent subgroups, hence is irreducible.
Yes, the projective symplectic group $G={\rm PSp}(2n,{\Bbb C})$ is an irreducible algebraic variety. Indeed, it is simple as an abstract group; see e.g. E. Artin, Geometric Algebra, 1957, Theorem 5.1. On the other hand, it is easy to see that the irreducible component of the identity in $G$ is a normal subgroup of $G$, and hence, coincides with $G$.
@LSpice: Yes, any connected semisimple group is generated by unipotent subgroups. Maybe we should understand the question as follows: Is the group ${\rm PSp}(2n,{\Bbb C})$ connected?
@MikhailBorovoi, yes, good point. I tend to think of the algebraic group $\operatorname{PGSp}_{2n}$ as "the (implicitly connected) adjoint group with root datum $\mathsf C_n$", so I glossed over that important issue.
OP: I guess that $n$ is an odd number (so $n+1$ is even)?
The symplectic group is connected. Otherwise there would be a special name for its unit component...
About the edit: the 10-dimensional component is the right one. The computation might detect another component consisting of matrices of determinant zero.
The two components are both 10-dimensional. One is of degree 12 and the other of degree 20.
I just checked. All the matrices in the component of degree 20 have zero determinant. Thanks for the hint. However, I still do not understand why this component shows up in the computation. It seems to me that those five equations select exactly symplectic matrices modulo scalar.
The symplectic group $G=\mathrm{Sp}(2n)=\mathrm{Sp}(V)$ is connected (say, in characteristic zero, as algebraic group), and hence so is its quotient $\mathrm{PSp}(2n)$. Let $K$ be the ground algebraically closed field, and $(V,\langle\cdot,\cdot\rangle)$ the given symplectic space.
Indeed, we have to check that every $g\in G$ belongs to the component of $1$. The Zariski closure of $\langle g\rangle$ is product of a unipotent (hence connected) group and a diagonalizable one. This reduces to the case when $g$ is diagonalizable, with eigenvalue decomposition $V=\bigoplus_{\lambda\in K^*}V_\lambda$. One sees that $\langle V_\lambda,V_\mu\rangle=0$ for $\lambda\mu\neq 1$. Hence, $$V_1\oplus V_{-1}\oplus\bigoplus_{\{\lambda,\lambda^{-1}\}\text{ of card 2}}(V_\lambda\oplus V_\lambda^{-1})$$
is an invariant orthogonal decomposition. This reduces to the case when this decomposition is trivial.
If $g=\pm 1$, there is nothing to do. Otherwise, say $g$ has eigenvalues $\lambda^{\pm 1}$ with $\lambda\neq\pm 1$; then one easily checks that there is a $g$-invariant orthogonal decomposition into 2-dimensional subspaces. This, in turn, reduces to the case $\dim(V)=2$. Then we're all set since then the symplectic group is just the connected group $\mathrm{SL}_2$.
PS: connectedness still holds for the stabilizer of an arbitrary alternating form: indeed it has a matrix block-triangular decomposition with diagonal blocks $\mathrm{Sp}(2n)$ and $\mathrm{GL}_m$, and upper block abelian unipotent of dimension $2mn$ (the unipotent radical).
Thank you very much to you all for the answer and the comments. I modified my question to better explain what is confusing me about the irreducibility of the symplectic group.
|
2025-03-21T14:48:30.016627
| 2020-03-06T18:12:07 |
354324
|
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"Gro-Tsen",
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|
Stack Exchange
|
Has Pillai conjecture been proven?
I found the paper https://hal.archives-ouvertes.fr/hal-00698687v9/document which claims the proof of Pillai conjecture.
However, it is not mentioned anywhere that it has been proved. It's stated still as an open problem (hence: the conjecture, not theorem).
My question is: Has Pillai conjecture been proven?
The picture is from the book Algebraic Number Theory and Fermat’s Last Theorem by Stewart and Tall.
This is not a proof, this is gibberish. Take this sentence: "We will define Pillai numbers : we note them: $b_a$ they depend of $a$ and heve the same proprieties than $b$ for $a= 1$." What does this even mean?
@Wojowu Thanks for pointing it out. So, what are the best-known results currently?
There is a page on OEIS (https://oeis.org/A074981), a paper by Prof M. Waldschmidt (https://arxiv.org/pdf/0908.4031.pdf) and a paper by Prof Michael A. Bennett (https://www.math.ubc.ca/~bennett/B-Pillai.pdf).
Where I can I find more result (the recent ones) on Pillai conjecture?
As, it has been pointed out in the comment, that the given paper is nothing but gibberish. I would like to know whether techniques used in the proof of Catalan conjecture by Prof Preda Mihăilescu (Pillai conjecture with $c=1$) can be applied to other values of $c$ or not? If yes, for which values?
For fixed $m,n,c$ you can prove finiteness using Baker's bound, which in principle can also be used to find all the solutions. If you let $m,n$ vary, then for $c>1$ Mihailescu's methods do not work - they rely too much on the properties of polynomials $x^k-1$ (they factor into cyclotomics, and the extensions their roots generate are abelian, and for no other $c$ in place of $1$ is it true for all $k$).
guess I will throw this in. For a specific triple of numbers, two primes $p,q$ and then a target $n, \; \; $ a youngster in Korea came up with a calculation that can often serve to prove the set of $p^a - q^b = n$ finite while identifying the largest $a,b$ that work. Maybe I will show the proof of the largest solution to
$$ 7^a - 3^b = 100 $$
We suspect that the largest solution is $343 = 243 + 100.$ Well, take $7^a - 343 = 3^b - 243.$ This becomes
$$ 343 ( 7^x - 1) = 243 ( 3^y - 1). $$ We are going to prove that we cannot accomplish this with $x,y \geq 1.$
Assuming $x,y \geq 1:$
Since $$ 7^x \equiv 1 \pmod {243}, $$ we find $$ 81 | x \Longrightarrow 27 | x. $$
$$ 7^{27} - 1 = 2 \cdot 3^4 \cdot 19 \cdot 37 \cdot 109 \cdot 811 \cdot 1063 \cdot 2377 \cdot 2583253 $$
This divides $7^x - 1.$ In particular, $811 | (7^x - 1),$ and so $811 | (3^y - 1.)$
jagy@phobeusjunior:~$ ./order 3 811
811 810 = 2 * 3^4 * 5
$$ 3^y \equiv 1 \pmod {811} \Longrightarrow 810 | y \Longrightarrow 81 | y. $$
$$ 3^{81} - 1 = 2 \cdot 13 \cdot 109 \cdot 433 \cdot 757 \cdot 3889 \cdot 8209 \cdot \mbox{BIG} $$
In particular, $3^{81} - 1$ is divisible by $3889,$ so $3^y - 1$ is divisible by $3889.$ In turn, this means that $7^x - 1$ is divisible by $3889.$
$$ 7^x \equiv 1 \pmod {3889} \Longrightarrow 1944 | x \Longrightarrow 243 | x. $$
jagy@phobeusjunior:~$ ./order 7 3889
3889 1944 = 2^3 * 3^5
We have shown $243 | x.$ However,
$$ 7^{243} -1 = 2 \cdot 3^6 \cdot 19 \cdot 37 \cdot \mbox{Many More}$$
This means that
$$ 729 | (7^x - 1) $$
This contradicts
$$ 343 ( 7^x - 1) = 243 ( 3^y - 1) $$ with $x,y \geq 1.$
I followed each step of your proof, but I have no idea how you guessed the particular succession of divisors of the orders of $3$ and $7$ modulo something that would eventually lead to a contradiction.
@Gro-Tsen sorry. I wrote three computer programs. By the time I got to this problem, I was pretty confident about what i was looking for. However, as you suspect, I did not predict those primes by hand, I did a few computer searches.
So in fact it is essentially about finding $n$'s which are powers of 3 or 7 such that $ 7^n-1$ has a high 3-valuation or $ 3^n-1$ has a high 7-valuation, correct?
|
2025-03-21T14:48:30.016977
| 2020-03-06T19:02:08 |
354327
|
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|
Stack Exchange
|
Atlas-like websites on specific areas of mathematics
In this post, we look for the existing atlas-like websites providing well-presented classifications or database about some specific areas of mathematics. Here are some examples:
GroupNames: https://people.maths.bris.ac.uk/~matyd/GroupNames
Finite groups of order ≤500, group names, extensions, presentations,
properties and character tables.
Atlas of Finite Group Representations: http://brauer.maths.qmul.ac.uk/Atlas/v3/
This ATLAS of Group Representations has been prepared by Robert
Wilson, Peter Walsh, Jonathan Tripp, Ibrahim Suleiman, Richard Parker,
Simon Norton, Simon Nickerson, Steve Linton, John Bray, and Rachel
Abbott (in reverse alphabetical order, because I'm fed up with always
being last!). It currently contains information (including 5215
representations) on about 716 groups [mainly finite simple groups or almost simple].
Atlas of subgroup lattice of finite almost simple groups: http://homepages.ulb.ac.be/~dleemans/atlaslat/
This atlas contains all subgroup lattices of almost simple groups $G$
such that $S≤G≤Aut(S)$ and $S$ is a simple group of order less than 1
million appearing in the Atlas of Finite Groups by Conway et al. Some
simple groups and almost simple groups or order larger than 1 million
have also been included, but not in a systematic way.
The L-functions and Modular Forms Database: https://www.lmfdb.org/
Welcome to the LMFDB, the database of L-functions, modular forms, and
related objects. These pages are intended to be a modern handbook
including tables, formulas, links, and references for L-functions and
their underlying objects [like field extensions and polynomial Galois groups].
The Inverse Symbolic Calculator: https://isc.carma.newcastle.edu.au/
The Inverse Symbolic Calculator (ISC) uses a combination of lookup
tables and integer relation algorithms in order to associate a closed
form representation with a user-defined, truncated decimal expansion
(written as a floating point expression). The lookup tables include a
substantial data set compiled by S. Plouffe both before and during his
period as an employee at CECM.
If you know such a website on any area of mathematics, please put it as an answer (with a short description).
You might be interested in Bridget Tenner's talk Fingerprinting Richard Stanley.
Also in the Notices article "Fingerprint Databases for Theorems" by Tenner and Billey.
@RussWoodroofe's reference: Billey and Tenner - Fingerprint databases for theorems (why reverse the author names?).
An Atlas of Even Primes would fill a much-needed gap in the literature.
@GerryMyerson: https://mathworld.wolfram.com/EvenPrime.html
@LSpice, I think I put Tenner first because I was replying to Timothy Chow's comment on Tenner's talk. (But you're right, I should've used the canonical order.)
The ISC is mentioned as a prime example of this. I would be interested in a currently operating version of the ISC.
@GeraldEdgar http://wayback.cecm.sfu.ca/projects/ISC/ISCmain.html
This catalogue of mathematical datasets could be of some interest to you - at least some of the entries are atlas-like websites. It includes several of the websites mentioned above, and I'm slowly adding more to it.
Although mere thanks in comments are discouraged, I cannot resist: very, very useful work, thank you very much for your initiative and effort!
It seems OP is not only collecting datasets but also working to standarize them, great work!
A classic: The On-Line Encyclopedia of Integer Sequences (OEIS).
I'd be very surprised if anyone here hasn't heard of this
@BlueRaja https://xkcd.com/1053/
I'm one of the lucky people today <3
Atlas of Lie Groups and Representations: info on lie groups, there is a nice article on the history of this atlas
Atlas of Finite Group Representations: info on finite groups actually the succesor of a book (already mentioned by OP but added it anyway to keep it symmetric with the first point)
Complexity zoo:list of complexity classes and what is know about them
Quantum algorithm zoo: list of quantum algorithms
Knot Atlas: list of knots
Encyclopedia of triangles center: list of triangle centers
Bilbao chrystallographic server: list of space groups and info although more physics oriented
Cantor's Attic: A wiki about large cardinal axioms, large countable ordinals, and infinity in general
Googology wiki: Info on googology
Digital Library of Mathematical Functions: a spiritual successor to a book too and there is also a nice article on the history
https://math.la.asu.edu/~jj/localfields/
https://polytope.miraheze.org/wiki/Main_Page
https://xlinux.nist.gov/dads/
https://www.2dcurves.com/
Does this also count as an atlas page? http://log.chez.com/text/logic/links_ordinals.html
@C7X I wouldn't call it an atlas page but I wouldn't say either that it less useful because it isn't, feel free to add it to the answer if you want, I have left some other pages as comment not because a particular criteria but only because while I want to mention them at the same time I want to avoid appearing in the feed of active answers.
https://errorcorrectionzoo.org/
Here's another one.
The Manifold Atlas Project: http://www.map.mpim-bonn.mpg.de/Main_Page
PolyDB is a database of polytopes. It can either be accessed through the website or directly from polymake.
Another one is the Catalogue of Lattices.
This data-base of lattices is a joint project of Gabriele Nebe, RWTH Aachen university (nebe(AT)math.rwth-aachen.de) and Neil Sloane. (njasloane(AT)gmail.com).
Our aim is to give information about all the interesting lattices in "low" dimensions (and to provide them with a "home page"!). The data-base now contains about 160,000 lattices!
I am trying to keep up with the area of symmetric functions, and various generalizations:
The symmetric functions catalog
The Information System on Graph Classes and their Inclusions:
http://www.graphclasses.org/
This is a database of graph classes with a java application that helps you to research what's known about particular graph classes.
$\pi$-base has examples of topological spaces and their properties.
Here are two more on general topology: http://hyperspacewiki.org/index.php/Main_Page https://topospaces.subwiki.org/wiki/Main_Page
The graded rings database
http://www.grdb.co.uk
A database of varieties (toric varieties, Fano varieties, Calabi-Yaus).
Here is a database of Vertex Operator Algebras and Modular Categories, though it is still in an early version.
The Blocks of Finite Groups wiki, which aims to classify the Morita equivalence classes of blocks with a given defect group. This is in part to understand Donovan's Conjecture better.
Maybe not exactly an "atlas", but there is a classification of some certain types of PDEs studied in analysis at
The Dispersive PDE Wiki: https://kvm16.pims.math.ca/DispersiveWiki/index.php?title=Main_Page
WikiWaves: https://wikiwaves.org/
The Database of Ring Theory (and Module Examples) features
over 100 examples of rings
tracking of over 100 different ring theoretic properties each can have.
Information on important subsets (i.e. zero divisors, Jacobson radical)
notions of dimension (composition length, Krull dimension, etc.)
Links to articles and/or books for each ring and property
Properties about the properties (being Morita invariant, passing to polynomial rings, etc.)
Sundry information about ring theory and theorems, errata in literature, diagrams of implications
I always welcome suggestions for additions: there is too much work for one person to do.
Update
The site now tracks examples of modules over the rings, too. Not all of the features tracked above are fully supported yet but that is the plan.
A classic by Stanisław Radiszowski:
= = = = = = Small Ramsey Numbers = = = = = =
https://www.combinatorics.org/ojs/index.php/eljc/article/view/DS1/pdf
Last update on 2017-03-03. All known Ramsey numbers or their estimates are provided for several different types of graphs and colorings.
There's the Reverse Math Zoo.
Pieter Belmans at the University of Bonn maintains a couple of such websites.
There is Fanography, "a tool to visually study the geography of Fano 3-folds."
And le superficie algebriche, "a tool for studying numerical invariants of minimal algebraic surfaces over the complex numbers". Johan Commelin also contributed to this one.
https://pbelmans.ncag.info/blog/atlas/
Regina and SnapPy are free program suits for research in low-dimensional topology (particularly in dimension 3), and are accompanied by large census databases of closed 3-manifolds, knots and links. Some highlights:
Knot tables up to 19 (!) crossings (Regina)
Closed, orientable 3-maifolds with at most 11 tetrahedra (Regina)
Knots and links with up to 14 crossings (SnapPy)
Various censuses of hyperbolic 3-manifolds (Regina, SnapPy)
The Cunningham Project seeks to factor the numbers $b^n \pm 1$ for
$b = 2$, $3$, $5$, $6$, $7$, $10$, $11$, $12$, up to high powers $n$.
(No relation.)
KnotInfo https://knotinfo.math.indiana.edu/
LinkInfo https://linkinfo.math.indiana.edu/
These are databases of all knots and links up to a certain number of crossings, along with many different computed invariants and properties.
The Atlas of Small Regular Polytopes (Michael Hartley) "contains information about all regular polytopes with $n$ flags where $n$ is at most 2000, and not equal to 1024 or 1536".
The website of Dimitri Leemans has
An Atlas of Abstract Regular Polytopes for Small Almost Simple Groups
An Atlas of Chiral Polytopes for Small Almost Simple Groups
An atlas of subgroup lattices of finite almost simple groups
The website of Marston Conder has many catalogs of information related to maps, hypermaps, polytopes, symmetric graphs, and surface actions.
Combinatorial Data (Brendan McKay) has complete collections of various classes of graph, up to some number of vertices, in graph6 format.
Combinatorial Catalogues (Gordon Royle) has class 2 graphs up to 9 vertices, trees up to 16, bipartite up to 14, 3-regular up to 22, and more.
Regular Graphs (M. Meringer) has information and shortcode files for "simple connected $k$-regular graphs on $n$ vertices and girth at least $g$ with given parameters $n,k,g$."
The House of Graphs is a searchable "Database of interesting graphs".
There is a database of matroids. For example, it includes a list of all $4,886,380,924$ isomorphism classes of matroids of rank 4 on 10 elements!
If you are interested in historical things,
John Jones' number fields database
and
William Stein's modular forms database (which also includes elliptic curve data, such as the Cremona database) used to be standard such websites. Now this information is subsumed into the
L-functions and modular forms database (LMFDB)
mentioned by the OP.
A database of number fields, by Jürgen Klüners and Gunter Malle. (Note this is not the same as the one mentioned in this answer.)
The site also provides links to similar databases.
Another classic, this time about prime numbers, it provides some theoretical results and a lot of data:
= = = = = = https://primes.utm.edu/ = = = = = =
There are many more pages and portals devoted to elementary number theory.
Enumerating Order Types for Small Point Sets with Applications
Database (maintained by Oswin Aichholzer, TU Graz)
Background paper (by O. Aichholzer, F. Auerhammer, H. Krasser (2002))
Many Internet pages provide (or used to) lotto design tables. In particular, the following paper seems solid:
= = = Lotto Design Tables (by P. C. Li and G. H. J. van Rees),
http://citeseerx.ist.psu.edu/viewdoc/download?doi=<IP_ADDRESS>.4298&rep=rep1&type=pdf
Table of absolutely maximally entangled states.
Abbreviated as AME(n,D), the table lists quantum states of n parties with D levels each that show maximal entanglement across every bipartition. These states are equivalent to pure $((n,1,n/2+1))_D$ quantum error correcting codes, and are in the case of n even also known as perfect tensors or multi-unitary matrices. They can be seen as a type of self-dual quantum codes.
http://www.polychora.de/wiki/images/ has renders of cross-sections, vertex figures and their duals, and cellets for many (possibly all) of the 1849 uniform polychora known at the time it was created.
|
2025-03-21T14:48:30.017903
| 2020-03-06T23:42:09 |
354348
|
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|
Stack Exchange
|
Reference request: proofs of integrals presented in Erdélyi's *Table of Integral Transforms*
I have asked related questions on MSE and received no answer, so: I am looking for proofs of the integrals presented in Erdélyi's Table of Integral Transforms vol. i-ii, specifically proofs of the various gamma integrals presented on page 297-299 of volume 2, for instance:
$$\int_{\mathbb{R}} \frac{dx}{\Gamma(\alpha + x)\Gamma(\beta - x)}= \frac{2^{\alpha + \beta -2}}{\Gamma(\alpha + \beta - 1)},~\Re(\alpha + \beta)>1.$$
That the proofs for the identities presented in the work were reviewed closely is mentioned in the preface, though I cannot find any separate references containing the proofs.
Link to e-PDFs of the work in question.
Titchmarsh’s Fourier integrals (1937, 7.6.4) has proof and attribution to Ramanujan.
Another approach appears as a comment on your question, so this is just a rip-off trying to make things tidier but surely there are other ways to Titchmarsh and this to prove it
$
\int_{\mathbb{R}}\frac{dx}{\Gamma(\alpha+x)\Gamma(\beta-x)}\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{\Gamma(\alpha+\beta-1)dx}{\Gamma(\alpha+x)\Gamma(\beta-x)}\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{(\alpha+\beta-2)!dx}{(\alpha+x-1)!(\beta-x-1)!}\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\binom{\alpha+\beta-2}{\alpha+x-1}dx\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{1}{2\pi i}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-2}}{z^{\alpha+x}}dzdx\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-2}}{z^{\alpha}}\frac{1}{2\pi i}\int_{\mathbb{R}}z^{-x}dxdz\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-2}}{e^{i(\alpha -1) \theta}}\frac{1}{2\pi }\int_{\mathbb{R}}e^{-i\theta x}dxd\theta\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-2}}{e^{i(\alpha -1) \theta}}\delta(-\theta)d\theta\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-2}}{e^{i(\alpha -1) \theta}}\delta(\theta)d\theta\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-2}}{e^{i(\alpha -1) 0}}\\
=\frac{2^{\alpha+\beta-2}}{\Gamma(\alpha+\beta-1)}
$
Thanks Michael Engelhardt for edits, for the second one I selected improve instead of approve because I had another error not sure if that means the second edit won't appear as accepted
You're welcome - everything's fine, this is a cute derivation.
|
2025-03-21T14:48:30.018051
| 2020-03-07T00:46:33 |
354349
|
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|
Stack Exchange
|
What is an orthogonal form?
I am reading the article: Algebro - Geometric applications of Schur S- and Q-polynomials
On page 179, they said about an orthogonal form $\psi$ on $W$. There is no definition of an orthogonal form on the article, also on google. How an orthogonal form is defined? Is this a different name of an inner product?
Usually an orthogonal form is a symmetric bilinear form, as opposed to an inner product or unitary form, which is usually a conjugate-symmetric semilinear (= linear in one variable) form.
|
2025-03-21T14:48:30.018113
| 2020-03-07T02:29:21 |
354351
|
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|
Stack Exchange
|
Bound for the inverse of a summation of rank-1 matrices
Given vectors $x_1,\dots,x_T \in R^d$ satisfying $\|x_i\|_2 = 1$, define $A_0 = I$ and $A_t = I + \sum_{i=1}^tx_ix_i^\top$ for $t \geq 1$. We are interested in the following quantity:
\begin{align}
S_{d,T} = \sum_{i=1}^T\|A_{i-1}^{-1}x_i\|_2.
\end{align}
If $d = 1$, it is easy to bound $S_{1,T}$ by $\log T$ as follows:
\begin{align}
S_{1,T} = \sum_{i=1}^T\frac{1}{i} = O(\log T).
\end{align}
Is it true that for $d\geq 2$, we still have the similar bound $S_{d,T} \leq O(poly(d)\log T)$, where $poly(d)$ represents some polynomial of $d$?
|
2025-03-21T14:48:30.018177
| 2020-03-07T02:31:47 |
354352
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354352"
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|
Stack Exchange
|
Nonlinear fixed-point equation with linear solutions?
Let $S$ be an $N\times N$ row-stochastic matrix and let $w'$ be the left Perron eigenvector of $S$ (i.e., $w$ is the stationary distribution of the Markov chain represented by $S$). Let $T$ be the matrix whose $in$-th entry is $S_{ni}w_{i}/w_{n}$. $T$ is also a row-stochastic matrix.
For any vector $\left\{ \hat{z}_{i}\right\} _{i=1}^{N}>0$, we look for $\left\{ \hat{w}_{i}\right\}$ that form the solution to the following system of equations:
$
\hat{w}_{i}=\sum_{n}T_{in}\hat{w}_{n}\frac{\left(\hat{w}_{i}/\hat{z}_{i}\right)^{-1}}{\sum_{k}\left(\hat{w}_{k}/\hat{z}_{k}\right)^{-1}S_{nk}}.$
Claim: the vector $(\hat{w}_i)$ is log-linear in $(\hat{z})$. That is, there exists a matrix $M$ such that the vector $\ln\hat{w}\equiv(\ln\hat{w}_i)$ is equal to $M\ln\hat{z}$.
I tried taking log of the RHS and taking a first-order approximation, which by definition creates a linear relationship between $\ln\hat{w}$ and $\ln\hat{z}$. Based on numerical solutions of the nonlinear equations, the first-order approximation seems to coincide with the numerical solution, but I don't know how to prove this.
|
2025-03-21T14:48:30.018267
| 2020-03-07T03:34:10 |
354354
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354354"
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|
Stack Exchange
|
Cohomological propriety
Let $k$ be a field of characteristic $p$, $\pi: \mathbf P^1_k\times\mathbf{A}_k^{n-1}\to\mathbf{A}_k^{n-1}$ with zero section $s$, $U\subset \mathbf P^1_k\times\mathbf{A}_k^{n-1} $ a Zariski open neighborhood of $D:=s(\mathbf{A}_k^{n-1})$, $\mathscr F$ a lisse $\mathbf{Q}_\ell$ sheaf on $U-D$, $j: U-s(\mathbf A^{n-1})\hookrightarrow U$ the open immersion of $\mathbf A^{n-1}$-schemes. If $\mathscr F$ is the restriction of a lisse sheaf $j_*\mathscr F$ on $U$, then cohomological purity tells us that $R^1j_*\mathscr F$ is lisse on $D$, locally isomorphic to the restriction of $j_*F$, with formation commuting with arbitrary base change $S’\to\mathbf A_k^{n-1}$, and $R^qj_*\mathscr F=0$ for $q>1$. If $\mathscr F$ is tamely ramified along $D$, then $(j,\mathscr F)$ is cohomologically proper in the language of Illusie in the appendix to Th. finitude; i.e. the formation of $Rj_*\mathscr F$ commutes with arbitrary base change $S’\to\mathbf A^{n-1}_k$.
Deligne-Laumon attach to the above situation a function $\Psi:D=s(\mathbf A^{n-1})\to\mathbf N$ given by
$$\Psi(s)=\operatorname{Sw}_{\overline s}(j_!\mathscr F|X_{\overline s}),$$
$\overline s$ being a geometric point centered on $s$. When this function is constant on $D$, their theorem says that $(U,j_!\mathscr F,\pi_U)$ is universally locally acyclic. (If $U=\mathbf P^1_k\times\mathbf{A}_k^{n-1}$, this implies that the $R^q\pi_!\mathscr F$ are lisse; it follows that the $R^q\pi_*\mathscr F$ are also, and both commute with arbitrary base change $S’\to\mathbf A^{n-1}_k$.)
For general $U$, even if $\Psi$ is constant on $D$, I do not believe the formation of the $Rj_*\mathscr F$ need commute with finite base change $S’\to\mathbf A^{n-1}_k$. My question is, are there additional hypotheses weaker than ‘$\mathscr F$ tamely ramified along $D$’ that will make this true?
|
2025-03-21T14:48:30.018403
| 2020-03-07T03:51:54 |
354355
|
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|
Stack Exchange
|
Rank of matrix over UFD polynomial ring
I have a matrix, $M$, size of approximately $20\times 25$, over a polynomial ring $\mathbb{Q}[x_1, \cdots, x_n]$ and a number $r$ such that I would like to test whether or not there exist values for the $x_i$ such that the $\text{rank }M\leq r$.
The most natural approach to me is to take the $r+1 \times r+1$ minors and consider the ideal that these minors generate. However, the issue is that if I wish to test rank $r = 6$, for example, the sheer number of minors and the following Groebner basis calculation take an exceedingly long time to compute (on the order of weeks in some cases). I am currently using SageMath software to do these computations.
I should state that I have already leveraged putting the matrix in a partial Smith normal form and the sparsity of matrix, when possible.
Are there any other algorithms I could employ in addition to (or in lieu of) the above strategy?
Rather than using all minors of a fixed size, often choosing a few randomly chosen ones is enough. If you get the empty scheme, then you are done. Otherwise, you might be able to find points which have a chance of having small rank...
How would you define "few". Do you have any reference for theorems that precisely state and prove your assertion that only choosing a few randomly is enough? I do not see any reason why often it would suffice to only choose random ones. I have not imposed any genericity conditions on the matrix whatsoever
I took @damiano's comment to be giving a one-sided test. That is, the assertion seems to be only that often a random selection of minors can show you that the rank cannot be made less than $r$.
@LouisDeaett Thank you for the clarification. I am still unsure as to why this is "often" the case, unless I impose the condition that the matrix is generic
Indeed, as @LouisDeaett says, I only meant it as a one-sided test. Still, you can squeeze out a little bit more, if you do have solutions satisfying the few minors you tested. Such solutions correspond to matrices with some minors forced to be zero. If you are lucky (which has often been the case in my experience), then you may find that the matrices actually have small rank.
|
2025-03-21T14:48:30.018583
| 2020-03-07T04:18:16 |
354356
|
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}
|
Stack Exchange
|
Finding a specific solution to $X^T\Sigma X = D$
I'm looking to solve for a specific $X$ in the following equation:
$$X^T\Sigma X = D,$$
where $\Sigma \succ 0$, $D$ is a diagonal matrix with strictly positive entries, and all matrices are square. It may also be worth noting that $\det(\Sigma) = \det(D)$.
I'm looking for a (known to exist) solution $X$ which can be written as $I-W$, where $W$ is permutation-similar to a strictly-triangular matrix. Using Sylvester's Law of Inertia, I'm able to solve for an $X$ (up to a unitary transformation) by finding the transformation which converts $\Sigma$ and $D$ to the same diagonal matrix (since they both have all positive entries). However, I don't know how to take this solution to one that matches the form I-W as described above.
Alternatively, I've also considered how one might solve this with the solution to a convex program, but haven't had much success so far.
One solution is obtainable via $X^T\sqrt \Sigma = \sqrt D$ (so $X^T = \sqrt D\Sigma^{-1}$). Here the $\sqrt \centerdot$ refers to the unique positive definite one (since $\Sigma$ has to be invertible). Then you can through in a unitary, $X^T\sqrt \Sigma = (\sqrt D)U$, and obtain all of them (I think).
And by "one", I meant "square root"; by "through", I meant "throw".
Is $det(\Sigma)=det(D)$ an additional constraint? How does it follow from $X^\top \Sigma X=D$, unless you seeking a a solution with a unit determinant?
|
2025-03-21T14:48:30.018701
| 2020-03-07T08:33:53 |
354363
|
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|
Stack Exchange
|
Paramodular newvectors and twists
In the book Local Newforms for GSp(4), Roberts and Schmidt have defined a theory of "new vectors" for smooth representations of $GSp_4$ over a nonarchimedean local field $F$ with trivial central character, using the paramodular congruence subgroups.
If $\pi$ is a generic unramified representation of $GSp_4(F)$ with trivial central char, and $\chi$ is a ramified quadratic character of $F^\times$, then $\pi \otimes \chi$ again has trivial central character; so there must be some vector in $\pi$ whose image in $\pi \otimes \chi$ is the new vector of the twist.
Can one write down an explicit element of $\mathbf{C}[GSp_4(F)]$ which sends the spherical vector of $\pi$ to the paramodular new vector of $\pi \otimes \chi$?
For $GL_2$ it's well known that the operator $\sum_{u \in (\mathcal{O}_F / \varpi)^\times} \chi(u)^{-1} \begin{pmatrix} 1 & u/\varpi \\ 0 & 1 \end{pmatrix}$ does the job, so I'm hoping for something like this. There is a paper by Andrianov "Twisting of Siegel modular forms" which seems related, but he doesn't use the paramodular subgroups.
I've stumbled across an answer to this old question of mine so I'm going to answer it myself. The following paper:
Johnson-Leung, Jennifer; Roberts, Brooks, Twisting of paramodular vectors, Int. J. Number Theory 10, No. 4, 1043-1065 (2014). (preprint version, official journal version)
addresses this exact question, giving an explicit (but lengthy) formula for a twisting map from newvectors of $\pi$ to newvectors of $\pi \otimes \chi$ (for any paramodular $\pi$, not necessarily unramified).
I just got a nudge about this post. Happy to discuss if it’s still of interest to you.
|
2025-03-21T14:48:30.018831
| 2020-03-07T11:14:11 |
354367
|
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"Mateusz Kwaśnicki",
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|
Stack Exchange
|
Convergence of a succession obtained by the Gram–Schmidt process
Let $H$ be an Hilbert space over $\mathbb{C}$.
Let $\{h_n\}_{n \in \mathbb{N}} \subset H$ be a sequence of linearly independent vectors in $H$ such that $h_n \to h \neq 0$ in norm topology.
We apply Gram–Schmidt process, without normalizing, to $\{h_n\}_{n \in \mathbb{N}} \cup \{h\}$ starting from $h$ and then $h_1, h_2,\dots$ obtaining the sequence $\{v_n\}_{n \in \mathbb{N}}$.
My question is if is it true that $v_n \to 0$ in norm topology or that $v_n \overset{w}{\to} 0$ in weak topology.
Thank you for all suggestions.
Yes, it is: the projection of $h_n$ onto the span of $h_1$ through $h_{n-1}$ is no farther away from $h_n$ than $h_{n-1}$.
thanks @MateuszKwaśnicki
|
2025-03-21T14:48:30.018910
| 2020-03-07T11:59:49 |
354370
|
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|
Stack Exchange
|
Uniform bound on $\lVert \chi_{\{u_n=0\}}\rVert_{W^{s,p}(\Omega)}$ for a bounded sequence $u_n$ in $H^1_0(\Omega)$?
Suppose I have a sequence $u_n \to u$ in $H^1_0(\Omega)$ on a smooth and bounded domain. For some $p>1$ and $s \in (0,\frac 12)$, is it possible to estimate the norm of the characteristic function of the zero level set of $u_n$, $$\lVert \chi_{\{u_n=0\}}\rVert_{W^{s,p}(\Omega)}$$
in terms of norms of $u_n$? In particular I am looking for a uniform bound for the above expression.
We know it belongs eg to $W^{\epsilon, 2}(\Omega)$ for $\epsilon < \frac 12$, I just want to know if it can be bounded uniformly.
For the set ${u_n=0}$ to make sense, you need something, say continuity of $u_n$, which does not follow from your assumption of regularity when the dimension is $\ge 2$.
@Bazin let us assume $n \leq 3$ and $u_n, u$ belong to $H^2(\Omega)$, so they are all continuous functions
|
2025-03-21T14:48:30.019121
| 2020-03-07T12:33:15 |
354372
|
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|
Stack Exchange
|
Bordism groups of $X$, Thom isomorphism and characteristic numbers
Recap: bordism group
An oriented singular $n$-manifold in $X$is a map $f:M^n\to X$ where $M$ is a finite disjoint union of $n$-dimensional smooth manifolds.
The empty set is an admissible oriented singular $n$-manifold.
Two oriented singular manifolds $f_i:M_i^n\to X$ $i=1,2$ are bordant in $X$ if exists a map $F:W^{n+1}\to X$ ($W^{n+1}$ smooth oriented) such that $M_1\sqcup -M_2\subset \partial W$ (orientation is induced), $F|_{M_1\sqcup -M_2} \equiv f_1\sqcup f_2$.
This defines an equivalence relation and the equivalence classes $[M^n,f]$ of oriented singular $n$-manifolds up to bordism define the oriented $n$-bordism group of $X$ denoted $\Omega_n(X)$.
The (abelian) group structure is given by the operation $[M_1,f_1]+ [M_2,f_2] = [M_1\sqcup M_2, f_1\sqcup f_2]$.
If we forget all the orientation requirement above, we get the (unoriented) n-bordism group of $X$ denoted as $\eta_n(X)$.
With $\Omega_n$ is intended $\Omega_n(pt)$ i.e. $X={*}.$
Question 1:Thom's theorem
Thom discovered an isomorphism of groups:
$$\phi:\Omega_n \to \pi_{n+k}(MSO_k)$$
where $k>n+1$ and $MSO_k$ is the Thom space of the universal $SO_k$ bundle $ESO_k\to BSO_k$.
Roughly $\phi([M])$ is constructed from an embedding $M\to \mathbb{R}^{n+k}$, taking the classifying map for the normal bundle $\tau_\nu:M\to BSO_k$ and extending it to a map $\mathbb{R}^{n+k}\to ESO_k$ that is constant in a neighbourhood of $\infty$. This yelds a map $\mathbb{S}^{n+k}\to MSO_k$.
Question 1: This works for $\Omega_n(pt)$, what kind of isomorphism do we have for $\Omega_n(X)$? How is it defined? References?
I hope there is a geometric definition similar to the one I sketched above (that clarifies the relation with the normal bundle of an embedding for example).
Question 2: Thom-Pontryagin theorem on characteristic numbers
We have the following theorem: $[M]= 0 \in \Omega_n$, i.e. $M$ bounds an oriented manifold iff all the Stiefe-Whitney numbers and all the Pontryagin numbers of $M$ vanish:
$$\langle w_{i_1}(M)\cup \cdots \cup w_{i_{k'}}(M), [M]\rangle = 0\ \ \ \ \ \langle p_{i_1}(M)\cup \cdots \cup p_{i_k}(M), [M]\rangle = 0 $$
For $\eta_n(X)$ things are similar: $[M,f]= 0\in \eta_n(X)$ iff
$$\langle w_{i_1}(M)\cup \cdots \cup w_{i_k}(M)\cup f^*(h), [M]\rangle = 0 \in \mathbb{Z}_2\ \ \ \forall h \in H^*(X,\mathbb{Z}_2) \text{ and SW classes}$$
i.e. all SW numbers of the map $f$ vanish.
Curiously, for $\Omega_n(X)$ the only statement I've found requires an additional assumption: if all torsion classes in $H_*(X,\mathbb{Z})$ have order exactly $2$, then $[M,f]= 0 \in \Omega_n(X)$ iff the SW and Pontryagin numbers of $f$ vanish. Also usually it's required $X$ to be (h-equivalent to) a finite CW complex.
Question 2a: how to prove the statement for $\eta(X)$ and $\Omega(X)$? What are some good references? Is there any geometric interpretation of why the vanishing of the characteristic numbers entails the extendability of the map $f$?
Question 2b: Can we get rid of the assumption on the torsion of $H_*(X)?$ Why is it needed?
Question 3: What important achievements regarding bordism theory has been reached after the 60s?
Note: the reference I have used here is Conner P.E and E.E. Floyd: Differentiable Periodic Maps. Bull. Am. Math. Soc. 68, 76-86 (1962).
Question 1 and the second part of 2b have more or less trivial answer. A1. Oriented bordism is a generalized homology represented by the spectrum MSO. You can find this in any decent textbook of cobordism. For example Rudyak. A2b') $H_(*,\mathbf{Z})$ has no torsion. I have a feeling that the first part of the question 2b should have a trivial answer, if we are looking for an example of the pair $([M,f],X)$ and not an example of $[M,f]$ for fixed $X$.
For question 2b, the answer is that elements of $H^n(X; A)$ determine bordism invariants
$\Omega_n^{\mathrm{SO}}(X)\to A$, and if $H^*(X)$ contains $p$-torsion for $p$ odd, these can't be interpreted as
Stiefel-Whitney or Pontrjagin numbers. A simple example is $\Omega_1^{\mathrm{SO}}(B\mathbb Z/3)$, the bordism
group of oriented 1-manifolds with a principal $\mathbb Z/3$-bundle. A circle with a nontrivial principal $\mathbb
Z/3$-bundle does not bound, but all Stiefel-Whitney and Pontrjagin numbers here vanish. In fact pushing the
fundamental class forward defines an isomorphism $\Omega_1^{\mathrm{SO}}(B\mathbb Z/3)\to H_1(B\mathbb
Z/3)\cong\mathbb Z/3$.
Question 2a is harder: the proofs are generally not geometric, but rather homotopical. One takes the spectrum
$\mathit{MSO}$ (built from the spaces $\mathit{MSO}_k$) and splits it as a wedge sum of other spectra we
understand. Such a splitting entails a lot of messy calculations with the Steenrod algebra and as far as I know
there isn't a geometric proof. For unoriented bordism, Thom split $\mathit{MO}$ splits as a wedge sum of shifts of
Eilenberg-Mac Lane spectra. The isomorphism is built out of the cohomology of $\mathit{BO}$, which ultimately means
it can be interpreted as sending a manifold to its Stiefel-Whitney numbers. I don't know what the best reference
is, but Thom's thesis is the original reference.
For oriented bordism, the analogous splitting result holds only after localizing. Thom showed that if you
rationalize, you can split $\mathit{MSO}$ into a sum of Eilenberg-Mac Lane spectra, and the map sends a manifold to
its Pontrjagin numbers.
If you complete at 2, the calculation is harder; there is
again a splitting, interpretable as sending a manifold to its Pontrjagin and Stiefel-Whitney numbers, and I think
this was shown by Wall, again homotopically rather than geometrically.
(At odd primes, one uses a generalized cohomology theory called Brown-Peterson cohomology to split $\mathit{MSO}$, and again the proof is homotopical.) I don't know of a good reference digging
into these calculations other than the original papers, but Manifold
Atlas has a good summary and list of references.
For question 3, some of the most exciting achievements have been the study of bordism categories, such as the
work of Galatius-Madsen-Tillmann-Weiss on the homotopy types of bordism categories; the cobordism hypothesis; and
some of the applications to topological field theory, homological stability, and so on. These results use and
strengthen Pontrjagin-Thom theory.
|
2025-03-21T14:48:30.019505
| 2020-03-07T14:51:29 |
354378
|
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|
Stack Exchange
|
Relative de Rham Cohomology groups of k-algebra
Let $A$ be a commutative unital $k$-algebra. Then we have de Rham complex given as:
$C_{\ast}(A)$ : $ 0 \rightarrow A \rightarrow \Omega_{A \lvert k}^{1} \rightarrow \Omega_{A \lvert k}^{2} \rightarrow ... $ and de Rham Cohomology group $H_{dR}^{\ast}(A \lvert k)$ is defined as the cohomology group of this complex. I want to know is there a definition of relative de Rham cohomology groups($H_{dR}^{\ast}(A\lvert k, I))$ of $A$ with respect to some ideal $I$? I don't know any reference of this definition. I need this notion because I want to relate these groups with the relative cyclic homology groups $HC_{\ast}(A\lvert k, I)$.
I am guessing that these groups is defined as the cohomology group of complex $C_{\ast}(A, I)$, where the complex $C_{\ast}(A, I)$ is defined by the following exact sequence of complex. $ 0 \rightarrow C_{\ast}(A, I) \rightarrow C_{\ast}(A) \rightarrow C_{\ast}(A/I)$. Tell me whether my guess is correct or not. It would be great if someone can give reference for this literature?
|
2025-03-21T14:48:30.019603
| 2020-03-07T15:42:24 |
354383
|
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"jack"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354383"
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|
Stack Exchange
|
Show that $(\sum_{k=1}^{n}x_{k}\cos{k})^2+(\sum_{k=1}^{n}x_{k}\sin{k})^2\le (2+\frac{n}{4})\sum_{k=1}^{n}x^2_{k}$
Let $x_{1},x_{2},\cdots,x_{n}>0$, show that
$$\left(\sum_{k=1}^{n}x_{k}\cos{k}\right)^2+\left(\sum_{k=1}^{n}x_{k}\sin{k}\right)^2\le \left(2+\dfrac{n}{4}\right)\sum_{k=1}^{n}x^2_{k}$$
This question was posted some time ago at MSE. A bounty was placed on it, but no complete solution was received.
The only solution there claims to solve the problem when $n \le 10^9$.
It's easy to see that the inequality can be proved when $(2+n/4)$ is replaced by $n$ (in fact, this follows directly from Cauchy-Schwarz inequality).
Also the LHS is equal to: $\sum_{k=1}^n x_k^2 + 2\sum_{i < j} x_ix_j\cos(i - j)$.
I'm looking for a proof or any reference of this result.
Any help would be appreciated.
It looks true, but very hard, depending on the estimates on irrationality measure of $\pi$. Do you see any evidence for it?
As a side question: if this is true, the same should then morally/logically also hold if we multiply all arguments with a real constant $\lambda$, i.e. consider $A=(\cos(\lambda(i-j)))_{i,j=1\ldots n}$ instead. Of course $\lambda$ must not be a rational multiple of $\pi$, though.
@FedorPetrov So far, only computer evidence. This problem is very hard indeed. I don't know any other reference for it apart from MSE.
I have added a theorem to my answer that shows that the $n/4$ dependence of the RHS is correct conditional on the irrationality measure of $\pi$ being less than 3.
However even if we knew that $\mu(\pi)$ was less than 3 this doesn't allow us to bound the constant since I think there still could be a finite set of rational approximations with $|\pi-p/q|<1/q^m$ for large m that could affect the value of $\sum_{k=1}^{\infty} \frac{1}{k^3|\sin(k)|}$.
It is a bit long for a comment.
Your question is about the matrix $A=(\cos((i-j)))_{i,j=1\ldots n}$, specifically, the maximum of the quadratic form $q(x)=(Ax,x)$ on the subset $M_+$ of the unit sphere $(x,x)=1$ consisting of the vectors with positive coordinates.
Since the sphere is compact, the maximum is attained, but in general it would be attained in the closure of the set $M_+$. So in fact, we have some subset $I\subset\{1,\ldots,n\}$ where the desired point has zero coordinates, and its complement where the coordinates are positive, and the correct Lagrange multiplier problem involves the function
$$
(Ax,x)-\lambda((x,x)-1)-\sum_{i\in I}\mu_i x_i .
$$
Taking derivatives gives us conditions
$$
Ax-\lambda x-\frac12\sum_i \mu_i e_i=0,\quad (x,x)=1, \quad x_i=0 \text{ for } i\in I,
$$
or in plain language, $x$ is a unit vector that is a combination of vectors $e_i$ for $i\notin I$, and such that $Ax$, up to a vector proportional to $x$, is a combination of vectors $e_i$ for $i\in I$.
I did not follow it through fully, but I convinced myself that the maximum is indeed contained in a nontrivial boundary point of $M_+$. In fact, it is possible to describe eigenvalues and eigenvectors of $A$ in a very cute way. Consider the matrix
$$
A(t)=(\cos((i-j)t))_{i,j=1\ldots n}
$$
Suppose that $t$ is not a rational multiple of $\pi$. First, I claim that this matrix has $n-2$ zero eigenvalues. To show that, I shall exhibit $n-2$ linearly independent elements in the kernel (here $e_i$ are standard unit vectors of $\mathbb{R}^n$):
$$
v_i=\sin(t)e_i-\sin((n-i)t)e_{n-1}+\sin((n-1-i)t)e_n,\quad i=1,\ldots,n-2 .
$$
Next, I claim that two remaining eigenvalues are
$$
\frac{n}{2}\pm\sum_{j=1}^k\cos((2j-1)t)
$$
for even $n=2k$ and
$$
\frac{n-1}{2}+\sum_{j=1}^{k-1}\cos(2jt),\quad
\frac{n+1}{2}-\sum_{j=1}^{k-1}\cos(2jt)
$$
for odd $n=2k+1$.
Moreover, the corresponding eigenvectors are as follows
$$
\sum_{j=1}^n \cos((k-j)t)e_j,\quad
\sum_{j=1}^n \sin((k-j)t)e_j
$$
for even $n=2k$
$$
\sum_{j=1}^n (\sin((k-j)t-\sin((k-1-j)t))e_j,\quad
\sum_{j=1}^n (\sin((k-j)t+\sin((k-1-j)t))e_j
$$
for odd $n=2k+1$. All these claims are checked by a rather direct calculation.
Now, if the maximum of $(Ax,x)$ were attained at an interior point, we would need to look at eigenvectors, and the maximum would be given by the respective eigenvalue. Vectors of the kernel do not interest us, clearly. From the explicit formulas above, the other two eigenvectors almost immediately start having negative coordinates too (some silly exceptions for small $n$ exist).
Not sure if it is easy to finish this, I might revisit it later.
It is possible to prove that the inequality holds for sufficiently large $n$ with $1/4$ replaced by $1/4+\epsilon$ for any $\epsilon>0$ .
(Update - see below for a stronger result conditional on the irrationality measure of $\pi$)
In fact we have:
Theorem
For any $\epsilon>0$ and large enough $n$, depending on $\epsilon$ only, the following inequality holds for any real $a_i\geq 0$
$$|\sum_{k=1}^n a_k e^{ik}|\leq (1/2+\epsilon) \sqrt n (\sum_{k=1}^{n}a_k^2)^{1/2}.$$
Proof
We know that $$|\sum_{k=1}^n a_k e^{ik}| = e^{ i\theta}\sum_{k=1}^n a_k e^{ik}$$ for some $\theta \in [0,2\pi]$. Hence
$$|\sum_{k=1}^n a_k e^{ik}| = \sum_{k=1}^n a_k e^{i(k+\theta)}=\sum_{k=1}^n a_k\cos(k+\theta).$$
Define a function $ \cos_{+}:\mathbb{R}\to[0,1]$ by $\cos_{+}(x)=\cos(x)$ if $\cos(x)\geq 0$ and $0$ otherwise.
Then using $a_k\geq0$ and Cauchy's Theorem,
$$\sum_{k=1}^n a_k\cos(k+\theta)\leq\sum_{k=1}^n a_k\cos_+(k+\theta)\leq(\sum_{k=1}^n a_k^2)^{1/2}(\sum_{k=1}^n {\cos_+}^2(k+\theta))^{1/2}.$$
Now the equidistribution theorem says that $\{\frac{i}{2\pi}\}$ is uniformly distributed in $[0,1]$. Hence by the Riemann integral criterion for equidistribution,
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} f\left(s_k\right) = \frac{1}{b-a}\int_a^b f(x)\,dx$$
with $f(x)={\cos_+}^2(2\pi x+\theta)$, $s_k=\{\frac{k}{2\pi}\}$ and $a=0$, $b=1$.
In other words,
\begin{equation}
\begin{split}
& \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n {\cos_+}^2(k+\theta)=\int_0^1 {\cos_+}^2(2\pi x+\theta)\,dx =\frac{1}{2\pi}\int_{\theta}^{2\pi+\theta} {\cos_+}^2(\phi)\,d\phi \\ & =\frac{1}{2\pi}\int_{-\pi/2}^{+\pi/2} \cos^2(\phi)\,d\phi=1/4.
\end{split}
\end{equation}
Hence
$$\frac{|\sum_{k=1}^n a_k e^{ik}|}{\sqrt n (\sum_{k=1}^n a_k^2)^{1/2} }\leq \left(\frac{1}{n}\sum_{k=1}^n {\cos_+}^2(k+\theta)\right)^{1/2}\rightarrow 1/2 \ \text{ as } \ n\rightarrow \infty$$
and the result follows.
$\blacksquare$
In addition I can prove the following estimate, conditional on the irrationality measure of $\pi$ being less than 3:
Theorem
If the irrationality measure for $\pi$, $\mu(\pi)$ is strictly less than 3 then $$\frac{|\sum_{k=1}^n a_k e^{ik}|^2}{ \sum_{k=1}^n a_k^2 }\leq\frac{n}{4}+D.$$
for some fixed constant $D$.
Proof
We know from the above that
$$|\sum_{k=1}^n a_k e^{ik}|^2\leq (\sum_{k=1}^n a_k^2) \sum_{k=1}^n {\cos_+}^2(k+\theta)$$ for some $\theta \in [0,2\pi]$.
Also, ${\cos_+}^2(x) = (\cos^2 x+\cos x|\cos x|)/2$.
Hence $$\sum_{k=1}^n {\cos_+}^2(k+\theta)=\sum_{k=1}^n (\cos^2 (k+\theta)+\cos (k+\theta)|\cos (k+\theta)|)/2\\=\frac{1}{2}\sum_{k=1}^n \cos^2 (k+\theta)+\frac{1}{2}\sum_{k=1}^n (\cos (k+\theta)|\cos (k+\theta)|)$$.
Clearly $\sum_{k=1}^n \cos^2 (k+\theta)\leq n/2+B$ for some constant $B$ so it remains to bound the other term which is more complicated.
Let $f(x) = |\cos{(x)}| \cos{(x)}$. Here it is noted by Andreas that the expression $|\sin{(x)}| \sin{(x)}$ can be written as a Fourier series, which we modify to provide a series for $f$,
$$
f(x) = \frac{8}{\pi}\sum_{m=0}^\infty \frac{(-1)^m}{4(2m+1)-(2m+1)^3} \cos((2m+1)x).
$$
Now we can sum
$$
S_n = \sum_{k=1}^{n} f{(k + \phi)} = \frac{8}{\pi}\sum_{m=0}^\infty \frac{(-1)^n}{4(2m+1)-(2m+1)^3} \sum_{k=1}^{n}\cos((2m+1)(k + \phi))
$$
where
$$
\sum_{k=1}^{n}\cos((2m+1)(k + \phi)) =
\frac{\sin(n(m + \frac12)) \cdot \cos((1 + n + 2 \phi)(m + \frac12)) }{ \sin(m + \frac12)}.
$$
Thus
$$
S_n = \frac{8}{\pi}\sum_{m=0}^\infty \frac{(-1)^n}{4(2m+1)-(2m+1)^3} \frac{\sin(n(m + \frac12)) \cdot \cos((1 + n + 2 \phi)(m + \frac12)) }{ \sin(m + \frac12)}.
$$
Taking absolute values and using the triangle inequality
$$
|S_n| \leq \frac{8}{\pi}\sum_{m=0}^\infty \frac{1}{|4(2m+1)-(2m+1)^3|\sin(m + \frac12)|}\\=\frac{8}{\pi}\sum_{m=0}^\infty \frac{|2\cos (m+\frac12)|}{|4(2m+1)-(2m+1)^3||\sin(2 m + 1)|}
\\ \leq \frac{16}{\pi}\sum_{m=0}^\infty \frac{1}{|(2m+1)^3-4(2m+1)||\sin(2 m + 1)|}
\\ =\frac{16}{3 \pi \sin 1}+\frac{16}{15 \pi \sin 3}+\frac{16}{\pi}\sum_{m=2}^\infty \frac{1}{|(2m+1)^3-4(2m+1)||\sin(2 m + 1)|}
\\ \leq \frac{16}{3 \pi \sin 1}+\frac{16}{15 \pi \sin 3}+\frac{16}{\pi}\sum_{m=2}^\infty \frac{1.3}{(2m+1)^3|\sin(2 m + 1)|}
$$
To complete the estimate we note that Theorem 5 of Max A. Alekseyev's paper "On convergence of the Flint Hills series" implies that if $\mu(\pi)<3$ then $\sum_{n=1}^{\infty}\frac{1}{n^3|\sin n|}$ converges, hence $\sum_{n=1}^{\infty}\frac{1}{(2n+1)^3|\sin (2n+1)|}$ converges also and we have
$$|S_n| \leq C$$ for some fixed constant $C$. Combining the two estimates we have on the assumption that $\mu(\pi)<3$,
$$\sum_{k=1}^n {\cos_+}^2(k+\theta)=\frac{1}{2}\sum_{k=1}^n \cos^2 (k+\theta)+\frac{1}{2}\sum_{k=1}^n (\cos (k+\theta)|\cos (k+\theta)|)\\ \leq n/4+(B+C)/2$$
and the result is proved. $\blacksquare$
Unfortunately although most irrational numbers have irrationality measure 2, and this is probably the true value of $\mu(\pi)$, the best upper bound for $\mu(\pi)$ is 7.103205334137 due to Doron Zeilberger and Wadim Zudilin - see here for their paper so we are a long way from being able to prove the inequality this way at least.
it is weaker than needed
@FedorPetrov Do you mean because it's only proved for sufficiently large n?
It is not proved for large $n$, what you prove is $n/4+o(n)$ which is a weaker upper bound than $n/4+2$.
@FedorPetrov Thank you for clarifying - yes I understand - you are absolutely right we don't get the OP's inequality, not quite anyway. I will update my answer to reflect your comment. I do have a more complicated method that I think will give the result precisely for large n and I'll post that if I have time later.
This was already noted in the linked question (Bounding a Fourier ...)
@FedorPetrov I have added a stronger result but conditional on the irrationality measure of $\pi$ being less than 3.
Too long for a comment. Let us consider for $X\in \mathbb S^{n-1}$,
$
\langle X,(e^{i \alpha k})_{1\le k\le n}\rangle_{\mathbb C^n}.
$
The question at hand is
$$
\max_{X\in \mathbb S^{n-1}}\vert\langle X,(e^{i \alpha k})_{1\le k\le n}\rangle_{\mathbb C^n}\vert\le \sqrt{2+\frac n4},
\tag{$\ast$}$$
for $\alpha =1$.
Of course, the above inequality does not hold when $\alpha =π$ since in that case ($\ast$) means
$$
\max_{X\in \mathbb S^{n-1}}\vert\langle X,((-1)^k)_{1\le k\le n}\rangle_{\mathbb C^n}\vert\le \sqrt{2+\frac n4}
$$
and choosing
$
X_{2l+1}=0, X_{2l}>0,
$
the lhs is $\sqrt{n/2}$.
The question for a given $\alpha$ is related to the closedness in $\mathbb C^n$ of the vector $(e^{i \alpha k})_{1\le k\le n}$ to a real-valued unit vector (i.e. in $\mathbb S^{n-1}$). In fact one may write
$$
\langle X,\underbrace{(e^{i \alpha k})_{1\le k\le n}}_{\mathbf e_\alpha \sqrt n}\rangle_{\mathbb C^n}=\frac14\Vert X+\mathbf e_\alpha\sqrt n\Vert^2_{\mathbb C^n}-\frac14
\Vert X-\mathbf e_\alpha\sqrt n\Vert^2_{\mathbb C^n},
$$
so that $(\ast)$ means that for $X\in \mathbb S^{n-1}$
$$
\vert\langle X,\mathbf e_\alpha \rangle_{\mathbb C^n}\vert\le\sqrt{\frac 2n+\frac 14,}\quad \text{i.e.}\quad \cos(\mathbb S^{n-1}, \mathbf e_\alpha )
\le\sqrt{\frac 2n+\frac 14}.
$$
Passing to the limit when $n\rightarrow+\infty$ we would get
$
\text{angle}(\mathbb S^{n-1}, \mathbf e_\alpha )\ge π/3.
$
Going back to the case $\alpha =1$ the subgroup $\{e^{ik}\}_{k\in \mathbb Z}$ is dense in the unit circle (it cannot be discreet), which implies that for a given $\epsilon >0$, you can find infinitely many $k\in \mathbb Z$ such that
$
\vert e^{ik}-1\vert <\epsilon.
$
This is probably not enough to violate your property, since what would be needed is a fixed proportion (more that $1/4$) of integers like this, but the property at hand seems related to diophantine properties of the above subgroup.
The condition $x_i>0$ is important, and you seem to have ignored it. If one allows all possible $x_i$, it is easy to get to values approximately $n/2$.
|
2025-03-21T14:48:30.020259
| 2020-03-07T17:41:57 |
354388
|
{
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"Sitan Chen",
"https://mathoverflow.net/users/153290",
"https://mathoverflow.net/users/35520",
"ofer zeitouni"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/354388"
}
|
Stack Exchange
|
L2 norm of the diagonal entries of a random rotation of a fixed matrix?
Let $X\in\mathbb{R}^{d\times d}$ be the diagonal matrix with $d/2$ entries equal to $1$ and $d/2$ entries equal to $-1$. Let $F_U \triangleq \frac{1}{d}\|\operatorname{diag}(U^{\dagger}XU)\|^2_F$ denote the sum of squares of the diagonal entries of $U^{\dagger}XU$. If $U$'s columns consisted of independent random complex unit vectors, then we know that $\mathbb{E}[F_U] = \frac{1}{d+1}$, and $\Pr\left[\left|F_U - \frac{1}{d+1}\right| > t \right] \le e^{-\Omega(t/d^{3/2})}$.
Now suppose $U$ is Haar-random unitary, in which case we still have that $\mathbb{E}[F_U] = \frac{1}{d+1}$. My question is: does $F_U$ concentrate comparably in this case?
Weingarten calculus tells us that $$\mathbb{E}[F_U^n] = \sum_{\substack{\sigma,\tau\in \mathbb{S}_{2n}: \\ \tau \text{ has only even cycles}}} d^{\kappa(\tau)}\cdot \operatorname{Wg}(\sigma\tau^{-1},d)\cdot \Pr_{s\in[d]^n}[\sigma \text{ fixes the word } (s_1,s_1,s_2,s_2,\ldots,s_n,s_n)],$$ where $\kappa(\cdot)$ denotes number of cycles, but it is not clear to me how to get a good estimate for this quantity for large $n$. Alternatively, perhaps a stretch, but is there some HCIZ-esque way of computing the MGF of $F_U$?
The concentration you state in the independent column case is very weak. Are you sure about the normalization? Shouldn't it be $e^{-ct^2d^2}$, which would be consistent with the variance of $F_U$? In any case, have you tried to apply Corollary 4.4.28 in Anderson-Guionnet-Zeitouni's book? I think this is what it gives.
Yes, I messed up the normalization. Because $U\mapsto \frac{1}{\sqrt{d}}|\text{diag}(U^{\dagger}XU)|_F^{1/2}$ is $O(1/\sqrt{d})$-Lipschitz, Corollary 4.4.28 will indeed give $e^{-\Omega(t^2d^2)}$ tails. Thanks very much!
|
2025-03-21T14:48:30.020408
| 2020-03-07T17:49:17 |
354390
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/12481",
"https://mathoverflow.net/users/44191",
"joro",
"user44191"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/354390"
}
|
Stack Exchange
|
When is $\lfloor C^n \rfloor \mod b$ efficiently computable?
For real irrational $C > 1 $ and natural $n,b$, define
$a(C,n,b)=\lfloor C^n \rfloor \mod b$
Q1 For which $C,b$ is $a(C,n,b)$ computable in time polynomial
in $\log{n}$?
Searching in OEIS suggests that for $C \in \{1+\sqrt{2},1+\sqrt{3},(1+\sqrt{5})/2\}$, $a(C,n,b)$ satisfy linear recurrence with constant coefficients and
so it is efficiently computable over the integers and all bases $b$.
In OEIS:
a(n) = floor(phi^n)
a(n) = floor((1+sqrt(2))^n)
a(n) = floor((1+sqrt(3))^n)
For natural $k$, $a(b^{1/k},n,b)$ is related to the base
$b$ representation of $b^{1/k}$ so it is probably hopeless.
Q2 Is $a(1+\sqrt{6},n,b)$ efficiently computable in some base $b$?
(We couldn't find linear recurrence for it)
Q3 Except linear recurrences, are there other islands of tractability
for algebraic $C$?
In comments @user44191 asked about specific constant near $1.75$.
We couldn't find linear recurrence, but got degree 2 relation
factoring into linear factors, which might be hint:
0 == (2*a(n + 2) - 3*a(n + 1) - 3*a(n - 1) - a(n - 3) + a(n) - 2) *
(a(n + 1) - a(n - 1) - a(n - 3) - a(n) - 1)
Computational bugs are possible.
For algebraic $\alpha$ such that $|\alpha| > 1, |\alpha'| < 1$ for all conjugates $\alpha'$, I suspect it should be easily computable; this is true for the examples given.
I should note that I assumed $\alpha$ is an algebraic integer. The above may be somewhat overly general; a sufficient restriction is that there is a unique conjugate of second-largest magnitude (which, necessarily, will be real). In that case, there will be a clear linear recurrence. It may still be possible to get a linear recurrence otherwise; I don't have any examples with 2 roots of the 2nd largest magnitude off the top of my head, but it might be worth investigating those.
An example to test: the unique positive root of $x^3 - 2x^2 + x - 1$; I think that computing $\lfloor C^n \rfloor$ is equivalent to determining whether $n \text{arg}(\alpha')$ has positive cosine, where $\alpha'$ is either nonpositive root.
@user44191 I edited with partial results about the test.
For the $\alpha$ I wrote above, there is a degree 6 relation that use only 4 terms that's true for sufficiently large $n$: if $b(n) = a(n + 3) - 2 a(n + 2) + a(n + 1) - a(n)$, then $\prod_{i = -2}^3 (b(n) - i) = 0$. This is because $b(n)$ is an integer and because $a(n) - \alpha^n$ is either very close to $0$ or very close to $-1$.
@user44191 Could there be a pattern in "c(n)=a(An+B)"?
|
2025-03-21T14:48:30.020586
| 2020-03-07T18:34:07 |
354392
|
{
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"authors": [
"Federico Poloni",
"Manfred Weis",
"https://mathoverflow.net/users/115044",
"https://mathoverflow.net/users/1898",
"https://mathoverflow.net/users/31310",
"pregunton"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/354392"
}
|
Stack Exchange
|
Name for matrix associated to smooth continuation
Is there an established name for the matrices that establish the conditions for a linear combination of $n$ functions $\lbrace f_1(x),\dots,f_n(x)\rbrace$ being the $n$-times smoothly differentiable continuation of a function $g(x)$ for $x=x_0$ after adding an appropriate constant:
$$\begin{pmatrix}\frac{d}{dx}f_1(x_0)&\dots&\frac{d}{dx}f_n(x_0)\\ \vdots & &\vdots\\ \frac{d^n}{dx^n}f_1(x_0) & \dots & \frac{d^n}{dx^n}f_n(x_0)\end{pmatrix}\begin{pmatrix}c_1 \\ \vdots\\ c_n\end{pmatrix} = \begin{pmatrix}g'(x_0)\\ \vdots\\ g^{(n)}(x_0)\end{pmatrix}$$
According to this Wikipedia page, it would be the derivative of what is sometimes called a "fundamental matrix".
@pregunton thank you very much, that ianswers my question
@pregunton Could you formulate it as an answer? Answering in comment is not a good practice on Stack Exchange sites. (In addition, you'll get internet points for it your answer.)
@FedericoPoloni My apologies. I expanded the comment into an answer.
No worries, and thanks!
According to this Wikipedia page, the matrix in the question is the derivative of what is sometimes called a fundamental matrix. A Google search of that term together with "Wronskian" gives quite a few relevant hits, so the name seems to be in common use.
|
2025-03-21T14:48:30.020705
| 2020-03-04T08:19:44 |
354122
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"Mtheorist",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354122"
}
|
Stack Exchange
|
Solving equations of motion of holomorphic BF theory - pure gauge in complex coordinates
In this paper by Bailieu and Tanzini, aspects of holomorphic BF theory are presented.
Holomorphic BF theory on a four dimensional Kahler manifold is discussed from page 5, and on page 8 the following equations of motion are derived from it:
$$
F_{\bar{m}\bar{n}}=0,~~~~~\epsilon^{mn\bar{m}\bar{n}}D_{\bar{n}}B_{mn}=0.
$$
The authors claim that these equations imply that classically, $A_m$ is undetermined, $A_{\bar{m}}$ is pure gauge, and $B_{mn}$ is holomorphic.
My question is, why is $A_{\bar{m}}$ definitely pure gauge?
Now, the pure gauge configuration $A_{\bar{m}}=-\partial_{\bar{m}}gg^{-1}$ certainly appears to be a solution of $F_{\bar{m}\bar{n}}=0$.
However, in general, flat connections can only be written as pure gauge in simply-connected regions of the underlying manifold. (This is because to gauge transform a connection, $A$, to zero, the group element, $U$, required is the parallel transport of $A$ along a curve, and to show that $U$ does not depend on the curve, the manifold must be simply-connected. Further details can be found in the answer to this question.)
The authors do not seem to be assuming that the Kahler manifold at hand is simply-connected.
Therefore, why is $A_{\bar{m}}$ pure gauge? Does this follow somehow because they use complex coordinates?
Cross-post at https://physics.stackexchange.com/questions/533613/solving-equations-of-motion-of-holomorphic-bf-theory-pure-gauge-in-complex-coo
|
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