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2025-03-21T14:48:31.898371
| 2020-08-24T08:59:26 |
369984
|
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"Dylan Wilson",
"Surojit Ghosh",
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|
Stack Exchange
|
A question on recognition of equivariant loop spaces
I have a question about equivariant loop space that has been bothering me, and that I have not been able to find an answer to in the obvious places.
We know from the work of Segal that to give a loop space structure on $X$ is equivalent to producing a simplicial space $X_\bullet$ in which $X_0$ is weakly contractible, $X_1$ is weakly equivalent to $X$ and the map $X_n \to (X_1)^n$, corresponding to the order-preserving inclusion $[1] \to [n]$ taking $0$ to $0$, is a weak equivalence. (see Proposition 1.5 of the article CATEGORIES AND COHOMOLOGY THEORIES by G. Segal)
For the $n$-fold loop spaces case one may see the work of Peter Cobb. The approach is in the same spirit as G. Segal's investigating of the infinite loop spaces via special $\Gamma$-spaces.
Can we have a (similar) description for the equivariant loop space $\Omega^V X$ where $X$ is $G$-space and $V$ is a $G$-representation?
Thank you so much in advance. Any help will be appreciated.
whoa! Thanks for pointing out Cobb's work... it looks like he discovered Joyal's category Theta_n and proved results like Berger did, but way back in 1974! (or am I misreading?) I don't have an answer to your question, but my suspicion is that only (reduced) permutation representations would be amenable to a combinatorial description like Cobb's
Thank you so much for your comment. Do you think that it follows from Shimakawa's work https://www.semanticscholar.org/paper/Infinite-Loop-G-Spaces-Associated-to-Monoidal-Shimakawa/c357074145e0c6f52054f29a0a84a130e1f0c9fa
|
2025-03-21T14:48:31.898497
| 2020-08-24T09:51:46 |
369988
|
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"GH from MO",
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}
|
Stack Exchange
|
The Fermat-Catalan conjecture with signature $(2,n,4)$, $n\ge4$
The Fermat-Catalan conjecture is that for coprime $x,y,z$ and positive integers $a,b,c$ with $1/a+1/b+1/c<1$, the generalized Fermat equation $x^a + y^b = z^c$ has only finitely many solutions. I'm considering signatures $(a,b,c)$ which are solved.
Table 1 of [BCDY] surveys known results and states that $(2,n,4)$, $n\ge4$ has been solved completely and that this is 'Immediate from Bennett–Skinner [BS], Bruin [Br3]'. [Br3] covers the case $n=5$. Fermat dealt with $n=4$.
This leaves $n=6, 9$ and prime $n\ge7$, but I can't see how [BS] is relevant to that. Can someone explain and/or point me to the relevant part of [BS].
[BCDY] 'Generalized Fermat equations: A miscellany', Bennett, Chen, Dahmen, Yazdani, International Journal of Number Theory, Vol. 11, No. 1 (2015)
[BS] 'Ternary Diophantine Equations via Galois Representations and Modular Forms', Bennett, Skinner, Canad. J. Math. Vol. 56(1), 2004 p23-54.
[Br3] 'Chabauty methods using elliptic curves', Bruin, J.reine angew. Math. 562 (2003), 27-49.
Note: This question was originally posted in MSE on 2020-07-03. It's had some upvotes, but no answers as of 2020-08-24.
This is a pretty good question. I suggest that you contact Bennett, and share his answer here.
[Br2] Theorem 1 covers the case $n=6$. So this leaves $n=9$ and prime $n\ge7$.
As suggested in a comment, I contacted Michael Bennett directly and he kindly explained the rest to me:
We have $x^2+y^n=z^4$ with $x,y,z$ coprime integers.
So $(z^2 - x)(z^2 + x) = y^n$.
The gcd of $(z^2 - x)$ and $(z^2 + x)$ is $1$ or $2$.
For a gcd of $1$ we have $x$ and $z$ of opposite parity, and can write
$z^2-x = u^n$ and $z^2+x = v^n$
so that $u^n+v^n = 2z^2$.
This is solved for coprime integer $u,v,z$ for $n\ge4$ by [BS] Theorem 1.1.
For a gcd of $2$ we have $x$ and $z$ both odd, and one of
$z^2-x = 2 u^n$ and $z^2+x = 2^{n-1}v^n$, or
$z^2+x = 2 u^n$ and $z^2-x = 2^{n-1}v^n$.
In either case, $u^n + 2^{n-2} v^n = z^2$.
This is solved for coprime integer $u,v,z$ for prime $n\ge7$ by [BS] Theorem 1.2.
This leaves the case $n=9$ with $y$ even.
Going back to the original equation, we have
$x^2+y^9=z^4$.
[Co, Section 14.4.1] gives complete parametrizations of
$x^2+w^3=z^4$
in terms of $s$ and $t$. In our case, $w$ is an even cube and from this and the parity constraints on $s$ and $t$ given in [Co] it follows that there exist coprime integers $s$ and $t$ with
$s t (s^3 - 16 t^3) (s^3 + 2 t^3)$ a cube.
The factors on the left hand side are pairwise coprime. ($s^3-16t^3$ and $s^3+2t^3$ could possibly have a common factor of $3$, but if they do, the whole expression is divisible by $9$ but not by $27$, and hence is not a cube.)
Since the factors are pairwise coprime, $s^3+2t^3$ is a cube.
This corresponds to a rational point on the curve $A^3+2B^3$=1 which is isomorphic to the elliptic curve $Y^2=X^3-1728$ via standard transformations. The latter curve has rank $0$ (and only the rational points corresponding to the point at infinity and $(X,Y)=(12,0)$). Tracing these back to $A^3+2B^3=1$, we find that $(A,B)=(1,0)$ or $(A,B)=(-1,1)$.
These points lead to either $t=0$ (which gives $z=0$ in $x^2+y^9=z^4$) or to $st = -1$ (which does not make $s t (s^3 - 16 t^3) (s^3 + 2 t^3)$ equal to a cube).
[Br2] 'The Diophantine Equations $x^2 \pm y^4 = \pm z^6$ and $x^2 + y^8 = z^3$', Bruin, Compositio Mathematica 118: 305-321, 1999.
[Co] 'Number Theory Volume II: Analytic and Modern Tools', Henri Cohen
Well done! Thank you.
|
2025-03-21T14:48:31.898725
| 2020-08-24T10:47:08 |
369990
|
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"Bazin",
"Fedor Petrov",
"Giorgio Metafune",
"Karl Fabian",
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|
Stack Exchange
|
Estimate for an Airy integral
Let me define for $x\in\mathbb R$,
$
F(x)=\int_{\mathbb R} e^{-π t^2}\cos(x t^3) dt.
$
I claim that $F(x)>0$ for all $x\in\mathbb R$.
Well, it is obvious for $x=0$ since $F(0)=1$ and also for $x$ near $0$ by continuity of $F$.
I guess that for $x$ large, a van der Corput method or a version of the stationary phase method should give the result. How can I prove this at "finite distance", in a situation where no asymptotic method could help?
Mathematica calculates the integral as
$$ \frac{2 \pi e^{\frac{2 \pi ^3}{27 x^2}} K_{\frac{1}{3}}\left(\frac{2 \pi ^3}{27 x^2}\right)}{3 \sqrt{3} \left| x\right| }$$
which is positve because the modified Bessel function $K_\nu(z)$ is.
The most elementary way is to exploit the periodicity and compare positive and negative contributions. Changing variable in the integral, with $\tau=xt^3$, and integrating by parts, you will find an analogous statement for
$$\int_0^\infty f(\tau)\sin(\tau)d\tau>0,$$
for some positive and decreasing function (also depending on $x$) $f$, which makes the claim evident since any positive contribute on $[2k\pi,(2k+1)\pi]$ is larger in absolute value than the successive negative contribution on $[(2k+1)\pi,(2k+2)\pi]$.
alternatively, we may use $\sin \tau d\tau=d(1-\cos \tau)$ that yields $\int_0^\infty f(\tau)\sin \tau d\tau=-\int_0^\infty f'(\tau)(1-\cos \tau) d\tau\ge 0$
so a double integration by parts, even better
yes, and may be done without change of variables (here this is a matter of taste, but when the change of variables is not so explicit, it looks more suitable), I posted the computation as an alternative answer
Very nice, thanks for both answers.
Similarly $\int_0^\infty f(t) \cos t, dt \ge 0$ if $f \ge0$ is decreasing and convex.
$$
\frac{F(x)}2=\int_0^\infty e^{-\pi t^2}\cos(x t^3) dt=\int_0^\infty \frac{e^{-\pi t^2}}{3xt^2}d\sin(x t^3)=-\int_0^\infty \left(\frac{e^{-\pi t^2}}{3xt^2}\right)'\sin(x t^3)dt\\=\int_0^\infty \frac1{3xt^2}\left(\frac{e^{-\pi t^2}}{3xt^2}\right)'d\left(1-\cos(x t^3)\right)=-\int_0^\infty \left(\frac1{3xt^2}\left(\frac{e^{-\pi t^2}}{3xt^2}\right)'\right)'\left(1-\cos(x t^3)\right)dt\\=
\int_{0}^\infty\frac{2e^{-\pi t^2}(2\pi^2t^4+5\pi t^2+5)}{9x^2t^6}\left(1-\cos(x t^3)\right)dt\geqslant 0.
$$
|
2025-03-21T14:48:31.898907
| 2020-08-24T10:50:17 |
369991
|
{
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"Asaf Shachar",
"Robert Bryant",
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|
Stack Exchange
|
A vector field whose flow has constant singular values
$\newcommand{\tr}{\operatorname{tr}}$
$\renewcommand{\div}{\operatorname{div}}$
Let $D \subseteq \mathbb{R}^2$ be the closed unit disk. Given a vector field $X$ on $D$, let $\psi_t$ be its flow.
Does there exist a (non-homothetic) divergence-free vector field $X \in \Gamma(D)$ whose associated flow $\psi_t \in \operatorname{diff}(D)$ has singular values which depend only on $t$ and not on the point in $D$? Equivalently, I want $|(d\psi_t)_p|^2$ to be independent of $p$.
By requiring that $X$ is not homothetic, I am excluding the possibility of $L_x g=\lambda g$ for some constant $\lambda$.
Write $f(t,p)=|(d\psi_t)_p|^2=\tr_g\big((\psi_t^*g)_p\big)$.
A necessary and sufficient condition on $X$ is that $$\frac{\partial }{\partial t}f(t,p)=\tr_g(\psi_t^*L_Xg)$$ would be independent of $p$, for all $t$. Can we write this as an explicit PDE on $X$?
Some partial information is obtained from differentiating twice at $t=0$:
$$
\frac{\partial^2 }{\partial t^2}\left|_{t=0}\right.\tr_g\big((\psi_t^*g)_p\big)=\operatorname{tr}_g(\mathcal{L}_X(\mathcal{L}_Xg))=2X(\operatorname{div}X)+2\|\nabla X\|^2+2\operatorname{tr}(\nabla X\circ\nabla X)=
$$
$$
2\|\nabla X\|^2+2\operatorname{tr}(\nabla X\circ\nabla X),
$$
so $\|\nabla X\|^2+\operatorname{tr}(\nabla X\circ\nabla X)$ must be constant.
$X(r,\theta)=\log r \frac{\partial}{\partial \theta}$ is an example on $D\setminus\{0\}$:
Its flow is $\psi_t: (r,\theta)\mapsto (r,\theta+t\log r)$;
$[d\psi_t]_{\{ \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta}\}}=\begin{pmatrix} 1 & 0 \\\ t & 1\end{pmatrix}$
is independent of $p$.
The vector fields $X$ that satisfy this condition are highly constrained, as there exist only a finite-dimensional family of $C^5$ solutions in a neighborhood of any given point in the plane. Here is how one can see this:
Using the standard coordinates, we have $g = dx^2 + dy^2$. Suppose that $X$ is defined on a simply-connected domain $D$ in the plane. Then, since $X$ is divergence-free, it can be written in the form
$$
X = u_y\,\frac{\partial}{\partial x} - u_x\,\frac{\partial}{\partial y}
$$
for some function $u$ on $D$. Define $\mathcal{L}_X^2g = \mathcal{L}_X(\mathcal{L}_X g)$ and, inductively $\mathcal{L}_X^{k+1}g = \mathcal{L}_X(\mathcal{L}^k_X g)$. As has been observed, we have $\mathrm{tr}_g(\mathcal{L}_X g) = 0$, since $X$ is divergence free. A direct computation gives
$$
\mathrm{tr}_g(\mathcal{L}^2_X g) = 2\left((u_{xx}-u_{yy})^2 + (2u_{xy})^2\right) = 8c^2
$$
for some constant $c\ge0$. If $c=0$, then either $X = a_0\,(y-y_0)\partial_x - a_0\,(x-x_0)\partial_y$ for some constants $a_0,a_1,a_2$, and the flow of $X$ is rotation around $(x_0,y_0)$ or $X = a_1\partial_x + a_2\partial_y$ and hence the flow of $X$ is simply translation.
Thus, from now on assume that $c>0$. Then the Hessian matrix of $u$ is of the form
$$
\begin{pmatrix} u_{xx} & u_{xy}\\u_{xy} & u_{yy}\end{pmatrix}
= \begin{pmatrix} h + c\,\cos v & c\,\sin v\\c\,\sin v & h-c\,\cos v\end{pmatrix}
$$
for some functions $h$ (unique) and $v$ (unique up to an additive constant that is an integral multiple of $2\pi$). The identities $(u_{xx})_y = (u_{xy})_x$ and $(u_{xy})_y = (u_{yy})_x$ imply that
$$
\mathrm{d}h = c\bigl(v_y\,\cos v-v_x\,\sin v\bigr)\,dx
+ c\bigl(v_x\,\cos v+v_y\,\sin v\bigr)\,dy,
$$
and, since $c\not=0$, the identity $\mathrm{d}(\mathrm{d}h) = 0$ gives
$$
(v_{xx}-v_{yy}+2v_xv_y)\,\cos v + (2v_{xy}-{v_x}^2+{v_y}^2)\,\sin v = 0.
$$
Thus, the Hessian of $v$ can be parametrized as
$$
\begin{pmatrix} v_{xx} & v_{xy}\\v_{xy} & v_{yy}\end{pmatrix}
= \begin{pmatrix}
q+p\,\sin v - v_xv_y & \tfrac12({v_x}^2-{v_y}^2)-p\,\cos v\\
\tfrac12({v_x}^2-{v_y}^2)-p\,\cos v & q-p\,\sin v + v_xv_y\end{pmatrix}
$$
for some functions $p$ and $q$. Now, a straightforward calculation yields that
$$
\mathrm{tr}_g(\mathcal{L}^3_X g) = 0
\quad\text{while}\quad
\mathrm{tr}_g(\mathcal{L}^4_X g) = 32\,c^4 - 8c^2\,(2h-u_xv_y+u_yv_x)^2.
$$
It follows that we must have $h = b + \tfrac12(u_xv_y-u_yv_x)$ for some constant $b$. Differentiating this equation and using the various formulae above allows one to solve for $p$ and $q$ in the form.
$$
p = \frac{P(\cos v, \sin v, v_x,v_y,u_x,u_y)}{\bigl(({u_x}^2-{u_y}^2)\cos v +2u_xu_y\sin v\bigr)}
$$
and
$$
q = \frac{Q(\cos v, \sin v, v_x,v_y,u_x,u_y)}{\bigl(({u_x}^2-{u_y}^2)\cos v +2u_xu_y\sin v\bigr)},
$$
where $P$ and $Q$ are certain polynomials in their arguments, which I won't write out here. Moreover, using these formulae, we find that
$$
\begin{align}
\mathcal{L}_X g
&= 2c\bigl(\sin v\,dx^2-2\,\cos v\,dx\,dy-\sin v\,dy^2\bigr)\\
\mathcal{L}^2_X g
&= 4c\bigl((c{+}b\,\cos v)\,dx^2+2b\sin v\,dx\,dy + (c{-}b\,\cos v)\,dx^2\bigr)
\end{align}
$$
and
$$
\mathcal{L}^3_X g = 4(c^2{-}b^2)\,\mathcal{L}_X g,
$$
from which it follows by induction that $\mathcal{L}^{k+2}_X g = 4(c^2{-}b^2)\,\mathcal{L}^k_X g$ for all $k\ge 1$, so that
$$
\mathrm{tr}_g(\mathcal{L}^{2k+1}_X g) = 0
\quad\text{and}\quad
\mathrm{tr}_g(\mathcal{L}^{2k+2}_X g) = 8c^2 \bigl(4(c^2{-}b^2)\bigr)^k
$$
for all $k\ge0$.
Moreover, because the equations express all of the second derivatives of $u$ and $v$ algebraically in terms of $u$, $v$ and their first derivatives, it follows that any solution $(u,v)$ that is $C^2$ is real-analytic. Thus, any $C^2$ solution $(u,v)$ of the equations above defines a vector field $X$ whose flow has singular values that depend only on $t$.
Now, it is easy to show that the equation $({u_x}^2-{u_y}^2)\cos v +2u_xu_y\sin v = 0$ cannot hold on any open domain. Moreover, given constants $(r_1,r_2,s,s_1,s_2)$ that satisfy $({r_1}^2-{r_2}^2)\cos s +2r_1r_2\sin s \not= 0$, and fixing constants $b$ and $c\not=0$, there is at most one solution on a neighborhood of $(x,y) = (0,0)$ of the above equations such that
$$
\bigl(u_x(0,0),u_y(0,0),v(0,0),v_x(0,0),v_y(0,0)\bigr) = (r_1,r_2,s,s_1,s_2).
$$
In fact, one finds that, in the case $b=0$, there exists such a local solution, as the PDE system is involutive (i.e., Frobenius). Unfortunately, there is good evidence that the 'generic' solution with $b=0$ cannot be written in terms of elementary functions. Sometimes, though, one can express a solution implicitly: As an example of a particular solution with $b=0$, one can show that when $x+i\,y$ is parametrized by a holomorphic parameter $w$ such that
$$
dx + i\,dy = \mathrm{e}^{w^2}\,dw,
$$
then, setting
$$
u_x + i\,u_y = \overline{w}\,\mathrm{e}^{w^2},
$$
one finds that, for all $t$, the time $t$ flow of the unimodular vector field $X = u_y\,\partial_y - u_x\,\partial_x$ has constant singular values. (This is most easily checked in the $w$ coordinate, where $X = 2i\left(w\,\partial_{\bar w} - \bar w\,\partial_w\right)$ and $g = \mathrm{e}^{w^2+\bar w^2}\,dw\circ d\bar w$.)
When $b\not=0$, it turns out that there are more restrictions on the initial conditions. In fact, an analysis shows that, when $b\not=0$, either $v$ is constant, or else the equation $(u_xv_x+u_yv_y)=4(c^2{-}b^2)$ must hold.
When $v$ is constant, the vector field $X$ is an affine vector field
whose flow is a $1$-parameter family of (unimodular) affine transformations of the plane, which is clearly a solution.
When $v$ is not constant and $b\not=0$, it turns out that one can integrate the equations explicitly. One finds that, up to a rigid motion in the plane, these solutions can be written in polar coordinates $(r,\theta)$ (where $g = dr^2 + r^2\,d\theta^2$) in the form
$$
X = a_1\left(r\,\frac{\partial}{\partial r}-2\theta\,\frac{\partial}{\partial\theta}\right) + (a_2 - 2b\log r)\,\frac{\partial}{\partial\theta}
$$
where $a_1, a_2, b$ are constants and $c^2=a_1^2+b^2$. Note that the case $a_1=a_2=0$ is, up to a constant multiple, the example provide by the OP.
@AsafShachar: I don't know about specifying terms on the boundary, other than that you can't do with it $b\not=0$. Unfortunately, there is good evidence that the general solution with $b=0$ (and there is a $6$-parameter family of them in a neighborhood of each point in the plane) cannot be written down in elementary terms. However, these solutions are all real-analytic, and one can effectively compute the Taylor series to arbitrary order, so even without integrating the equations, one can tell whether an integral curve of the vector field is a circle, so there is hope to answer the question.
@AsafShachar: Yes. The point is, though, that constancy of the trace for $k=2$ and $k=4$ is sufficient, already, to guarantee that a $C^5$ solution is, in fact, real-analytic, so you don't need to assume it.
@AsafShachar: One more comment: Since $X$ is divergence free, if it were 'inward pointing' on the boundary of a compact domain $D$ where it is differentiable, it would have to be tangent to $\partial D$. If $D$ is the interior of a circle, this would force the boundary to be a union of flow lines of $X$ and hence those flow-lines would have constant curvature. I believe that the only local solutions that have flow lines of constant nonzero curvature have $b\not=0$ and have the form that I gave above with $a_1=0$, and these are not smooth at the origin.
Thank you for your last comment, and thanks again for this thorough and extremely beautiful analysis. I have now succeeded (after two days effort) verifying most of your computations. I am very interested in fully understanding everything you wrote, and I still have some questions about some parts: (1) Can you please elaborate on why the equation $
({u_x}^2-{u_y}^2)\cos v +2u_xu_y\sin v = 0$ cannot hold on an open domain? (2) Can you elaborate on the analysis of the case $b\not=0$? How do you deduce that either $v$ is constant, or else the equation $(u_xv_x+u_yv_y)=4(c^2{-}b^2)$ must hold?
(3) Can you describe the explicit integration process which led you to the last formula in polar coordinates? I will appreciate any further elaboration you feel convenient to provide. I understand that I might be asking too much here for an MO setting, but I really do want to understand more, and I find your analysis intriguing. If you feel that the answer is already too long, and do not want to answer further in the comments (I deleted my previous comments to reduce the clutter), I am of course ready to continue the discussion outside of MO. Thanks again for the effort and patience.
@AsafShachar: I can explain (1) easily, but (2) and (3) are harder and will have to wait until I have more time to write things out. First, we can't have $u_x=u_y=0$ on an open set since $c\not=0$. Away from the closed set where $u_x=u_y=0$, we would have $$ \cos v = \pm\frac{2u_xu_y}{u_x^2+u_y^2}\quad\text{and}\quad \sin v = \mp \frac{u_x^2-u_y^2}{u_x^2+u_y^2},,$$ and this would give a formula for $\mathrm{d}h$ in terms of $u_x$, $u_y$, and $h$. But then, using this formula, $\mathrm{d}(\mathrm{d}h)=0$ gives a contradiction, since $c\not=0$.
|
2025-03-21T14:48:31.899598
| 2020-08-24T12:31:11 |
369995
|
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"url": "https://mathoverflow.net/questions/369995"
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|
Stack Exchange
|
The double dual of the unitization of a $C^*$-algebra
I am studying the proof that if $A$ is a $C^*$-algebra such that $A^{**}$ is a semidiscrete vN algebra, then $A$ has the completely positive approximation property (CPAP). I was able to handle the unital case, but I am stuck in the non-unital setting: The authors (N. Brown and N. Ozawa) suggest that one should prove that if $A^{**}$ is semidiscrete then so is $(\tilde{A})^{**}$ and then conclude by proving that if $\tilde{A}$ has the CPAP then so does $A$.
My problem is this: I can't prove that the double dual of the unitization will be semidiscrete. I cannot understand the double dual of the unitization in relevance to the double dual of $A$ at all. The authors state that $(\tilde{A})^{**}\cong A^{**}\oplus\mathbb{C}$ and mention that it is furthermore true that if $B$ is any $C^*$-algebra with a (closed, two-sided) ideal $J$, then $B^{**}\cong J^{**}\oplus(B/J)^{**}$. First of all, does $\cong$ mean as vector spaces or as $C^*$-algebras? How can one prove this isomorphism? Extra bonus question: If all double duals involved are endowed with their ultraweak topologies, is $\cong$ a homeomorphism?
Believe it or not, these are $*$-isomorphisms as C${}^*$-algebras. If $J$ is a closed two-sided ideal of $B$ then $J^{**}$ is a weak* closed two-sided ideal of $B^{**}$, and every weak*-closed two-sided ideal of a von Neumann algebra is a direct summand. I suppose these are good exercises. The supremum in $B^{**}$ of an approximate unit for $J$ will be a central projection $p$ such that $pB^{**} = J^{**}$.
Bonus, any $*$-isomorphism of von Neumann algebras is automatically a weak* homeomorphism. That is because it preserves order and hence must be normal.
Thank you very much for your answer. Much appreciated, I will try the exercises you are pointing out, even though they seem a little difficult to me.
Hint: a weak*-closed ideal would itself be a von Neumann algebra and hence must have a unit.
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2025-03-21T14:48:31.899763
| 2020-08-24T13:10:44 |
370000
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Stack Exchange
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Does every map $K(\mathbb{Z}, n) \to K(\mathbb{Z}/m, n + k)$ factor through $K(\mathbb{Z}/m, n)$?
Recall that a cohomology operation is a natural transformation $H^n(-; \pi) \to H^{n+k}(-; G)$ defined on CW complexes.
Does every cohomology operation $H^n(-; \mathbb{Z}) \to H^{n+k}(-; \mathbb{Z}/m)$ factor through $H^n(-; \mathbb{Z}/m)$?
The cohomology operations $H^n(-; \mathbb{Z}) \to H^n(-; \mathbb{Z}/m)$ are all multiples of the map induced on cohomology by the quotient map $\mathbb{Z} \to \mathbb{Z}/m$. In particular, if the above question has a positive answer, then for any such cohomology operation $\theta$, we have $\theta(mx) = 0$.
One can rephrase the above question in terms of Eilenberg-MacLane spaces:
Does every map $K(\mathbb{Z}, n) \to K(\mathbb{Z}/m, n + k)$ factor through $K(\mathbb{Z}/m, n)$?
I would actually be content with a positive answer to the following broader question:
Does every map $K(\mathbb{Z}, n) \to K(\mathbb{Z}/m, n + k)$ factor through $K(\mathbb{Z}/r, n)$ for some $r$?
I initially asked question 1 in the Homotopy Theory chatroom. Piotr Pstrągowski's comments, see here, explain why there is a positive answer for $m = 2$.
Maybe for context one should point out that for stable cohomology operations, the answer is yes. That is, any map of spectra $\Sigma^n H\mathbb Z \to \Sigma^{n+k} H\mathbb Z / m$ commutes with multiplication by $m$ (which is null on $\Sigma^{n+k} H\mathbb Z / m$), and so descends, via the cofiber sequence $\Sigma^n H\mathbb Z \xrightarrow m \Sigma^n H\mathbb Z \to \Sigma^n H\mathbb Z / m$, to a map $\Sigma^n H\mathbb Z / m \to \Sigma^{n+k} H\mathbb Z / m$.
sorry- I deleted my previous answer because I can't seem to make it work without basically just doing the whole computation anyway... maybe one needs to extract something from the Cartan seminar? It would be nice if there was a clean argument like Tim's though... Messing around it seems like $\mathbb{Z}/m[B\mathbb{Z}]\to\mathbb{Z}/m[B\mathbb{Z}/m]$ has an $\mathbb{E}_{\infty}$-$\mathbb{Z}/m$-retract? (which would do it) But I don't really trust that I didn't obscure some error while doing that... if I end up trusting that, I will update the post.
Along with $m=2$, there is also a positive answer for $n=1$ :)
@JohnGreenwood nice! but I guess there's a negative answer for n=k=0 !
@DylanWilson well played, sir
The answer is (also) yes when $m=p$ is an odd prime, by Theoreme 2 in
Cartan, H. Détermination des algèbres $H_*(\pi, n; Z_p)$ et $H^*(\pi, n; Z_p)$, $p$ premier impair.
Séminaire Henri Cartan, Tome 7 (1954-1955) no. 1, Exposé no. 9, 10 p.
http://www.numdam.org/item/SHC_1954-1955__7_1_A9_0/
which gives $H^*(K(\pi, n); Z/p)$ as a free graded commutative algebra on a sum of copies of $Hom(\pi, Z/p)$ (indexed by certain words of the "first kind") plus a sum of copies of $Hom({}_p \pi, Z/p)$ (indexed by certain words of the "second kind"). Here ${}_p \pi$ denotes the subgroup of elements of exponent $p$. Applying this with $\pi = Z$ and $\pi = Z/p$ shows that $H^*(K(Z/p,n); Z/p) \to H^*(K(Z,n); Z/p)$ is surjective.
Cartan first proves the dual statement in homology (Theoreme Fondamental on page 9-03), showing that $H_*(K(Z,n); Z/p) \to H_*(K(Z/p,n); Z/p)$ is injective.
Thanks for your answer. This gives me some hope that it may be true in general, although I am worried about powers of primes.
I Steenrod's 1957 Colloquium Lectures, published as
Steenrod, Norman E.
Cohomology operations, and obstructions to extending continuous functions.
Advances in Math. 8 (1972), 371–416.
he ends Section 17 with:
There are certain elementary cohomology operations which are taken for
granted but must be mentioned in order to state the main result. These
are: addition, cup products, homomorphisms induced by homomorphisms
of coefficient groups, and Bockstein coboundary operators associated
with exact coefficient sequences $0 \to G' \to G \to G'' \to 0$. Then
the main result becomes:
The elementary operations and the operations $Sq^i$, $\beta_2$, $P_p^i$, $\beta_p$ generate all reduced power operations by forming
compositions.
Thereafter, at the end of Section 21, he writes:
Using the full strength of Cartan's result, Moore [18] has shown that
all cohomology operations, whose initial coefficient groups are finitely generated, are generated by the cohomology operations listed
at the end of Section 17.
Here, reference [18] is: "J. Moore, Seminar notes 1955/1957, Princeton University."
Moore's result implies that the answer to Question 1 is "yes", because any Bockstein operation will vanish on any integral cohomology class. Unfortunately, I do not know if Moore's seminar notes are available somewhere.
There was a copy of Moore's notes in the Stanford library when I was a graduate student there. Maybe one could request it via interlibrary loan. (Also I think I have a photocopy in my office, though I haven't been there in a long time...)
Dmitri Pavlov has a nice scan of these notes (better quality than what I could get from my 2002 scan): https://dmitripavlov.org/scans/moore-algebraic-homotopy-theory.pdf
Caveat to my previous comment: The above notes might not be what Steenrod was referencing. (They're from the right time period, at least.)
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2025-03-21T14:48:31.900092
| 2020-08-24T13:26:40 |
370003
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Stack Exchange
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Weak Hopf algebra structure on twisted group algebra
A (normalized) $2$-cocycle on a finite group $G$ with values in $S^1$ is a map
$\sigma:G\times G\rightarrow S^1$ such that $$\sigma(g,h)\sigma(gh,k)=\sigma(h,k)\sigma(g,hk)$$ and $$\sigma(g,e)=\sigma(e,g)=1.$$ Let $\{\delta_g\}$ denote the canonical basis of $l^2(G)$. Define, $\lambda_{\sigma}(g)(\delta_h)=\sigma(g,h)\delta_{gh}.$ The twisted group von Neumann
algebra $L_{\sigma}(G)$ is the von Neumann subalgebra of $\mathcal{B}(l^2(G))$ generated by $\lambda_{\sigma}(G).$ My question is the following:
$\textbf{Question:}$ Can one provide a natural weak Kac algebra structure on $L_{\sigma}(G)$?
Edit: Note that group $C^*$-algebra has a natural Kac algebra structure if we define $\Delta(g)=g\otimes g$, $\epsilon(g)=1$ and $S(g)=g^{-1}$. In my question I have used "natural" in this spirit. In other words, the formulae for comultiplication, antipode and counit should involve group and the cocycle.
For definition of weak Hopf (and Kac) algebra see:
[BNS] Böhm, Gabriella; Nill, Florian; Szlachányi, Kornel (1999). "Weak Hopf algebras. I. Integral theory and $C^*$-structure". Journal of Algebra. 221 (2): 385–438.
[NV] Dmitri Nikshych, Leonid Vainerman; Algebraic versions of a finite-dimensional quantum groupoid.
Probably not. Take $G = \Gamma \times \hat\Gamma$ for some finite Abelian grouo $\Gamma$ and the cocycle $(g, \chi) \mapsto \langle g, \chi\rangle$. Then, $L_\sigma(G) = L \Gamma \rtimes_{\alpha} \hat{\Gamma}$, where $\alpha$ is the dual action. That equals $M_{|\Gamma|}(\mathbb{C})$ by Sakai's duality, but i think there are no nontrivial comultiplication structures on matrices.
I think there is a natural comultiplication structure on $M_n(\mathbb{C})$ given by matrix units which also makes it a coalgebra structure together with a counit given by $\epsilon(e_{ij})=\delta_{ij}$. However, since matrix algebra is simple this coalgebra structure is not compatible with the algebra structure of the matrices and hence it is not a bialgebra (and hence not a Hopf algebra). But, it is not clear to me why matrix algebra can not be made into a weak Hopf algbera. @AdriánGonzález-Pérez
@SebastienPalcoux Why it should follow that $n=1$?
@Sebastein I think it is appropriate to keep "Can one provide" as explained in my first comment. I am doubtful about your guess.
My guess was false according to the paper of Nikshych and Vainerman cited above: « Then we show that for any f-d C*-algebra $A$ there exists a unique weak Kac algebra structure (non-trivial when $A$ is non-abelian) on the full matrix algebra $M_n(C)$ (where $n = dim A$) with Cartan subalgebras isomorphic to $A$. This gives a classification of weak Kac algebras simple as C*-algebras ».
@AdriánGonzález-Pérez: see Theorem 3.2.2 in the second paper cited above.
@ Sebastien Yes I have also realised that your guess can not be correct. In fact, as I have already indicated earlier that you can construct $\Delta, \epsilon $ and $S$ using matrix units and using these by hand one can easily show that matrix algebra has a natural weak Kac algebra structure.
What about defining a weak Kac algebra structure on any finite dimensional C*-algebra, component-wise, using what is known for $M_n(\mathbb{C})$?
I was not aware of that result, thanks for the reference. My first comment is invalid.
@SebastienPalcoux Yes, this will do the job. Thanks.
@SebastienPalcoux But the information of co-cycle will be lost.
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2025-03-21T14:48:31.900588
| 2020-08-24T13:38:22 |
370004
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Stack Exchange
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Counting polygons in arrangements
For an arrangement of lines $\cal{A}$ in the plane, an
inducing polygon $P$ is a simple polygon satisfying:
(a) every edge $e$ of $P$ lies on some line $\ell$ of $\cal{A}$, and
(b) every line $\ell \in \cal{A}$ is collinear with an edge $e$ of $P$.
If $P$ has $k$ edges and $\cal{A}$ has $n$ lines, $k \ge n$.
Note that several edges of $P$ might lie on the same line of $\cal{A}$.
It is known that if the lines in $\cal{A}$ are in general position
in the sense that no two lines are parallel and no three lines meet in a point,
then $\cal{A}$ has an inducing polygon.1
My questions concern counting the inducing polygons.
Q. Over all arrangements $\cal{A}$ of $n$ lines in general position,
what are upper and lower bounds on the number of inducing polygons
for $\cal{A}$, and which arrangements achieve those bounds?
To clarify (thanks MaxAlekseyev): Let $\cal{A}$ be a specific arrangement of $n$ lines
in general position. $\cal{A}$ supports a certain number of incongruent
inducing polygons. What are max and min of this number, over all arrangements of $n$ lines?
Other possibly easier questions suggest themselves, e.g.:
Does any arrangement ever have more than one convex inducing polygon?
My original aim was to find a minimum area inducing polygon,
which is likely difficult.
1Scharf, Ludmila, and Marc Scherfenberg.
"Inducing polygons of line arrangements."
In International Symposium on Algorithms and Computation,
pp. 507-519. Springer, Berlin, Heidelberg, 2008.
Springer link.
I assume you enumerate polygons up to some equivalence relation - what is it?
@MaxAlekseyev: Good question. I was thinking congruency, but one could imagine other equivalence relations.
Congruency won't work - e.g. in you top leftmost example rotating one of the lines would produce infinitely many non-congruent polygons.
@MaxAlekseyev: But for a fixed arrangement, congruency makes sense.
Ok, I've misread your question - it's not about polygons induced by all arrangements of $n$ lines, but about those induced by a single arrangement.
@MaxAlekseyev: Yes. Your questions clarified matters. Thanks.
@MattF. Could you elaborate?
Going to the geometric dual,
lines map to points with polar coordinates $(\varphi,\,r)$ where $\varphi$ is the angle of the normal, pointing away from the origin, with the positive $x$-axis and $r$ the distance of the origin to the line
pairs of intersecting lines in the euclidean plane map to line-segments connecting the respective dual points.
point-segment arrangemnts in the dual plane can be interpreted as planar embeddings of graphs and simple arrangements of lines yield a complete graph.
That simple arrangement of lines yield a complete graph implies that they can always be represented by a single polygon: any Hamilton cycle through the points in the dual plane will do.
The other questions seem to be answered by results about cell complexes, some of which arein the cited wikipedia article like e.g."Although a single cell in an arrangement may be bounded by all n lines, it is not possible in general for m different cells to all be bounded by n lines. Rather, the total complexity of m cells is at most $Θ(m^{2/3}n^{2/3} + n)$,[11] almost the same bound as occurs in the Szemerédi–Trotter theorem on point-line incidences in the plane"
$ILP$ formulation
If a one-to-one correspondence between lines and polygon sides is desired, then a integer linear programming formulation may yield solutions that can be subjected to desirable optimization criteria:
the binary variables correspond to the edges created by splitting the lines at the intersection points, the constraints being that the variables of collinear edges sum to $1$ and that in every intersection of two lines the sums of the variables corresponding to their adjacent edges are equal, i.e.
if $l_{1i},l_{i1},l_{2,i},l_{i2}$ are the binary variable corresponding to the edges of lines $L_1$ and $L_2$ that intersect in point $(x_i,y_i)$, then $l_{1i}+l_{i1}=l_{2,i}+l_{i2}$ must be satisfied.
Imposing subtour elimination constraints may answer the existence of a single polygon with bijection between its edges and the lines.
Nice idea to explore the dual! However, I don't believe there is a 1-1 correspondence between inducing polygons and Hamiltonian cycles in the dual: (a) an inducing polygon must be simple (non-self-intersecting); (b) several edges of an inducing polygon can be collinear with one line.
@JosephO'Rourke I just provided an ILP formulation for the case when no mare than to lines can intersect in a point; that should aid in investigating the problem; maybe directly or by its dual.
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2025-03-21T14:48:31.900912
| 2020-08-24T14:40:53 |
370007
|
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Stack Exchange
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Construction of differential of a smooth map between diffeological spaces with respect to internal tangent spaces
According to page 5 , definition 3.1 of https://arxiv.org/pdf/1411.5425.pdf, there is a notion of Internal Tangent Space of a Diffeological space $X$ at a point $x \in X$.
Basically if $x \in X$, they defined the internal tangent space $T_xX$ as the colimit of the following diagram(composite of functors) in Vect, the category of finite dimensional vector spaces:
where
$ DS_0$ is the category with objects all open connected neighbourhoods of $0$ in $ \mathbb{R}^n$ for all $ n \in \mathbb{N}$ and morphisms are the smooth maps between them sending $0$ to $0$.
$(X,x) \in Diff_*$, the category of pointed diffeological spaces.
$ DS_0/(X , x)$ is the category with objects the plots $ P: U \rightarrow X$ such that $ U$ is connected , $ 0 \in U$ and $ P(0)=x$ and morphisms are the commutative triangles
where $ p, q \in Obj(DS_0/(X , x))$ and $ f$ is a smooth map such that $f(0)=0$.
$F$ is the usual forgetful functor.
On Objects: $\phi(U)= T_0U$, the tangent space of $U$ at $0$ and
On Morphisms: $ \phi(f)= f_{*}$ where $ f_{*}: T_oU \rightarrow T_0V$ is the differential of the smooth map $ f: U \rightarrow V$ such that $ f(0)=0$.
Now after defining the Tangent space of $X$ at $x$ it is natural to ask about the differential of a smooth map.
So given a smooth map $f:X \rightarrow Y$ between diffeological spaces I want to construct a morphism $f_*: T_xX \rightarrow T_{f(x)}Y$ in $C$.
Now $f$ induce a functor $\bar{f}: DS_0/(X,x) \rightarrow DS_0/(Y,f(x))$ in a natural way that is
On Objects: $p:U \rightarrow X \mapsto f \circ p : U \rightarrow Y$
On Morphisms $\lambda \mapsto \lambda$
"Now given such a map $\bar{f}: DS_0/(X,x) \rightarrow DS_0/(Y,f(x))$ I am tying to construct $f_*: T_x X \rightarrow T_{f(x)}Y$ in $C$ in a natural way."
In Page 6 of https://arxiv.org/pdf/1411.5425.pdf it is mentioned that given such $\bar{f}$ one can construct a functor $T:Diff_* \rightarrow Vect$ which is a left Kan extension of $\phi: DS_0 \rightarrow Vect$ along the inclusion functor $i:DS_0 \rightarrow Diff_*$.
But I can't completely understand how the above paragraph is related to my question of explicitly constructing a morphism $f_{*}: T_xX \rightarrow T_{f(x)}Y $ in $C$.
So I thought that the construction might be followed from a following category theoretic reason which I stated below:
Let $I$ , $J$ and $B$ be small categories. Let $C$ be a cocomplete category. Let us consider the following :
where $f :I \rightarrow J$ is a functor such that $ \beta \circ f= \alpha$ and $g: B \rightarrow C$ be any functor.
Now let $g \circ \alpha = F$ and $ g \circ \beta = G$. Let $L_F$ and $L_G$ be the Colimits of the diagram $F$ and $G$ respectively in the category $C$.
My question is the following:
For the given functor $f:I \rightarrow J$ such that $\beta \circ f =\alpha$, is there any natural choice of morphism $f_*: L_F \rightarrow L_G$ in $C$ between the colimits?
I cannot see immediately any such choice.
Note: Here in this post I used the same notation for convenience for different objects, but I feel the meanings are evident from the context.
Apology in advance if the question is not upto the standard of this forum.
I got my answer. Its easily following from the simple observation which I mentioned at the end of the question.(which can be easily proved using the universal property of colimits). I personally feel that this question is not upto the standard of this forum. So if someone also feel the same and suggest me in the comments section that I should delete this question then I am happy to do that. Sorry for the inconvenience caused.
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2025-03-21T14:48:31.901158
| 2020-08-24T14:59:48 |
370008
|
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|
Stack Exchange
|
Period map on non-Kähler manifold
Is there a theory of period map on non-Kähler manifolds that has Hodge decomposition? Any reference is helpful. Thank you.
@user145520 Thank you for your comment. Is there any intuition or some ideas for you to believe it is wrong?
Have you checked Angella's book on non-Kähler geometry? He did some research on $\partial\bar\partial$ manifolds maybe provide some information which you need.
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2025-03-21T14:48:31.901351
| 2020-08-27T19:45:11 |
370269
|
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Stack Exchange
|
Is Carlitz's paper correct about the number of similarity classes of commuting matrices?
L. Carlitz has a paper, Classes of pairs of commuting matrices over a finite field, that computes the number of simultaneous similarity classes of of pairs of commuting matrices in $\operatorname{Mat}_n(\mathbb F_q)$. Two pairs $(A,B)$ and $(A',B')$ are called simultaneously similar if there is $U\in \operatorname{GL}_n(\mathbb F_q)$ such that $A'=UAU^{-1}$, $B'=UBU^{-1}$.
However, from the proof (specifically, to go from equation (4) to (5) on p. 193), it seems that he implicitly uses the statement that $(A,B)$ and $(A,B')$ belong to different similarity classes whenever $B\neq B'$. But it is possible that there is $U\in \operatorname{GL}_n(\mathbb F_q)$ that commutes with $A$ and such that $UBU^{-1}=B'$. In this case, they belong to the same class.
Is that it? If this paper turns out to be wrong, is there any result about the same question?
Could you point out where this happens in the proof? I scanned the paper and failed to find it.
The equation (4) computes the number of solutions of AX=XA for A in a fixed similarity class, and concludes in (5) that if we sum up such numbers over all possible similarity classes of A, we get the answer. This assumes that if X and X' are different, then (A,X) and (A,X') always contribute to different simultaneous similarity classes.
It is wrong. See the correction published in AMM 71 (1964), issue 8, page 900. Unfortunately this was published in the "Mathematical Notes" section, making it hard to find (as these notes don't get DOIs of their own).
Ah, I see. @darijgrinberg Doesn't that provide an answer to the question in the title?
(The relevant correction is a tiny note at the very end of @darijgrinberg's link: "J. Towber has kindly pointed out to the writer that there is an error in the paper: …. The error occurs in equation (5) of the paper. The results of the paper remain valid if we redefine $Q(n)$ as equal to the number of pairs of $n\times n$ matrices $(A_i, B_i)$, with elements in …
… $\operatorname{GF}(q)$, where $A_i$ runs through a complete set of nonsimilar matrices, and, for each $A_i$, $B_i$ commutes with $A_i$. \
Thus the problem of determining the number of classes of pairs of commuting matrices remains open.")
@darijgrinberg This does answer the question and is all I need. Could you put it as an answer?
It is wrong. See the correction published in AMM 71 (1964), issue 8, page 900. (You have to scroll down to the bottom of the last page to find this correction.)
Unfortunately this was published in the "Mathematical Notes" section, making it hard to find (as these notes are not individually indexed much of the time, and don't get separate DOIs). I've only managed to find it by looking at the back references to Carlitz's original paper (always a good first step if you suspect something is wrong in a paper; whoever cited it might too have noticed), and realizing that one of these back references also cites a correction.
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2025-03-21T14:48:31.901585
| 2020-08-27T20:47:09 |
370271
|
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"Steven Landsburg",
"Yifeng Huang",
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}
|
Stack Exchange
|
Leray spectral sequence for the complement of an arrangement of high-codimensional varieties
Assume $X$ is a smooth complex varieties and $D_1,\dots,D_r$ are smooth irreducible divisors that "intersect like hyperplanes", i.e. their union is locally given by a product of linear functions. Then the equation (2.3-1) of Looijenga's paper Cohomology of M3 gives a spectral sequence that computes the cohomology of $X-D_1-\dots-D_r$.
Note that in the case $X=C^n$ where $C$ is a smooth curve, and the divisors in the list are the big diagonals $\Delta_{ij}=\{(x_1,\dots,x_n)\in C^n: x_i=x_j\}$, the spectral sequence above is consistent with Theorem 1.1 of Kriz 1994, i.e. they predict the same spectral sequence converging to $H^*(F(C,n))$, where $F(C,n)=X-\bigcup \Delta_{ij}$ is the unordered configuration space.
Is there a version of Looijenga's (2.3-1) for arrangements of smooth irreducible closed subvarieties that may have codimension higher than one? The examples of subvarieties with particular interest are the big diagonals in $X=V^n$ where $V$ is smooth variety of dimension at least $2$, and the preimages of coordinate projections (e.g. $\{P\}\times V^{n-1}\subseteq X$, where $P$ is a point of $V$).
Yes, these spectral sequences exist quite generally. There is Dan Petersen's paper "A spectral sequence for stratified spaces" and also my paper "Lattice spectral sequences and cohomology of configuration spaces"-- these also handle cases where the subvarieties are not smooth.
Do you have $\neq$ where you mean $=$ in the definition of the big diagonal?
@StevenLandsburg Yes, my bad.
@PhilTosteson I see, so I can say that when talking about big diagonals and preimages of projections, we have a similar spectral sequence to the divisor case because the "combinatorics of intersections" (i.e. order complex) are the same. Does your Theorem 1.8 have a simpler version in the case where all (sub)varieties are smooth and they intersect like linear subspaces? P.S. Peterson's formula for compactly supported cohomologies does involve simpler terms, but I find that taking Poincare dual involves some algebra (in particular, enough to deduce the Orlik--Solomon relations in divisor case)
Yes, there's some simplification because when $Z_p$ is a smooth subvariety the cohomology groups H^*(X,X-Z_p) can be computed using a tubular neighborhood + excision. The cohomology order complex will also simplify if the poset is the same as the poset of a hyperplane arrangement. Also, I don't talk about product structures/comparing with the Leray spectral sequence in the paper. Feel free to email me if you want more info.
@PhilTosteson I now see how it is done (see the answer I just posted). Do you know of a reference that shows that this is a sequence of mixed Hodge structures?
Compatibility with mixed Hodge structure is proven in Subsection 3.2 of my paper that Phil mentioned.
I think I figure out the answer to my question after seeing @PhilTosteson 's comments.
Short answer:
If an arrangement of smooth complex subvarieties intersect "like a hyperplane arrangement" and every multi-wise intersection is connected, then we have a spectral sequence whose $E_1$ page is the same as the $E_2$ in Bibby's "Cohomology of abelian arrangements" Theorem 4.1, except that the generator $g_Y$ (corresponding to a subvariety $Y$ of complex codimension $c$) now has bidegree $(2c,1)$ and Hodge type $(c,c)$. If the ambient variety is smooth projective (or more generally, such that $H^i$ is pure of weight $i$), then the spectral sequence degerates at $E_2$ page.
Long answer:
Consider an arrangement $\mathcal A$ of closed subspaces $Y_1,\dots,Y_r$ in a topological space $X$, and denote $M_\mathcal A = X - Y_1 - \dots Y_r$. Consider the poset $P$ consisting of all nonempty intersections of $Y_i$'s, ordered by inclusion. Form $\hat P:=P\cup \hat 1$ by adjoining a top element $\hat 1$ corresponding to $X$. Then Theorem 1.8 of Tosteson's "Lattice spectral sequences ..." gives an exact sequence converging to $H^*(M_\mathcal A;\mathbb Z)$, whose $E_1$ page is a direct sum over all $Z\in \hat P$ of terms involving
(1) the relative cohomology $H^i(X,X-Z;\mathbb Z)$ for $Z\in \hat P$.
(2) a combinatorial datum associated to $Z$ and $\hat P$, recorded as the reduced homology $\tilde H_{j-2}$ of the simplicial complex of the sub-poset $(Z,\hat 1):=\{Z':Z<Z'<\hat 1\}$.
In the case where $X$ and all $Z\in P$ are connected smooth complex varieties, tubular neighborhood + excision + Thom isomorphism implies a canonical isomorphism $H^i(X,X-Z;\mathbb Z) \cong H^{i-2\mathrm{codim}_{\mathbb C}Z}(Z;\mathbb Z)$. (The assumption of being complex varieties can be relaxed; it was to ensure that the normal bundle of Z in X has a canonical orientation.)
In the case where $P$ is isomorphic to a poset arising from a hyperplane arrangement, one can apply Folkman 1966 "The homology groups of a lattice" and get
$$ \tilde H_{j-2}((Z, \hat 1);\mathbb Z) = \mathbb Z^{\mu(Z,\hat 1)} = A_Z(\mathcal A)$$
if $j$ is the "rank" of $Z$ (i.e. the distance between $Z$ and $\hat 1$ in the poset), and 0 otherwise. Here, $\mu$ is the Moebius function on the poset, and $A_Z(\mathcal A)$ is the subgroup of the Orlik--Solomon algebra generated by $g_I$ with $\bigcap I = Z$.
At the end, if we apply this theorem to the $n$-th ordered configuration space of a $d$-dimensional complex variety, the $E_1$ page is the algebra given in Totaro 1996 Theorem 4, except that Totaro's $E_{2d}$ page corresponds to Tosteson's $E_1$ page, and the bidegree $E_{2d}^{0,2d-1}$ (where the Orlik--Solomon generator lives) corresponds to the bidegree $E_1^{2d,1}$. The only nontrivial differential is on page $2d$ in Totaro's sequence, and page $1$ in Tosteson's sequence.
|
2025-03-21T14:48:31.901992
| 2020-08-27T22:09:05 |
370274
|
{
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|
Stack Exchange
|
Theory of weak enrichment in higher categories
Has there been work towards a general theory of weak enrichment in higher categories? To be more pointed, has there been any work towards trying to make sense of statements such as
There is a (weak) $(n+1,r+1)$-category $\mathcal V\mathbf{Cat}$ of categories enriched in a monoidal $(n,r)$-category $\mathcal V$.
generalising the fact that there is a $2$-category of categories enriched in an ordinary monoidal category, and how I imagine there is a (weak?) $3$-category of categories enriched in a monoidal bicategory.
I know there's work by Gepner and Haugseng (and more in this direction) which makes sense of an $(\infty,1)$-category of categories enriched in a monoidal $(\infty,1)$-category, though from my (limited) understanding there is little care here for the non-invertible "enriched natural transformations" which would exist in the corresponding $(\infty,2)$-category of enriched categories, and a lot of the work is put into ensuring that the equivalences of enriched categories correspond to the correct notion of fully faithful and essentially surjective functors.
Besides these, and other texts that address special cases (e.g., using a notion of weak enrichment to create a theory of weak $n$-categories), I haven't found anything that addresses a more general notion of enrichment in higher monoidal categories. Perhaps I just don't know how to look for them, but it also leads me to wonder: is there reason to study weak enrichment in this way? Or conversely, is there no good reason to be concerned with this? (e.g., problems where this would come in handy don't come up often / don't exist, or the $(\infty,1)$-category theory of enrichment is sufficient for most practical purposes, etc.)
Edit: Harry Gindi pointed out that Gepner and Haugseng do provide some work in the direction of higher categorical structure. In particular (this is their example 7.4.11), if $\mathcal V$ is a $\Bbb E_2$-monoidal (presentable) $(\infty,1)$-category, then the $(\infty,1)$-category $\mathcal V\mathbf{Cat}$ of $\mathcal V$-enriched categories will be monoidal also; moreover, if $\mathcal V$ is closed, then so is $\mathcal V\mathbf{Cat}$. Especially, the latter is self-enriched to provide a "$\mathcal V$-$(\infty,2)$-category" of $\mathcal V$-enriched categories (in that between $\mathcal V$-enriched categories is a $\mathcal V$-enriched category of functors).
While this provides $\mathcal V\mathbf{Cat}$ with a nice self-enrichment in instances where the enriching category is nice, this is somehow giving more than I asked for given less than I have. For instance, if $\mathcal V$ is just an ordinary monoidal $1$-category, then $\mathcal V\mathbf{Cat}$ will always be a $2$-category, though with additional assumptions on $\mathcal V$ (take it to be a Bénabou cosmos, for instance) it can also be given an internal hom (making it a "$\mathcal V$-enriched $2$-category").
If $\mathcal V$ is a general monoidal $(\infty,1)$-category, is it possible that $\mathcal V\mathbf{Cat}$ will to naturally an $(\infty,2)$-category?
Edit 2 (for clarification): While Harry's answer does a great job of elaborating on the work of Gepner and Haugseng, it doesn't really answer my question (as far as I know). I'm not trying to inductively define $(n+1,r+1)$-categories through iterated weak enrichment; rather, I'm just trying to see if there are constructions which, given for example a (nice?) monoidal $n$-category $\mathcal V$, produces the $(n+1)$-category $\mathcal V\mathbf{Cat}$ of categories enriched in $\mathcal V$. I mentioned Gepner and Haugseng's work because it's a step in this direction, but I don't think they were trying to address this generality.
Since my post is getting a bit long, I'll keep a summary of the questions I'm really trying to ask:
Is there existing work towards a general theory which describes the $(n+1,r+1)$-category of categories enriched in a monoidal $(n,r)$-category?
If not, is this because there is no merit to this endeavour in the first place, or is it because this is just a difficult thing to accomplish in general?
Gepner and Haugseng have shown in fact that the induced $E_{n-1}$ monoidal structure on V-enriched categories is closed as long as the $E_n$-monoidal category $V$ is closed, so in fact, it does provide an internal hom. The thing that they don't have is a fully worked out theory of lax and oplax natural transformations, which are very special and specific to the $(\infty,n)$-category of $(\infty,n-1)$-categories. The point is that the objects of (op)lax natural transformations are not enriched functors. They are highly combinatorial, but this is something very special and specific.
(At least when $V$ is presentably $E_n$-monoidal, which is true in most cases of interest).
This is a good question, if a bit open ended. One thing it illustrates is that, in the best of futures, there will be many works, by different authors with different perspectives, developing this theory. Gepner--Haugseng have certainly done a lot, but my impression is that there is more to do. I hope that an answer will teach me about other work that I am not aware of.
@HarryGindi Ah, I wasn't so careful reading the last section of their paper so I missed that, thanks. I'm not currently concerned with op/lax transformations, for reasons like that you mentioned. If $\mathcal V$ is not closed / not $\Bbb E_2$-monoidal, would it be unreasonable to expect $\mathcal V\mathbf{Cat}$ to be an $(\infty,2)$-category?
If you take a look at Gepner-Haugseng Corollary 5.7.12, the first thing you'll notice is that the functor $\mathbf{Cat}^{(-)}_\infty:\mathbf{MON}^{\operatorname{lax}}_∞\to \mathbf{CAT}_\infty$ is lax monoidal and sends $\mathcal{O}$-algebras to $\mathcal{O}$-monoidal categories for any symmetric operad $\mathcal{O}$. Another way to say this is that enrichment takes an $\mathcal{O}\otimes E_1$-monoidal category to an $\mathcal{O}$-monoidal category.
Now here's a neat thing to note: If we restrict this to the subcategory of presentably monoidal $\infty$-categories, we get the stronger statement that the functor $\mathbf{Cat}^{(-)}_\infty$ carries $\mathbf{Mon}^{\operatorname{pr},\operatorname{lax}}_\infty$ to $\mathbf{Pr}^L_\infty$ and is lax monoidal, sending the tensor product of presentably monoidal $\infty$-categories to the tensor product of presentable $\infty$-categories.
In particular, we now get the same statement as before, namely that for any symmetric operad $\mathcal{O}$, the functor $\mathbf{Cat}^{(-)}_\infty$ carries presentably $\mathcal{O}\otimes E_1$ monoidal categories to presentably $\mathcal{O}$-monoidal categories. Ok, so how does this answer your question?
Well, presentably $\mathcal{O}$-monoidal categories are closed, namely for each arity, we have an n-fold tensor product $\otimes^n:C^{\otimes n}\to C,$ and if we choose a family of $n-1$ objects $(c_1,\dots,c_{\hat{i}},\dots,c_n)$ (omitting the ith index), we obtain a colimit-preserving functor $C\to C$, which now admits an adjoint, the internal hom (if you aren't symmetric monoidal, these can all vary in complicated ways).
Now applying this to $\mathbf{Cat}^{\mathcal{V}}_\infty$, for $\mathcal{V}$ presentably $E_n$-monoidal, we see that it is canonically enriched over itself. Moreover, the assignment $\ast \mapsto \mathbb{1}_{\mathcal{V}}$ extends to a lax monoidal functor $\mathcal{S}\to \mathcal{V}$, which we can use to understand the 'underlying $(\infty,2)$-category $\widetilde{\mathbf{Cat}}^{\mathcal{V}}_\infty$'.
You can then iterate this procedure to produce an $E_{n-k}$ presentably-monoidal $(\infty,k)$-category of $k$-fold iterated enriched categories.
So I'm a bit confused what you're talking about. You do indeed get all of the correct kinds of natural transformations. There is a problem in the case where you choose $\mathcal{V}$ to be $\mathbf{Cat}_{\infty,n}$, but this is not an enriched story.
It is expected (or already proven, depending on who you ask) that there is another special biclosed $E_1$ presentably-monoidal structure on $\mathbf{Cat}_{\infty,n}$ for each $n\leq \omega$. This is called the Gray tensor product, or the lax Gray tensor product. This is not an enriched tensor product at all. It can't commute with or distribute over the Cartesian product. It is sui generis, and its right adjoints classify functors with lax or oplax natural transformations between them. This is an extremely important construction, but it has very little to do with enrichment. It looks like the easiest way to construct it is actually by first inducing it on $\mathbf{Cat}_{\infty,\omega}$, then inducing it on each finite $n$ by localization.
Thank you for your detailed answer! This probably fully answers my question, but there's one part that isn't clicking with me. Namely, does Gepner and Haugseng's work really apply to enrichment in monoidal $(n,r)$-categories? I can see how this procedure makes sense of creating an $(\infty,n+1)$-category of categories enriched in the $(\infty,n)$-category of $n$-fold iterated enriched categories, but what if my starting $(n,r)$-category is not of this form?
More precisely, does this encompass the theory of enrichment in a monoidal bicategory? I'm sorry if this is a dumb question...
@shibai (n,r)-categories are a presentable localization of (∞,r)-categories. They are also still cartesian-closed symmetric monoidal, so yeah, this would give you a way to write down some model of (n,n)-categories if you so-desired. But I was mentioning this before: Considering n-categories merely as enriched categories isn't quite right. There are other structures that are crucial to the whole theory that don't come merely from iterated enrichment.
If anyone is interested also, Rune and Hongyi have a recent paper doing all of this again for enriched operads, and it's pretty cool!
I think there's something wrong with my understanding. I was not trying to construct a theory of $n$-categories via iterated enrichment. For example, a monoidal bicategory need not be cartesian closed, or even braided, but nonetheless you can still make sense of enrichment in it, and the resulting collection of such enriched categories ought to be a $3$-category of sorts. Certainly specialising to the bicategory $\mathbf{Cat}$ and its cartesian product would produce the tricategory $\mathbf{Bicat}$, but this is only a very special case.
For an explicit example, could this framework address enrichment in the monoidal bicategory $\operatorname{Alg}R$ for $R$ a commutative ring, where the objects are $R$-algebras, $1$-cells are bimodules, and $2$-cells are bimodule homomorphisms, where the monoidal product is given by $\otimes_R$?
@shibai Rune sometimes hangs around the homotopy chat on MO. Maybe ping him and ask him about it. Enriching in bicategories is something that people haven't really done yet in homotopy theory, as far as I'm aware. If anyone knows the answer, it will probably be Rune.
My paper with David does not consider enrichment in monoidal bicategories (or monoidal $(\infty,2)$-categories for that matter).
@RuneHaugseng Thank you for clarifying! Might you know of some progress that has been done in this direction perhaps, or if this direction is even worth garnering attention?
|
2025-03-21T14:48:31.902694
| 2020-08-28T00:47:23 |
370276
|
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|
Stack Exchange
|
Algebraic/étale representation varieties
There has been extensive study of representation varieties $R(S, G)$ which are moduli spaces of representations of the fundamental group of a compact Riemann surface $S$ into a reductive group $G$. Concretely $R(S, G)$ is the quotient $\operatorname{Hom}(\pi_1(S), G)/G$ for conjugation action of $G$. It inherits an analytic structure from $G$. $R(S, G)$ can be identified with the moduli space of flat $G$-connections over $S$. This can in fact be done for arbitrary compact manifolds $M$ in place of $S$, and possibly in even greater generality.
Since Riemann surfaces can arise as complex points of smooth projective algebraic curves, it is natural to ask if algebraic counterparts of this construction and identification exist as well, so:
Fix a field $k$. Take a scheme $S$ and a reductive algebraic group $G$ over $k$. Consider the functor $\underline{\rho}^{ét}_{S,G} = \operatorname{Hom}(\pi^{ét}_1 (S), G)$, see if it is representable by a (formal - see R. van Dobben de Bruyn's comment below) scheme $\rho^{ét}_{S,G}$, take its GIT-quotient by the conjugation action by $G$, and call the resulting (formal) stack $R^{ét}_{S,G}$.
Or perhaps it'd be more appropriate to take one of the constructions of the fundamental group scheme of $S$, e.g., by Nori or Vakil-Wickelgren, instead of $\pi^{ét}_1 (S)$, and consider the analogous stack $R_{S,G}$ when it exists.
Some basic questions are:
(1) Can this be done always or only for certain $k, S$ and $G$? In which cases does one obtain a scheme rather than a stack? When and in what form does identification with the moduli space of flat $G$-connections over $S$ still hold?
(2) If $S$ is a smooth projective algebraic curve $C$ over $\mathbb{C}$, how are $R_{C,G}(\mathbb{C})$ and $R^{ét}_{C,G}(\mathbb{C})$ related to the representation variety of the associated Riemann surface, $R(C(\mathbb{C}), G(\mathbb{C}))$?
(3) If $k$ is a number field or a $p$-adic field, $S = \operatorname{Spec}(k)$ and $G = \operatorname{GL}_n$, what is $R^{ét}_{S,G} := \operatorname{Hom}(\operatorname{Gal}(\overline{k}|k), \operatorname{GL}_n)//\operatorname{GL}_n$? How are its $\overline{\mathbb{Q}}_l$-points related to the deformation spaces of $\overline{\mathbb{Q}}_l$-valued Galois representations (via connected components over residual representations?) What is $R_{S,G}$ in this case?
(4) What is a good reference for all this?
Because the étale fundamental group is profinite, it is more natural to ask for $\operatorname{Hom}(\pi_1^{\operatorname{\acute et}}(S),G)$ to be represented by a formal scheme or rigid analytic space. The coefficients of a local system are usually taken $\ell$-adic, so the same should apply to the space of such local systems.
|
2025-03-21T14:48:31.902898
| 2020-08-28T02:00:09 |
370278
|
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|
Stack Exchange
|
Why some operations on tensors don't give a tensor?
I asked the following question on math.stackexchange but no one seemed to have an authorative answer so I'm posting here hoping that experts will see it.
The gradient is a tensor $\nabla f:\mathbf{V} \to \mathbf{R}$ where the partial derivatives are evaluated at some point $(x_0, y_0, z_0)$. And evaluation of this linear form at some vector $v=(v_1,v_2,v_3)$ gives
$$
(\nabla f)(\mathbf{v}) = \partial_x f v_1 + \partial_y f v_2 + \partial_z f v_3
$$
Furthermore in going to a new coordinate system these partial derivatives transform in the expected way.
But what about a function $g:\mathbf{V} \to \mathbf{R}$ which is defined only using the partial derivative of $f$ in the $x$ direction.
$$
g(\mathbf{v}) = \partial_x f v_1 + \partial_x f v_2 + \partial_x f v_3
$$
As I understand this is not considered a tensor because in moving to a new coordinate system it does not transform correctly.
This has confused me endlessly. The strict definition considers a map such as $f:\mathbf{V}\times\mathbf{V} \to \mathbf{R}$ a tensor if linearity holds in each parameter. The function $g$ above certainly satisfies that. It seems to me that this definition is not used and that the definition of a tensor that is actually used consists of two parts.
linearity in each parameter (i.e. multilinear form),
and the algebraic structure of the coefficients is maintained in coordinate transformation
Because once we have calculated $\partial_x f$ it is just a scalar and we just hit $(\partial_x f,\partial_x f,\partial_x f)$ with the usual transformation for a covariant vector to get the new coefficients for $g$ in the new coordinate system. That these new coefficients don't have the right algebraic structure doesn't make multilinearity of $g$ go away. Is this at all correct?
One reason for a lack of authoritative answer could be want of a definition. Any of your $g$s, which is on $V^3$, not $V$, is (canonically identified with an element of $(V^*)^{\otimes3}$, so a mathematician might say (as you do) that it is a tensor. What $g$ you get, and hence with which element it is identified, depends on the coördinates used to take the partial derivative of $f$, so a physicist might care to argue that it is not a tensor. To answer your question of who is right, we need a definition. What definition do you want to use?
Indeed, a physicist would balk at more than one of the premises. First, that "$\partial_{x} f$ is just a scalar". It's not, it's the first component of a vector, which is not the same thing. But, let's indulge and interpret this as "in some fixed coordinate system, I have calculated $h=\partial_{x} f $, and I declare this $h$ that I have obtained a scalar." But then, secondly, $(h,h,h)$ is not a vector - after all, it doesn't change upon transformation. You don't get to "hit it with the usual transformation", you must show that it behaves according to that transformation - which it doesn't.
Your $g(\mathbf{v}) = \partial_x f v_1 + \partial_x f v_2 + \partial_x f v_3$ formula is not a well defined function $g:\mathbf{V} \to \mathbf{R}$.
The behaviour of tensors (1-forms in your case) under coordinate changes is a necessary condition for your g functions to be well defined (that is: independent of the coordinate system adopted to describe the vector space V=R^3).
Is $\mathbf V$ an abstract vector space or is it $\mathbb{R}^n$?
@KonstantinosKanakoglou That g is well defined is exactly the whole point of my post. See the comment by Stinking Bishop in https://math.stackexchange.com/questions/3801095/why-some-operations-on-tensors-dont-give-a-tensor where he elaborates on this.
@DeaneYang V is meant to be R^3.
@user782220, i think that the comment from the other site, essentially says the same thing (although not in a clear manner): that your g is not actually a function. So $(\partial_x f , \partial_x f , \partial_x f)$ is not really a 1-form (a (0,1) tensor). Does this answer your question?
For me it’s easier to understand if you work with an abstract vector space. Then there are no coordinates already defined. A tensor, such as the gradient, is well defined without any use of coordinates.. You get the formula you wrote only after you choose coordinates, say, by choosing a basis. But since the second formula requires coordinates to define and changes meaning in different coordinates, it can’t be a tensor.
|
2025-03-21T14:48:31.903224
| 2020-08-28T03:39:54 |
370280
|
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|
Stack Exchange
|
Primality testing using Chebyshev polynomials
Can you provide a proof or a counterexample for the claim given below?
Inspired by an alternative definition of the Frobenius primality test which is given in this paper I have formulated the following claim:
Let $n$ be an odd natural number that is not a perfect square and let $c$ be the smallest odd prime number such that $\left(\frac{c}{n}\right)=-1$ , where $\left(\frac{}{}\right)$ denotes a Jacobi symbol . Let $T_m(x)$ be the mth Chebyshev polynomial of the first kind , then $n$ is prime iff $T_n\left(1+\sqrt{c}\right) \equiv 1- \sqrt{c} \pmod{n}$ .
The test runs in polynomial time , you can try it here. I have verified this claim for $n$ up to $10^{10}$ .
The Android app that implements this test can be found on Google Play.
But $\left(\frac cn\right)$ is $-1$ means that $c$ is not a square $\mod n$, then what does $1-\sqrt c\ (\mathrm{mod}\ n)$ mean?
@მამუკაჯიბლაძე Please see Definition 1.1. in the linked paper.
I see, thanks. Still I am not sure how to make this rigorous. Is it equality in $\mathbb Z/n\mathbb Z[t]/(t^2-c)$ or something?
@მამუკაჯიბლაძე I don't know.
I believe the relevant place in your code is Mod(x,Mod(x^2-c,n)). Does this (for Sage) mean what I wrote? I guess so but I am not sure.
@მამუკაჯიბლაძე You are right.
"only if" is certainly true, following from the congruence $T_p(x) \equiv x^p \mod p$ for $p$ an odd prime. For "if" I have no clue, but in the most naive possible random model ( $T_n(1+\sqrt{c} )$ is a random element in $\mathbb Z[\sqrt{c}]/n$ for $n$ composite) it has a probability $1/n^2$ of failure for each $n$, hence based on your data we expect it is true for all $n$.
|
2025-03-21T14:48:31.903491
| 2020-08-28T03:48:35 |
370281
|
{
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"Chilperic",
"Stefaan Vaes",
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|
Stack Exchange
|
Is the set of all ICC amenable groups countable?
Is the set of all ICC amenable groups countable?
If "yes", then in general, the classes of all countable ICC groups that give rise to the same von Neumann algebra (factor) -- are these classes always countable, do we know?
Is it useful to consider this as an equivalence relation?
How about the same question with the crossed product construction of factors, and equivalence classes of (measure space with a group action) setups?
No, there are continuum many non-isomorphic simple locally finite groups (there are continuum many locally finite fields, take their $PSL_2$).
Thank you. Shouldn't you post it as an answer so that the question doesn't stay near the top?
Done. Just to help the reader, recall that "icc" means the FC-center, defined as the union of all finite conjugacy classes, is reduced to ${1}$. That is, all conjugacy classes other than ${1}$ are infinite.
The set of countable ICC groups doesn't exist, so I think you're asking about isomorphism classes.
There are continuum many non-isomorphic locally finite fields (e.g., take, for $S$ any set of primes, the invariants by $\prod_{p\in S}\mathbf{Z}_p$, which has Galois group $\prod_{p\notin S}\mathbf{Z}_p)$ in the algebraic closure of $\mathbf{F}_p$.
It's known that non-isomorphic fields give rise to non-isomorphic groups $\mathrm{PSL}_2$. Hence, taking $\mathrm{PSL}_2$ of such fields yield continuum many non-isomorphic locally finite (hence amenable) groups, which are infinite simple, hence icc.
(Actually there are continuum many non-isomorphic solvable icc finitely generated groups as well, by another argument.)
As for the second question, the equivalence relation on the class of countable ICC groups given by $G \sim \Gamma$ if and only if $L(G) \cong L(\Gamma)$ is very interesting and usually called $W^*$-equivalence of $G$ and $\Gamma$.
On page 45 of [Con82], Connes conjectures that the $W^*$-equivalence class of an ICC property (T) group is a singleton (up to isomorphism). Such groups are now called $W^*$-superrigid. This conjecture is wide open. At the moment, there is no counterexample, but also no example: there is no known $W^*$-superrigid property (T) group.
In Section 4 of [Pop07] it is proven that an ICC property (T) group is $W^*$-equivalent with at most countably many nonisomorphic groups.
In [IPV10] we introduced the first family $W^*$-superrigid ICC groups. These examples are given by a generalized wreath product construction and do not have property (T).
For all countably infinite abelian groups $\Gamma$, the group von Neumann algebra $L(\Gamma)$ is the unique diffuse abelian von Neumann algebra. One can deduce that for all countably infinite abelian groups $\Gamma_1,\Gamma_2$, the ICC group $\Gamma_1 * \Gamma_2$ is $W^*$-equivalent to the free group $\mathbb{F}_2$. More generally, by Theorem 4.6 in [Dyk92], whenever $\Gamma_1$ and $\Gamma_2$ are infinite amenable groups, we have that their free product $\Gamma_1 * \Gamma_2$ is $W^*$-equivalent to $\mathbb{F}_2$.
[Con82] A. Connes, Classification des facteurs. In Operator algebras and applications, Part 2 (Kingston, 1980), Proc. Sympos. Pure Math. 38, Amer. Math. Soc., Providence, 1982, pp. 43–109.
[Pop07] S. Popa, Deformation and rigidity for group actions and von Neumann algebras. In International Congress of Mathematicians (Madrid 2006), Eur. Math. Soc., Zürich, 2007, pp. 445-477.
[IPV10] A. Ioana, S. Popa and S. Vaes, A class of superrigid group von Neumann algebras. Ann. of Math. 178 (2013), 231-286.
[Dyk92] K. Dykema, Free products of hyperfinite von Neumann algebras and free dimension. Duke Math. J. 69 (1993), 97–119.
Thanks very much, Stefaan and YCor. I was aware of some of this work on cardinality of these classes. I was wondering more about thinking of them along the results we have for countable Borel equivalence relations. Does Silver's dichotomy hold -- either countable number of classes or an uncountable number of singletons? Is it possible to have a group action so that the classes are orbits? That sort of thing.
I think that in your question, your are only mentioning countability of equivalence classes. Maybe you can post a new and more precise question about these other points that you have in mind.
|
2025-03-21T14:48:31.903795
| 2020-08-28T05:05:06 |
370283
|
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}
|
Stack Exchange
|
What sets can be unraveled?
A set $X\subseteq\omega^\omega$ is unravelable iff there is a possibly larger set $A$ and a clopen set $Y\subseteq A^\omega$ (with respect to the product topology coming from the discrete topology on $A$) such that winning strategies for the game on $A$ with payoff set $Y$ can be converted to winning strategies for the game on $\omega$ with payoff set $X$ in a particularly simple way (see Martin's paper A purely inductive proof of Borel determinacy for the precise definition). By Gale–Stewart, unravelability implies determinacy; however, it is not the only route to proving the determinacy of a pointclass.
My question is whether there is a known upper bound on the complexity of an unravelable pointclass. To make this somewhat precise:
Is it consistent with $\mathsf{ZFC}$ that every set in $\mathcal{P}(\mathbb{R})^{L(\mathbb{R})}$ is unravelable?
The strongest results I can find are vastly weaker than this, namely that under large cardinal assumptions $\Pi^1_1$ sets are unravelable (Neeman 1,2; I remember seeing the same result for Borel-on-$\Pi^1_1$ sets, but I can't find a reference for it at the moment). On the other hand, I don't see an easy proof that this is anywhere close to the most unravelability we can expect, nor can I find this stated in the literature.
I emailed Itay Neeman, and he told me the following:
As far as I know it's open. I don't think anything is known about
unraveling beyond what you can get from my methods. These give the
Suslin operation on $\Pi^1_1$ sets, and various iterations of that. In
terms of the large cardinal hierarchy it's still all just using
measures. Anything above that is open I think.
In particular I don't think it's known if $\Sigma^1_2$ sets can be
unraveled.
For now I think that settles this question. (Itay later said that it's unclear whether we "should" expect unravelability of $\Sigma^1_2$ sets from large cardinals or not in the first place. So I think my main takeaway at this point is that unravelability is not best thought of as yet another tameness property, where the slogan "large cardinals make definable things tame" is expected to prevail.)
The opposite of settled, but that makes it more interesting, no?
@AndrésE.Caicedo Indeed. I'm curious: do you have a sense as to whether unravelability of $\Sigma^1_2$ or more is something we should expect from large cardinals? (I asked Itay, and his response was that it was unclear at present but I'm curious if anyone else has ideas.)
|
2025-03-21T14:48:31.903989
| 2020-08-28T05:51:18 |
370285
|
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"Cusp",
"Ian Agol",
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|
Stack Exchange
|
Weil-Petersson metric with respect to covering
Let $S$ be a closed oriented surface of genus $g\geq 2$. Consider the Teichmuller space $T(S)$. Let $d_t$ be the Teichmuller metric and $d_{WP}$ be the Weil-Petersson metric on $T(S)$. Let $P:S_1\rightarrow S$ be a covering map. It is a well known result that if $P$ is finite then the embedding $(T(S),d_t)\rightarrow (T(S_1),d_t)$ is an isometric embedding (see Section 7). My question is
Q) Is the map $((T(S),d_{WP})\rightarrow (T(S_1),d_{WP})$ an isometric embedding?
My idea is the following. As the WP pairing and the (co)metric on $T(S)$ is defined by $\pi_1(S)$ invariant quadratic differentials, they are naturally $\pi_1(S_1)$ invariant. Hence the above map is an isometry. But for this argument, I don't need $P$ to be finite. I am not sure whether this argument is correct or I am making some silly mistakes.
Any kind of suggestion/reference will be extremely helpful. Thanks in advance.
See the remark on p. 2 of this paper: https://arxiv.org/abs/0806.2304
For a covering, the embedding of Teichmuller space induced by a covering increases distance by the square root of the degree of the cover.
@IanAgol Thank you very much for the reference. I forgot about the normalization.
|
2025-03-21T14:48:31.904106
| 2020-08-28T06:14:39 |
370288
|
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"Gerry Myerson",
"Gordon Royle",
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"sort": "votes",
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}
|
Stack Exchange
|
Finding a special plane graph with some requirements on the faces
Is there a plane graph such that
(1) the outer face has degree 3, i.e, is a triangle,
(2) every inner face has degree 5, and
(3) any two degree 5 faces share at most one commong edge.
If you have a plane graph with a pair of inner faces of degree 5 that share more than one edge, you can simply "insert" a dodecahedron into one (or both) of the faces to alleviate the situation. So an example should be easy to construct.
To supplement Gordon's answer based on my comment, here's a planar drawing:
Using Timothy Budd's suggestion, it is easy to find such a graph by gluing two dodecahedra together. Here is the SageMath code to make this graph.
d1 = graphs.DodecahedralGraph()
d2 = graphs.DodecahedralGraph()
h = d1.disjoint_union(d2)
h.merge_vertices([(0,0),(1,0)])
h.merge_vertices([(0,1),(1,1)])
h.merge_vertices([(0,2),(1,2)])
h.add_edge((0,3),(1,3))
h.relabel()
Here is the picture
If you need to check that everything is satisfied then the below code finds all the 5-faces as sets of vertex-pairs (hence the nested "Set" stuff) and then checks all the pairwise intersections.
f5s = [Set([Set(e) for e in f]) for f in h.faces() if len(f) == 5]
[len(x.intersection(y)) for x in f5s for y in f5s if x != y]
This is a plane graph?
@GerryMyerson It is planar, else Sage would complain if I tried to get its faces, Timothy Budd has provided the sensible plane drawing...
OK, very good..
|
2025-03-21T14:48:31.904243
| 2020-08-28T06:18:45 |
370290
|
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|
Stack Exchange
|
Procedure to color the edges of a circulant graph
From the first theorem in this paper, it is clear that a cayley graph on abelian group for all generating sets of even order is class $1$, that is can be edge colored in exactly $\Delta$ colors. But, this procedure and proof is not clear for me in the case of circulant graphs.
In the proof of the theorem $2.3$ of the above paper, the author proves the theorem for abelian groups by first proving the theorem for any graph defined on any generating set on the groups $H\rtimes G$. But, how can a graph defined on a cyclic group be written in this form? Specifically, say we have a cyclic group of order $2p$, where $p$ is $\it{odd}$. Then, though the group is isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_2$, i cannot see how a graph defined on the whole group $\mathbb{Z}_{2p}$ is isomorphic to a cartesian product of graphs of the form $\Gamma(S:\mathbb{Z}_p)\times\mathbb{Z}_2$, as expected from the paper. Am I missing something? Thanks beforehand.
The 14-cycle can be coloured with two colours, but there is no reason why it must be coloured with only two colours. Using more colours on the 14-cycle may leave you more flexibility in colouring the two 7-cycles in such a way that fewer colours are required overall.
@GordonRoyle thanks! so does this imply for a general algorithm we use all the colors on all the cycles (according to their lengths)?
Have you tried looking at the proof of the statement you claim, to see if it gives an algorithm?
@verret yes. it is the stong's paper. i quite dont get the algorithm, if any, described there. And surely, this is non-trivial, for an algorithm for any circulant graph seems not easy. And your downvote seems unjustified!
So the paper you already know about (which you should probably have explicitly included in your question) has the answer to your question. What you then asking?
@verret no, the paper shows that it is possible to 1-factorize the graph. But, the procedure is not clear. The paper is here
Right, and that proof can be turned into an algorithm. It's not very complicated, the whole proof with preliminaries is less than a page. So again, what is the question?
@verret edited the post. It is not clear to me how the proof holds for graphs on cyclic groups. Would be helpful if you could elaborate
Which specific part of the proof do you not see how it holds in this particular case?
@verret the part when the paper says that any graph for any gen. set on a group of the form $H\rtimes G$ can be edge colored using the edge coloring of the graph on $S\times {s}\quad s\in G\quad S\subset H$.
|
2025-03-21T14:48:31.904452
| 2020-08-28T07:13:07 |
370294
|
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"Nicholas Kuhn",
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|
Stack Exchange
|
Invariants of a $kG$-module via its comoposition series, when does $M^P \supsetneq N^P$ hold for a $p$-group for $N\subseteq M$ maximal?
Let $G$ be a finite group, $k$ a field, $M$ a $kG$-module, $M^G$ the invariants of $M$ under $G$, $P$ a Sylow $p$-subgroup of $G$ where $p = \text{char}(k)$, $N$ a maximal submodule of $M$ and $S$ the trivial simple $kG$-module. One can show that
\begin{align*}
M^G \supsetneq N^G \Leftrightarrow M^P \supsetneq N^P \text{ and } M/N \cong S.
\end{align*}
Additionaly if $M^G\supsetneq N^G$ then $\text{dim} M^G/N^G = 1$. Using this one could compute the invariants $M^G$ of $M$ via the composition series of $M$. Now my question is, if there is any way to determine $M^P \supsetneq N^P$ for a $p$-group?
Oh yes, thanks!
Standard homological algebra tells you that this happens if and only if there is a nonzero map map of modules from S to M/N that lifts back to M. There isn't much more to say than that, though if you are assuming that M/N=S, then there is such a nonzero map if and only if M is the direct sum of N and S.
Keep in mind that the dimension of the invariants is just counting the number of occurrences of the trivial module as a SUBmodule.
|
2025-03-21T14:48:31.904571
| 2020-08-28T07:46:51 |
370295
|
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|
Stack Exchange
|
Relative canonical class of blowing-up a flag ideal
Let $X$ be a smooth complex projective variety of dimension $n$. Consider a flag ideal $I$ on $X\times \mathbb{P}^1$, namely,
$$
I=I_0+I_1t+I_2t^2+\cdots+I_{N-1}t^{N-1}+(t^N)\,,
$$
where $t$ is the variable on $\mathbb{C}\subset \mathbb{P}^1$, $I_j$ are coherent ideal sheaves on $X$.
We assume that the flag ideal $I$ is special in the sense that each $I_j$ is reflexive, corresponding to an effective divisor $D_j$. Moreover, we assume that each $D_j$ is snc.
Now let $Y$ be the normalised blowing-up of $X\times \mathbb{P}^1$ along $I$. Then how can we compute the relative canonical class $K_{Y/X\times\mathbb{P}^1}$ in terms of all these $D_j$?
Even in the simplest case of deformation to the normal cone (i.e. when $N=1$), this does not seem to be clear to me.
In the case $N=1$, where the ideal is $I = \mathcal{O}(-D) + (t)$ for some snc divisor $D$ on $X$, $I$ is a global complete intersection of codimension 2 ($V(I) = D \times \{0\}$), with normal bundle $\mathcal{N} = \mathcal{O}(D)|_D \oplus \mathcal{O}_D$. If $Y := \mathrm{Bl}_I X$ and if $\pi\colon Y \to X$ is the projection, then the exceptional locus $E = \pi^{-1}(D \times \{0\})$ can be identified as $E \simeq \mathrm{Proj}_D \mathcal{N}$, a $\mathbb{P}^1$-bundle over $D \times \{0\}$. In particular, if $D = \sum_i D_i$ (with the $D_i$ smooth and irreducible) then $E = \sum_i E_i$ where the $E_i = \pi^{-1}(D_i)$ are also smooth and irreducible.
Now, we know that $K_Y - \pi^* K_{X \times \mathbb{P}^1} = \sum_i a_i E_i$. The coefficient $a_i$ only depends on the valuation corresponding to $E_i$ (see Rmk. 2.23 of Kollár-Mori's Birational Geometry of Algebraic Varieties), so it can be computed on any neighborhood of the generic point of $E_i$. This essentially reduces to the case where $D$ is smooth, and so Lem. 2.29 of the above reference shows $a_i=1$ for all $i$.
The above argument fails when $N>1$ since $I$ is no longer a complete intersection. However, if we assume that $I_i = \mathcal{O}(-D_i)$ for each $i$, where the $D_i$ are smooth divisors on $X$ so that $\sum_i D_i$ is snc, there might be a tractible description of $E$ in terms of the strata of $\sum D_i$ (the components of intersections $\cap_{i \in I} D_i$, where $I \subseteq \{0, \dots, N-1\}$). For example, if $X = \mathbb{A}^2$ and $I = (x) + (yt) + (t^2)$, the exceptional locus has 2 components: a $\mathbb{P}^1$-bundle over $V(x) \times \{0\}$ and a divisor centered over $(0, 0)$. At the risk of extrapolating from 1 example: one could ask if with the assumptions of this paragraph, and setting $Z_m = \bigcap_{i=0}^m D_i$, whether $K_Y - \pi^* K_X = \sum_{m=0}^{N-1} (m+1) E_m$, where $E_m$ is an exceptional divisor centered at $Z_m$. Note that $m+1 = \mathrm{codim}(Z_m \subset X \times \mathbb{P}^1) -1$.
Edit: Macaulay2 code for the example.
i1 : k = ZZ/9973
o1 = k
o1 : QuotientRing
i2 : XtimesP1 = k[x, y, t]
o2 = XtimesP1
o2 : PolynomialRing
i3 : I = ideal(x, y*t, t^2)
2
o3 = ideal (x, y*t, t )
o3 : Ideal of XtimesP1
i4 : Y = reesAlgebra(I)
o4 = Y
o4 : QuotientRing
i5 : E = I*Y
2
o5 = ideal (x, y*t, t )
o5 : Ideal of Y
i6 : primaryDecomposition E
2 2
o6 = {ideal (t, x, w ), ideal (x, t , y*t, y )}
1
o6 : List
i7 : describe Y
XtimesP1[w , w , w ]
0 1 2
o7 = -----------------------------------------
2
(y*w - t*w , t w - x*w , y*t*w - x*w )
1 2 0 1 0 2
```
|
2025-03-21T14:48:31.904799
| 2020-08-28T07:52:32 |
370296
|
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|
Stack Exchange
|
What are the rational solutions of $y^2 = 4x^n + z^{n-1}$?
Let $n>3$ be an integer. What are the rational solutions of
$$y^2 = 4x^n + z^{n-1}$$
?
This is not elliptic-curves nor hyperelliptic-curves. It is a surface rather than a curve. Nevertheless the usual attack to this kind of problems is as in ABC conjecture or Fermat's last theorem, which involves considering an associated Frey curve. I don't know how deep one can go with this equation.
We consider two cases: $n$ is odd or $n$ is even. If $n=2k+1$ then put
$$
x=uz, \quad y=vz^{k},
$$
where $u, v$ are non-zero rational parameters. Thus $v^2z^{2k}=4u^{2k+1}z^{2k+1}+z^{2k}$ and after division by $z^{2k}$ we get linear equation in $z$ which can be easily solved, i.e., $z=(v^2-1)/(4u^{2k+1})$. We thus get
$$
x=\frac{v^2-1}{4u^{2k}},\quad y=v\left(\frac{v^2-1}{4u^{2k+1}}\right)^{k}.
$$
If $n=2k$ one can use exactly the same type of reasoning, i.e., we put
$$
y=ux^{k},\quad z=vx.
$$
After necessary simplifications, we get $x=v^{2k-1}/(4-u^2)$ and thus
$$
y=u\left(\frac{v^{2k-1}}{4-u^2}\right)^{k},\quad z=v^{2k}/(4-u^2).
$$
Are there other solutions besides yours?
Some solutions are omitted. If $n$ is odd and $x=0$ then we need to take $v=\pm 1$. However, then we get $y=0$ from our parametrization. This is not a real problem because we can parametrize $y^2=z^{n-1}$ easily. The same situation occurs in the case $n$ even and $v=0$. In both cases the reason for this is simple: the maps constructed are rational and not defined everywhere.
So you gave rational parametrization of two families of rational surfaces?
@joro Exactly. That is what I have done.
|
2025-03-21T14:48:31.904938
| 2020-08-28T07:52:40 |
370297
|
{
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|
Stack Exchange
|
number of integer points inside a triangle and its area
Let $T$ be a triangle in $\mathbb{R}^2$ defined by $y = \alpha x$, $y = \beta$ and $x = \gamma$ where
$\alpha, \beta, \gamma \in \mathbb{R}_{>0}$. I am interested in obtaining an estimate for the number of integral points inside (either within or on the boundaries) of $T$ which I denote $N$. A simple computation yields
$$
N = Area(T) + E
$$
with
$$
|E| \ll |\gamma - \frac{\beta}{\alpha}| + |\beta - \alpha \gamma|
$$
in other words $E$ is bounded by the sum of the two side lengths. Are there ways to get better upper bound than this? Any comments are appreciated.
In general I don't think you can do much better. Consider the triangle with end points $(1,1), (n,1), (n,(n+1)/n), n \in \mathbb{N}$ for example. The area is equal to $1/2$, but there are $n$ integral points on the boundary. This is essentially the worst case though.
The question reduces to estimates of the sums $\sum_{k=1}^n {\gamma k}$. For fixed irrational $\gamma$ and large $n$ this is about $n/2$.
Better upper bound on the deviation from the area? Certainly not: if you move $\beta$ and $\gamma$ by almost $1$ from some integer plus epsilon to the next integer minus epsilon, the number of integer points does not change but the area changes pretty much by the sum of the sides that is your current bound. Perhaps you wanted to ask something else or I misunderstood the question?
The bound on $|E|$ can certainly be improved: $|E|\le|E|$; so, we get the "perfect" (but completely useless) upper bound, $|E|$, on $|E|$.
To make the problem meaningful, we need to specify the terms in which the upper bound is to be expressed. Counting the unit squares intersecting the boundary of the triangle, it is easy to see that
$$|E|\ll s:=a+b+1,$$
where $a$ and $b$ are the lengths of the horizontal and vertical sides of the right triangle (in terms of your $\alpha,\beta,\gamma$, we have $a=|\gamma-\beta/\alpha|$ and $b=|\beta-\alpha\gamma|$; your bound is missing "$+1$"). Note that $s>1$.
Let us show that $s$ is the best (up to a constant factor) upper bound on $|E|$ in terms of $s$. Indeed, take any real $s>1$ and consider the right triangle $T$ with vertices $(0,0),(0,a),(a,a)$, where $a:=(s-1)/2>0$, so that $a+a+1=s$. Then the area of $T$ is $A=a^2/2$ and
the number $N$ of integral points that are either inside $T$ or on the boundary of $T$ is the number of pairs $(i,j)$ of integers such that $0\le i\le j\le a$. The latter triple inequality can be rewritten as $0\le i\le j\le n$, where $n:=\lfloor a\rfloor\ge0$, so that $n\le a<n+1$. So, $N=(n+1)(n+2)/2$ and $A<(n+1)^2/2$. Hence,
$$|E|=|N-A|=N-A \\
>\frac{(n+1)(n+2)}2-\frac{(n+1)^2}2 \\
=\frac{n+1}2\ge\frac{2(n+1)+1}6>\frac{2a+1}6=\frac s6.$$
Thus, the best upper bound on $|E|$ in terms of $s$ is $\,\asymp s$, as claimed.
Working a bit harder, one can show that $|E|\le s$. So, the best upper bound on $|E|$ in terms of $s$ is between $s/6$ and $s$.
Since the question is posed in terms of the parameters $\alpha,\beta,\gamma$, I think one hopes for a bound in terms of these, rather than in terms of $a,b$.
@GerryMyerson : I have inserted expressions of $a,b$ in terms of $\alpha,\beta,\gamma$. However, the expression of the upper bound in terms of $a,b$ seems more transparent.
|
2025-03-21T14:48:31.905178
| 2020-08-28T08:36:33 |
370302
|
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"Joseph O'Rourke",
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|
Stack Exchange
|
A claim on partitioning a convex planar region into congruent pieces
Let us define a perfect congruent partition of a planar region $R$ as a partition of it with no portion left over into some finite number n of pieces that are all mutually congruent (ie any piece can be transformed into another piece by an isometry. We consider only cases where each piece is connected and is bounded by a simple curve).
Note: It is known there are convex planar regions - indeed quadrilaterals - which do not allow perfect congruent partition for any n ([1] proves a stronger result).
Claim: If a convex polygonal $R$ allows a perfect congruent partition of itself into $N$ non-convex pieces each with finitely many sides, then $R$ also allows a perfect congruent partition into $N$ convex pieces with finitely many sides. In other words, allowing the pieces to be non-convex polygons does not improve the chances of a convex planar region achieving a perfect congruent partition into $N$ pieces.
I know no proof, no counter example. One can consider replacing 'congruent' with 'similar' in the above question. Some more related thoughts are in [2].
References:
https://www.research.ibm.com/haifa/ponderthis/challenges/December2003.html
https://arxiv.org/abs/1002.0122
A bit added on March 15th, 2024: Are there convex polygonal regions with even number of edges that has this property of being perfect congruent partitionable only into 2 non-convex pieces?
If you do not demand the regions to be simple polygons and congruence includes rotations, the counterexample is simple:take the regular hexagon and build something on its sides in the pattern aabbcc. Then you can split it into two congruent pieces by connecting the center to the "first" a,b,c and then to the "second" but no convex partition (i.e, splitting with a single line) into two congruent pieces will be possible. If you mean something more restricted, just say what exactly.
Thanks. The partition is to be into simple polygonal regions. Made a couple of edits to the question.
This picture looks like a counterexample with $N=2$ and $R$ a convex pentagon:
This should work more generally starting from an $n \times (n+1)$ rectangles
for any integer $n>1$, removing two congruent right triangles that are
not isosceles (the picture shows $n=3$ with $2:5$ triangles).
Replacing the heavy lines with more complicated polygonal convex paths
yields convex polygons with more sides.
Beautiful! Nice to use a pentagon that cannot be partitioned into two congruent convex pieces, as that would require a chord to create identical angles.
Really elegant example; thanks!! Let me add a small extension to the question, basically to bound this property: to find that convex shape with such a perfect congruent partition into 2 non-convex pieces that 'wastes' the highest fraction of its area when given the best partition into 2 convex pieces.
|
2025-03-21T14:48:31.905506
| 2020-08-28T09:02:04 |
370303
|
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"Ariyan Javanpeykar",
"Evgeny Shinder",
"Francesco Polizzi",
"Nguyen",
"Nick L",
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|
Stack Exchange
|
Variety of negative Kodaira dimension contains a projective line
Does a smooth projective variety over an algebraically closed field of negative Kodaira dimension contain a projective line?
I do not want to assume any conjectures.
What is your precise definition of "projective line"? Just a smooth rational curve, or you also require that it has degree $1$ with respect to some fixed polarization?
@FrancescoPolizzi smooth rational curve
I remark that it is true in dimension at most $3$, over $\mathbb{C}$. (This probably isn't the easiest way to see it and is not a complete solution, but it was too short for a comment. )
In dimension $2$ it follows from the fact that all of the minimal surfaces with negative Kodaira dimension are covered by smooth rational curves (they are either $\mathbb{P}^1$-bundles or $\mathbb{P}^2$).
In dimension $3$ it follows from the existence and classification of extremal contractions for smooth $3$-folds due to Mori. Let $X$ be a smooth $3$-fold with Kodaira dimension $-\infty$ then $X$ is uniruled (Theorem 6.1.8 (FV)). Hence by a result of Miyoaka and Mori there is a non-empty Zariski open $U \subset X$ such that each point in $U$ is contained in an irreducible curve $C \subset X$ with $K_{X}.C <0$ (Theorem 6.1.4 (FV)), in particular $K_{X}$ is not nef.
(by (FV) I mean the book "Fano varieties" by Iskovskikh and Prokhorov).
Hence by the contraction theorem, there is some extremal contraction. You may consult the list of extremal contractions which can occur. In all cases; other than when the image is a point, some fibre of the extremal contraction clearly contains smooth rational curves (For conic bundles and del Pezzo bundles one needs that a general fibre is smooth, which is true). If the image is a point, $X$ is Fano $3$-fold with $b_{2}=1$. Then since Fano $3$-folds are classified we may go through each of the $17$ prime Fano $3$-folds and check there is a smooth rational curve in each.
In dimension 3, a variety with negative Kodaira dimension is uniruled.
I used that fact in my answer, although it is not clear to me that uniruledness immediately gives a smooth rational curve in $X$ as asked by the OP.
@NickL Let's say that a proper variety is "covered by rational curves" if, for every proper closed subset $Z\subsetneq X$, there is a rational curve passing through a point in $X\setminus Z$. Note that this is clearly a birational invariant of the class of proper varieties. Moreover, if $X\to Z$ is a proper surjective morphism and $X$ is covered by rational curves, then so is $Z$, right? This is what abx probably had in mind (and is simpler than what you wrote).
I am sorry, but I don't see any relationship between what you said and smooth rational curves. Could you explain what you mean?
I wonder if a single uniruled parametrization may ever suffice in this question. Namely, do there exist uniruled parametrizations with images of all components of all fibers being singular rational curves? It's easy to arrange such parametrization with general curve being singular, but these typically break up to have smooth components.
|
2025-03-21T14:48:31.905735
| 2020-08-28T09:33:49 |
370306
|
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|
Stack Exchange
|
Number of critical points of sum of two functions
I ran into the following "simple" question and I am wondering whether there are any references, which might help me. I am coming from statistics, so I am not so aware which branch of math could have already dealt with this question.
Let $S\subset\mathbb{R}^D$ be compact and assume that $f,g: S\rightarrow \mathbb{R}$ are $C^2$ Morse functions (so basically requiring finitely many critical values). Further assume that $f+g$ has only finitely many isolated critical points.
Denote with $C_f$ and $C_g$ the number of critical points of $f$ and $g$.
Is it possible to find a bound on $C_{f+g}$ depending only on $C_f, C_g$ and $D$ and maybe $S$?
For polynomials this is easy and simply Bezout's Theorem, yet I need something like it for more general functions.
Suppose that $f$,$g$ are also stable, meaning that different critical points have different critical values. This is a generic situation, i.e., probability 1 event. Consider the function $h_t=f+tg$. For $t$ very small $h_t$ has the same number of critical points as $f$ and for $t$ large it has the same number of critical points as $g$.
Is the following example helpful? This is inspired by the fact that no bound on the "degree" of the two functions $f$ and $g$ is assumed. As far as I understand Bezout's theorem, this would make a bound difficult even for polynomial functions.
Here let $A,B > 1$ be large and $0 < \epsilon < 1$ be small. Let $D = 1$, and $S = [0,1]$ be the closed unit interval. Define $f: x \mapsto Ax + \epsilon \cos(Bx)$ and $g: x \mapsto Ax$. Then $f'(x) = A - B\epsilon \sin(Bx)$ and $g'(x) = A$, so that neither function has critical points in $[0,1]$ provided just that $A > B \epsilon$. However their difference $f-g$ has derivative $(f-g)'(x) = - \epsilon \sin(Bx)$. By taking suitably large $A,B$ this function has arbitrarily many critical points in $[0,1]$.
Perhaps it is worth pointing that really only $B$ needs to be large in the argument above. In particular, by first picking $B$ large and then a small $\epsilon > 0$ in terms of it, one is free to pick a small $A$ as well. Therefore the functions $f,g$ can be taken to have arbitrarily small first and second derivatives.
Hey, thanks. That is a nice, simple example. Still trying to bound moments of the critical value of a "simple" random field obtained through basis functions, yet I might need to use Kac-Rice-formulas to do so.
|
2025-03-21T14:48:31.905911
| 2020-08-28T10:05:09 |
370309
|
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|
Stack Exchange
|
Commutative algebras associated to simple Lie algebras
In Section 2 of the article https://www.sciencedirect.com/science/article/pii/S0021869307000385, the authors study the center $Z=Z_Q$ of certain preprojective like algebras associated to the simply laced Dynkin quivers/simple Lie algebras and show that the dimension of the center is equal to the number of roots of the simple Lie algebra. My knowledge of this stuff is very limited and Im not sure whether there is more research on this topic.
Question 1: Are explicit quiver and relations known for the centers $Z_Q$ associated to the simple Lie algebra?
In type $A_n$, the relations are very easy: $Z_{A_n}=k[x,y]/(p_n(x,y))$ , where $p_n(x,y)$ are all homogenous monomial of degree $n$. For example $Z_{A_3}=k[x,y]/(x^3,x^2y,xy^2,y^3)$.
Question 1.5: Since those algebras have a Dynkin type classification, it seems likely that they come from another classification problem from homological algebra/representation theory. Is something in this direction known? Did they appear elsewhere?
In type $A_n$ the natural basis $e_i$ of $Z_{A_n}$ realises the root poset via the poset of submodules of the form $e_i Z_{A_n}$.
For example for $n=3$ and $B:=Z_{A_3}$ those modules are $B$, $xB$, $yB$, $x^2B$, $y^2B$, $xyB$.
Question 2: Is there such a natural basis $e_i$ in the other Dynkin types too such that the submodule lattice realises the root poset?
Those algebras are over $\mathbb{C}$ and thus one can not calculate their quiver and relations using the GAP package QPA (or is there a trick to calculate quiver and relations at least over the rationals for centers of quiver algebras using QPA?).
But here is the quiver and relations of Dynkin type $D_4$ over the finite field with 17 elements (the field should probably have high enough characteristic to cause no problems) directly from the QPA output:
$Z_{D_4}=K[a1,a2,a3]/I$, where
I=[ (Z(17)^0)*a2^2+(Z(17)^10)a3^2+(Z(17)^7)a1^2a2+(Z(17)^0)a1^2a3, (Z(17)^0)a1a2^2, (Z(17)^0)a1a2a3, (Z(17)^0)a1a3^2, (Z(17)^0)*a2^3,
(Z(17)^0)a2^2a3, (Z(17)^0)a2a3^2, (Z(17)^0)*a3^3, (Z(17)^3)*a2^2+(Z(17)^12)a2a3+(Z(17)^0)*a3^2+(Z(17))a1^2a2+(Z(17)^0)*a1^4, (Z(17)^0)a1^3a2 ]
Of course those relations do not need to be in an optimal way. One might expect that there are "type $D_n$" versions of the homogeneous monomial polynomials that give the relations in that case. Are there any good candidates for such polynomials (that might even exist in types like $G_2, F_4 , B_n$ etc.) ? A good presentation of $I$ might be also key to find a canonical/good basis of $Z_Q$.
I was not able yet to find a "canonical" basis of $Z_{D_4}$ that realises the root poset. But $Z_{D_4}$ has 4-dimensional socle and that fits with the fact that the root poset of $D_4$ has 4 minimal elements.
Question 3: Is there such a canonical basis at least for $D_4$ and what does it look like?
Assume we have a canonical basis (as in type $A_n$) and let $M$ be the direct sum of all submodules $e_i Z_Q$. Let $C_Q:=End_{Z_Q}(M)$.
Question 4: Is it true that $C_Q$ has finite global dimension? What is the global dimension in that case?
I am especially interested in Dynkin type $A_n$. Here the basic version of $M$ is given by choosing one submodule of each dimension. I can show that $C_{A_n}$ at least has Cartan determinant equal to one, so it seems likely that it has finite global dimension. For $n=2$, the global dimension is 2, for $n=3$ and $n=4$ it was 3.
Furthermore in type $A_n$ the algebra $C_Q$ was even a quasi-hereditary quadratic algebra for $n=2,3,4$. I guess it might even be Koszul.
Question 5: Is $C_{Q}$ for $Q=A_n$ a quasi-hereditary Koszul algebra?
Here quiver and relations for $C_{A_3}$:
Quiver( ["v1","v2","v3"], [["v1","v2","a1"],["v1","v2","a2"],["v2","v1","a3"],["v2","v3","a4"],["v2","v3","a5"],["v3","v2","a6"]] )
[ (Z(17)^0)a1a3, (Z(17)^0)a2a3, (Z(17)^8)a1a5+(Z(17)^0)a2a4, (Z(17)^8)a3a1+(Z(17)^0)a4a6, (Z(17)^8)a3a2+(Z(17)^0)a5a6 ]
|
2025-03-21T14:48:31.906169
| 2020-08-28T10:10:53 |
370310
|
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|
Stack Exchange
|
Disjoint paths in temporal graphs
Given a graph $G=(V,E)$ and a pair of source-destination nodes $s$ and $t$. Time is divided in periods with the total number of periods denoted by $T$. Each edge $e$ is either operational or broken at each period $\tau$. A path is good at period $\tau$ if all its edges are operational at period $\tau$. Suppose we cannot find a path that is good for the entire $T$ periods. We seek a pair of node-disjoint paths such that at least one of them is good at each period $\tau$. What is the hardness of this problem? Is it related to some known problem? Any suggestion in developing algorithm to find the good path pair?
Is the given G guaranteed to contain a good path pair?
@RayButterworth Not necessarily. In fact we can formulate the entire problem as (1) decide the existence of such path pair (2) find them.
https://www.sciencedirect.com/science/article/pii/S1572528612000035 looks like a bad news (NP-complete even for $T=2$, see Lemma 2). The killer seems to be the requirement that the paths have to be vertex disjoint. Do you really need it?
@fedja Thank you very much for the reference (the result is not surprising to me). Is there any nice algorithm if the disjointness constraint is relaxed?
Only if $T$ is small, but then it is trivial. If $T$ is allowed to be comparable to $|V|$, you can enforce the vertex disjointness by introducing $|V|$ extra periods in each of which one intermediate vertex is blocked but all other edges remain open. What are the actual numbers $|V|,|E|,T$ you are thinking of?
@fedja I don't impose particular constraint on the relationship among $V$, $E$, and $T$. But the algorithm needs to work under sufficient large values of them. It is meaningful to consider the case $T\sim V$.
@Ichen, let us see if I understand you (if I do, I may have an answer for you): say you have two nodes N1 and N2. Now you would like to find a path from N1 to N2 that is broken for all t in T. But because that may well be impossible, you would like to search for two ROUTES P1 and P2 (ie two paths) such that for each t, either one is not broken. Right? Would you be happy to find a path that at each t, wherever you are between N1 and N2, enables you to continue your journey as long as you get to N2?
@MircoA.Mannucci. Thank you for your comment. Yes, you reformulation is correct, but I fail to catch what you mean by "at each t, I can continue my journey". Do you mean that I will wait some time? This is an interesting direction that I havn't thought of, but definitely pertinent to my problem and what I want to achieve. Any thought is welcome.
Not exactly. What I think you have in mind is something like this: let us say that you have a road map, and you want to go from A to B. There are many paths which you can drive through, but problem is, at each t there may be an edge broken. Now, what I was suggesting is: suppose you go from A to A1 in the first instant. You had a plan, and wanted to go from A1 to A2, but alas, the road from A to A2 is broken. Well, suppose you have an option to go to A3 instead, because at that time is ok. Now you keep choosing unbroken edges, as long as you reach B you are ok, right?
@MircoA.Mannucci With that last interpretation it would be a trivial dynamic programming problem. However, If, indeed, Ichen has in mind something different from what was formally posted, it might help a lot to figure out what it really is. NP-hard problems are no fun to search a decent algorithm for. :-)
@fedja, agree 100% with you. I think Ichen's question is quite interesting, but he needs a bit more details, as far as what he is actually looking for
Thank you both. Let me clarify my problem. I want to go from $s$ to $t$. I seek a path such that no matter when I leave $s$, I can reach $t$ (suppose that the time to traverse the path is negligible). Such path may not exist. So I seek two paths such that no matter when I leave $s$, I can reach $t$ via one of them.
Yep, that agrees with the formal problem you posted and that is at least as difficult as 3-SAT (with linear reduction, which is a really bad sign). So, you'll have to think of whether you can settle with something weaker. I'll try to think if there is a quick reduction of your problem to some other well-studied NP-complete problem (perhaps, the same SAT. Then, at least, you'll be able to try some existing state of art algorithms on it instead of inventing everything from scratch :-)
@fedja Totally agree. I also thought of 3-SAT. The problem we are facing is how we can do in this case, given the global objective to have resilient $st$-paths. Any related thought or formulation is welcome.
|
2025-03-21T14:48:31.906752
| 2020-08-28T11:18:23 |
370312
|
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|
Stack Exchange
|
Non-uniruled connected smooth fibers implies flat
Let $f:X\to Y$ be a surjective morphism of connected smooth projective varieties over an algebraically closed field.
Assume all fibers are connected smooth and none are uniruled. Is $f$ flat?
In particular if all fibers are abelian varieties is $f$ flat? Can the general fiber be e.g. a point and the special fiber be an elliptic curve?
All fibers are smooth, or the generic fiber is smooth?
@AmorFati all fibers are smooth
"Can the general fiber be e.g. a point and the special fiber be an elliptic curve?"
Welcome new contributor. That certainly cannot happen. Abyhankhar proved that for a birational morphism to a smooth target the irreducible components of fibers are point or ruled varieties. That suggests an approach to your general question. Consider the rational transformation from $Y$ to the Chow variety of $X$ determined by the fibers of $f$. If the morphism is not flat, then the closure of the graph of this rational transformation has some uniruled fibers . . .
|
2025-03-21T14:48:31.906861
| 2020-08-28T11:45:20 |
370313
|
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Stack Exchange
|
When can we decompose a multivariable p-adic power series into product of single variable power series?
Is there any known result of decomposing multivariable power series over $p$-adic field into product of single variable power series ?
For example, consider the following power series in $n$ variables:$$ f(x_1,~x_2, \cdots, x_n)=\sum_{j_1,~j_2,\cdots, j_n=0}^{\infty} a_{j_1,~j_2, \cdots, j_n} \prod_{k=1}^{n} (x_k-c_k)^{j_k}.$$
Now we want to express $f(x_1,~x_2, \cdots, x_n)$ in the following way:
$$ f(x_1,~x_2, \cdots, x_n)=\left( \sum_{i_1=0}^{\infty} a_{i_1} x_1^{i_1}\right) \cdot \left( \sum_{i_2=0}^{\infty} a_{i_2} x_2^{i_2}\right) \cdots \left( \sum_{i_n=0}^{\infty} a_{i_n} x_n^{i_n}\right).$$
When and under which condition is it possible?
Is there any results or notes or resources available in this regard ?
Thanks,
A necessary condition is given by $$\frac{\partial f}{\partial x_i} \frac{\partial f}{\partial x_j} = \left( \frac{\partial}{\partial x_i} \frac{\partial}{\partial x_j} f \right) f$$ for all $i \neq j$. My guess is that this is also sufficient (at least in characteristic zero, which you are assuming anyway).
@RP_, Thank you very much. Yes , I checked for small degrees, your condition comes true. But I didn't see how did you get the necessary condition ? Can you be more explicit about the derivation of the necessary condition ?
Well, just write $f = f_1(x_1)f_2(x_2) \cdots f_n(x_n)$ and compute both sides of my equation. For simplicity, let $i=1, j=2$. Then both sides are clearly equal to $f'_1(x_1)f_1(x_1)f'_2(x_2)f_2(x_2)f_3(x_3)^2 f_4(x_4)^2 \ldots f_n(x_n)^2$.
@RP_, Thanks. So it seems that it is also sufficient condition as you said in p-adic case (char $0$).
Just to be sure: by "necessary condition" I meant that if the power series can be written as a product, then it must satisfy the equation (so if it doesn't satisfy it, it is not a product of $n$ univariate power series). From what I wrote above I can't draw the conclusion that the condition is also sufficient. I have thought about it but I don't see how to prove that it is sufficient. I thought I could maybe prove it but unfortunately I was too optimistic.
@RP_, I got your point. Thank you for nice effort
@RP_, can you please look into my another question in mathstackexchange.com ? It is here here
@RP_, While I am trying to prove the sufficiency part, I just want to discuss one thing with you. Once we have the decomposition into single variable power series, without loss of generality, we can assume each factor (each single variable power series) in the product has distinct zeros. Second, finite product and direct sum are equivalent. So we can think $f(x_1,x_2, \cdots, x_n)=\oplus_{i=1}^{n} f_i(x_i)$ instead of $f(x_1,x_2, \cdots, x_n)=\prod_{i=1}^{n} f_i(x_i)$ also. Is it ?
Some night thoughts on your question which are too long for a comment. For simplicity, I will look at the two variable case. Firstly, there is a very simple discrete criterion for when a function of two variable splits in the way you are interested in: Let $f$ be a function from $X \times Y$. Then is can be represented as a product of two functions of one variable if and only if $$ f(x_1,y_1)f(x_2,y_2)=f(x_1,y_2)f(x_2,y_1) $$ for all possible values of the variables.
This is purely set theoretical situation but one can ask the same question in various contexts (continuous, smooth functions, or power series) and a small additional argument is required to show that if a function splits in the above sense, then the factors automatically have the required smoothness.
With respect to a differential condition, I will cheat and suppose that $f$ is a smooth real-valued function on the plane. (I know nothing about the $p$-adic case but hope that this might give you some pointers). Then, as above,
$f f_{xy}=f_xf_y$ is a necessary condition for splitting and the question is whether it is sufficient. This is certainly true (using logarithms) if $f$ has no zeroes. In cases (like yours) where it can only have isolated zeroes I imagine that one could use a localisation argument to prove the sufficiency but I haven’t examined the details.
Thank you very much . In the last para 3rd line, what do mean by $\text{as above}$ ? Are you referring the comment by @RP_ ?
Yes, indeed I was.
ok,thank you very much
I didn't understand the argument how the condition is sufficient using logarithm ? Can you explain it ? Second, using localization we will get new functions, the rational functions or Laurent(type) power series. How this conclude the sufficiency ? Any clue please ?
If you use the tranformation $g= \ln f$, the differential equation takes on a very simple form. For the second question, one can use this fact to get a factorisation in a neighbourhood of any point where $f$ is non zero, then piece them together to get one away from the zeroes. But as I said, I haven’t thought this through so if you want to use this “fact” anywhere you will have to bite the bullet and find out whether it can be solidified to a precise proof.
Thank you very much for your nice help.
Can you please look into my another question in mathstackexchange.com ? It is here here
Sorry—I looked but don’t have the expertise to be able to help you.
Thank you very much for your effort.
While I am trying to prove the sufficiency part, I just want to discuss one thing with you. Once we have the decomposition into single variable power series, without loss of generality, we can assume each factor (each single variable power series) in the product has distinct zeros. Second, finite product and direct sum are equivalent. So we can think $f(x_1,x_2, \cdots, x_n)=\oplus_{i=1}^{n} f_i(x_i)$ instead of $f(x_1,x_2, \cdots, x_n)=\prod_{i=1}^{n} f_i(x_i)$ also. Is it ?
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2025-03-21T14:48:31.907187
| 2020-08-28T12:03:05 |
370314
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Stack Exchange
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Dimension of derived Artin stacks and perfect complexes
I am interested in the concept of dimension of derived and $n$-Artin stacks. Take for example the definition 4.10 of From HAG to DAG: derived moduli stacks. in which they define the dimension of a tangent derived stack as an alternating sum of the of cohomology dimensions of the cotangent complex $\mathbb{R} \Omega_{F}^{1}$:
Definition 4.10 If $x:$ $i$Spec $\mathbb{C} \longrightarrow F$ is a
point of a strongly geometric $D$ -stack, then we say that the
dimension of $F$ at $x$ is defined if the complex $\mathbb{R}
\Omega_{F, x}^{1}$ has bounded and finite dimensional cohomology. If
this is the case, the dimension of $F$ at $x$ is defined by $$
\mathbb{R} D i m_{x} F:=\sum(-1)^{i} H^{i}\left(\mathbb{R} \Omega_{F,
x}^{1}\right) $$
However I don't understand some issues. Why the the complex $\mathbb{R} \Omega_{F}^{1}$ is concentrated in degrees $(-\infty,1]$? For any derived Artin stack $F$, locally of finite presentation, the cotangent complex is perfect. I expected it to be only concentrated in degrees $[-p,1]$ being $p$ bounded in some way by the highest dimension of the schemes which form the atlas for $F$ (or by the topological space $X$ on you define your $F$). Otherwise what is the utility of those higher terms?
QUESTION 1: Can you have non-zero cohomology for arbitrary high terms in your
perfect complex, disregarding the dimension of your scheme or
topological space on the perfect complex is defined?
This leads to the more general question
QUESTION 2: Can you have examples of derived stacks built over a scheme or topological space $X$ in which the derived degrees of freedom (i.e. its dimension as a derived stack) can grow arbitrarily high independently of the dimensions of $X$? I can easily imagine that freedom for $n$-stacks considering for example the moduli stack of $n$-vector bundles over a projective variety (with $n$ arbitrarily high) but I cannot find such examples in the derived side for derived stacks.
Maybe more complicated Artin stacks, like the (higher derived ) stack of perfect complexes
on a scheme $X$ can do the job?
Assuming you're using cohomological gradings, the negative terms in the cotangent complex arise from derived structure or singularities, rather than descent data. A non-LCI ring of finite type will have cotangent complex in degrees $(-\infty,0]$. I think an example for both of your questions is given by CDGAs of the form $A:=k[V[r]]$ for $r>0$; since $H_0A=k$, the underlying space is a point, but the dimension is $(-1)^r(\dim V)$.
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2025-03-21T14:48:31.907376
| 2020-08-28T12:12:46 |
370315
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Stack Exchange
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Infinite coproducts and monoidal structure
It is a well-known fact that on categories admitting all finite coproducts one can define a monoidal structure where the monoidal product is exactly the coproduct and the monoidal unit is the initial object.
For coproducts there is nothing that prohibits me from working with a countable (or even uncountable) number of objects. Now I was wondering if there exists any general theory on how this can be generalized to the monoidal context? (In general I have barely found anything on the extension of monoidal products to an infinite number of objects.)
This question popped up while I was studying monoidal posets (see Seven Sketches in Compositionality by Spivak & Fong: http://math.mit.edu/~dspivak/teaching/sp18/7Sketches.pdf). On one hand I am interested in this question from an educational point of view because I would in general like to know how the concepts from monoidal category theory generalize to ''infinite products''. But on the other hand I am also interested from a professional point of view because not all operations I consider come from coproducts (e.g. the addition on $\mathbb{R}^+$) and hence I cannot resort to a description in those terms.
You can construct infinitary coproducts from finite coproducts and directed colimits in a straightforward fashion (it is say, in MacLane's book). My suggestion is to define infinitary monoidal products in the same way.
Does this answer your question? $\infty$-ary tensor product on a category
This indeed gives an idea of what the infinitary construction would be (at least in abstract terms). I assume that the definition of infinitary (lax) monoidal functors will also be a suitable modification of the finitary definition from Leinster's Higher operads, higher categories?
Yes, that's right.
This paper might be of interest.
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2025-03-21T14:48:31.907545
| 2020-08-28T13:29:58 |
370318
|
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Stack Exchange
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intuition about perverse sheaves
firstly, I would know if my very basic intuition on perverse sheaves is correct .
secondly, I would have some clarification in what perverse sheaves behaves better than regular sheaves .
my intuition is :
In smooth cases perverse t-structures is approximately the same as ordinary t-structures (up to translation by the dimension of the space)
But in the singular case the perverse sheaves behave much better than the ordinary sheaves so one can actually imagine that the perverse sheaves are the correct category,
it is a more correct category than the category of ordinary sheaves and the reason for this is that the construction of the perverse sheaves is actually the gluing of the ordinary t structures on all strata.
The intuition for this construction is, a stratification of a space is supposed to divide the space into smooth elements which makes it possible to “correct the singularity” and therefore it makes sense to think that the correct object is perverse sheaves and not ordinary sheaves
In addition it is possible to prove that the perverse sheaves are stacks which means that they behave just like sheaves.
I would be grateful is someone could said me if this is a correct explanation\intuition(and by the way my intuition on stratification is correct?) ?
and if someone have another intuition on perverse sheaves I am absolutely open for other intuition !
thanks in advance !!
A couple of points. 1) When $X$ is smooth, a locally constant sheaf is perverse up to translation. But the converse isn't true (expect in trivial cases). 2) If you understand a bit about $D$-modules, then on the $D$-module side, under Riemann-Hilbert, perverse sheaves look very natural.
"In smooth cases perverse t-structures is approximately the same as ordinary t-structures" - This depends on what you mean by "smooth cases". For example, the perverse t-structure for (algebraically) constructible complexes on $\mathbb A^1 = \mathbb C$ is interesting - not just a shift of the usual t-structure - even though all the strata and links are themselves smooth. E.g. If $j:\mathbb C^\times \hookrightarrow \mathbb C$ is the open inclusion, $Rj_\ast \mathbb{Q}_{\mathbb{C}^\times}[1]$ is perverse, but it has ordinary cohomology in degrees -1 and 0.
When I said in the smooth cases I am meaning smooth sheaves on essentially smooth and equidimensional. scheme ...
@Anonyme I've edited to make sure that you have at least one top-level tag on your question. This is considered good practice and is also a good way to maximize the number of relevant users who see your question. Since you already had 5 tags (the maximum number), I had to delete one (in this case "t-structures", which is only watched by 3 people). Feel free to re-adjust the tags as you see fit, but I do recommend having at least one top-level tag.
My current favorite reference is "The Decomposition Theorem and the Topology of Algebraic Maps". A very nice example if you know a little toric geometry is the section on the application of the decomposition theorem to toric varieties. In toric geometry, everything in sight consists of T-orbits and closures of T-orbits. For a very down to earth example of cases in which intersection homology works more nicely than regular homology or cohomology is that intersection is a purely combinatorial invariant of the polytope from which you construct the variety, unlike regular cohomology.
One difficulty with this question is that there is no way to give a short answer. So I will second Marc Besson's suggestion. Here is a link to the article https://arxiv.org/abs/0712.0349
If you are looking for a more topological intuition, I would recommend MacPherson's lecture notes, which were mentioned in this other question.
|
2025-03-21T14:48:31.907948
| 2020-08-28T13:55:27 |
370319
|
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|
Stack Exchange
|
Short research articles
I am a masters student. I am interested in short articles which have counter examples and very few references. I want to write a short and interesting article.
For example; One of the best known shortest and best academic paper articles I read is Counterexample to Euler's Conjecture on Sums of Like Powers by L. J. Lander and T. R. Parkin (Bull. Amer. Math. Soc. 72 (1966) p 1079, doi:10.1090/S0002-9904-1966-11654-3).
It has only one reference. It's really fascinating.
Is there any short articles in Mathematics and especially in Analysis/Complex Analysis? I am also looking at Counterexamples books for learning something and I searched open problems in Wikipedia and looked at undiscovered, newly valued and current topics.
So can you share these type of articles you read? I am curious about a good and interesting short article topic. What do you recommend to me about it? You can also give some references.** Thanks for your ideas and answers.**
See https://mathoverflow.net/q/7330/12357
it's very nice. thank you very much.
There is a journal called Examples and Counterexamples, https://www.journals.elsevier.com/examples-and-counterexamples/ (I haven't seen it, so I can't do better than to refer you to the website)
William Chen, Counterexamples to Knaster's conjecture, Topology Vol. 37. No. 2, pp. 401-405, 1998.
@GerryMyerson thanks a lot:) the link is very useful. i will research journal and article.
On a lighter note: The shortest papers ever published.
[a bit too long for a comment]
I understand from the question that the aim is to find a research project based on the search for a counterexample. By construction, this will mean showing that some existing paper in the literature is mistaken. That is typically not a productive way to start a project in a new field, simply because (a) if the author of that paper is clueless then there is not much gained in showing them wrong by finding a counterexample,$^\ast$ while (b) if the author is an expert you are facing an uphill battle if you are just entering a field.
Typically, a more productive way to enter a field is to try to generalize/extend work of others, basically by exploring corners of the field they left untouched (or didn't bother to explore). You may find that this leads you to uncover an error/oversight in the paper you started from, but that would then be a byproduct of your research and not the primary motivation.
$^\ast$ many questions here on MathOverflow can be readily dismissed by finding a counterexample, but that rarely becomes something worthy of a publication
it's perfect comment and very nice perspective. i see it:)
Finding a counterexample to a conjecture might not be hard. James Davis's counterexample to Conway's climb-to-a-prime conjecture comes to mind (although I don't think he got a formal publication out of it).
@TimothyChow: As that example shows, there exist significant conjectures for which finding a counterexample isn’t hard. But finding such conjectures is hard.
@PeterLeFanuLumsdaine : I may be misreading Carlo Beenakker's answer, but it sounded like he was talking about counterexamples to a published theorem and I was trying to say that it's easier to find counterexamples to a published conjecture. I would not normally describe a person who conjectures something that turns out to be false as "clueless" or even "mistaken."
"By construction, this will mean showing that some existing paper in the literature is mistaken." That's not at all what a counterexample is! There are many papers whose main contribution is an ingenious counterexample to a well-known published conjecture or folklore question. These can be quite lengthy and occasionally highly technical, and it is uncommon that they only have 1 or 2 references like the OP suggests.
For counterexamples in analysis, a good start would be "Counterexamples in Analysis" by Gelbaum and Olmsted, there's even a Dover edition.
Thank you:) I will research its details.
Also Theorems and Counterexamples in Mathematics by the same authors, https://www.springer.com/gp/book/9780387973425
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2025-03-21T14:48:31.908401
| 2020-08-28T15:00:42 |
370327
|
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|
Stack Exchange
|
Extension of positive functionals II
This is a follow-up to
Extension of positive functionals.
Assume that $X=R^n$ with the canonical order (I indicate with $K$ the positive cone, $x \in K$ iff $x_i \ge 0$ for all $i$) and let $L:M \to R$ is a positive functional defined on a subspace of $X$. Then it should be true that $L$ has a positive extension to $X$. Actually, there is an old post here
https://math.stackexchange.com/questions/18593/extending-a-positive-linear-functional-in-finite-dimensions
where the author asks for a simpler proof of the result, but the original proof is not shown.
If $M$ contains an interior point of $K$ the extension follows from Krein-Rutman and if $M \cap K=\{0\}$ (or even if $(Ker L) \cap K=\{0\}$) from Bauer-Namioka. What to do with in the remaining cases? One could add $\epsilon$-times a strictly positive functional but I do not know how to control the norm of the extensions to let $\epsilon \to 0$.
So the question is about a proof of the above result, if this is true as it should be. Finally, I do not know what happens for more general (closed) cones in $R^n$. Thank you to all
It seems like in an arbitrary positive (with respect to a cone $K$) functional on $M$ has a positive extension to the whole space if and only if there exists a positive constant $c$ such that $d(x,K)\ge c d(x,M\cap K)$ for every $x\in M$ with the norm of the extension controlled by $c^{-1}|L|$, but I haven't checked all details yet. So, for a cone in $\mathbb R^d$ that is an intersection of finitely many half-spaces it is always possible but in general there are counterexamples.
@ Fedja. Thank you very much. I am not sure whether the condition on the distances you wrote above is necessary, but it is surely sufficient. You are right, for polyhedral norms in finite dimension the extension is always possible and I found some old papers by Klee and Mirkil proving that. Everything seems quite obvious, the proofs are not difficult but puzzling because there are small problems at several points and one feels silly after a while. I could give a reference (or even an indication of the proof) for somebody interested.
I answer myself but maybe somebody is interested and I give a sketch of the proof based on more general ideas by Mirkil and Klee. The main point is the following
Lemma 1. Let $f_1, \cdots, f_k$ be vectors in $X=R^n$ and $G=\{\sum_{i=1}^k a_if_i, \ a_i \geq 0 \}$. Then $G$ is closed.
This is obvious if the vectors are linearly independent. The proof in the general case is by induction on $k$ and holds in any Banach space.
The following is an easy consequence
Lemma 2. If $M$ is a subspace of $X$, then $M+K$ is closed.
For the proof one considers the quotient map $q$ from $X \to X/M$ (or the orthogonal projection onto $M^{\perp}$) and applies Lemma 1 in $X/M$ to $q(K)$ with $f_i=q(e_i)$ (here $(e_i)$ is the standard basis of $X$). Then $q(K)$ closed in $X/M$ gives $M+K=q^{-1}(q(K))$ closed in $X$.
The proof of the extension of any positive map $f:M \to R$ goes as follows.
Write
$f(x)=(x, x_0)$ for some $x_0 \in M$. We look for $\bar x \in K$ such that $\bar x-x_0 \in M^{\perp}$, so that $x \to (\bar x, x)$ is a positive extension. If such a $\bar x$ does not exist, then $x_0 \not \in K+M^\perp$ which is closed. Hahn-Banach then gives $y$ such that $(x_0,y) <(k+g,y)$ for every $k \in K, g\in M^\perp$.
If $k=0$, then $(x_0,y) < (g,y)$ for any $g \in M^\perp$ yields $y \in M$. But then $(x_0,y)<(k,y)$ for every $k \in K$ gives $(k,y) \ge 0$ for every $k \in K$, hence $y \in K$. Finally, $y \in M\cap K$ but $(x_0,y)<(0,y)=0$, against the positivity of $f$.
Two remarks
The same proof works for polyedral cones even in infinite dimension. One has to work with the dual cone in $X'$.
The condition $d(x,M \cap(-K)) \le Cd(x,-K)$, see the comment by @fedja, would give a positive extension as an application of the Hahn-Banach theorem. However, I cannot verify it for a general subspace $M$ and I do not know if this is necessary. Please, let know.
|
2025-03-21T14:48:31.908690
| 2020-08-28T15:41:31 |
370330
|
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|
Stack Exchange
|
Objects and morphisms in inverse limits of toposes?
Certain Galois toposes can be written as $\lim_{i \in I} \mathbf{PSh}(G_i)$ where $(G_i)_{i \in I}$ is an inverse system of discrete groups. (The limit is a strict limit in the 2-category of Grothendieck toposes and geometric morphisms.)
What are the objects and morphisms in this topos?
If all groups $G_i$ in the diagram are finite and the transition maps are surjective, then I believe the objects of the topos are sets $S$ together with a $G$-action, with the special property that the action morphism $G \to \mathrm{End}(S)$ factors through one of the groups $G_i$. The morphisms are the functions such that $\phi(g\cdot x) = g\cdot \phi(x)$ for all $g\in G$. Can you do something similar in general?
Edit: One of the standard references for this is "Continuous fibrations and inverse limits of toposes". The idea there is to replace the diagram of Grothendieck toposes by a corresponding diagram of Grothendieck sites, with a special kind of transition maps (called continuous fibrations). Then a "limit Grothendieck site" is constructed and the inverse limit topos is the topos of sheaves on this site. But I don't know how to follow this process in practice, even in the case described above where all groups are finite and the transition maps between the groups are surjective.
I'm pretty sure you take the colimit in the category of cocomplete and finitely complete categories (and cocontinuous, lex functors) of the corresponding diagram of toposes and the inverse image parts of the geometric morphisms between them induced by the diagram of groups. Then the adjoint functor theorem gives right adjoints to all the functors in the cocone, giving geometric morphisms as maps in a cone over the original diagram of Grothendieck toposes.
Thank you! Do you know how to compute the colimit in this category? For example, is it different from taking the colimit in the 2-category of categories?
Concerning @DavidRoberts's comment, the relevant literature is Sketches of an Elephant, B1.4.
@IvanDiLiberti I'm afraid I don't understand how to go from the results in B1.4 to David Roberts's statement.
I am sorry, I meant B3.4.
@JensHemelaer Your description of the classifying topos of a profinite group is not correct (the category you describe does not have infinite sums, for example). See SGA4, IV.2.7. Also, the inclusion Top ⊂ Cat preserves filtered limits, so objects/morphisms in such a limit of toposes are just compatible families of objects/morphisms.
@MarcHoyois Thanks for the reference, it gives precisely the concrete description that I was looking for. Further, the topos that is associated to a pro-group in SGA4 agrees with taking the inverse limit in the category of Grothendieck toposes, by Remark 2.7 in "Prodiscrete groups and Galois toposes". I'm also interested in your last statement: is the inclusion Top ⊂ Cat the one that sends a geometric morphism to its right adjoint part (direct image functor)?
@JensHemelaer Yes, the morphisms in Top are the pushforward functors. And "limit" should be understood in the 2-categorical sense, so a compatible family of objects includes isomorphisms satisfying the cocycle condition.
The question was solved thanks to comments by Marc Hoyois.
The classifying topos associated to a pro-group is described by Grothendieck and Verdier in SGA4, Exposé IV, 2.7 (link). Grothendieck and Verdier assume that the transition maps $\pi_{ij} : G_j \to G_i$ are surjective.
The classifying topos of a pro-group $(G_i)_{i \in I}$ as defined there, agrees with the inverse limit topos $\varprojlim_{i \in I} \mathbf{PSh}(G_i)$, by Remark 2.7 in "Prodiscrete groups and Galois toposes".
As suggested by მამუკა ჯიბლაძე below, here the construction of Grothendieck and Verdier in more detail:
The objects of $\varprojlim_{i \in I} \mathbf{PSh}(G_i)$ are sets $E$, written as directed union $\bigcup_{i \in I} E_i$ where each $E_i$ comes with a $G_i$-action, and such that $E_i = \{ x \in E_j : gx = x,~\forall g \in \ker(G_j \to G_i)\}$ and the action of $G_i$ on $E_i$ is the one induced by the action of $G_j$ on $E_j$ (this last part is left implicit in SGA4).
If we write $E_j$ as a direct union of orbits, then $E_i$ consists of the orbits isomorphic to $G_j/H$ with $H$ containing $\ker(G_j \to G_i)$. In this way, we get another description of the objects of $\varprojlim_{i \in I} \mathbf{PSh}(G_i)$. They are isomorphic to direct sums of quotients $G_i/H$, where two quotients $G_i/H$ and $G_j/H'$ are the same if there is a $k \leq i,j$ and a subgroup $H'' \subseteq G_k$ such that $\pi_{ik}(H'') = H$ and $\pi_{jk}(H'')=H'$.
It would be convenient to add the description itself, it is not so difficult: in the profinite case these are just continuous $\left(\varprojlim_iG_i\right)$-sets. In the general case (assuming all homomorphisms $G_j\to G_i$ surjective), one has a set $E$ covered by subsets $E_i$ with $G_i$-actions such that $E_i$ is the fixed set of the action of $\ker(G_j\to G_i)$ on $E_j$ for each $i\to j$.
@მამუკაჯიბლაძე Good idea, but I realize that at the moment I don't understand the description yet. Shouldn't there be an additional condition that says that the actions of the different groups $G_i$ are compatible with each other? Consider for example the pro-group coming from $\mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$. Then the total set $E$ has a $\mathbb{Z}/4\mathbb{Z}$-action and it has a subset $E' \subseteq E$ fixed by $2\mathbb{Z}/4\mathbb{Z}$. But then according to SGA4 you have another piece of data involving a $\mathbb{Z}/2\mathbb{Z}$-action on $E'$?
I would bet they mean the canonical action: if $G$ acts on $E$ and $N$ is a normal subgroup of $G$, this gives rise to an action of $G/N$ on $E^N$. It is this one they must have in mind there.
|
2025-03-21T14:48:31.909097
| 2020-08-28T16:18:04 |
370332
|
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|
Stack Exchange
|
Runs of consecutive numbers all of which are Murthy numbers
Murthy numbers, in a given base, are positive integers, such as 2009 in base 10, which are not relatively prime to their reversal, that is, the number written backwards (in base 10 such numbers are AO71249 in the OEIS).
In base 10, numbers from 8432 to 8440 are all Murthy numbers. Are there arbitrarly long runs of consecutive numbers in base 10 all of which are Murthy numbers? In other bases?
Let $m+1, \dots, m+n$ be a sequence of $n$ consecutive Murthy numbers such that each $m+i$ shares with its reversal $\overline{m+i}$ a prime factor $p_i\equiv 3\pmod4$ such that 10 is a quadratic nonresidue modulo $p_i$. We will show how to construct such a sequence of $n+1$ consecutive Murthy numbers.
Define $t := p_1\cdots p_n(10^{\frac12\mathrm{lcm}(p_1-1,\dots,p_n-1)}+1)$ and notice that the product $p_1\cdots p_n$ divides both $t$ and its reversal. (Any smaller $t$ with this property will also do the job.)
The new sequence will have the form:
$$t\cdot (10^k + 1) 10^l + m, \dots, t\cdot (10^k + 1) 10^l + m+n,$$
where integers $k,l$ (larger than the length of $t$ and $m+n$) are to be determined.
First we notice that the last $n$ numbers in this sequence are Murthy since $t\cdot (10^k + 1) 10^l + m+i$ shares with its reversal the same prime factor $p_i$.
So it remains to enforce Murthyness on $t\cdot 10^k + m$. Let $q\equiv 3\pmod{4}$ be a prime having $10$ as a primitive root. We require that both $t\cdot (10^k+1)10^l + m$ and its reversal are divisible by $q$, that is
$$\begin{cases}
t\cdot 10^{k+l} + t\cdot 10^l + m \equiv 0\pmod{q},\\
\overline{m}\cdot 10^{d+k+l} + \overline{t}\cdot (10^k+1)\equiv 0\pmod{q},
\end{cases}
$$
where $d$ is the difference in decimal lengths between $t$ and $m$. This system can be solved by first eliminating the terms $10^{k+l}$ and expressing $10^l$ in terms of $10^k$, and then obtaining a quadratic equation w.r.t. $10^k$. If it's not solvable, we can change the value of $q$ to make it solvable. Then the values of $k,l$ are obtained by taking discrete logarithms (thanks to $10$ being a primitive root modulo $q$).
Example. For $m=8434$ and $n=3$, we have $p_1=7$, $p_2=3$, $p_3=11$, and we can take $t=1617$. Then the system is solvable for $q=29$ with solutions $(10^l,10^k)\equiv (12,20)\pmod{29}$ or $(10^l,10^k)\equiv (8,16)\pmod{29}$. Correspondingly, $(l,k)\equiv (21,12)\pmod{28}$ or $(l,k)\equiv (5,16) \pmod{28}$. The latter produces the following sequence of consecutive Murthy numbers:
$$1617\cdot (10^{16}+1)\cdot 10^5 + 8434 + i,\qquad i=0,1,2,3.$$
P.S. We should have infinite supply of primes having 10 as a primitive root by Artin's conjecture.
|
2025-03-21T14:48:31.909290
| 2020-08-28T16:24:49 |
370333
|
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|
Stack Exchange
|
Number of reduced decompositions of the longest element of the Weyl group
Let $R$ be a reduced root system, $W$ the associated Weyl group, and $w_0 \in W$ the longest element of $W$. In general $w_0$ admits more than one reduced decomposition into a product of reflections, a number which we denote by $d_R$. Where can one find a list of values of $d_R$ for low-dimensional root systems?
For example are the explicit values of $d_R$ known for the exceptional root systems?
For Types A and B(=C) there are product formulas for these numbers: see the famous paper http://math.mit.edu/~rstan/pubs/pubfiles/56.pdf and https://math.stackexchange.com/questions/2271510/the-number-of-reduced-expressions-for-the-longest-element-of-b-n. I'm pretty sure for Type D there is not a product formula (as Stanley mentions there is a big prime, 193, in the factorization of the number of reduced words of the longest word in Type $D_4$). As for exceptionals I don't know of a list but this is in principle something a computer can do.
In the linked stackexchange webpage the B-series formula in the answer gives, for low values of $n$, a product with negative limits. Is this an error or is this an example of some convention I am not familar with? For example, what is the value for $B_2$?
Maybe the formula Zach wrote is not quite right. For $B_2$ the answer should be 2. It is the same as the number of linear extensions of the root poset (poset of positive roots whereby $\alpha \leq \beta$ if $\beta-\alpha$ is a nonnegative sum of simple roots). This poset is the same as the shifted trapezoid shape $(2n-1,2n-3,...,1)$ poset. This number also happens to be the same as the number of SYTs of $n\times n$ square shape.
You can see Corollary 5.2 of the paper of Stanley linked above for another way of writing the product formula for the Type B # of reduced decompositions (he lists it as a conjecture but it has been proven).
The basic thing that's going on here is that there's an Edelman-Greene style bijection in Types A and B (and also the non-Weyl types $I_2(m)$ and $H_3$- sometimes these types are called the 'coincidental types'). The other types don't have such a bijection.
This is easy to do in SageMath. E.g. the following code
G = WeylGroup("F4")
w = G.long_element_hardcoded()
print(w)
rw = w.reduced_words()
len(rw)
outputs 2144892. If you want to look at some of these reduced words just examine the list rw. To create a list for classical types of different rank do
res = {}
for n in range(2,5):
G = WeylGroup(["A", n])
w = G.long_element_hardcoded()
print("Calculating rank ", n)
res[n] = len(w.reduced_words())
Note 2144892 = 2^2 x 3 x 47 x 3803, again suggesting lack of product formula.
I created http://www.findstat.org/StatisticsDatabase/St001585. If you have enough power to do E6 that would be great!
@MartinRubey I will try some computer in Prague's Charles University to which I have remote access. But I am pessimistic. This algorithm actually generates and stores all the reduced words so it eats all available memory rather quickly.
Counting reduced words is the same as counting directed paths between two vertices in a certain directed graph (the weak order graph), so it's possible you could write code by hand that does this faster than what's in Sage.
@SamHopkins Actually, SageMath implements this via BraidOrbit and includes a TODO that promises counting the number of reduced words to be much faster. It's a bit outside of my expertise and I have enough work on my plate to dig into this right now. If no better answer shows up, feel free to ping me in a week or two.
@SamHopkins, indeed, this is much faster! I added corresponding code to the findstat entry. $D_5$ is now easy. Unfortunately, FindStat has a limitation on the size of statistic values, so anything beyond $A_6$ is too large.
It would be nice if this number would have a (large) natural (that is, predictable) factor.
@MartinRubey I gave it a shot but the $E_6$ calculation doesn't fit into 16 GB of RAM and I don't have access to more.
Did you try the code from findstat? This should not take much memory.
@MartinRubey Of course. The problem is that the weak poset doesn't fit into the memory. Is it possible to reimplement the chain polynomial of weak poset using weak_order_ideal method of the WeylGroup object?
Around the time Fomin and I wrote this
paper, Tao Kai Lam applied the technique to type $D_n$. It
emerged that it was "natural" to weight a reduced
decomposition $\rho$ by $2^{d(\rho)}$, where $d(\rho)$ is
the number of simple reflections in $\rho$ that correspond
to the $n-2$ "nonbranch nodes" in the Coxeter diagram for
$D_n$. Using this weighting, there is a
nice product formula for the number of weighted reduced
decompositions of the longest element, which I unfortunately
have forgotten. I hope someone can redo this work.
Is this weighted sum related to linear extensions in some way?
@SamHopkins: I have a hazy recollection that it is a power of 2 times the number of SYT of the shifted shape $(2n-2,2n-4,\dots,4,2)$ (for which there is a nice product formula).
That would make a lot of sense! In https://conservancy.umn.edu/bitstream/handle/11299/159973/Williams_umn_0130E_14358.pdf Williams calls the poset you're talking about the "flattened root poset" of $D_n$.
Actually, the result you mention might follow from this paper of Billey and Haiman: https://math.berkeley.edu/~mhaiman/ftp/schubert/schubert.pdf (see also the discussion on pg. 46 of the thesis of Williams linked above)
Name of this paper: Schubert polynomials and the nilCoxeter algebra.
|
2025-03-21T14:48:31.909834
| 2020-08-28T16:44:35 |
370336
|
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|
Stack Exchange
|
Two versions of Sylvester identity
MathWorld presents the following two versions of Sylvester's determinant identity, relating to an $n\times n$ matrix $\mathbb{A}$:
First:
$$
|\mathbb{A}||A_{r\,s,p\,q}| = |A_{r,p}||A_{s,q}| - |A_{r,q}| |A_{s,p}|
$$
where $r$ and $s$ ($p$ and $q$) are sets that indicate which rows (columns) of $\mathbb{A}$ are to be deleted (correcting MathWorld's typo).
Second:
$$
|\mathbb{A}|\left[ a_{k\,k}^{(k-1)}\right]^{n-k-1}
=
\left|
\begin{matrix}
a_{k+1\, k+1}^{(k)} & \cdots & a_{k+1\, n}^{(k)} \\
\vdots & \ddots & \vdots \\
a_{n\, k+1}^{(k)} & \cdots & a_{n\, n}^{(k)} \\
\end{matrix}\right|
$$
where
$$
a_{i\, j}^{(k)}
=
\left|
\begin{matrix}
a_{11} & \cdots & a_{1\,k} & a_{1 \, j} \\
\vdots & \ddots & \vdots& \vdots \\
a_{k\ 1} & \cdots & a_{k\,k}& a_{k\, j} \\
a_{i\ 1} & \cdots & a_{i\, k} & a_{i\, j} \\
\end{matrix}
\right|
$$
for $k<i$, $j \leq n$.
Would anyone help me prove that these two versions are indeed equivalent.
Note: As has been pointed our below, MathWorld's claim is obviously (using a counterexample) incorrect; the second version implies only a special case (when r,s,p,q are single numbers) of the first 'version'.
Why do you think they should be equivalent?
For the special case that $r,s,p,q$ are single elements, it is shown in these notes (page 7) how the first identity (known as the Desnanot-Jacobi identity) follows from the second identity.
Apply the second identity to the matrix
we thus arrive at the first identity, illustrated graphically as
source
Update: Since I could not find the first identity in the literature, for the more general case when $r,s,p,q$ each contain more than a single element, I tried to check it for an example. I took $n=6$, $r=1,2$, $s=5,6$, $p=1,2$, $q=5,6$. For the $6\times 6$ matrix $A$ I took
$$A=\left(
\begin{array}{cccccc}
0 & 1 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 1 & 0 \\
2 & 0 & 2 & 2 & 0 & 0 \\
2 & 1 & 0 & 0 & 2 & 1 \\
2 & 0 & 2 & 1 & 0 & 2 \\
1 & 0 & 0 & 0 & 2 & 2 \\
\end{array}
\right)$$
The left-hand-side of the first identity is 0,
$\det A \det \left(
\begin{array}{cc}
2 & 2 \\
0 & 0 \\
\end{array}
\right) = 24\cdot 0 = 0$,
but for the right-hand-side I find a nonzero answer:
$$ \det \left(
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
2 & 0 & 2 & 2 \\
2 & 1 & 0 & 0 \\
\end{array}
\right) \det \left(
\begin{array}{cccc}
2 & 2 & 0 & 0 \\
0 & 0 & 2 & 1 \\
2 & 1 & 0 & 2 \\
0 & 0 & 2 & 2 \\
\end{array}
\right) - \det \left(
\begin{array}{cccc}
2 & 0 & 2 & 2 \\
2 & 1 & 0 & 0 \\
2 & 0 & 2 & 1 \\
1 & 0 & 0 & 0 \\
\end{array}
\right) \det \left(
\begin{array}{cccc}
0 & 0 & 1 & 1 \\
1 & 0 & 1 & 0 \\
2 & 2 & 0 & 0 \\
0 & 0 & 2 & 1 \\
\end{array}
\right)=$$
$$\qquad\qquad= (-4) \cdot 4 - (-2)\cdot (-2)=-20\neq 0$$
Incidentally, I did find a determinantal identity of a somewhat similar form in Tao's blog (last equation, Karlin's identity). But it is not quite of the form of the first identity in the OP.
So unless I have made a mistake, my conclusion is that the first identity in the OP only holds when $r,s,p,q$ are single elements, but not more generally.
Thanks for a neat (special-case) proof. How would one proceed with a fully general proof? Now, I would be quite happy to see how the special case of FIRST version (with r, s, p, q each being a single row/column) implies the more general (r,s,p,q being sets) case. Or is it better to ask that as a separate question?
Thanks again. I believe that you are correct, which implies that MathWorld is making a false statement - what an incredible carelessness for such a popular and well respected source. This also means that the second formulation implies the special case (one row, one column) of the first one (as you have shown), but their equivalence is then out of the question.
|
2025-03-21T14:48:31.910081
| 2020-08-28T17:10:56 |
370337
|
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|
Stack Exchange
|
Frobenius algebras from symmetric polynomials
Let $K$ be a field of characteristic 0 (maybe it works for more general fields) and $K[x_1,...,x_n]$ the polynomial ring in $n$ variables. Let $e_1,e_2,...,e_n$ denote the elementary symmetric polynomials in $n$ variables (see for example https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial ). (One might also choose other symmetric polynomials such as https://en.wikipedia.org/wiki/Complete_homogeneous_symmetric_polynomial or https://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial for this problem ).
For $n,t \geq 2$ let $A_{n,t}:=K[x_1,...,x_n]/(e_1^t,...,e_n^t)$.
$A_{n,t}$ is by definition a Frobenius algebra if there is a unique polynomial up to scalars $p \in A_{n,t}$ with $p x_i=0$ for $i=1,...,n$. This is a purely combinatorical condition.
For example for $n=t=2$, we have $A_{2,2}$ has vector space basis $1,x_1,x_2,x_1^2,x_1 x_2,x_1^3,x_1^2 x_2,x_1^4$ and indeed the unique longest polynomial in $A_{2,2}$ is $p=x_1^4$. Note that we have in $A_{2,2}$ $x_1^4=-x_2^4$ and unique means here really in the algebra $A_{2,2}$ and not as a polynomial in the polynomial ring.
I was able to prove by force that $A_{2,t}$ is a Frobenius algebra for $t \geq 2$.
Question 1: Is $A_{n,t}$ for general $n,t$ a Frobenius algebra?
In case this is true in general, there is probably a nice proof that avoid heavy computation (and might work for the other classes of symmetric polynomials such as the power sum symmetric polynomials).
Question 2: Did those algebras appear in the literature already? Do they have other nice properties such as being Hopf algebras in special cases (maybe allowing fields of certain characteristics).
Question 3: Is there a generalisation of question 1 in case question 1 is true?
For example small experiments suggest that even the algebras $K[x_1,...,x_n]/(e_1^{t_1},...,e_n^{t_n})$ are Frobenius as long as $t_1 \geq 2$ and $t_i \geq 1$. But maybe one can also do more, for example taking certain polynomials of the $e_i$.
The problem gets complicated quickly for larger $n$. Here is the case $n=3, t=2$ with the computer over the rationals($x_1=x,x_2=y,x_3=z$):
$A_{3,2}$ has vector space dimension 48 and longest polynomial for example $x^8y$.
A vector space basis via the computer is given as follows:
[ [(1)*v1], [(1)*x], [(1)y], [(1)z], [(1)x^2], [(1)xy], [(1)xz], [(1)y^2], [(1)yz],
[(1)x^3], [(1)x^2y], [(1)x^2z], [(1)xy^2], [(1)xyz], [(1)y^3], [(1)y^2z],
[(1)x^3z], [(1)x^2y^2], [(1)x^2yz], [(1)x^4+(2)x^3z+(-2)xy^3], [(1)xy^3],
[(1)x^3y+(1)xy^3], [(1)xy^2z], [(1)y^4], [(1)x^4z], [(1)x^3y^2], [(1)x^3yz],
[(1)x^5+(2)x^4z+(-4)x^3y^2+(-4)x^3yz+(-2)x^2y^3],
[(1)x^4y+(2)x^3y^2+(2)x^3yz+(1)x^2y^3], [(1)x^2y^2z], [(1)x^2y^3+(1/2)xy^4],
[(1)xy^4], [(1)x^4yz], [(1)x^3y^3], [(1)x^5y+(2)x^4y^2+(2)x^4yz+(1)x^3y^3],
[(1)x^6+(4)x^5y+(2)x^5z+(4)x^4y^2+(4)x^4yz+(2)x^3y^3], [(1)x^2y^4],
[(1)x^4y^2+(2)x^3y^3+(1)x^2y^4], [(1)x^5z+(2)x^2y^4], [(1)x^6z+(-2)x^5y^2],
[(1)x^5y^2], [(1)*x^7+(3/2)x^6y+(3)x^5y^2], [(1)x^4y^3],
[(1)x^6y+(2)x^5y^2+(4)x^4y^3], [(1)x^7y], [(1)x^6y^2],
[(1)*x^8+(3/2)x^7y+(3)x^6y^2], [(1)x^8y] ]
In this commutative situation, a Frobenius algebra is the same as an
artinian Gorenstein ring. In general, if $\theta_1,\dots,\theta_n$ are
homogeneous elements of positive degree of $A=K[x_1,\dots,x_n]$ and if
$K[x_1,\dots,x_n]/(\theta_1,\dots,\theta_n)$ is artinian (i.e., a
finite-dimensional vector space in this situation), then
$A/(\theta_1,\dots,\theta_n)$ is Gorenstein (in fact, a complete
intersection, which is stronger). Since $A/(e_1,\dots,e_n)$ is artinian
the same is true for $A/(e_1^t,\dots,e_1^t)$, so the Gorenstein property
follows. In fact, the socle polynomial $p$ is given by $\prod_{1\leq
i<j\leq n}(x_i-x_j)^t$.
Thanks, do you have a reference for this?
@Mare: I don't have a specific reference, but this is standard commutative algebra and should be included in such texts as https://www.springer.com/gp/book/9780387942681 and https://www.cambridge.org/core/books/cohenmacaulay-rings/938BC2204D8A7C99E2CEBA1695A692A4.
|
2025-03-21T14:48:31.910320
| 2020-08-28T17:54:37 |
370339
|
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|
Stack Exchange
|
Reference request on a characterization of ellipses
One up-vote, 35 views, and no comments and no answers have resulted from this reference request that I posted on math.stackexchange.com . This was actually inspired by a probability exercise, and at this point I'm half-way expecting someone to figure out how that happened before anyone finds the answer. But we'll see.
(Note that there are two links to m.s.e. in this present question. The one that appears above is a link to the reference request; the one below is to an earlier posting.)
In this question I conjectured that a simple proposition about ellipses holds. In the accepted answer, "Chrystomath" proves it.
Is this in some published refereed source?
quote:
Consider a closed bounded set with non-empty interior in the plane. Suppose it is strictly convex, i.e. every point between two of its points is one of its interior points. This entails that a line that intersecting its boundary but not its interior intersects it at only one point. Call such a line a tangent line. It follows that for every tangent line, there is exactly one other tangent line parallel to it. Suppose that for every line parallel to those two and between them, the midpoint of the intersection of that line with our closed bounded convex set is on the line connecting the two points of tangency.
[i.e. for EVERY such pair of tangent lines]
Does it follow that our closed bounded set is the convex hull of an ellipse?
The probability exercise was stated like this: Suppose $X,Y$ are real-valued random variables and for some numbers $a,b$ with $a\ne 0$ we have $\operatorname E(Y\mid X) = aX+b.$ Prove or disprove that from that alone it follows that for some numbers $c,d$ with $c\ne0$ we have $\operatorname E(X\mid Y) = cY+d.$ That was the exercise. It is well known that if $(X,Y)$ is jointly normally distributed, i.e. so distributed that every linear combination $pX+qY$ is normally distributed,$,\ldots\qquad$
$\ldots,$then both of the conditional expectations are as above (with $c=1/a$ ONLY IF $\operatorname{cor}(X,Y)\in{\pm1}$), and that the level sets of the joint density are ellipses. And it's easy to see that with any joint density whose level sets are ellipses we get those two straight lines. Thus counterexamples are to be found only among joint densities whose level sets are not ellipses. But is EVERY instance where they're not ellipses a counterexample? $\qquad$
Yes, this characterization is a theorem proven by Blaschke in "Kreis und Kugel" (1916). The theorem has a higher dimensional version, characterizing ellipsoids as the unique strongly convex bodies with the property that all centroids of codimension 1 sections that are parallel to a fixed plane lie on a line. See the paper "Characterizations of ellipsoids by section-centroid location" by Meyer and Reisner.
BEGIN QUOTE $${}$$ In [2] Blaschke proved the following result which he relates to Brunn: Let $K$ be a convex body in $\mathbb R^n$ with the following property: the midpoints of every bundle of parallel chords of $K$ lie in a plane, then (and only then) $K$ is an ellipsoid. $${}$$ END QUOTE $${}$$That's the very first sentence in that paper. (The "only then" part is trivial.) $\qquad$
Here's a small data point that supports my preference for the more traditional way of using hyphens, which calls for using them somewhat more often than is often done today: "centroids of codimension 1 sections" For a second I thought: What is a centroid of codimension 1? Then I realized that "codimension 1" is an adjective qualifying the word "sections". So "centroids of codimension-1 sections" is how I would write it. Richard Borcherds once expressed the fear that such a thing could be mistaken for a minus sign, but I don't think that's true with proper typesetting.
|
2025-03-21T14:48:31.910597
| 2020-08-28T19:50:03 |
370349
|
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|
Stack Exchange
|
Factorizing $n$-th Wendt determinant $W(n)$
It has been conjectured that the $p$-th Wendt determinant $W(p)$, where $p$ is prime, has all its prime factors that are greater than $p$ congruent to $1$ (mod $p$). Furthermore its prime factors that are less than $p$, when fully exponeniated to divide $W(p)$ are also congruent to $1$ (mod $p$), if they exist. However the occurrences of $W(p)$ with $p$ prime such that it has prime factors less than $p$ is uncommon (see OEIS A336688).
Are such occurrences finite or is there an example greater than $W(331)$?
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2025-03-21T14:48:31.910664
| 2020-08-28T20:04:38 |
370351
|
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|
Stack Exchange
|
How much reduction in expected variance can we get from a single bit?
Consider the following protocol:
Alice has a number $X$, chosen according to a known distribution $\mathcal D$ (e.g., $X\sim U[0,1]$).
She can send a bit to Bob, giving him more information about $X$ (e.g., she can send $Y=\mbox{Bernoulli}(X)$. In turn, Bob estimates $X$ (e.g., as $\widehat X = Y$).
I'm interested in lower bounding the expected (over the choice of $X$) variance of any such protocol.
In the above example, we have $\mbox{Var}[\widehat X | X] = X(1-X)$ and thus
$$\mathbb E[\mbox{Var}[\widehat X]] = \mathbb E[\mbox{Var}[\widehat x | X]] = \int_0^1x(1-x)dx = 1/6.$$
For this specific distribution (uniform), this protocol seems to yield the lowest possible expected variance for any unbiased estimation.
Can we use information-theoretic arguments to lower bound the variance for different distributions $\mathcal D$ for any protocol?
I looked into applying conditional differential entropy, but this seems to apply to a specific choice of protocol (how to choose which bit to send).
Formally, as posed, Bob can just estimate $\widehat X=EX$ without even looking at what Alice sent him (and she can just send him a bit independent of $X$ to avoid tempting him to do anything), thus achieving variance $0$. Apparently you wanted to ask something else, but I'll leave it to you to figure out what exactly it was.
It seems a better goal to bound from below the conditional variance of $X$ given $\hat{X}$.
Is Alice's alphabet restricted to ${0,1}$ or can they send any message with Shannon entropy 1? (Not sure if this matters yet)
This is a specific instance of one-shot lossy source coding which is still open in general. The best work I know of is in this preprint by Elkayam and Feder where they distill it down to the open problem of identifying a convex minimizer $\tilde{D}(z,Q_Y)$.
https://arxiv.org/abs/2001.03983
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2025-03-21T14:48:31.910827
| 2020-08-28T20:26:31 |
370353
|
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|
Stack Exchange
|
Question about Correspondences from Mumford’s Complex Projective Varieties
I study David Mumford's Algebraic Geometry I - Complex Projective Varieties
and have some problems to understand a step in the proof of Lemma 6.7 (b).
Firstly, the general setting & preparations around the problem I intend to present below (pages 97/98):
Some remarks on used notations: For a rational map $Z: X \dashrightarrow \mathbb{P}^m$
Mumford uses the notation "$Z$" ambiguously. On the one hand for rational map itself,
on the other hand he associates to the rational map a closed subvariety
$Z \subset X \times \mathbb{P}^n$ which he calls "correspondence", noted also by $Z$:
(See also Def 2.15 page 29)
I think that if $Z: X \to \mathbb{P}^m= \operatorname{Proj} \mathbb{C} [Y_0,...,Y_m]$
is moreover a regular map defined by
$x \to [f_0(x):...f_m(x)]$ then $Z \subset X \times \mathbb{P}^m$ coinsides with
vanishing locus $V(..., f_i Y_j - f_j Y_i,... ) \subset X \times \mathbb{P}^m$.
First question is does this closed subvariety $V$ coinside exactly with the closure of graph
$\Gamma_Z$ of $Z$ of we assume that $Z$ is a regular map?
Another probably not well known notation Mumford uses is $Z[S]$ for closed $S \subset X$.
Here the definition (see Cor. 2.26 page 35):
Having the background now we now come to my actual problem:
PROBLEM: I not understand following argument from proof on 6.7 (b) (p 98):
For $H:=V(l)= V(\sum \alpha_i Y_i)$ and $H_i := V(Y_i)$, we observe that because of
$$((Y_i/l) \circ Z) = Z^*(H_i) -Z^*(H)$$
all $f_i=(Y_i/l) \circ Z$ have over $X \backslash \operatorname{Supp} Z^*(H)$
no poles. (because over this subset the divisor $(f_i)$ is positive)
But why does this observation imply that over $X \backslash \operatorname{Supp} Z^*(H)$
the correspondence $Z$ (considered again as subset of $X \times \mathbb{P}^n$) is contained in
$V(..., f_i Y_j - f_j Y_i,...) \subset X \times \mathbb{P}^m$? That is
in the locus of zeros of $f_i Y_j - f_j Y_i$ for $0 \le i,j \le m$?
In other words why the condition that $f_i$ have no poles over
$X \backslash \operatorname{Supp} Z^*(H)$ suffice, to conclude that
$Z \subset V(..., f_i Y_j - f_j Y_i,...)$? Why is this assumption
on the $f_i$ it neccessary?
Let $U$ be the open subset of $X$ where $Z$ defines a regular map. Let's consider the map: $U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H) \longrightarrow \mathbb{P}^n$ defined by:
$$ x \longrightarrow \left[ \dfrac{Y_0 \circ Z}{l \circ Z}(x), \ldots, \dfrac{Y_m \circ Z}{l \circ Z}(x) \right] $$
Let us denote by $\tilde{Z} \subset (U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H)) \times \mathbb{P}^m$ the correspondance defining this map. Using the remark you made about the equtions for the graph of a regular map $U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H) \longrightarrow \mathbb{P}^m$, we find that over $(U \cap X \ \backslash \ \mathrm{Supp} \ Z^*H)$, $\tilde{Z}$ is included in the zero locus of $(f_iY_j-f_jY_i)_{0 \leq i,j \leq m}$ where $f_i = \dfrac{Y_i \circ Z}{l \circ Z}$.
Since $Y_i$ is the projection on the $i$-th coordinate and the denominator $l \circ Z$ does not vanish on $X \ \backslash \ \mathrm{Supp} \ Z^*H$ (by definition of $Z^*H$), the regular map $U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H) \longrightarrow \mathbb{P}^n$ defined by:
$$ x \longrightarrow \left[ \dfrac{Y_0 \circ Z}{l \circ Z}(x), \ldots, \dfrac{Y_m \circ Z}{l \circ Z}(x) \right] $$
is equal to the regular map $Z : U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H) \longrightarrow \mathbb{P}^m$. As a consequence, we have :
$$ Z \cap \bigg(U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H) \times \mathbb{P}^n \bigg) = \tilde{Z}.$$
As $Z$ is assumed to be irreducible, $Z \cap \bigg(X \ \backslash \ \mathrm{Supp} \ Z^*H \times \mathbb{P}^n \bigg)$ is also irreducible and we deduce from the above equality that $Z$ is also included in the vanishing locus of $(f_iY_j-f_jY_i)_{0 \leq i,j \leq m}$ over $X \ \backslash \ \mathrm{Supp} \ Z^*H$.
The first part I not understand. You wrote that since all $\dfrac{Y_i \circ Z}{l \circ Z}$ have no poles on $X \ \backslash \ \mathrm{Supp} \ Z^*H$, the map $X \ \backslash \ \mathrm{Supp} \ Z^*H \longrightarrow \mathbb{P}^{m}, x \mapsto\left[ \dfrac{Y_0 \circ Z}{l \circ Z}(x), \ldots, \dfrac{Y_m \circ Z}{l \circ Z}(x) \right]$ is regular. But what, if there exist a $x \in X \ \backslash \ \mathrm{Supp} \ Z^*H$ with $\dfrac{Y_i \circ Z}{l \circ Z}(x)=0$ for all $i$. then this map to $\mathbb{P}^{m}$ is a priori not well defined.
What we definitely can deduce from the information that $\dfrac{Y_i \circ Z}{l \circ Z}$ have no poles, that $\dfrac{Y_i \circ Z}{l \circ Z} \in k[X \ \backslash \ \mathrm{Supp} \ Z^*H] = H^0(X \ \backslash \ \mathrm{Supp} \ Z^*H, O_X)$, ie that $\dfrac{Y_i \circ Z}{l \circ Z}$ live in coordinate ring of $X \ \backslash \ \mathrm{Supp} \ Z^*H$.
So if we take affine pieces $U \subset X \ \backslash \ \mathrm{Supp} \ Z^*H$ and $V \subset \mathbb{P}^{m}$ with wlog $V \cong \mathbb{A}^{m}$ and that $V$ contains the image of $U$, then the restriction map $U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H) \to \mathbb{A}^{m}$ given by $x \mapsto \left( \dfrac{Y_0 \circ Z}{l \circ Z}(x), \ldots, \dfrac{Y_m \circ Z}{l \circ Z}(x) \right)$ is well defined. Did you possibly had this in mind?
I made some further clarifications. Hope I reached the textbook standards...
Ok I see, choosing open, small enough $U \subset X$ our restricted map becomes regular and the subsequent arguments I understand.
One general aspect about correspondences still confuses me:
Assume we have two arbitrary closed subvarieties $X \subset \mathbb{P}^n$ and
$Y \subset \mathbb{P}^m$ and a rational map
$f: X \dashrightarrow Y$. Can the associated correpondence
$Z:= Z_f \subset X \times Y$ be described explicitly?
What we know is that there exist an open $ U \subset X$ such that
the restriction $f \vert _U: U \to Y$ is a regular map given by
$ x \mapsto [f_0(x):...: f_m(x)]$ with $f_i \in k[U]$.
Then as I remarked above
the corresponence $Z_U$ of the restriction $f \vert _U$ is defined as
closed subvariety $V(..., f_i Y_j - f_j Y_i,... ) \subset U \times Y$. That's what
we know.
But what about the correspondence
$Z:=Z_f\subset X \times Y $? Does it have an explicit description?
Clearly $Z_U \subset Z$ and the closure $\overline{Z_U}$ considered as
subset of $X \times Y $ is contaned in $Z$ as well. But is $\overline{Z_U}
\subset Z$ in general a proper inequality? If yes, why?
One other question, one other post? I think I properly answered that one.
Yes sorry, you are right, of course you answer the initial question properly. I will post my concern from last comments as a separate question.
|
2025-03-21T14:48:31.911200
| 2020-08-28T20:30:21 |
370354
|
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|
Stack Exchange
|
Which classes in $\mathrm{H}^4(B\mathrm{Exceptional}; \mathbb{Z})$ are classical characteristic classes?
Let $G$ be a compact connected Lie group. Recall that $\mathrm{H}^4(\mathrm{B}G;\mathbb{Z})$ is then a free abelian group of finite rank. Let us say that a class $c \in \mathrm{H}^4(\mathrm{B}G;\mathbb{Z})$ is classical if it is in the image of restriction along a map $G \to J$ where $J$ is a classical Lie group, i.e. a quotient of $\mathrm{Spin}(N)$, $\mathrm{SU}(N)$, or $\mathrm{Sp}(N)$. The classical classes are essentially the Chern and (fractional) Pontryagin classes of linear representations of $G$. Let us furthermore say that a class is virtually classical if it is a $\mathbb{Z}$-linear combination of classical classes.
Remark: Not every virtually classical class is classical. Indeed, $\mathrm{H}^4(\mathrm{B}G;\mathbb{Z})$ has a cone of positive classes, namely the ones that determine negative-definite pairings under the canonical map $\mathrm{H}^4(\mathrm{B}G;\mathbb{Z}) \to \mathrm{H}^4(\mathrm{B}G;\mathbb{R}) \cong \mathrm{Sym}^2(\mathfrak{g})^G$. Restriction along homomorphisms preserves positivity. Suppose $c_1,c_2$ are nonzero classical classes on $G_1,G_2$. Then $c_1 \ominus c_2 \in \mathrm{H}^4(BG_1 \times BG_2)$ does not have a definite sign, and so is not classical, but it is virtually classical.
Question 0: If $G$ is simple, is every virtually classical class classical?
When $G$ is already a classical Lie group, then all classes are classical, although it is certainly a fun game to understand the restriction maps between various groups. For $G = \mathrm{G}_2$, the seven-dimensional representation $\mathrm{G}_2 \to \mathrm{Spin}(7)$ is an isomorphism of $\mathrm{H}^4$, and so all classes in $\mathrm{H}^4(B\mathrm{G}_2; \mathbb{Z})$ are classical.
Question 1: For the other exceptional Lie groups $G = \mathrm{F}_4, \mathrm{E}_6,\mathrm{E}_7,\mathrm{E}_8$, which classes in $\mathrm{H}^4(\mathrm{B}G;\mathbb{Z})$ are classical? Which are virtually classical?
This is easy if the answer to Question 0 is "yes", because I'm pretty sure that the smallest (positive) classical class is achieved by the smallest-dimensional representation. (In terms of the smallest positive class, the smallest classical classes are, I think, $3, 6, 6?,30$.) But I could imagine a situation where there is some funny cancellation and there are virtually classical classes smaller than any classical class.
Edit: I'm pretty sure that the answer to Question 0 must be "no", modulo my mistakes in calculations. First, recall that the Dynkin index of a map $G \to J$ of simple simply connected Lie groups is the numerical value of the induced map $\mathbb{Z} = \mathrm{H}^4(BJ) \to \mathrm{H}^4(BG) = \mathbb{Z}$. The inclusions $D_4 \subset F_4 \subset E_6 \subset E_7 \subset E_8$ all have Dynkin index $1$, and the adjoint map $G \to \mathrm{Spin}(\mathfrak{g})$ has Dynkin index the dual Coxeter number $h^\vee$.
The Dynkin index of a complex representation $V : G \to \mathrm{SU}(N)$ is called the Chern class $c_2(V)$ (except maybe it is $-c_2$). For a Spin representation $V : G \to \mathrm{Spin}(N)$, $N \geq 5$, it is the fractional Pontryagin class $\frac{p_1}2(V)$, and for a symplectic representation $G \to \mathrm{Sp}(N)$ it is the quaternionic Pontryagin class, which I will call $q_1(V)$. (Wikipedia calls it also $p_1(V)$, which is confusing.) The maps $\mathrm{Spin}(N) \to \mathrm{SU}(N)$ and $\mathrm{SU}(N) \to \mathrm{Sp}(N)$ each have Dynkin index $2$, whereas the maps $\mathrm{Sp}(N) \to \mathrm{SU}(2N)$ and $\mathrm{SU}(N) \to \mathrm{Spin}(2N)$ each have Dynkin index $1$. (Note that, like a topologist, I write $\mathrm{Sp}(N)$ for the group of quaternionic automorphisms of $\mathbb{H}^N$. Representation theorists sometimes call this group $\mathrm{Sp}(2N)$, because maps into it are $2N$-dimensional symplectic representations.)
In any case, the adjoint representation of $\mathrm{E}_7$ has fractional Pontryagin class $\frac{p_1}2(\mathbf{133}) = h^\vee(\mathrm{E}_7) = 18$, so $18 \in \mathrm{H}^4(B\mathrm{E}_7)$ is classical. The smallest representation of $\mathrm{E}_7$ is the $56$-dimensional symplectic representation $\mathbf{56} : \mathrm{E}_7 \to \mathrm{Sp}(28)$. I claim that $q_1(\mathbf{56}) = 12$, which means, I think, that $6$ is virtually classical but not classical.
To justify the claim, I need to compute $q_1(\mathbf{56})$. I claimed already that the map $\mathrm{Sp}(28) \to \mathrm{SU}(56)$ that forgets the symplectic structure has Dynkin index $1$, which is to say $q_1(\mathbf{56}) = -c_2(\mathbf{56})$.
Now note that $\mathrm{E}_6 \subset \mathrm{E}_7$ has Dynkin index $1$, so it suffices to compute $c_2(\mathbf{56})|_{\mathrm{E}_6} = c_2(\mathbf{56}|_{\mathrm{E}_6}) = c_2(\mathbf{27} \oplus \overline{\mathbf{27}} \oplus \mathbf{1} \oplus \mathbf{1}) = 2 c_2(\mathbf{27})$.
So my claim amounts to the claim that $c_2(\mathbf{27}) = 6 \in \mathrm{H}^4(B\mathrm{E}_6)$. I can compute this by restricting further along the Dynkin-index-$1$ map $\mathrm{F}_4 \subset \mathrm{E}_6$, where $\mathbf{27} = \mathbf{26} \oplus 1$. So I want to compute $c_2(\mathbf{26}) = -2\frac{p_1}2(\mathbf{26}) \in \mathrm{H}^4(B\mathrm{F}_4)$.
So my claim amounts to the claim that $\frac{p_1}2(\mathbf{26}) = 3 \in \mathrm{H}^4(B\mathrm{F}_4)$. To compute this, restrict further along the Dynkin-index-$1$ map $\mathrm{D}_4 \subset \mathrm{F}_4$, where $\mathbf{26}$ splits as a sum of the three $8$-dimensional representations $\mathrm{D}_4 = \mathrm{Spin}(8) \to \mathrm{Spin}(8)$ (plus some trivial representations).
In summary, I know that $12$ and $18$ are both classical classes for $\mathrm{E}_7$, and so $6$ is a virtually classical class for $\mathrm{E}_7$. I have a hard time believing that it is actually classical. Well, I have an OK time believing that I made a mistake in the calculations, especially if I am wrong about the relation between Chern and quaternionic Pontryagin classes. And I don't see how some large representation of $\mathrm{E}_7$ will have small characteristic classes, but I don't have a proof that this cannot happen...
Comment: It is interesting that the numbers $3, 6, 6, 30$ for $F_4,E_6,E_7,E_8$ are also the largest square-free factors of the dual Coxeter numbers of these groups $(9, 12, 18, 30)$. Furthermore, if I ignore the prime $p=2$, then these are the only simple simply connected groups with torsion in their cohomology, and only at these primes (this is a result of Serre, I think). Ignoring the prime $p=2$ might make sense, since the characteristic classes for the classical series are all the same up to a factor of $2$: after inverting $2$, I could have just asked to understand the "Chern subring" (in degree $4$) of $\mathrm{H}^\bullet(BG)$.
This isn't going to particularly help you resolve your question, but as regards the Comment, I think it was Borel who closed the book on determining torsion primes, particularly in this paper: https://projecteuclid.org/euclid.tmj/1178244298. There is earlier joint work with Serre, though: https://link.springer.com/article/10.1007%2FBF02564557.
@jdc Thank you!
|
2025-03-21T14:48:31.911713
| 2020-08-28T21:18:16 |
370355
|
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|
Stack Exchange
|
GIT and singularities
Let $G$ be a complex reductive group acting on a complex affine variety $X$ and let $X // G = \operatorname{Spec}\mathbb{C}[X]^G$ be the GIT quotient.
Is there a relationship between the singular locus of $X$ and that of $X // G$?
Of course, $X//G$ can be highly singular while $X$ is smooth. But, for example, I was wondering if (or under what conditions) singular points of $X$ are mapped to singular points of $X // G$.
Edit. Spenser's nice comment below shows that the answer to the latter is no. But perhaps a better and more precise question is: If $X // G$ is non-singular at $y$, is there a non-singular $x \in X$ mapping to $y$? In other words, do all fibres of $X \to X // G$ at non-singular points contain a non-singular point of $X$? I'm willing to assume irreducibility or other nice properties.
I think that singular points can be mapped to smooth points. Take $xy = 0$ and let $\mathbb{C}^*$ act by $z\cdot(x, y) = (zx, y)$. The quotient is $\mathbb{C}$.
That is still not true. The GIT quotient of the conjugation action on the locus of non-invertible matrices is the affine space given by the coefficients of the characteristic polynomial (in particular, the constant coefficient equals $0$). The fiber over the origin in this affine space is the nilpotent cone inside the locus of non-invertible matrices, and every nilpotent matrix gives a singular point of this locus.
@JasonStarr I'm trying to understand your counterexample in the 2 x 2 case, and it doesn't seem to work (I may be wrong). The non-invertible matrices are the locus of $ad - bc = 0$, so the only singular point is the zero matrix. Hence, any non-zero nilpotent matrix is a non-singular point of the fibre above zero. What am I missing?
I was wrong! There are, indeed, nilpotent matrices that are regular.
This is clearly false for finite groups (take the union of the two coordinate axes in $\mathbb A^2$, with the involution that switches the two axes. For a connected example, embed the cyclic group into $\mathbb G_\mathrm{m}$, and consider the induced action.
@Angelo. That is precisely how I made the normal counterexample below. I started with a normal counterexample for the cyclic group of order 2 and induced an example for the multiplicative group.
This is an answer to the revised question. It is the simplest counterexample that I can think of where the reductive group is smooth and connected, where $X$ is normal and affine, and where $Y=X//G$ is smooth, even though there are fibers of the quotient map that are contained in the singular locus of $X$.
Let $Y$ be $\text{Spec}\ k[x,y,z]$, i.e., affine $3$-space. Let $G$ be the multiplicative group of units, $G=\text{Spec}\ k[u,u^{-1}]$. Let $X$ be $\text{Spec}\ k[x,y,z,s,t]/\langle f \rangle$ where $f$ is the polynomial, $$f=s^2+t(xz-y^2).$$ Let the action of $G$ on $X$ be defined by $$\mu:G\times_{\text{Spec}\ k} X \to X, \ \ \mu(u,(x,y,z,s,t)) = (x,y,z,us,u^2t). $$ The ring of $G$-invariant polynomials is the subring, $$k[X]^G = k[x,y,z].$$ The quotient map is just the usual projection, $$q:X\to Y, \ \ q(x,y,z,s,t) = (x,y,z).$$ For the dense Zariski open $U = D(xz-y^2)\subset Y$, the inverse image $q^{-1}(U)$ is a $G$-torsor over $U$.
The singular locus of $X$ is the single $q$-fiber, $q^{-1}(0,0,0)$. Even though the origin is a smooth point of $Y$, every point of this $q$-fiber is a singular point of $X$.
|
2025-03-21T14:48:31.912215
| 2020-08-28T21:28:34 |
370356
|
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|
Stack Exchange
|
Applications of the co-area formula
Kirchheim [2] generalized the classical area formula to the case of Lipschitz mappings into metric spaces. Ths paper is well known and widely cited. The area formula is a special case of the co-area formula. The classical co-area formula has plenty of applications.
On the other I am quite surprised that the co-area formula for mappings into metric spaces [1,3] is not that widely cited and I did not see too many applications. Are you aware of interesting applications of this general version of the co-area formula (i.e. for Lipschitz mappings into metric spaces)?
For the record, the theorem is:
Theorem: Suppose $X$ is an $\mathcal{H}^n$-$\sigma$-finite metric space, $A \subset \mathbb{R}^{n+m}$ is measurable and $f\colon A \to X$ is Lipschitz. Then, for any integrable $g$ on $A$,
\begin{equation}
\int_{A} g(x) C_n(\text{md}(f,x)) \ d\mathcal{H}^{n+m}(x) = \int_{X} \left( \int_{f^{-1}(\{y\}) \cap A} g d\mathcal{H}^{m} \right) d\mathcal{H}^n(y) \ .
\end{equation}
Here $\text{md}(f,x)$ is the so-called metric derivative of $f$ at $x$, which is a seminorm on $\mathbb{R}^{n+m}$, and $C_n(\text{md}(f,x))$ is its coarea factor -- some sort of $n$-Jacobian.
[1] M. Karmanova,
Rectifiable sets and coarea formula for metric-valued mappings.
J. Funct. Anal. 254 (2008), 1410–1447.
[2] B. Kirchheim, Rectifiable metric spaces: local structure and regularity of the Hausdorff measure. Proc. Amer. Math. Soc. 121 (1994), no. 1, 113–123.
[3] L. P. Reichel,
The coarea formula for metric space valued maps. Ph.D. thesis, ETH Zurich, 2009.
(Available online).
@ABitTooCurious Any reference to that?
|
2025-03-21T14:48:31.912356
| 2020-08-28T22:05:22 |
370357
|
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|
Stack Exchange
|
Examples of geodesically convex functions on (subsets of) $GL_n (\mathbb{R})$
I'm reading this paper [1]. To get some intuition on what a geodesically convex function is like, I'm trying to find some example in $GL_n(\mathbb{R})^+$ that is geodesically convex (near $I$).
Let me first copy some definitions from [1].
Definition 2 [1]. A function $f : \mathcal{M} \rightarrow \mathbb{R}$ is said to be geodesically convex if for any $x, y \in \mathcal{M}$, a geodesic $\gamma$ such that $\gamma(0) = x$ and $\gamma(1) = y$, and $t \in [0, 1]$, it holds that
$$f(\gamma(t)) \le (1 − t) f(x) + t f(y).$$
It can be shown that an equivalent definition is that for any $x, y \in \mathcal{M}$,
$$ f(y) \ge f(x) + \left< g_x, Exp_x^{-1} (y) \right>_{x}, $$
where $g_x$ is a subgradient of $f$ at $x$, or the gradient if $f$ is differentiable, and $\left< \cdot , \cdot \right>_x $ denotes the inner product in the tangent space at $x$ induced by the Riemannian metric.
Definition 3 [1]. A function $f : \mathcal{M} \rightarrow \mathbb{R}$ is said to be geodesically $\mu$-strongly convex if for any $x, y \in \mathcal{M}$,
$$ f (y) \ge f (x) + \left< g_x, Exp_x^{-1} (y) \right>_{x} + \frac{\mu}{2} d^2 (x,y), $$
[where $d$ is the distance between $x$ and $y$ induced by the Riemannian metric.]
Based on the definition of geodesic convexity and strong geodesic convexity, I'm guessing, in $GL_n (\mathbb{R})^+$, the function $f (x) = d^2(x,I)$ ($d (\cdot, \cdot)$ is the distance induced by some Riemann metric) is geodesically convexity near $I$.
I'm trying to use the (relatively common) Riemann metric $\left< \cdot, \cdot \right>_x$ on $GL_n (\mathbb{R})$ such that $ \left< A, B \right>_x = \left< x^{-1} A, x^{-1} B \right>_I $ and $ \left< A, B \right>_I = tr (A^\top B) $.
It seems, in the above mentioned setting, verifying $ f (x) = d^2 (x,I)$ fits into the convexity (or strong convexity) definitions may not be an easy task.
Are there some (clean) examples that fits into the definitions above on $GL_n (\mathbb{R})^+$ (near $I$)?
[1] H. Zhang and S. Sra, First-order Methods for Geodesically Convex Optimization, JMLR, 2016.
|
2025-03-21T14:48:31.912516
| 2020-08-29T00:09:42 |
370360
|
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|
Stack Exchange
|
What is the approximation of $\log(|\zeta'(\frac{1}{2}+it)|)$ in Dirichlet polynomial if it is exists?
I have done some search many times on web to find any approximation of $\log|(\zeta'(s))|$ in Dirichlet polynomial but I didn't got it, Probably that $\log(|\zeta'(s)|$ dosn't have a Dirichlet polynomial approximation yield probably for $\log(|\zeta'(\frac{1}{2}+it)|)$ also dosn't have an approximation in Dirichlet polynomial but I do not have complet confidence for that , Now my question here is: What is the approximation of $\log(|\zeta'(\frac{1}{2}+it)|)$ in Dirichlet polynomiall if it is exists ?
Edit $s$ is a complex variable ,note that $0<t\leq T$ , $T$ is large enough, I have added this detail because I missed it when I posted the question.
Related question:
(Series representation for $\log(|\zeta(\frac{1}{2}+it)|)$)
Note The motivation of this question is the mean -value estimate of derivative of Riemann zeta function over non trivial zero of Riemann zeta function
A Dirichlet polynomial is a function of the form
$\sum_{1\le n \le X} a_n n^{-s}$, where $a_n$ are complex numbers and
$s = \sigma + i t$ with $\sigma$ and $t$ real. It is an analytic
function of $s$.
You ask whether certain functions can be approximated by a Dirichlet
polynomial. The specific functions you mention are of the form
$\log(|f(s)|)$ and $\log(|f(\frac12 + i t)|)$. You don't say what
range of $s$, or of $t$, you want to have an approximation, but let's
assume the ranges are reasonably large because you want to use that
approximation to prove something.
Then the answer is 'no' because the functions you want to approximate are
real-valued, but a Dirichlet polynomial (or any analytic function)
cannot have a very small imaginary part throughout a large region.
That explains why $\log(|f(s)|)$ cannot be approximated (by any analytic
function, so in particular a Dirichlet polynomial). A Dirichlet polynomial
cannot have very small imaginary part at $\frac12 + i t$ for a wide range
of $t$, so that is why $\log(|f(\frac12 + i t)|)$ cannot be approximated
(take the imaginary part of each term in the sum, and recognize it as
a sum of trig functions).
It is irrelevant that the function $f$ happens to be $\zeta'$.
The main reason why you cannot find a Dirichlet series approximation (even one that holds "most of the time" and "within a certain error") is because $\zeta'(s)$ does not have an Euler product.
|
2025-03-21T14:48:31.912670
| 2020-08-29T02:24:20 |
370366
|
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|
Stack Exchange
|
The isomorphism class of the 1-representation of a complete quiver
Let $Q$ be a quiver with vertex set $Q_0$ and the arrows $Q_1$. A quiver self $Q$ is said to be complete if it has no loops and for every arrow in $Q_1$ the opposite arrow is also in $Q_1$.
A representation $(V_i,f_i)_{i \in Q_1}$ is said to be a $1$-representation if $V_i = V_j$ for each $i$ and $f_i$ is the identity linear transformation.
My questions:
How to find the isomorphism class of the $1$-representation of a quiver Q?
How to find whether a representation of $Q$ belongs to this class?
is there any combinatorial description of elements of this class?
is this class of particular importance in the theory of finite quivers?
Kindly share your thought and thank you for your valuable time.
"A representation $(V_i, f_i)_{i \in Q_1}$ …" Shouldn't the $V_i$ and $f_i$ be indexed separately, by $Q_0$ and $Q_1$ respectively?
Since your $1$-representation consists only of identity maps, it'll be uniquely determined by the dimension vector. So since the dimension at the vertices are all the same, they're indexed by $\mathbf{N}$?
|
2025-03-21T14:48:31.912771
| 2020-08-29T05:21:37 |
370369
|
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|
Stack Exchange
|
Maximizing a vector after a series of matrix multiplications
Problem Statement
Let's say we have a set of $n\times n$ matrices $X=\{M_1,\ldots,M_r\}$ and weights of these matrices $\{w_1,\ldots,w_r\}$ along with a set of "initial vectors" $\{v_1,\ldots,v_s\}$. We then define the weight of a vector $u\in\mathbb{R}^n$ by
$$w(u)=\min\{x:\big(\exists i_1,\ldots,i_k,j:x=w_{i_1}+\cdots+w_{i_k},u=M_{i_1}\cdots M_{i_k}v_j \big)\}\textrm{ or }w(u)=\infty\textrm{ if this set is empty}$$
Finally, we introduce a function $f:\mathbb{R}^n\to\mathbb{R}$ which we wish to maximize. We can then ask questions like "what is the maximum value of $f$ over all vectors of weight at most some $y$?", or "given $x$, what is the minimum weight of a vector $u$ satisfying $f(u)\geq x$?". We can also ask for the most efficient algorithm to answer these questions, or if a closed form exists.
I am very curious if this general question has been studied at all. Are there any known results in this area? Any references or pointers would be greatly appreciated.
Motivation
An example of a problem which can be formulated as above is the $UwU$ problem. In this problem, Alice wants to send Bob as many $UwU$'s as possible. She starts by manually typing "$UwU$" and afterwards has two options at each step: copy everything she has already written (which takes $kp$ seconds for $p>0,k>1$) or paste the last thing she copied (which takes $p$ seconds). This can be formulated using the matrix set $COPY=\big(\begin{smallmatrix}
1 & 0\\
1 & 0
\end{smallmatrix}\big)$ and $PASTE=\big(\begin{smallmatrix}
1 & 1\\
0 & 1
\end{smallmatrix}\big)$ and $w_1=pk$ and $w_2=p$ and a set of initial vectors consisting just of $v_1=\big(\begin{smallmatrix}
1\\
0
\end{smallmatrix}\big)$. We then want to maximize the first coordinate of a vector by multiplying by $COPY$ and $PASTE$ while keeping the cost/sum of weights low.
By generalizing this problem to the format above, I am hoping to see if anyone has studied this class of problems, and to uncover any research that may have already been done in this area.
What do you mean by the weight of a matrix? Is this just some arbitrary positive number?
Yes. The weight of a matrix is just a positive real number we assign to the given matrix.
|
2025-03-21T14:48:31.912944
| 2020-08-29T06:55:06 |
370371
|
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|
Stack Exchange
|
Constructing a model structure without knowing the class of weak equivalences
I need to prove the existence of a model structure but I am still unable to formulate a definition of the class of weak equivalences. I have the following informations:
The underlying category is locally finitely presentable
I have the set of generating cofibrations and every object is cofibrant
I have a conjectural set of generating trivial cofibrations
I have a cylinder functor but, unfortunately, it is not a left adjoint (so I cannot use Olschok's theorems)
This model structure is left determined and probably not simplicial
I can prove that there exist objects which are not fibrant (so Isaev's theory cannot be used either).
Does it exist a method for this specific situation ? I tried to relate
this category to other categories by adjunctions but the acyclicity
condition is never satisfied. Of course, that would be much simpler if
I could characterize conjecturally the class of weak equivalences.
I'm going to say something trivial that you know (it's just to determine what options are available), but you have all acyclic cofibrations, and you can determine acyclic fibrations from the generating cofibrations, right? Then weak equivalences should just be the morphisms that can be written as compositions of the two. So standard theorems about cofibrantly generated model categories could apply. But this may not be a very useful description of weak equivalences of course.
@NajibIdrissi I agree with you but the problem is precisely that I cannot figure out from that what is the definition of a weak equivalence. That is the problem. I will find out eventually, I asked the question just in case someone would know a generic method.
Do you know what the fibrant objects ought to be?
@AlexanderCampbell Yes I have a conjectural description.
In some model structures (e.g. Joyal) a full description of the weak equivalences is tricky. It might be valuable to concentrate on weak equivalences between fibrant objects -- from what you say, these should be the homotopy equivalences with respect to your cylinder. Perhaps this can be reformulated into something more easily checkable. And for instance, if you choose a fibrant replacement functor $F$, then you can describe the weak equivalences as those maps sent to w.e.'s (between fibrant objects) by $F$. Otherwise I'm not sure what to say with this level of information.
May I suggest to have a look to my last paper https://arxiv.org/abs/2005.02360 ? It gives a lots of criterion to get left,right and "weak" model categories from assumption of this kind. there is an emphasize on Cisinski-Olschok type construction, but many results go beyond this. Note that because you are in the case where every object is cofibrant, a left semi-model structure would be the same as a Quillen model structure.
@SimonHenry I will. Thanks.
|
2025-03-21T14:48:31.913157
| 2020-08-29T08:54:09 |
370380
|
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|
Stack Exchange
|
Polyharmonic Maass forms are automorphic forms on $\mathrm{SL}_2(\mathbb{R})$
Let $G=\mathrm{SL}_2(\mathbb{R})$, $K=\mathrm{SO}(2)$, and $f$ be a holomorphic modular form of weight $k$ for $\Gamma$ a Fuchsian group of the first kind. In Borel's book, 'automorphic forms on $\mathrm{SL}_2(\mathbb{R})$', the lifted function $\tilde{f}$ is considered, where
$$
\tilde{f}(g)=j(g,i)^{-k}f(gi), \quad g \in G.
$$
In this book, a Lemma says that $f$ satisfies
$$
C\tilde{f}=(\frac{k^2}{2}-k)\tilde{f},
$$
where $C$ is the Casimir operator $C=\frac{1}{2}H^2+EF+FE$, $H=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, E=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, F=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$. This implies that $\tilde{f}$ is an automorphic form for $\Gamma$.
My first question is
Why the above lemma is true?
To ask the second question, let me define the polyharmonic maass forms. A polyharmonic maass form $f$ of weight $k$, depth $m$ for $\Gamma$ is a smooth function on $\mathbb{H}$ with the following properties:
(1). $f|_k \gamma (z) = f(z) \quad$ for $\gamma \in \Gamma$.
(2). $\Delta_k^m f =0 \quad$ where $\Delta_k=-y^2(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})-iky(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y})$.
(3). moderate growth at the cusps
J.C.Lagarias and R.C.Rhoades' paper 'Polyharmonic Maass forms for $\mathrm{PSL}(2,\mathbb{Z})$' says that the finite dimensionality of the space of polyharmonic maass forms follows from the finite dimensionality of the space of automorphic forms. Hence it seems that there is a natural embedding from polyharmonic maass forms to automorphic forms.
I guess that the same lift induce the natural linear embedding. If this true, then for a polyharmonic maass form $f$, there is a polynomial $P$ in $C$ such that
$$
P(C)\tilde{f}=0.
$$
Is my suggestion true?
If yes, what polynomial $P$ satisfies the above condition?
Please restrict to one question per post (standard MO policy). Your first question concerns Lemma 5.15 in Borel's book. The proof is included in the book, right below the lemma (it takes about one page).
GH from MO // The proof in the Borel's book is wrong. I resolved the questions, using different approach. Anyway, thank you for the comment.
|
2025-03-21T14:48:31.913338
| 2020-08-29T10:14:53 |
370384
|
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|
Stack Exchange
|
Is there a 'certainty' principle?
Heisenberg's uncertainty principle is a restriction on which probability distributions can describe the position and momentum of a quantum particle.
In mathematical terms it says that if $\psi\in L^2$ is normalized, and we define $f,g\in L^1$ by $f(x)=|\psi(x)|^2$ and $g(k)=|\hat\psi(k)|^2$ then we have
$$V(f)V(g)\geq\frac14$$
where $V$ is the variance of the probability distribution with the given density function.
There are various other uncertainty principles, including the Entropic uncertainty principle and Hardy's uncertainty principle. Define $f,g\in L^1$ to be compatible if there exists $\psi\in L^2$ such that $f(x)=|\psi(x)|^2$ and $g(k)=|\hat{\psi}(k)|^2$. Then each uncertainty principle states a condition which compatible $f$ and $g$ must obey.
I noticed a curious fact, which holds true of everything I could find in the literature calling itself an 'uncertainty principle'. For fixed $f$ the restriction on $g$ is always a convex set. For example the set of $g$ satisfying $V(g)\geq\frac1{4V(f)}$ is convex because variance is a concave function on the space of probability distributions.
This does makes sense with the name 'uncertainty principle'. Intuitively, mixing probability distributions cannot produce a result that is more 'certain' than all of them.
However, playing with the Discrete Fourier Transform as a toy model, I noticed that the set of $g$ compatible with a given $f$ need not be convex.
Randomly sampled $g$ compatible with $f = (0.46,0.46,0.08)$ and $f = (0.46,0.46,0.07,0.01)$:
Note that these sets are nonconvex, and don't even contain the maximally-uncertain uniform distribution (the centre point of the simplex of possible distributions). So the uncertainty of distributions in these sets is bounded above as well as below.
In the case of distributions on $\mathbb R$, can we even find a single $f$ for which we can prove the set of compatible $g$ is not convex?
Is there a 'certainty principle' that, for $f$ within some class, puts an upper bound on the variance or entropy of compatible $g$?
Sorry but what exactly is the set of $g$ "compatible" with $f$? Thanks.
That is, I think that given $f$ you obtain a statistic, e.g. $V(f)$, and you want to consider sets like "all $g$ whose Fourier transform has variance $7$". But I'm not sure.
@usul Define $f,g\in L^1$ to be compatible if there exists $\psi\in L^2$ such that $f(x)=|\psi(x)|^2$ for almost all $x$ and $g(k)=|\hat{\psi}(k)|^2$ for almost all $k$. I'll edit the post to make that more explicit, thanks!
The Paradoxical Certainty Principle: With all certainty, there is no certainty principle.
Thanks for the clarification! I would think that someone must know quite a bit this about, given $f$, the structure of the sets of feasible/compatible $\psi$, $\hat{\psi}$, and $g$.
By request, I add a comment as an answer with some additional details; but what I meant is really straightforward. The simplest realization is as follows: take any $\psi\in L^2$. Split its support into finitely many parts to obtain a representation $\psi=\sum_{k= 0}^N\psi_k$ where $\psi_0$ is small in $L^2$ (the infinite tail) and $\psi_k$ for $k>0$ are small (less than $\varepsilon$) in $L^1$ (short intervals). Now multiply each $\psi_k$ with $k>0$ by $e^{2\pi i Mkx}$ with $M$ chosen so that $\sup_{\lvert y\rvert>M,1\le k\le N}\lvert\widehat\psi_k(y)\rvert\le \frac{\varepsilon}N$ (it exists by Riemann–Lebesgue). Then the Fourier transform of the resulting function at any point $y$ will be bounded by $\lvert\widehat\psi_0(y)\rvert+3\varepsilon$. The first part doesn't influence anything because its $L^2$-norm is small and the rest is uniformly small and, therefore, spread wide.
If $\psi\in L^1\cap L^2$, then no special treatment of $\psi_0$ is needed. Also, you can get the true uniform smallness by splitting into countably many parts and choosing the phase shifts inductively instead of just using an arithmetic progression. And so on, and so forth.
Edit: Now about convexity. Take $f$ to be the characteristic function on $[0,1]$ and consider $g(k)$ where $k\in\mathbb Z$ (in this case the point values are continuous functionals). Clearly, every sequence with all zeroes and one $1$ is admissible ($\psi(x)=e^{2\pi ik_0x}$ on $[0,1]$). Thus, if the convexity had held, we would be able to construct a function on $[0,1]$ that is identically $1$ (or, at least, as close to that as we would like) such that $g(0)=g(1)=\frac 12$ and all other $g(k)=0$. However, that would be just a two-term polynomial with equal coefficients, so it would vary quite a bit in absolute value on $[0,1]$. This proves at least that sometimes convexity does not hold. I suspect that this trick can be generalized quite a bit but the details are elusive yet.
The original post included a quote of (I guess) a comment by @OscarCunningham on a comment of yours, both apparently now deleted. For the benefit of posterity, I inlined this as a description of what was happening, rather than a quote of deleted comment. I hope that was OK. (EDIT: Oops, I see that the comments you were quoting were on a different answer, not the main post. Sorry! Feel free to revert my edit if it was inappropriate.)
@LSpice Nah, it is fine either way. I'd rather think of the second half of the question now (if I figure it out, I'll edit the post anyway) :-)
Nice proofs! Took me a while to understand the second one. For fellow novices: The Nyquist–Shannon sampling theorem implies that if $\psi$ is supported on $[0,1]$ then $\hat\psi$ is determined by its values on $\mathbb Z$.
I agree it would be nice if we could generalize to more $f$. The compact support looks to me to be essential though.
@OscarCunningham "The compact support looks to me to be essential" Nope, it isn't. The construction is stable enough: just replace $f$ by anything that is close to the characteristic function of $[0,1]$ in $L^1$ and, instead of specifying the values at points, specify the averages over short intervals around them. So, the convexity fails quite often. My question is rather if it ever holds, i.e., if there exists a single $f\in L^1$ with integral $1$ such that the (closure of) the corresponding $g$ is convex.
With $D_x=\frac{d}{i dx}$, the Heisenberg uncertainty principle in its most classical form can be deduced from the equality
$$
2\Re \langle \hbar D_x u, ix u \rangle_{L^2(\mathbb R)}= \langle \bigl[\hbar D_x, ix\bigr] u, u \rangle_{L^2(\mathbb R)}=\hbar\Vert u\Vert_{L^2(\mathbb R)}^2,
$$
which implies
$
\Vert \hbar D_x u\Vert_{L^2(\mathbb R)}\Vert xu\Vert_{L^2(\mathbb R)}\ge
\frac\hbar 2\Vert u\Vert_{L^2(\mathbb R)}^2,
$
where the constant $\hbar/2$ can be proven sharp by testing on a Gaussian function.
So much for the lowerbound. Maybe a "certainty principle" would mean that we want to deal with the upperbound (?) We have
$$
\Vert \hbar D_x u\Vert_{L^2(\mathbb R)}\Vert xu\Vert_{L^2(\mathbb R)}\ge
\Re \langle \hbar D_x u, ix u \rangle_{L^2(\mathbb R)}=
\frac\hbar 2\Vert u\Vert_{L^2(\mathbb R)}^2,
$$
but it is true that the left-hand-side could be much larger than the rhs: take for instance
with $\omega$ smooth, valued in $[0,1]$, equal to 1 for $\vert x\vert\ge 2$, to 0 on $\vert x\vert\le 1$, $\lambda \ge 1$,
$$
u_\lambda(x)=(x^2+1)^{-1/2}\omega(x/\lambda),\quad
\Vert u_\lambda\Vert_{L^2(\mathbb R)}^2 \le π,
$$
$$
u'_\lambda(x)=-\underbrace{x(x^2+1)^{-3/2}\omega(x/\lambda)}_{\text{bounded in $L^2$}}+\underbrace{\frac1\lambda \omega'(x/\lambda) (x^2+1)^{-1/2}}_{\substack{
\text{with limit $0$ in $L^2$}\\\text{since $\omega'$ has support $[\lambda, 2\lambda]$}
}},
$$
$$
x u_\lambda(x)=\frac{x}{\sqrt{x^2+1}} \omega(x/\lambda),\quad
\Vert xu_\lambda(x)\Vert_{L^2(\mathbb R)}={+\infty}.
$$
As a consequence, the upperbound is $+\infty$.
I don't think this answers the question. The fact that there's no bound on $\Vert \hbar D_x u\Vert_{L^2(\mathbb R)}$ in terms of $\Vert xu\Vert_{L^2(\mathbb R)}$ doesn't mean that we can't bound $\Vert \hbar D_x u\Vert_{L^2(\mathbb R)}$ by some other function of $u$.
@OscarCunningham Given any $f$ (i.e., $|\psi|$), we can put some "random phase" on it so that the Fourier transform will be uniformly arbitrarily small, which (due to the conservation of the $L^2$ norm), will force it to spread as far as you wish, so no estimate from above on some "concentration quantity" is possible. The convexity question is more interesting one. I would say that the set of $g$ given $f$ is almost never convex, but I don't have a proof.
@fedja That would certainly answer the second question. But I can't put together the proof from what you wrote. Could you add it as an answer with some additional details?
You might find the von Neumann-Koopman mechanics of interest. Here, classical mechanics is formulated in the same formal language of Diracs transformational theory which superseded both the Wave Mechanics of Schrodinger and the Matrix Mechanics of Heisenberg.
Observables, as in Quantum Mechanics, are represented by self-adjoint operators on the Hilbert space of KvN wave functions. However, unlike quantum mechanics, these operators commute and so are simultaneously measurable. This means that the uncertainty principle of Heisenberg disappears to be replaced by the usual deterministic laws of classical Newtonian mechanics - aka, a 'certainty principle'.
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2025-03-21T14:48:31.914063
| 2020-08-29T10:58:53 |
370387
|
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|
Stack Exchange
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Subharmonic function in unbounded regions
The harmonic majorization for a subharmonic function $h$ is well-known for bounded regions $\Omega \subset \mathbb{C}$:
$$h \le 0 \text{ in }\partial \Omega \Longrightarrow h \le 0 \text{ in }\Omega.$$
I know this is related to maximum principle.
I need a reference for the same result for unbounded regions of the complex plane. I think one should further assume that the function is bounded on $\overline{\Omega}$. I saw this somewhere but I am not able to find a reference.
Yes, this is called the Phragmen-Lindelof Principle: For every region on the Riemann sphere, if $h$ is subharmonic and bounded from above, and
$$\limsup_{z\to\zeta}h(z)\leq 0$$ for all $\zeta\in\partial\Omega$, except finitely
many points, then $h\leq 0$ in $\Omega$. If your domain $\Omega$ is an unbounded domain in $C$, just include $\infty$ to this finite exceptional set.
There are many improvements of this, for example, finite exceptional set can be replaced by a set of zero capacity. Boundedness from above can be replaced by a weaker condition $h(z)<o(\log|z|),\; z\to\infty$. This can be replaced by a
weaker growth condition,
if something is known about the shape of the unbounded domain near infinity. For example, if the portion of $\Omega$ near $\infty$ is contained in a sector of opening $<\pi/\alpha$, then instead of boundedness one can impose the growth condition $h(z)<o(|z|^\alpha)$.
Refs. Ransford, Potential theory in the plane,
Levin, Lectures on entire functions,
Hayman, Kennedy, Subharmonic functions.
In fact, the proof is very simple. Suppose $h$ is bounded from above and $h(z)\leq 0$ on $\partial\Omega$, where $\Omega$ is an unbounded domain.
Here $\partial$ is with respect to $C$, so it does not include $\infty$.
Suppose for simplicity that $\Omega$ does not intersect the unit disk. Consider $u(z)=h(z)-\epsilon\log|z|$, where $\epsilon>0$. Then $\limsup_{z\to\zeta}u(z)\leq 0$ for $\zeta\in \partial^*\Omega$, the boundary with respect to the Riemann sphere, so it includes $\infty$. By the usual Maximum principle we conclude that $u(z)\leq 0$ on $\Omega$.
Passing to the limit for fixed $z$ as $\epsilon\to 0$, we obtain $h(z)\leq 0$.
To obtain the result under other conditions, you use other auxilliary functions in place of $\log|z|$.
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2025-03-21T14:48:31.914236
| 2020-08-29T11:47:41 |
370389
|
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|
Stack Exchange
|
Where does the Laplace transform come from?
The Gelfand transform on the commutative Banach *-algebra $L^1(\mathbb{R})$ is just the Fourier transform.
Q. What can we say concerning the Laplace transform?
this is answered at https://en.wikipedia.org/wiki/Gelfand_representation#Examples
We can evaluate a power series on the real line, or we can evaluate it on the unit circle in the complex plane. The latter gives us Fourier series. We can also evaluate it on the whole complex plane. These three options also show up in the continuous transforms, giving the Laplace transform on the real line, the Fourier transform, and the Laplace transform on the entire complex plane. The function being transformed is the analogue of the coefficients of the power series, and the function you get after transforming is the analogue of the function you get when you evaluate the power series.
Set $A = L^1([0, \infty))$, equipped with the structure of a Banach $*$-algebra via convolution. The spectrum of this algebra is the half plane $\text{Re}(z) \geq 0$, and the Gelfand transform is the Laplace transform.
Cool, it is concerning with one-sided Laplace transform. And for two-sided Laplace transform?
This is more or less what Carlo was referring to with his wikipedia link. (But I guess it may as well be posted as an answer.)
@AliBagheri One way to do it is to take the convolution algebra of the multiplicative group of positive real numbers, equipped with its Haar measure. Here the Gelfand transform is what is more classically known as the Mellin transform, and the Mellin transform is a change of variables away from the two sided Laplace transform. Maybe it's possible to get straight to the two sided Laplace transform via another choice of measure or something - I'm not sure.
@YemonChoi Yeah, I posted my answer within a minute of Carlo posting his comment. I thought for a moment about deleting it, but I noted that I got reasonably far in a PhD with an emphasis on operator algebras before I learned this fact. The question is borderline for MO, but I guess it's still open and I think this is the right answer, so :shrug:
Sure, TBH I miss the days of MO when we would all share more of the things we happened to learn early on and other people didn't (the difference between Banach and Cstar algebras is one reason why examples such as $L^1({\bf R}_+)$ are baked into my "training" while e.g. I can never remember the proof of Kaplansky density :) )
|
2025-03-21T14:48:31.914433
| 2020-08-29T12:55:48 |
370392
|
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|
Stack Exchange
|
Feynman rule in finite volume or at periodic boundary condition
I am wondering, whether the Wick theorem for free particles (in the form: $n$-point correlators split into $2$-point correlators) also hold for particles in a finite volume or for particles on a circle (periodic boundary conditions)? Or are there controlled extra terms?
I was curious to derive a set of Feynman rules for a quantum mechanical particle (not yet a field) in some potential etc, in either of the two settings.
PS: I might add I would be also interested to see any finite or periodic analogon of the harmonic oscillator, for which again Wicks theorem holds, so I can use it as the underlying free theory for deriving Feynman rules.
I am very sure that many people have thought about this before, but I have a hard time finding literature on it. Any hint would be much appreciated :-)
Greetings,
Simon
also posted at https://physicsoverflow.org/43194 (it is best practice to disclose cross-postings, in order to avoid duplication of efforts)
I thought that Wick's theorem holds for any quadratic Hamiltonian; a finite volume or periodic boundary conditions would be realized by a potential term, without spoiling the quadratic nature, so it should not invalidate Wick's theorem.
Carlo, that would be great - do you have a reference? I would somehow expect that, but the proofs I found where too involved for myself to easily include additional potential terms (e.g. the binomial formula for V(x+deltax) does not hold if I include a cut-off)?
see, for example, page 170 of these notes --- the only assumption is a bilinear Hamiltonian in the fermionic (or bosonic) fields.
The algebraic steps involved in a Wick decomposition are fairly simple: Introducing sources with respect to which you can take derivatives, and completing the square. Off the top of my head, I don't see how spatial boundary conditions would interact nontrivially with these operations - the boundary conditions will influence the propagator, not the combinatorics of Gaussian integrals. Of course, in practice, I'd quickly run through these steps for any particular model in question to make sure.
|
2025-03-21T14:48:31.914601
| 2020-08-29T13:34:09 |
370393
|
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Stack Exchange
|
Can convex programming capture affine curves or surfaces?
A friend of mine asked me a question that relates to algebraic geometry and convex programming.
The question was "What is the minimal size of a program that capture (the intended object is not necessarily one slice or a projection of the program (perhaps each slice captures at most one point of the variety) and the program could have arbitrary large but finite number of dimensions compared to the dimension of the space in which the variety at hand lives) the portion of a given algebraic curve presented by monomials that lies within a box around the origin?".
I am wondering can convex programs or quantified convex programs capture exactly one-dimensional curves or two dimensional surfaces etc. in affine spaces?
Or is there a characterization of what curves or surfaces can be captured?
Perhaps the question is trivial for well known and elementary reasons but in that case I just do not know.
|
2025-03-21T14:48:31.914705
| 2020-08-29T13:39:15 |
370394
|
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"Giorgio Metafune",
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Stack Exchange
|
Optimization problem with definite integral inequality constraints
Question: How can we prove that there exists a real constant $c\ge 1$ such that the following inequality holds for all integers $d>1$ and all real numbers $r\in\left[1,\sqrt{d}\right]$?
$$\int_{-1}^1 \left(\sqrt{r^2-x^2}\right)^{d-1} dx\le c\cdot \frac{r^d}{\sqrt{d-1}}$$
(Furthermore, is it also possible to find an upper bound for the minimum value of $c\in [1,\infty)$ such that the above inequality holds for all $d>1$ and all $r\in\left[1,\sqrt{d}\right]$?)
I get that your integral is less than $2r^{d-1}$ for all $r \ge1$. Just estimate the integrand with $r^{d-1}$.
Just estimate the LHS by $r^{d-1}\int_{-\infty}^{\infty}\exp{-(d-1)\frac{x^2}{2r^2}}dx=\sqrt{2\pi}\frac{r^d}{\sqrt{d-1}}$.
Thank you @GiorgioMetafune. This way, whan $r$ is small (say constant), we get $c$ depending on $d$, and we do not get that the inequality holds for a constant $c$ and for all $d>1$ and all $r\in\left[1,\sqrt{d}\right]$.
Thank you once again @fedja! This is precisely consistent with what I was expecting to obtain!
Ah, I misunderstood the questione, sorry.
@fedja how much can be the LHS different from this estimate?
Looks like just constant number of times, i.e., there is a lower bound of the same type in the range you are interested in.
Than you @fedja!
Mathematica answers
NMaximize[{Integrate[(r^2 - x^2)^(d/2 - 1/2), {x, -1, 1},
Assumptions -> d > 1 && r >= 1]/r^d*Sqrt[d - 1],
r >= 1 && r <= Sqrt[d] && d > 1 && d \[Element] Integers},{r, d}]
$$\{2.43959,\{r\to 1.01254,d\to 28\}\} $$
and
NMaximize[{Integrate[(r^2 - x^2)^(d/2 - 1/2), {x, -1, 1},
Assumptions -> d > 1 && r >= 1]/r^d*Sqrt[d - 1],
r >= 1 && r <= Sqrt[d] && d > 1}, {r, d}, AccuracyGoal -> 4,PrecisionGoal -> 4]
$$ \{2.50662,\{r\to 149.294,d\to 611671.\}\}$$
Addition. The command of Maple confirms it by
restart;Digits := 20;DirectSearch:-Search((d, r) -> int((r^2 - x^2)^(1/2*d - 1/2),x = -1 .. 1, numeric, epsilon = 1/1000)*sqrt(d - 1)/r^d, {1 <= d, 1 <= r, r <= sqrt(d)}, maximize);
$$[ 2.5066284493892445574, \left[ \begin {array}{c}
280247431.41221862419\\ 1059.2785342279385942
\end {array} \right] ,317]
$$ It seems the supremum is attained as $d \to \infty$ and $r \to \infty$.
Thank you! However I would like to prove it mathematically.
@Penelope Benenati: But this is math.
Elementary asymptotic estimates give $\sqrt{2\pi}=2.506628274631$ for the constant.
Do you mean "asymptotic" w.r.t. $d,r\to\infty$? if so, how can you first mathematically prove that "the supremum is attained as $d,r\to\infty$"?
@Penelope Benenati: I think the asymptotics of the integral under consideration as $d\to\infty$ can be found by the Laplace's method. That asymptotics depends on $r$. According to numeric calculations, its maximum should be attained for $r\approx \sqrt d$.
Thank you for your answer. How can we prove that the supremum for $c$ is attained exactly as $d\to\infty$?
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2025-03-21T14:48:31.914906
| 2020-08-29T14:06:43 |
370396
|
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Stack Exchange
|
Name and properties of $\mathrm{lcm}(\{1,\,\cdots,\,n\})$
one of the most prominent functions of the first $n$ natural numbers is the factorial $n!$ that denotes their product.
Today however I wondered whether the least common multiple $\mathrm{lcm}(n):=\mathrm{lcm}(\{i\in\mathbb{N}\,|\,1\leqq i\leqq n\})$ of the first $n$ natural numbers has already been the subject of mathematical studies.
The first elements of their sequence are $1,\, 2,\, 6,\,12,\,60,\,60,\,420,\,840,\,2520,\,2520,\,27720,\, \dots$ and it is A003418 – in the OEIS, as Christian Bernert pointed out in his comment.
Questions:
is there already an established name for $\mathrm{lcm}(n)$?
is there an agreed upon notation?
does it have an extension for non-integral complex values, analogous to the gamma function?
what can be said about bounds on, resp. growth-rate of the number of consecutive equal values apart from their relation to prime-gaps?
It is not hard to see that $\mathrm{lcm}(n)$ equals the procuct of maximal prime powers less or equal n
This is a long- and well-studied sequence and I could not believe it is not yet in OEIS. And of course it is: http://oeis.org/A003418
(Also the comments there should answer some of your questions.)
@ChristianBernert maybe because entered it differently; I will edit my question accordingly.
This is similar to primorial.
The quantity $\operatorname{lcm}(n)$ is equal to $e^{\psi (n)}$, where $\psi$ is the second Chebyshev function. This function is well studied, and the prime number theorem is equivalent (indeed, usually proved using this equivalence) to the fact that $\psi (x)\sim x$.
|
2025-03-21T14:48:31.915059
| 2020-08-29T14:11:28 |
370397
|
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Stack Exchange
|
K-Theory of $C^{*}(X)$
I'm new to K-Theory for $C^{*}$-algebra and $C^{*}$-algebra of groups.
If $X$ is the group of finite support bijections of natural numbers then what is the K-Theory of $C^{*}(X)$?
I was planning to calculate $C^{*}(X)$ first and then its projections and their homotopy classes but I failed to determine $C^{*}(X)$ in the first place.
Here is the definition of the $C^{*}$-algebra of a group:
https://pages.uoregon.edu/ncp/Courses/2016ShanghaiCrPrdFiniteGps/Slides/Lecture1_Print_NoP.pdf
Can anybody help me?
Thanks a lot.
The group you describe should be the infinite symmetric group $S_{\infty}$. The $K$-theory of its $C^*$-algebra has been determined by Kerov and Vershik in
The K -functor (Grothendieck group) of the infinite symmetric group
https://link.springer.com/article/10.1007/BF02104985
The main result can be summarised as follows: Let $\mathcal{A}$ be the ring of symmetric polynomials in infinitely many variables. This is isomorphic to $\mathbb{Z}[a_1, a_2, \dots]$, where $a_i$ is the $i$th elementary symmetric function in an infinite number of arguments. Then
$$
K_0(C^*S_{\infty}) \cong \mathcal{A}\,/\, (a_1 - 1)\mathcal{A}\ .
$$
The isomorphism sends the irreducible representation $\pi_{\lambda}$ corresponding to the Young diagram $\lambda$ to the Schur function corresponding to $\lambda$. This is also an isomorphism of rings, where the ring structure on the left hand side comes from the observation that
$$
K_0(C^*S_{\infty}) \cong \lim_n K_0(C^*S_n)
$$
and multiplies two representations $\pi_1 \colon S_n \to GL(V)$ and $\pi_1 \colon S_m \to GL(W)$ to
$$
Ind_{S_n \times S_m}^{S_{n+m}}(\pi_1 \otimes \pi_2)\ ,
$$
where $S_n \times S_m$ sits inside $S_{n+m}$ with $S_n$ permuting the first $n$ elements and $S_m$ permuting the other $m$.
Thank you for your answer sir, but I don't understand the first paragraph: The group is infinite symmetric group, how do you know that? And second what is its $C^{*}$-algebra? Sorry if my questions are so dumb.
@PegLegScott please don't take this the wrong way, but it really doesn't look like you have enough background to be studying K-theory of C*-algebras. Why not start a little further back?
You wrote you consider the group of finite support bijections of $\mathbb{N}$. Such a bijection is the identity outside of a finite set and permutes the elements in that finite set. Therefore you can identify the group of finite support bijections with the union over all groups $S_n$. Does that help?
As for the group $C^$-algebra, I guess the answer is: It is what it is. You can view it as the direct limit over all $C^$-algebras $C^(S_n)$, which reveals it as an AF-algebra, if that helps. The algebras $C^(S_n)$ are of course finite-dimensional and can be seen as direct sums of matrix algebras.
@UlrichPennig Thank you very much sir. I learned a lot from your answer. It is clear now for me.
Karen Strung has good notes about C*-algebras in the context of the classification programme. They also treat K-theory and can be found here: https://strung.me/karen/CStarIntroDraft.pdf . Maybe these are useful.
I appreciate it, sir.
|
2025-03-21T14:48:31.915275
| 2020-08-29T15:07:49 |
370399
|
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|
Stack Exchange
|
Artin reciprocity via Shimura varieties
The point of Shimura varieties, as far as I've understood it, is that for a given Shimura datum $(G,D)$, there exist models, by which I mean that for congruence subgroups $\Gamma$ there exists a Shimura variety $X(\Gamma)$ defined over some number field. Hence we get a action of the absolute galois group $G_{\mathbb{Q}}$ on
$$V:=\lim_{\Gamma} H^*_{ét}(X(\Gamma),\mathbb{Q}_{\ell}).$$
However, from an adelic point of view, we also get a (continuous) action of $G(\mathbb{A}_f)$ on the shimura variety, and so in fact $V$ is a representation of $G_{\mathbb{Q}}\times G(\mathbb{A}_f)$. The point now (from a Galois representation/ Langlandian point of view) is that for every representation $\rho:G_{\mathbb{Q}}\rightarrow \mathbb{\mathbb{Q}_\ell}^\times$, we can associate a representation of $G(\mathbb{A}_f)$ as
$$\text{Hom}(\rho,V).$$
The main difficulty (as far as I understand it) is now to show that we can generate sufficently many representations that way to prove Langlands Program. My question is how this argument looks like in abelian case, i.e. what happens when $G=\text{GL}_1$? The Shimura varieties of tori are relatively simple to understand, namely we know that the Shimura variety associated to $\text{GL}_1$ is of the form
$$\mathbb{Q}^\times \backslash \mathbb{A}_f^\times/K$$
for an open compact subgroup $K\subset \mathbb{A}_f^\times$ and is a finite étale scheme over some number field. How do we finish the proof from there on to get Artin's reciprocity? Or do we need to restrict ourself to the local case to even be able to complete the proof?
If $K$ is 'everything 1 mod N' for some N, then the canonical model of $\mathbf{Q}^\times_+ \backslash \mathbf{A}^\times_{\mathrm{f}} / K$ is exactly $\mu_N / \mathbf{Q}$, the scheme of $N$-th roots of unity. Any open compact $K$ will contain one of these, so $GL_1 / \mathbf{Q}$ Shimura varieties all look like quotients of $\mu_N$ for some $N$.
Hence the answer to "How do we finish the proof from there on to get Artin's reciprocity?" is "we prove the Kronecker--Weber theorem" [i.e. every abelian extension of $\mathbf{Q}$ is contained in a cyclotomic field].
Perhaps it's disappointing that Shimura varieties don't tell you how to prove Kronecker--Weber. But they do something much more important: they tell you how to generalize Kronecker--Weber, pointing you towards a much more general (mostly conjectural) picture of which Kronecker--Weber is just one small part.
Thank you for the nice answer. Does the situation change if we want to study $G_{\mathbb{Q}_p}^{ab}$? Harris and Taylor proved local Langlands with Shimura varieties, so is there hope for $\text{GL}_1$ in the local case?
"Harris and Taylor proved local Langlands [for $GL_n$] with Shimura varieties" is in the same category of statements as "Michaelangelo painted the Sistine Chapel with a paintbrush". Yes, Shimura varieties for unitary groups were involved; but it's very far from that to imply that once you know the standard properties of Shimura varieties, the local Langlands conjecture is somehow obvious.
|
2025-03-21T14:48:31.915508
| 2020-08-29T15:43:55 |
370403
|
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|
Stack Exchange
|
Conditions on the velocity ensuring that a flow moves points along the boundary of a manifold
Let
$\tau>0$;
$d\in\mathbb N$;
$v:[0,\tau]\times\mathbb R^d\to\mathbb R^d$ be Lipschitz continuous in the second argument uniformly with respect to the first with $v(\;\cdot\;,x)\in C^0([0,\tau],\mathbb R^d)$;
$X^{s,\:x}$ denote the unique element of $C^0([s,\tau],\mathbb R^d)$ with $$X^{s,\:x}(t)=x+\int_s^tv(r,X^{s,\:x}(r))\:{\rm d}r\;\;\;\text{for all }t\in[s,\tau]\tag1$$ for $(s,x)\in[0,\tau]\times\mathbb R^d$ and $$T_t(x):=X^x(t):=X^{0,\:x}(t)\;\;\;\text{for }x\in E$$ for $t\in[0,\tau]$.
Now let $M$ be a $d$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary and $\partial M$ denote the manifold boundary of $M$.
I would like to know which assumption on $v$ we need to impose in order to ensure that $$T_t(\partial M)=\partial M\;\;\;\text{for all }t\in[0,\tau]\tag2.$$
Phrased differently, I want to find a condition on $v$ ensuring that $T_t$ moves a "particle" $x$ "along the boundary" $\partial M$ for all $t\in[0,\tau]$:
I've read that we need to assume that $$\langle\left.v\right|_{[0,\:\tau]\times\partial M},\nu_{\partial M}\rangle=0\tag3,$$ where $\nu_{\partial M}$ denotes the unique outer unit normal field on $\partial M$. However, I don't see why $(3)$ implies the $(2)$. And is $(3)$ an additional assumption at all? It seems like $(3)$ should always hold: Let $(s,x)\in[0,\tau]\times\partial M$. The claim $\langle v(s,x),\nu_{\partial M}(x)\rangle=0$ is equivalent to $v(s,x)\in T_x\:\partial M$, where $T_x\:\partial M$ denotes the tangent space of $\partial M$ at $x$. Now, if $$\gamma(h):=X^{s,\:x}(s+h)\;\;\;\text{for }h\in[0,\tau-s],$$ then $\gamma(0)=x$ and $\gamma'(0)=v(s,x)$. Thus, $v(s,x)\in T_x\:\partial M$. Am I missing something?
What is $E$ supposed to be?
Do you see why its true when M is a disk in a plane? Does your 'proof' given at end of your question hold in this case?
@JackLee Sorry, I'm used to consider flows on a Banach space $E$. Here $E=\mathbb R^d$.
@PiyushGrover Please clarify: Do you mean that $(3)$ doesn't hold if $E=\mathbb R^2$ and $M={x\in\mathbb R^2:\left|x\right|\le1}$?
And $\Omega=M$ I suppose? You'd make it a lot easier for those of who try to answer your questions if you'd proofread them before posting them, and make sure all of your notations are consistent and well defined.
No, I meant your proof claiming that (3) is trivially satisfied. Lets take $v$ to be a constant vector field pointing in horizontal direction, parallel to x axis. Clearly (3) doesn't hold for most points on boundary of $M$.
@JackLee Yes, sorry. Fixed that now.
@PiyushGrover If $v$ is a constant, the wholse disk would be moved in that direction. So, I guess you're right, but where is the mistake in my proof?
In the situation you described, if $M$ is properly embedded (i.e., topologically embedded and closed), the flow of a vector field takes $\partial M$ to itself if and only if the vector field is everywhere tangent to the boundary. For a proof of the "if" direction, see Lemma 9.33 in my Introduction to Smooth Manifolds. (That lemma shows that the flow takes $M$ to itself if the vector field is everywhere tangent to the boundary, but if you look closely at the proof, it also shows that the flow preserves $\partial M$.)
The condition that $v$ is tangent to the boundary at a point $x\in \partial M$ is equivalent to the condition that $v$ is orthogonal to the unit normal vector there. (This is essentially the definition of a normal vector.)
If $M$ is not closed, the result is not true. For example, consider the case in which $M$ is the closed unit disk in $\mathbb R^2$ with one boundary point removed, and $v$ is a vector field that generates rotations. Then there is no nonzero time such that $T_t(\partial M)\subset\partial M$.
To answer the question in your comment: Once we know that $T_t(\partial M)\subset \partial M$ for each $t$, here's how to show that in fact equality holds.
First suppose $\partial M$ is connected. Because $T_t$ is a diffeomorphism from $\mathbb R^d$ to itself and $\partial M$ is closed in $\mathbb R^d$, it follows that $T_t(\partial M)$ is closed in $\mathbb R^d$ and therefore also (relatively) closed in $\partial M$. On the other hand, since $T_t$ restricts to a diffeomorphism from $\partial M$ to itself, $T_t|_{\partial M}$ is an open map (as a map between $(d-1)$-dimensional manifolds), and therefore $T_t(\partial M)$ is (relatively) open in $\partial M$. By connectivity, therefore $T_t(\partial M) = \partial M$.
If $\partial M$ is not connected, just apply the argument above to each connected component.
Thank you for your answer. In the "if-direction": Is your suggestion to consider $$\mathcal T:={t\in[0,\tau]:X^x(t)\in\partial M}$$ for some fixed $x\in\partial M$ and prove that this set is equal to $[0,\tau]$? You're doing this by showing that it is closed and open, but you've assumed that $D$ is a regular domain of a smooth manifold. Here $D=\partial M$, but $\partial M$ is not topologically closed (hence not "properly embedded") in $\mathbb R^d$ or am I missing something? ($\partial M$ is closed in $M$ though, if that's relevant)
@0xbadf00d: You're right -- for this to be true, you have to add the hypothesis that $M$ is properly embedded. I'll edit my response.
Meanwhile, I've figured out how we can prove that $T_t(\partial M)\subseteq\partial M$ for all $t\in[0,\tau]$ as long as $\partial M$ is closed. However, why can we infer from this result that it's not only "$\subseteq$", but "$=$"?
@0xbadf00d: See my addition above.
Can we infer from $T_t(\partial\Omega)=\partial\Omega$ for all $t\in[0,\tau]$ that $T_t(\Omega)=\Omega$ for all $t\in[0,\tau]$?
I'd still be interested in the question from my last comment.
What's $\Omega$?
I'm sorry, I'm always confusing $\Omega$ and $M$. So, $\Omega$ should be $M$. BTW, if we are able to show that $T_t(∂ M)\subseteq\partial M$ (as you did in your answer), aren't we able to use the same argumentation to hsow $T_t^{-1}(\partial M)\subseteq ∂M$? In the same manner, showing $T_t(M)\subseteq M$ and $T_t^{-1}(M)\subseteq M$ should be doable in the same way. All we would need to do is whowing that $T_t^{-1}$ solves a similar system as $(1)$. This is clearly the case when $v$ doesn't depend on time. I've asked for the general case here: https://math.stackexchange.com/q/3842350/47771.
A similar argument shows that $T_t(\text{Int}(M)) = \text{Int}(M)$, and therefore $T_t(M) = M$. You first show $T_t(M)$ is closed in $\mathbb R^d$ as above, and therefore $T_t(\text{Int}(M)) = \text{Int}(M)\cap T_t(M)$ is relatively closed in $\text{Int}(M)$. And $T_t(\text{Int}(M))$ is open because $T_t|_{\text{Int}(M)}$ is a diffeomorphism onto its image.
In your last comment, I think we first need to show that $T_t(\operatorname{Int}:M)\subseteq\operatorname{Int}:M$ for all $t$ (are you denoting the manifold interior or the topological interior by $\operatorname{Int}:M$?). How do you infer this from $(3)$?
It's still not clear to me how you infer the invariance of the (manifold, I guess) interior. Maybe I've misunderstood you, but if I got you right, you say that $(3)$ implies this. And it's actually intuitively plausible that $(3)$ ensures that a trajectory originating from the manifold cannot leave the boundary. Please take a look at the question I've asked for this: https://mathoverflow.net/q/372810/91890.
|
2025-03-21T14:48:31.916075
| 2020-08-31T17:27:28 |
370539
|
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|
Stack Exchange
|
Hilbert module over $C_0(\Lambda)$ as space of continuous sections of HIlbert bundle
Let $\Lambda$ be a manifold and $p:H\to\Lambda$ a continuous Hilbert bundle with $H(\lambda):=p^{-1}(\lambda)$. Suppose $\Gamma_0^0(\Lambda)$ is the space of continuous sections vanishing at infinity of $H$. I proved that $\Gamma_0^0(\Lambda)$ has the structure of a $C_0(\Lambda)$ module (with $C_0$ being the space of continuous functions vanishing at infinity). Define $H_{\lambda}=\Gamma_0^0(\Lambda)/\overline{K_{\lambda}}$ where $$K_{\lambda}=\text{span}\{f\varphi:\varphi\in \Gamma_0^0(\Lambda)\text{ and } f(\lambda)=0\}$$
I am interested in showing that $H_{\lambda}$ is isomorphic to $H(\lambda)$. Does anyone know some way to prove this or some reference with a similar proof? I've encounterd problems because $\overline{K_\lambda}$ is not necessarily the kernel of the evaluation map from $\Gamma_0^0$ to $\mathbb{C}$. There are several references pointing to properties similiar to this one, in fact in Mathematical Quantization by Nik Weaver there is a similar result regarding $C(\Lambda)$ modules in the case $\Lambda$ is a compact manifold but there is no proof. I've already checked the references provided by Weaver and didn't find a proof.
The reference I cited in my book is Fell and Doran, Representations of ${}^*$-Algebras, Locally Compact Groups, and Banach ${}^*$-Algebraic Bundles, vol. 1 (1988). Did you check there? I don't have a copy handy but I remember this treatment being very complete.
If that doesn't work, you could look at my paper with Chris Phillips, Modules with norms which take values in a C${}^*$-algebra. Theorem 8 gives a more general result about Banach modules/Banach bundles. The full proofs are not given here either, but there are specific references to Takahashi's dissertation, Fields of Hilbert Modules (Tulane, 1971) which should fill all the gaps.
|
2025-03-21T14:48:31.916226
| 2020-08-31T17:54:20 |
370541
|
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|
Stack Exchange
|
Interaction of plethysm with other operations
The plethysm $s_{\nu}[s_{\mu}]$ of two symmetric functions is the character of the composition of Schur functors $S^{\nu}(S^{\mu}(V))$. We know that this operation is linear and multiplicative in its first argument. But is there a way to develop
$s_{\nu}[s_{\mu} + s_{\lambda}]$;
$s_{\nu}[s_{\mu}s_{\lambda}]$;
in terms of plethysms $s_{\nu}[s_{\mu}]$ and $s_{\nu}[s_{\lambda}]$ ? I think I have heard of a formula for the first one, but I don't find it anymore!
If a symmetric function $f$ satisfies $\Delta\left(f\right) = \sum_{i=1}^k g_i \otimes h_i$ (where $\Delta$ is the comultiplication of the Hopf algebra of symmetric functions), then $f\left[u + v\right] = \sum_{i=1}^k g_i\left[u\right] h_i\left[v\right]$ whenever $u$ and $v$ are elements of a $\lambda$-ring (e.g., symmetric functions). Thus, any formula for $\Delta s_\nu$ (for example, the classical $\Delta s_\nu = \sum_{\lambda, \mu} c^\nu_{\lambda, \mu} s_\lambda \otimes s_\mu$) will give you a formula for $s_\nu\left[u + v\right]$.
The same applies to $f\left[uv\right]$, but this time you need the second comultiplication (i.e., the internal comultiplication, whose structure constants are the Kronecker coefficients) instead of $\Delta$.
In principle one can develop (1) using the coproduct in the ring of symmetric functions. By the Littlewood–Richardson rule, $\Delta(s_\nu) = \sum_{\alpha}\sum_\beta c^\nu_{\alpha\beta} s_\alpha \otimes s_\beta$ where $c^\nu_{\alpha\beta}$ is a Littlewood–Richardson coefficient, and correspondingly
$$s_\nu[s_\lambda + s_\mu] = \sum_{\alpha}\sum_\beta c^\nu_{\alpha\beta} s_\alpha[s_\lambda] s_\beta[s_\mu].$$
Here the sum is over all partitions such that $|\alpha|+|\beta| = |\nu|$.
Somewhat similarly,
$$s_\nu[s_\lambda s_\mu] = \sum_{\alpha}\sum_{\beta} k^\nu_{\alpha\beta} s_\alpha[s_\lambda] s_\beta[s_\mu]$$
where the sum is over all partitions $\alpha$ and $\beta$ of $|\nu|$ and $k^\nu_{\alpha\beta}$ is the Kronecker coefficient, most easily defined as the inner product $\langle \chi^\nu, \chi^\alpha \chi^\beta \rangle$ in the character ring of the symmetric group. Equivalently the$k^\nu_{\alpha\beta}$ are the structure constants for the internal product, usually denoted $\star$, on the ring of symmetric functions. These formulae can be found in MacDonald's textbook: see (8.8) and (8.9) on page 136, and hold replacing $s_\lambda$ and $s_\mu$ with arbitrary symmetric functions.
In practice, at least in my experience, this usually leads to a mess. One special case that's worth noting is when $\nu = (n)$, in which case the Littlewood—Richardson coefficient is non-zero only if $\alpha = (m)$ and $\beta = (n-m)$ for some $m \in \{0,1,\ldots, n\}$ and we get
$$s_{(n)}[s_\lambda + s_\mu] = \sum_m s_{(m)}[s_\lambda] s_{(n-m)}[s_\mu].$$
This is the symmetric function version of $\mathrm{Sym}^n (V \oplus W) = \sum_{m=0}^n \mathrm{Sym}^m V \otimes \mathrm{Sym}^{n-m} W$ for polynomial representations of $\mathrm{GL}_d(\mathbb{C})$. There is a corresponding rule for exterior powers and so for $s_{(1^n)}$.
This also gives one indication that (2) is even harder: one related question was asked on MathOverflow. Example 3 on page 137 of MacDonald gives the special case for $\nu = (n)$, when $\chi^{(n)}$ is the trivial character, and so $\langle \chi^{(n)}, \chi^{\alpha}\chi^{\beta}\rangle = \langle \chi^{\alpha}, \chi^\beta\rangle = [\alpha=\beta]$. Hence
$$s_{(n)}[s_\lambda s_\mu] = \sum_{\alpha} s_\alpha[s_\lambda] s_\alpha[s_\mu]. $$
Great care is needed when extending these rules to arbitrary symmetric functions. For instance, $s_\nu[-f] = (-1)^{|\nu|} s_{\nu'}[f]$ for any symmetric function $f$ and, as Richard Stanley points out in a comment below, the expression $s_\nu[f-f]$ should be interpreted as a plethystic substitution using the alphabets for $f$ and $-f$, not as $s_\nu[0]$; correctly interpreted, it can be expanded using the coproduct rule and the rule for $s_\nu[-f]$ just given.
It is not true that $s_\nu[0] = s_\nu[f-f]$. This illustrates the subtlety of plethysm notation. The expression $f-f$, at least when $f$ is a sum of monomials, means $f+(-f)$, where the $+$ refers to a union of alphabets and $-f$ to a negative alphabet.
The expansion of $s_{(n)}[s_\lambda+s_\mu]$ can be interpreted nicely in terms of (very slightly generalized) combinatorial species: $s_{(n)}$ is the (cycle index series of the) species of sets with $n$ elements, so the expansion says: any set of $n$ items, some of which are $\lambda$ime coloured and some are $\mu$agenta coloured, is a set of $n-m$ $\lambda$ime coloured items together with a set of $m$ $\mu$agenta coloured items. In general, combinatorial species are quite a good language to phrase plethystic identities in.
@RichardStanley: Really? Can you give a counterexample? I'm pretty sure that $f\left[u-u\right] = f\left[0\right]$ for any symmetric function $f$ (a consequence of the defining axiom of the antipode in a Hopf algebra).
What a minefield. I think darij is correct. By P2 in the survey article by Loehr and Remmel, $g \mapsto p_m \circ g$ is an algebra homomorphism. Hence $p_m[-u] = -p_m[u]$. Using $\Delta[p_m] = p_m \otimes 1 + 1 \otimes p_m$, we get $p_m[u-u] = p_m[u]1[-u] + 1[u]p_m[-u] = p_m[u] + p_m[-u] = p_m[u] - p_m[u] = 0$. By P1 in the survey article, for any $h$, the map $f \mapsto f \circ h$ is an algebra homomorphism. Hence $p_\mu[u-u] = \prod p_{\mu_i}[u-u] = 0$. Since the $p_\mu$ span, P3 implies that $f[u-u] = 0$ for all $f$.
@darijgrinberg: oops, you are right. I was thinking of the ambiguity of notation such as $p_2(1-q)$, which could mean either $p_2(1-q,0,0,\dots)=(1-q)^2$ or $p_2(1,0,0,\dots)-p_2(q,0,0,\dots)=1-q^2$.
@RichardStanley: Yeah, but this is (I think) due to the presence of variables inside the plethystic bracket. (Which is why I prefer to use $\lambda$-rings when formalizing plethysm; this way, both options can be obtained by choosing the appropriate $\lambda$-ring structures.)
|
2025-03-21T14:48:31.916589
| 2020-08-31T19:37:01 |
370550
|
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|
Stack Exchange
|
Finding approximate eigenvectors: quantitative results
Let $X$ be a complex Banach space and $T \colon X \to X$ be a bounded operator. For every $x \in X \setminus \{0\}$, denote by $Y_x$ the smallest closed $T$-invariant subspace of $X$ containing $x$. By basic spectral theory, the restriction of $T$ to $Y_x$ has an approximate eigenvalue. So we have the following:
For every $x \in X$, and every $\varepsilon > 0$, there exists a polynomial $P_{x, \varepsilon} \in \mathbb{C}[X]$ such that, letting $y = P_{x, \varepsilon}(T)(x)$, we have $\|y\| = 1$ and $dist(T(y), \mathbb{C}y) \leqslant \varepsilon.$
General question: Are there known bounds on the degree of the polynomial $P_{x, \varepsilon}$, depending on $X$, $T$, $x$ and $\varepsilon$?
I would be interested in any reference of a work in this direction. But I am actually interested in a special case of this:
Specific question: Suppose $T$ is an isomorphism between $X$ and a proper subspace $Y$, and fix $\varepsilon > 0$. Does there exists a $d \in \mathbb{N}$ (depending on $X$, $T$, and $\varepsilon$) such that for every $x \in X \setminus Y$, one can find a polynomial $P_{x, \varepsilon}$ as above, with degree at most $d$?
|
2025-03-21T14:48:31.916692
| 2020-08-31T19:54:43 |
370552
|
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|
Stack Exchange
|
Who introduced the abstract definition of a DGA?
Differential graded algebras, or DGAs, are a basic object of study in many areas of modern mathematics. While they were present (implicitly at least) since the start of modern differential geometry, I would like to know where the abstract definition of a DGA was first written down, and by whom?
Search for the earliest appearance of "differential graded algebra" and "DGA" on MathSciNet. The earliest hit is DGA in a review of a 1954 paper of Cartan where DGA (or more precisely, the redundant term "DGA-algebra") is defined: "Sur les groupes d'Eilenberg-Mac Lane $H(\Pi,n)$ I. Méthode des constructions," Proc. Nat. Acad. Sci. U.S.A. 40 (1954), 467–471. In the same year there is an entry from the Cartan seminar: Séminaire Henri Cartan de l'Ecole Normale Supérieure, 1954/1955. Algèbres d'Eilenberg-MacLane et homotopie.
I don't think "DGA-algebra" is redundant, as the A there stands for "associative" (while in DGA, it stands for "algebra").
@AleksandarMilivojevic fair enough, I was not reading the review closely.
In Cartan's usage, the A stands for "augmented", not "associative". A DGA-algebra is a differential graded augmented algebra.
@JohnRognes Then it probably means "augmented" everywhere I've seen DGA-algebra. Thank you for the correction.
|
2025-03-21T14:48:31.916809
| 2020-08-31T20:18:39 |
370553
|
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|
Stack Exchange
|
About a type of permutations
How many permutations are there on the set $\{1,2, \cdots, n\}$ ($n\geq 3$), such that any three elements are not in increasing or decreasing order? For example, for $n=3$ we have $(1,3,2), (2,1,3), (2,3,1)$ and $(3,1,2)$. For $n=4$, we have $(2,1,4,3), (2,4,1,3), (3,1,4,2)$ and $(3,4,1,2)$.
I believe that, for $n\geq 5$, there are no such permutations.
https://en.m.wikipedia.org/wiki/Erdős–Szekeres_theorem
@SamHopkins corrected link: https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Szekeres_theorem
The statement is that every sequence of length $\ge (r-1)(s-1)+1$ has either an increasing (= non-decreasing) sequence of length $\ge r$ or a decreasing subsequence of length $\ge s$. In particular every sequence of length $\ge (r-1)^2+1$ has a monotonic subsequence of length $\ge r$.
@SamHopkins Thank you!
@YCor Thank you!
|
2025-03-21T14:48:31.916897
| 2020-08-31T21:12:47 |
370555
|
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|
Stack Exchange
|
Dyck paths of Dynkin type
(The conjecture is a homological algebra question, but question 2 is a pure combinatorics question given that the conjecture is true)
A Dyck path of length $n$ is a list of positive integers $[c_1,c_2,...,c_n]$ with $c_i -1 \leq c_{i+1}$ for all $i$ and $c_i \geq 2$ for $i \neq n$ and $c_n=1$.
(One can show that those sequences really correspond to the classical Dyck paths via the area sequence and the number of Dyck paths of length $n$ is $C_{n-1}$ when $C_n$ denotes the Catalan numbers). Dyck paths can get naturally identified with the Nakayama algebra $A_D$ with a linear quiver having Kupisch series $[c_1,c_2,...,c_n]$, see for example https://arxiv.org/abs/1811.05846 .
Let $D=[c_1,c_2,...,c_n]$ be a Dyck path of length $n$.
We define the Cartan matrix $C_D$ of $D$ as the $n \times n$ upper triangular matrix with entries 0 or 1 as follows:
In the $i$-th row $C_D$ has entries equal to one in position $(i,i)$, $(i,i+1)$,...,$(i,i+c_i-1)$ and all other entries are zero.
Define the Coxeter matrix $\phi_D$ as $-C_D^{-1} C_D^T$ and the coxeter polynomial $p_D$ as the characteristic polynomial of this matrix.
We say that a Dyck path is of Dynkin type $Q$ in case the corresponding Nakayama algebra $A_D$ is derived equivalent to $KQ$.
Conjecture: A Dyck path $D$ is of Dynkin type $Q$ if and only if the algebra $A_D$ has coxeter polynomial $p_Q$.
This is true for type $A_n$ as was proven in What are the periodic Dyck paths? by Gjergji Zaimi (together with the fact that bouncing Nakayama algebras are exactly those of Dynkin type $A_n$, which can be proved by using special tilting modules). With the help of a computer it is also true for all exceptional types $E_6, E_7$ and $E_8$ and true for $D_i$ for $i=4,5,6,7,8,9$ (which is why I made it a conjecture now), but maybe there is a nice uniform proof that works for all types.
In the theorem of page 23 in http://prospero.dmat.usherbrooke.ca/ibrahim/publications/Alg%C3%A8bres_pr%C3%A9inclin%C3%A9es_et_cat%C3%A9gories_d%C3%A9riv%C3%A9es.pdf one can find a homological characterisation when a (Nakayama) algebra/Dyck path is of Dynkin type $D_n$.
We have for the Dynkin types: $p_{A_n}=x^n+x^{n-1}+x^{n-2}+....+x^2+x+1$, $p_{D_n}=x^n+x^{n-1}+x+1$, $p_{E_6}=x_1^6+x_1^5-x_1^3+x_1+1$, $p_{E_7}=x_1^7+x_1^6-x_1^4-x_1^3+x_1+1$, $p_{E_8}=x_1^8+x_1^7-x_1^5-x_1^4-x_1^3+x_1+1$. See for example Table 1.1. in the book "Notes on Coxeter Transformations and the McKay Correspondence" by Rafael Stekolshchik.
The bouncing Dyck paths are exactly the Dyck paths of Dynkin type $A_n$ and there are $2^{n-2}$ many. A classification/enumeration seems not so easy in type $D_n$, which leads to the following question:
Question 2: How many Dyck paths are there for a given $n$ with Coxeter polynomial equal to $p_{D_n}$ (call this sequence $a_n$)? How many Dyck paths are there for a given $n$ that are of Dynkin type $D_n$ (call this sequence $b_n$)?
Note that in case the conjecture is true, we have $a_n=b_n$ and calculating $a_n$ is a purely elementary problem. The sequence $a_n$ starts for $n=4,5,6,7,8$ with 1,6,13,29,65 and does not appear in the oeis. Here are the Dyck paths with Coxeter polynomial equal to $p_{D_n}$ for $n=4,5,6,7$. Maybe someone sees a pattern what they might be:
$D_4$:
[ [ 3, 3, 2, 1 ] ]
$D_5$:
[ [ 3, 3, 2, 2, 1 ],
[ 2, 3, 3, 2, 1 ],
[ 3, 3, 3, 2, 1 ],
[ 4, 3, 3, 2, 1 ],
[ 3, 4, 3, 2, 1 ],
[ 4, 4, 3, 2, 1 ] ]
$D_6$:
[ [ 3, 3, 2, 2, 2, 1 ],
[ 3, 3, 3, 2, 2, 1 ],
[ 3, 4, 3, 2, 2, 1 ],
[ 3, 3, 2, 3, 2, 1 ],
[ 2, 2, 3, 3, 2, 1 ],
[ 3, 2, 3, 3, 2, 1 ],
[ 2, 3, 3, 3, 2, 1 ],
[ 4, 3, 3, 3, 2, 1 ],
[ 2, 4, 3, 3, 2, 1 ],
[ 5, 4, 3, 3, 2, 1 ],
[ 3, 3, 4, 3, 2, 1 ],
[ 3, 5, 4, 3, 2, 1 ],
[ 5, 5, 4, 3, 2, 1 ] ]
$D_7$:
[ [ 3, 3, 2, 2, 2, 2, 1 ],
[ 3, 3, 3, 2, 2, 2, 1 ],
[ 3, 4, 3, 2, 2, 2, 1 ],
[ 3, 3, 2, 3, 2, 2, 1 ],
[ 2, 3, 3, 3, 2, 2, 1 ],
[ 4, 3, 3, 3, 2, 2, 1 ],
[ 3, 3, 4, 3, 2, 2, 1 ],
[ 3, 5, 4, 3, 2, 2, 1 ],
[ 3, 3, 2, 2, 3, 2, 1 ],
[ 3, 3, 3, 2, 3, 2, 1 ],
[ 3, 4, 3, 2, 3, 2, 1 ],
[ 2, 2, 2, 3, 3, 2, 1 ],
[ 3, 2, 2, 3, 3, 2, 1 ],
[ 2, 3, 2, 3, 3, 2, 1 ],
[ 4, 3, 2, 3, 3, 2, 1 ],
[ 2, 2, 3, 3, 3, 2, 1 ],
[ 3, 2, 3, 3, 3, 2, 1 ],
[ 2, 4, 3, 3, 3, 2, 1 ],
[ 5, 4, 3, 3, 3, 2, 1 ],
[ 2, 2, 4, 3, 3, 2, 1 ],
[ 3, 2, 4, 3, 3, 2, 1 ],
[ 2, 5, 4, 3, 3, 2, 1 ],
[ 6, 5, 4, 3, 3, 2, 1 ],
[ 3, 3, 2, 4, 3, 2, 1 ],
[ 2, 3, 3, 4, 3, 2, 1 ],
[ 4, 3, 3, 4, 3, 2, 1 ],
[ 3, 3, 5, 4, 3, 2, 1 ],
[ 3, 6, 5, 4, 3, 2, 1 ],
[ 6, 6, 5, 4, 3, 2, 1 ] ]
Here are the Dyck paths of Dynkin type $E_6$:
[ [ 2, 3, 3, 2, 2, 1 ],
[ 4, 3, 3, 2, 2, 1 ],
[ 4, 4, 3, 2, 2, 1 ],
[ 3, 3, 3, 3, 2, 1 ],
[ 4, 4, 3, 3, 2, 1 ],
[ 2, 3, 4, 3, 2, 1 ],
[ 4, 3, 4, 3, 2, 1 ],
[ 2, 4, 4, 3, 2, 1 ],
[ 3, 4, 4, 3, 2, 1 ],
[ 4, 4, 4, 3, 2, 1 ],
[ 5, 4, 4, 3, 2, 1 ],
[ 4, 5, 4, 3, 2, 1 ] ]
(there are 12 Dyck paths with Coxeter polynomial $p_{E_6}$, 54 with Coxeter polynomial $p_{E_7}$ and 133 with Coxeter polynomial $p_{E_8}$.)
Note that it is in general not true that having the same Coxeter polynomial as $KQ$ implies that a finite dimensional algebra is derived to $KQ$( there even non-derived equivalent examples with the same Cartan matrix), so the conjecture might be special to Nakayama algebras in case it is true.
Is this related to https://mathoverflow.net/q/366009 ?
@MaxAlekseyev It is not really related. Also the arxiv paper is much older than this MO link and is also not directly related.
It seems to me that the values of the Coxeter polynomials for $D_n$ that you give do not match the product formula. For the given values, I have $p_{D_4} = \phi(6)\phi(2)^2$, $p_{D_5} = \phi(8)\phi(2)$, $p_{D_6} = \phi(10)\phi(2)^2$, $p_{D_7} = \phi(12)\phi(4)\phi(2)$,
@MartinRubey Thank you very much. It seems I forgot a factor $\Phi_2$ in the formula. I gave also a reference for the formula.
Yes, I saw that reference, but it still doesn't fit. In $p_{D_7}$, the factor $\phi(3)$ is missing, and $p_{D_5}$ and $p_{D_7}$ only have one factor $\phi(2)$.
@MartinRubey Thanks again. I deleted the cyclotomic factorisation as it is probably not needed here. Instead, I just wrote down the polynomials directly.
OK, then the correct expression would be the product over all divisors of $2(n-1)$ which are not divisors of $n-1$, times $\phi(2)$.
I very much hope that my code has a bug: the numbers of Dyck paths of type $D_n$ which only return to the axis with their final step are $1, 4, 5, 6, 7, 8, 11, 10$ and $1473$ for $n$ from $4$ to $12$. The numbers of all Dyck paths of type $D_n$ would be $1, 6, 13, 29, 65, 145, 323, 705, 3837$.
More data for $D_n$: the numbers of paths are $1, 6, 13, 29, 65, 145, 323, 705, 3837, 3329, 7169, 15361, 32769, 69633$, the numbers of paths which only return to the axis with their final step are $1, 4, 5, 6, 7, 8, 11, 10, 1473, 12, 13, 14, 15, 16$. I have no idea what makes $D_4$, $D_{10}$, $D_{12}$ special.
Put differently, the sequence would like to be $a_4=3, a_5=6$ and $a_{n+2} = 1 + 4(a_{n+1} - a_n)$, but isn't.
@MartinRubey Thanks, those numbers make it look like there is no really nice solution (at least not as nice in type $A_n$).
Also, maybe I am making an error calculating by hand, but are you sure that [3,3,3,2,1] is in $D_5$?
@GjergjiZaimi Im pretty sure, I also send you an email.
@MartinRubey, empirical observations: $D_4$ is special because it has to be: there are $2^{n-2}$ paths of type $A_n$, so that only leaves one path which could be in $D_4$. For $D_{10}$ the two exceptional paths are $[3, 4, 5, 5, 4, 4, 3, 3, 2, 1]$ and $[5, 6, 5, 6, 5, 5, 4, 3, 2, 1]$; in general the $D_n$ paths which return to the axis only on the final step are $[n-1, n-1, n-2, \ldots, 2, 1]$; $[3, n-1, n-2, \ldots, 2, 1]$; and $[k, k-1, \ldots, 3, 3, n-k+1, n-k, \ldots, 2, 1]$ for $k \in [3, n-1]$ (and for $n=4$ they all coincide).
|
2025-03-21T14:48:31.917328
| 2020-08-31T22:57:41 |
370559
|
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|
Stack Exchange
|
Algebras derived equivalent to a hereditary algebra
Let $A=KQ/I$ be a quiver algebra with relations in $I$ having only coefficients 1 or -1. This implies that $A=FQ/I$ is defined over any other field $F$ (possibly of even another characteristic).
Question: Is $A$ derived equivalent to a hereditary algebra $KQ$ over a field $K$ if and only if $A$ is derived equivalent to $FQ$ over another field $F$?
This is true in case $Q$ is of Dynkin type, but even here I know no elementary reason (is there a nice argument in this case?).
|
2025-03-21T14:48:31.917395
| 2020-08-31T23:22:02 |
370560
|
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|
Stack Exchange
|
Square-integrability in lemma 4.30 of Folland's "A Course in Abstract Harmonic Analysis"
This question was originally posted on MSE (https://math.stackexchange.com/q/3796602/793374), but nobody has found a correct answer in about two weeks, so I decided to repost it here:
In lemma 4.30 of Folland's "A Course in Abstract Harmonic Analysis" (Second Edition) one needs to show the square-integrability of the function $f$ defined below and I don't understand how Folland deduces it from the inequality below.
For context, $G$ is a locally compact abelian Hausdorff group and $dx$ is a Haar measure on $G$ (note that Folland defines Radon measures to be outer regular and inner regular on open sets). Furthermore $f$ is continuous, bounded and a linear combination of functions of positive type (this is encoded in the notation $f \in \mathcal{B}(G)$). Here is what i have tried so far:
With Plancherel's theorem we see (as in Folland's proof) that $$(L^1(G) \cap L^2(G), \|\cdot\|_2) \to \mathbb{C}, \ k \mapsto \int_G f(x) \cdot k(x) \, dx$$ defines a bounded linear functional which extends to a bounded linear functional $F \in L^2(G)^*$ by the BLT theorem (I removed the complex conjugation for linearity; this should not make a difference in the argumentation).
Now Riesz's theorem yields an $r \in \mathcal{L}^2(G)$ such that $F$ is given by integration against $r$, i.e. $$F(k) = \int_G r(x) \cdot k(x) \, dx \ \text{ for all } k \in L^2(G).$$
In particular we have $$\int_G f(x) \cdot k(x) \, dx = \int_G r(x) \cdot k(x) \, dx \ \text{ for all } k \in L^1(G) \cap L^2(G).$$
With this we can show that the set $N := \{x \in G: r(x) \neq f(x)\}$ is locally null with respect to the Haar measure $dx$ since for any Borel set $A \subseteq N$ with finite Haar measure we can set $$k(x) := 1_A(x) \cdot \frac{|f(x) - r(x)|}{(f(x) - r(x)) + 1_{G \setminus N}(x)}$$ to obtain a function $k \in L^1(G) \cap L^2(G)$, so $$0 = \int_G (f(x) - r(x)) \cdot k(x) \, dx = \int_A |f(x) - r(x)| \, dx,$$
i.e. $A \cap N = A$ has Haar measure $0$.
To conclude $f \in L^2(G)$ we now need to show that $N$ has Haar measure $0$ and this problem can be reduced further: The set $$R := \{x \in G: r(x) \neq 0\} = \bigcup_{n \in \mathbb{N}} \{x \in G: |r(x)| \geq \tfrac{1}{n}\}$$ is $\sigma$-finite since $r \in \mathcal{L}^2(G)$, so $R \cap N$ is again $\sigma$-finite and locally null. Hence $R \cap N$ has Haar measure $0$ and we only need to show that the set $$M := (G \setminus R) \cap N = \{x \in G: r(x) = 0 \neq f(x)\}$$
has Haar measure $0$.
If I recall correctly, Folland adopts a style/approach where most of the arguments are stated in a way that works for groups with a sigma-finite Haar measure, and he has some comments earlier in the book on how one can reduce to this case or adapt the arguments. Is this "hand-waving" the part that concerns you?
In the following attempt, I am shameless using the fact that $f$ is continuous and bounded. My philosophy on $L^p$-spaces is shaped heavily by Banach space theory rather than measure theory, and most of my experience is with $\sigma$-finite measure spaces, so I apologize if I have missed some subtleties or conversely if I have belaboured some easy points.
Pick a compact $K\subset G$ and set $k=1_K\cdot f$; this certainly belongs to every $L^p(G)$ since $f$ is continuous and bounded, and since Haar measure is finite on compact sets.
Then, using the inequality that you quote from Folland,
$$ \int_K f\overline{f} \,dx \leq {\Vert \phi \Vert}_2 \left( \int_K |f|^2 \right)^{1/2}$$
so that
$$
\int_K |f(x)|^2\,dx \leq {\Vert\phi\Vert}_2^2
$$
(I think this is what someone was suggesting on MSE.) Now we are done provided we can justify the following claim.
Claim: Let $h\geq 0$ be a non-negative continuous bounded function on $G$, and let $\mu$ be a Radon measure on $G$. Then
$$ \int_G h\,d\mu = \sup_K \int_K h\,d\mu $$
where the supremum is over all compact $K\subseteq G$.
(Note that I am not assuming that $G$ is $\sigma$-compact.)
Proof of claim.
If the RHS is infinite there is nothing to prove; so we may assume it is finite, and denote this supremum by $C$. Clearly the LHS is $\geq C$ so we only need to establish the converse inequality.
Given $r \in (0,1)$, let $E_r= \{ x\in G \colon h(x) > r \}$. This is open, so by inner regularity of $\mu$ on open sets, there is an increasing sequence of compact subsets $K_1 \subseteq K_2 \subseteq \dots \subseteq E$ with $\mu(K_n) \nearrow \mu(E_r)$. But then, using our assumption,
$$
r\mu(K_n) \leq \int_{K_n} h\,d\mu \leq C \qquad\hbox{for all $n$}
$$
and so we have $\mu(E_r) \leq C/r<\infty$.
Since $E_r$ has finite measure and $\sup_n\mu(K_n)=\mu(E_r)$, the set $E_r \setminus \bigcup_{n\geq 1} K_n$ has measure zero and we have
$$
\int_{E_r} h\,d\mu = \lim_n \int_{K_n} h\,d\mu \leq C.
$$
By taking $r\searrow 0$ along some decreasing sequence in$(0,1)$, it follows that $\int h\,d\mu \leq C$, as required.
Thanks for your answer! One more question: Is boundedness of $h$ actually required for the claim you prove? I don't see it used anywhere (since the two integral limits should follow from monotone convergence), but I might be missing something.
I think you are right: I was making the proof up as I went along, and in an earlier version I wanted to play safe by assuming h is bounded. It also seems that lower semi-continuity of h is sufficient
@YemonChoi Nice answer! A more general claim than the one you use and prove can be found in Folland's book "Real analysis" proposition 7.12 (which indeed holds for general lower semicontinuous functions).
|
2025-03-21T14:48:31.918217
| 2020-09-01T02:13:28 |
370567
|
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|
Stack Exchange
|
Upper bound $\sum_{i=1}^m \sum_{j=1}^n p_{i,j}(1-p_{i,j})$
Let
$$p_{i,j} = \frac{\sum_{l=i}^{i+j-1} {l-1 \choose i-1} {m+n-l \choose m-i}}{{m+n \choose n}}$$
I am interested in approximating/upper bounding the sum
$$ \sum_{i=1}^m \sum_{j=1}^n p_{i,j}(1-p_{i,j})$$
Putting it through Sympy's Concrete module showed no simplification. Does the summation have any reasonable closed form bounds?
I've been able to prove a few bounds for individual cases, say $m=1$ which was trivial but had a closed form solution.
For $m=1$, $i$ could only be $1$. $p_{i,j}$ simplifies to:
$$p_{1,j} = \frac{\sum_{l=1}^{j} 1}{{n+1 \choose n}}=\frac{j}{n+1}$$
Then
$$\sum_{j=1}^n \frac{j}{n+1}(1-\frac{j}{n+1})$$ can be trivially evaluated and upper bounded using the equation for $\sum_i^n i$ and $\sum_i^n i^2 $
|
2025-03-21T14:48:31.918307
| 2020-09-01T03:53:21 |
370570
|
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|
Stack Exchange
|
Algebraic geometry reference for differential geometer
I am a graduate student in differential geometry and would like to learn more about algebraic geometry recently. Are there any recommended textbook/reference/lecture notes which is easier for a differential geometer to approach? Or are there any standard textbooks that can build my way up to algebraic geometry instead of some AG books for DG students?
My background for algebra is only an honor course I took in undergraduate which includes group, ring, field, and module.
Your help is very much appreciated!!
Among the classic references, Griffiths and Harris's Principles of algebraic geometry is one of the more accessible ones to more (complex) analytically minded geometers. But maybe you're looking for something even more specifically aimed at differential geometers? (Note: this is a pretty serious book, so will take some time. But then algebraic geometry is a very rich subject, so it's not so easy to learn it without a substantial commitment.)
Not being a geometer of any kind, I found my brief attempts to use G+H as a gateway rather like running into a brick wall. (A more sober and informed take on the book can be found here https://projecteuclid.org/euclid.bams/1183545213 )
Given the background you list, you might find it more interesting/accessible to start with complex algebraic curves a.k.a. Riemann surfaces, and then one text which seems quite readable is Fulton's book, a PDF version of which can be found on his website at http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf Alternatively, maybe Miranda's book Algebraic Curves and Riemann Surfaces but I don't know too much about the content of that one.
@R.vanDobbendeBruyn Thank you for your recommendation! I am working on Kähler manifolds, so GH's book may actually be a good reference for me.
@YemonChoi Thank you for the links provided, the pdf seems to be a very nice way to start learning AG. I will definitely read that first as an introduction to the field.
Griffiths and Harris is a classic, but I kind of agree with Yemon Choi, that it is a difficult read. For basic facts about Kähler manifolds and Hodge theory, I would recommend Wells, Differential Analysis on Complex Manifolds as an alternative.
@DonuArapura Great! I have Wells' book in my bookshelf, I can start reading it now. Thanks!!
Take a look at Mumford's "Algebraic Geometry I: Complex Projective Varieties." It was game changing for me as a beginning graduate student struggling with Hartshorne's.
@auniket Great, I will definitely take a look at that too. Thanks!
|
2025-03-21T14:48:31.918489
| 2020-09-01T04:04:37 |
370572
|
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|
Stack Exchange
|
Indexed character tables for wreath products in Sage and GAP
I am trying to obtain character table for the Hyperoctahedral group $\mathcal{H}_n$ in Sage using GAP.
This group arises as the wreath product $\mathcal{C}_2 \wr \mathcal{S}_n$, so of course I can write (calling gap from sage)
n=4
group = gap.WreathProduct(gap.SymmetricGroup(2), gap.SymmetricGroup(n))
character_table_1 = gap.CharacterTable(group)
However, knowing that conjugacy classes and irreps of $\mathcal{H}_n$ are indexed by pairs of partitions (of $k$ and of $n-k$), I would like to see this in the output. Specifically, I would like to be able to write something like
Irr(character_table)[ [[1],[2,1]] ] # (1)
To obtain the character corresponding to the given pair of partitions. Fortunately, this is possible:
n=4
cyclic_group_character_table = gap.CharacterTable(gap.SymmetricGroup(2))
character_table_2 = gap.CharacterTableWreathSymmetric(cyclic_group_character_table, n)
However, unlike the first character table, this one doesn't store the underlying group, so when I get the character using (1), I can't apply it to elements of group (and also, another package I am using expects to have the group stored in the character table).
Now, it seems like the exact thing I need is implemented in GAP, namely the function CharacterTableWithStoredGroup, which (I think) should give me the old character table but with the underlying group stored within it. I tried it out:
group = gap.WreathProduct(gap.SymmetricGroup(2), gap.SymmetricGroup(n))
character_table_3 = gap.CharacterTableWithStoredGroup(group, character_table_2)
But the function returns fail. If I understand it correctly, according to the documentation this should work.
My question is why does the code fail? and is there some other way to get the object I'm after?
Have you tried running the corresponding commands directly in GAP (invoked as "sage -gap" from the command line)? That would show whether the problem is with GAP itself or with the Sage interface to it.
Thanks for your comment!
Just tried it. Doesn't work unfortunately, still returning `fail' which apparently indicates "GAP cannot identify the classes of G up to automorphisms of tbl"
|
2025-03-21T14:48:31.918636
| 2020-09-01T05:27:00 |
370574
|
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|
Stack Exchange
|
Request for resources or techniques for bounding the infinity norm of an infinite product convolved with a simple function
I'm attempting to bound an expression of the form.
$$
\lVert(\prod_{i=1}^{\infty} \phi_i) * s \rVert_{\infty}
$$
Where $\phi_i$ are bounded periodic step functions which can be replaced by smoothed versions if need be, and for any sufficiently large integer n, $\exists m$ such that $i \geq m$ implies $\phi_i(x) = 1$ for $x \in (-n, n).$
Young's convolution inequality is clearly insufficient. I also attempted to deal with the problem using the Fourier transform as I have a nice closed form solution to the Fourier transform of $\phi_i$ and $s$.
I'm trying to get a sense for the standard techniques and references that would be applicable to this problem generally, and any lesser known results that could be useful.
p.s. Thank you for any help you can provide.
Edit: In particular I'm trying to bound this by a constant, and $\phi_i(x) = 1 - f_i(x)$
$$f_i(x) = \begin{cases}
1 & \text{if the nearest integer of $x$ has property $p_i$} \\
0 & \text{otherwise} \\
\end{cases}$$
$p_i$ is normally a divisibility criteria such as $p |n \text{ and } p^k \nmid n.$
Note: Most of the formulation of the problem is up in the air, and as such i'm not really looking for a particular solution but techniques to bound this type of convolution generally.
As written, it is not even clear whether it is a product of convolutions or a convolution of the product. In general, if you want something better than wild guesses about what might work, post all details (what exactly the functions are, what bound you would consider satisfactory, etc.). Then someone will think of it :-)
|
2025-03-21T14:48:31.918770
| 2020-09-01T09:28:41 |
370586
|
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|
Stack Exchange
|
On the width of the Catalan monoid and the rank of K-groups of the Furstenberg transformation group
The semigroup algebra of the Catalan monoid is isomorphic to the incidence algebra of $P_n$, where $P_n$ is the poset consisting of subsets of { 1,...,n } where for two subsets $X \leq Y$ if and only if $X$ and $Y$ have the same cardinality and if X= {x_1 < ... < x_k } and Y= {y_1 < ... < y_k } we have $x_i \leq y_i$ for $i=1,...,k$. This is for example proved in https://arxiv.org/pdf/1806.06531.pdf. Recall that the width of a poset is the maximum size of an antichain.
I noted that the width of the poset $P_n$ starts with 2, 3, 4, 6, 8, 13, 20, 32, 52, 90, 152 for $n=1,...,11$ and this leads to the sequence https://oeis.org/A084239 of the rank of K-groups of the Furstenberg transformation group of the $n$-torus. See table 1 in https://arxiv.org/pdf/1109.4473.pdf .
Question 1: Is this true for all $n$? Is there a deeper explanation? Does the width have a homoogical interpretation for the Catalan monoid?
The poset $P_n$ has $n+1$ connected components, for each of the $k$-subsets and one can restrict to find antichains in those subsets and put them together. But I am more curious whether there is a deeper connection to the $K$-group sequence or is this just random?
One might also ask about other nice properties of $P_n$. I noted that appending a minimum and maximum to $P_n$, one obtains a lattice.
The width of $P_n$ is equal to the maximal number of covers an element can have in the distributive lattice of order ideals $L(P_n)$ of $P_n$.
Question 2: Does the incidence algebra of $L(P_n)$ have an algebraic meaning in relation to the Catalan monoid?
Question 3: Is the Coxeter matrix of $L(P_n)$ periodic?
Question 3 has a positive answer for $n=1,2,3,4$ and the periods are given by 6,12,30,42 in that case.
(Small values suggest that also the Coxeter matrix of $P_n$ might be periodic, but that might be not so good evidence since the connected compoenent have not many points for small $n$.)
You might look at the answer to this question https://mathoverflow.net/questions/193913/is-there-a-standard-name-for-this-poset which gives another interpretation of this poset in terms of well known geometry.
The fact that this algebra is an incidence algebra was o think first proved by Hivert and Thiery. The proof you linked to is just the shortest one.
The poset $P_{n,k}$ that's discussed Benjamin Steinberg is the interval of Young's lattice between the empty and a rectangular shapes partition. Using geometry, Stanley shows these posets are Sperner in http://math.mit.edu/~rstan/pubs/pubfiles/42.pdf. This makes computing the width easy (it's then the maximum size of a rank of the poset).
Again, if we break your $P_{n}$ into the $P_{n,k}$, then the widths are given by https://oeis.org/A067059 by Stanley's Spernicity result. https://oeis.org/A084239 is the sum of the diagonals of https://oeis.org/A067059
You can translate it in a way that doesn't mention $K_0$ or $K_1$. Suppose that $A$ is the $n\times n$ matrix with $a_{i,j}=1$ if $i=j$ or $j=i+1$. Then the claim is that the largest antichain in $P_{n,r}$ is the same as the dimension of the kernel of $\bigwedge^r A-I$.
@SamHopkins Thanks, that seems to be the explanation to answer question 1. If you want you can turn it into an answer.
Oh, I see, the fact that the rank of the kernel of $\bigwedge^r A-I$ and the number of half sized partitions agree is proven in section 6.2 of the Reihani and Milnes paper referenced in the Oeis link.
The proof of proposition 6.30 in the Reihani-Milnes paper, is essentially very close to Stanley's proof of Spernicity for $P_{n,r}$. In particular the observation that A084239 is the sum of the diagonals of A067059 seems to be just as hard as the computation of the latter. I wonder if there is a more direct route.
The distributive lattice $L(k,n-k) := [\varnothing,(n-k)^k]$, the interval between the empty partition and the rectangular shape $(n-k)^k$ in Young's lattice, is the same as the poset of subsets of $[n]:=\{1,2,\ldots,n\}$ of size $k$ ordered by $\{x_1<\cdots<x_k\}\leq \{y_1<\cdots<y_k\}$ iff $x_i \leq y_i$ for all $i=1,\ldots,k$. We also have $L(k,n-k)=J([k]\times[n-k])$, the distributive lattice of order ideals of the product $[k]\times[n-k]$ of two chains. Finally, and perhaps most importantly, $L(k,n-k)$ is the Bruhat order on the cells of the Grasmannian $\mathrm{Gr}(k,n)$.
Your $P_n$ is a disjoint union of these $L(k,n-k)$.
In "Weyl groups, the Hard Lefschetz Theorem, and the Sperner property", Stanley proved that $L(k,n-k)$ is Sperner, i.e., the maximum size of an antichain of this poset is the maximum size of one of its rank. (He proves more, namely, the strong Sperner property, and in a more general context of parabolic quotients of Weyl groups, using some basic geometry of generalized flag manifolds.)
The maximum size of a rank of $L(k,n-k)$ is easily seen to be given by the OEIS sequence https://oeis.org/A084239. Since https://oeis.org/A084239 is a sum of diagonals of https://oeis.org/A084239, this explains your observations about $P_n$.
A lot is known about $L(k,n-k)$, because of its connection to representation theory/geometry. For instance, $[k]\times[n-k]$ is a so-called "minuscule poset," which implies a lot of nice properties for $J([k]\times[n-k])$: see this paper of Proctor.
Similarly, your observation about the Coxeter transform being periodic seems to be proved by Yildirim in this paper in the more general context of minuscule posets.
EDIT: Ah, sorry, the paper of Yildirim addresses the periodicity of the Coxeter transform for $L(k,n-k)=J([k]\times[n-k])$. For $J(P_n)$, I bet what you observed only happens for small values of $n$.
|
2025-03-21T14:48:31.919129
| 2020-09-01T09:46:24 |
370587
|
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|
Stack Exchange
|
Toeplitz operators for other measures then Lebegue
In the standard setting there is a lot known about Toeplitz operators i.e that the compression of a multiplication operator restricted to the Hardy space.
Are there any results when one has a arbitrary probability measure on the unit circle instead of Lebesgue? and furthermore when the dimension of the boundary is increased to say $n=2$. Results on multiplication operators seems to generalize to any finite measure.
If anyone has a good reference to what in known in this case let me know!
|
2025-03-21T14:48:31.919189
| 2020-09-01T10:03:12 |
370588
|
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|
Stack Exchange
|
On $\Psi$-generating paths in the Bruhat order of a Weyl group
Let $W$ be a Weyl group with roots $R$ and positive roots $R^+$. Let $v\in W$ of length $r$. We call $\mathbb{m}=(\alpha_1,\ldots,\alpha_r)\in(R^+)^r$ a Bruhat path from $1$ to $v$ if $1\lessdot s_{\alpha_r}\lessdot s_{\alpha_{r-1}}s_{\alpha_r}\lessdot\cdots\lessdot s_{\alpha_1}\cdots s_{\alpha_r}=v$ where $\lessdot$ means the covering relation in the Bruhat order. Let $\mathbb{m}=(\alpha_1,\ldots,\alpha_r)$ be a Bruhat path from $1$ to $v$, then we write $x_{\mathbb{m}}=x_{\alpha_1}\otimes\cdots\otimes x_{\alpha_r}\in V_W^{\otimes r}$ where $V_W$ is the Yetter-Drinfeld $W$-module defined by generators $x_\alpha$ where $\alpha\in R$ and relations $x_{-\alpha}=-x_\alpha$ for all $\alpha\in R$. Let $\mathbb{B}_r$ be the braid group with generators $\sigma_1,\ldots,\sigma_{r-1}$. This group acts on $V_W^{\otimes r}$ by $\sigma_i x=\Psi_{i,i+1}x$ where $1\leq i\leq r-1$, $x\in V_W^{\otimes r}$, and where $\Psi_{i,i+1}$ is the braiding of the $i$th and the $(i+1)$th tensor factor. Let $t\colon\mathbb{S}_{r}\to\mathbb{B}_r$ be the Matsumoto section, then Bazlov conjectures that
$$
\left(\sum_{\sigma\in\mathbb{S}_r}t(\sigma)\right)x_{\mathbf{m}_0}=\sum_{\text{all Bruhat paths $\mathbf{m}$ from $1$ to $v$}}x_{\mathbf{m}}
$$
for some distinguished (not necessarily unique) Bruhat path $\mathbf{m}_0$ from $1$ to $v$. A Bruhat path $\mathbb{m}_0$ from $1$ to $v$ which satisfies the above identity is then called a $\Psi$-generating path from $1$ to $v$.
While I believe that this conjecture is not really true (although I would appreciate a proof as well), I would like to ask a simpler
Question. Let $\mathbf{m}_0$ be a $\Psi$-generating path from $1$ to $v$. Is then necessarily $\mathbf{m}_0=(\beta_1,\ldots,\beta_r)$ for some reduced expression $v=s_{\beta_1}\cdots s_{\beta_r}$?
Second question. Do you know any further literature except that is related to this problem?
|
2025-03-21T14:48:31.919308
| 2020-09-01T10:40:11 |
370589
|
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|
Stack Exchange
|
Definition of Euler system of cyclotomic units
I am not sure about my understanding of Euler system of cyclotomic unit. This is what I have learnt:
Let $F=\mathbb{Q}(\mu_m)$.
Let $\mathcal{I}(m)$ = {positive square free integers divisible only by primes $\mathit{l}\equiv{1}$ (mod $m$)}.
Let $\mathit{r}\in\mathcal{I}(m)$ i.e $\mathit{r} = \mathit{l_1}\mathit{l_2}...\mathit{l_r}$ such that $\zeta_r = \prod\limits_{l\mid r} \zeta_l$, then
An Euler system over the field $\mathbb{Q}(\mu_m)$ is defined to be a map $\alpha\colon\mathcal{I}(m)\rightarrow\overline{\mathbb{Q}}^\times$ such that $\forall\mathit{r}\in\mathcal{I}(m)$ and each prime $\ell|\mathit{r}$ we have:
$\alpha(r)$ = $\prod\limits_{j}\left((1-\zeta_m^j\zeta_r)(1-\zeta_m^{-j}\zeta_r)\right)^{a_j}$ $\quad$ $j,a_j\geqslant 1$ $\quad$ $\longrightarrow$ $\in\mathbb{Q}(\mu_{mr})^\times$ = $F(\mu_r)^\times$
$\alpha(r)$ is a unit
if $\ell\nmid r$ then $\quad$ $N_{F(\mu_{rl})/F(\mu_r)}\alpha(rl)$ = $\alpha(r)^{\mathrm{Frob}_\ell - 1}$
$\alpha(r\ell)\equiv \alpha(r)$$\quad$ modulo all primes of $F(\mu_{r\ell})$ above $\ell$.
How could this be related to what Rubin explains:
Rubin explains that if we fix a collection $\{\zeta_m\colon m\geqslant1\}$ such that $\zeta_m$ is a primitive $m$-th root of unity and $\zeta_{mn}^n$=$\zeta_m$$\quad$$\forall{m,n}$ then for any $m\geqslant1$ and prime $\ell$ we have the relation:
$N_{\mathbb{Q}(\mu_{ml})/\mathbb{Q}(\mu_m)}(1-\zeta_{ml})$ =
a) $1-\zeta_m$, if $l\mid m$
b) $(1-\zeta_m)^{\mathrm{Frob}_l-1}$, if $l\nmid m$
Now we define
$\tilde{C}_{m,p} = N_{\mathbb{Q}(\mu_{mp})/\mathbb{Q}(\mu_m)} (1-\zeta_{mp})\in\mathbb{Q}(\mu_m)^\times$
and
$\tilde{C}_m = N_{\mathbb{Q}(\mu_m)/\mathbb{Q}(\mu_m)^+}(\tilde{C}_{m,p})$
Then the distribution relation above makes the collection $\{\tilde{C}_{m,p},\tilde{C}_m\}$ into an Euler System.
Can you put the reference of the paper that you are using? Also you should fix the typos...
The choice of notations is unnecessarily confusing here: in the first definition, $r$ is varying and $m$ is fixed (but $r$ has to lie in a set that depends on $m$); in the second definition, there is no $r$, and $m$ is varying.
|
2025-03-21T14:48:31.919463
| 2020-09-01T10:44:48 |
370590
|
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"Dominic van der Zypen",
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|
Stack Exchange
|
$1$-factorizability for linear hypergraphs with infinite edges on $\omega$
Let $H=(V,E)$ be a hypergraph. We say that $M\subseteq E$ is a matching if the members of $M$ are pairwise disjoint, and $M$ is said to be perfect if $\bigcup M = E$. Moreover, $H$ is $1$-factorizable if $E$ is the disjoint union of perfect matchings.
Let $[\omega]^\omega$ denote the set of infinite subsets of $\omega$. We say that the hypergraph $(\omega, E)$ is linear if $\varnothing \neq E\subseteq[\omega]^\omega$ and $E$ has the following property:
$e_1\neq e_2\in E$ implies $|e_1\cap e_2|\leq 1$.
We say $(\omega, E)$ is maximal if for $E_1\subseteq [\omega]^\omega$ with $E\subseteq E_1$ and $E\neq E_1$ we have that $(\omega, E_1)$ is not a linear hypergraph.
Is there a $1$-factorizable maximal linear hypergraph $H = (V,E)$ with $|E|=\aleph_0$ and $\bigcup E = \omega$?
What about a countably infinite projective plane? If we regard the points as vertices and the lines as edges, then it's a "maximal linear hypergraph" with $\aleph_0$ vertices and $\aleph_0$ edges, and it doesn't even have two disjoint edges, let alone a perfect matching.
That's right - thanks @bof for the question and for your remark. By regular I meant $|e| =\aleph_0$ for all $e\in E$. And I should exclude the projective plane (and similar hypergraphs) from the question
Does anyone else use "regular hypergraph" this way? As far as Iknoe, hypergraphs where all hyperedges have the same size are called "uniform", and "regular" means that all vertices have the same degree.
A countably infinite affine plane, regarded as a hypergraph with points as vertices and lines as edges, is a maximal linear hypergraph with $\aleph_0$ vertices and $\aleph_0$ edges, and it is $1$-factorizable; each maximal family of parallel lines is a perfect matching.
I think this is analogous the Erdos-Faber-Lovasz conjecture for infinite hypergrphs
@vidyarthi Thanks for your comment. Erdos-Faber-Lovasz is about coloring, though, and this is about 1-factorizability. But maybe you know a link between colouring and 1-factorizability that I am not aware of? I would be interested in this for sure!
No, i just thought of extending the fact that any regular graph with degree $d$ is $d$-edge colorable iff it is 1-factorizable to linear hypergraphs. May be we have to specify the degree of each vertex in the hypergraph to have a proper equivalence between factorizability and colorability
|
2025-03-21T14:48:31.919634
| 2020-09-01T10:52:10 |
370591
|
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|
Stack Exchange
|
Doubt regarding invariance of discrete absolute value under automorphism
I have been reviewing some basic algebraic number theory and $p$-adic analysis, and the following thought crossed my mind: if $F/ \mathbb Q$ be a finite Galois extension, and $\eta$ be a $\mathbb Q$-automorphism of $F$, then it is easy to show that $\eta$ leaves the ring of integers of $F$ invariant, that is, $\eta\mathcal{O}_F = \mathcal{O}_F$. I was wondering what happens about $p$-integers: more precisely, if $v$ be a non-archimdedean (hence discrete) place of $F$ and $p$ be the unique rational prime satisfying $|p|_v<1$, with $| \cdot |_v$ normalized such that $|p|_v = 1/p$, can it be said that for any $\alpha \in \mathcal{O}_{v, F} := \{x \in F: |x|_v \leq 1\}$, $|\alpha|_v=|\eta\alpha|_v$?
It seems that this question only reduces to checking whether $\eta\mathcal{O}_{v, F} \subset \mathcal{O}_{v, F}$ for any $\mathbb Q$-automorphism $\eta$ of $F$: for if this were true and $p^r:=|\alpha|_v$ with $r \in \mathbb Z_{\leq 0}$, then $p^r\alpha \in \mathcal{O}_{v, F}$ and $p^{r-1} \alpha \not\in \mathcal{O}_{v, F}$, whence $p^r\eta\alpha = \eta(p^r\alpha) \in \mathcal{O}_{v, F}$, $p^{r-1}\eta\alpha = \eta(p^{r-1}\alpha) \not\in \mathcal{O}_{v, F}$, so that indeed $|\eta\alpha|_v = p^r$. Is my reasoning for this reduction correct?
On the other hand, it is not clear to me why $\eta\mathcal{O}_{v, F} \subset \mathcal{O}_{v, F}$ - if we show this, the above argument will imply that any automorphism of $F$ is also an isometry with respect to non-archimedean valuations of $F$. However, I can see an intuitive reason for this happening: namely, the invariance (i.e. constancy) of ramification indices of all primes of the finite Galois extension $F$ lying over a fixed prime of $\mathbb Q$. I would really appreciate if someone could give me any hint as to how to complete the argument, if the assertion under consideration is true, - and a counterexample otherwise. In the former case, is my intuition of invariance of ramification indices correct (in the other case, why is it flawed)?
Pretty sure that the claim is false. Let $F = \mathbb Q \left[i\right]$ and $\xi = 2+i$ and $\tau = 2-i$. Then, $1/\tau$ is an "integer" for the $\xi$-valuation, but $\overline{1/\tau} = 1/\xi$ is not.
On the other hand, $\left{x \in F \mid p^k x \text{ is an algebraic integer for some } k \geq 0 \right}$ is a subring of $F$ and (pretty obviously) invariant under any $\mathbb{Q}$-algebra automorphisms of $F$.
I see, thank you. Could you please tell me where the ramification-index-intuition fails? Thanks. Also changed notation.
Afraid I can't, as I don't have that intuition myself. But I think the underlying ur-problem is that while only one $p$ corresponds to a given $v$, there are usually many $v$s corresponding to a given $p$.
As noted in other comments, this is false for a number field $F/\mathbb Q$. However, it is true if you work in the completion, i.e., a finite Galois extension $F/\mathbb Q_p$. There is then a unique extension of the absolute value to $F$, so $\operatorname{Gal}(F/\mathbb Q_p)$ preserves the absolute value. Actually, for Galois extensions of number fields, Galois acts transitively on the primes lying over $p$, and each prime lying over $p$ determines a unique $p$-adic absolute value on $F$.
|
2025-03-21T14:48:31.919860
| 2020-09-01T11:36:47 |
370593
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370593"
}
|
Stack Exchange
|
Cover by $K$-invariant affine open sets
Let $X$ be a non-singular complex algebraic variety (quasi-projective if necessary) and $K$ a connected compact Lie group acting on $X$ real algebraically, i.e. the action map $K \times X \to X$ is real algebraic and for each $k \in K$ the map $k : X \to X$ is complex algebraic.
Question. Is there a cover of $X$ by $K$-invariant affine open sets?
The reason I'm asking is that I'm thinking about whether the $K$-action extends to a complex algebraic action of the complexification $K_{\mathbb{C}}$. If $X$ is affine, the action does extend, so a positive answer to the above question would settle the general case.
No, that is not true. Let $X$ be a complex Abelian variety of complex dimension $g>0$, and let $K$ be the underlying real Lie group of $X$ considered as a product of $2g$ copies of the real Lie group $U(1)$. The only $K$-invariant open subsets are the empty set and all of $X$.
...and Jason's example also gives a counterexample to the OP's more general question: this action of K does not extend to an algebraic action of K_C, at least if one complexifies to a split algebraic torus.
Homogenous varieties under a connected reductive group are another class of examples.
|
2025-03-21T14:48:31.919981
| 2020-09-01T11:42:36 |
370594
|
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"authors": [
"Dieter Kadelka",
"Mehmet Onat",
"Todd Eisworth",
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}
|
Stack Exchange
|
Regularity and ultrafilter
I read the following result in an article.
Let $X$ be a regular space. Let $\mathcal{M}$ be free closed
ultrafilter on $X$. Set $\mathcal{U=}\left\{ U:U\text{ is open and there
exists a }F\in \mathcal{M}\text{ such that }F\subset U\right\} $. $\mathcal{U%
}$ is contained in an open ultrafilter $\mathcal{W}$. By regularity, $\bigcap
\overline{\mathcal{W}}=\emptyset $.
Why the regularity implies $\bigcap \overline{\mathcal{W}}=\emptyset $?
Only to clarify terminology: By closed/open ultrafilter you mean a maximal filter in the set of closed/open subsets of $X$?
By regularity each point will have an open neighborhood whose closure is disjoint to a set in $\mathcal{M}$. The complement of this closure is in $\mathcal{W}$, and the closure of this complement omits the original open set around the point.
@DieterKadelka, yes, I mean a maximal filter in the set of closed/open subsets of $X$.
Maybe that this answer is essentially the not elaborated answer of Todd Eisworth. Assume that $x \in \bar{W}$ for all $W \in \mathcal{W}$. Since $\mathcal{M}$ is free there is $M \in \mathcal{M}$ with $x \not\in M$. By regularity of $X$ there are open $U,V \subset X$ with $x \in U$, $M \subset V$ and $U \cap V = \emptyset$. Therefore $V \in \mathcal{U}$, but $U \not\in \mathcal{U}$, in particular $V \in \mathcal{W}$ and $x \not\in \bar V$, since $U \cap V = \emptyset$ and $x \in U$ open.
|
2025-03-21T14:48:31.920225
| 2020-09-01T11:44:54 |
370595
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Chris",
"Donu Arapura",
"guest0803",
"https://mathoverflow.net/users/116075",
"https://mathoverflow.net/users/164620",
"https://mathoverflow.net/users/4144",
"https://mathoverflow.net/users/519",
"naf"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/370595"
}
|
Stack Exchange
|
What can be said about a projective morphisms that admit decomposition theorem like smooth morphisms?
Let $f\colon X\to Y$ be a surjective morphism of smooth projective varieties. If the decomposition theorem for $f$ is given by $$Rf_*\mathbb{C} \simeq \bigoplus_i R^if_*\mathbb{C}[-i],$$ what are the necessary conditions the morphism $f$ must satisfy? Is there an example where such a morphism is not smooth but the decomposition theorem nonetheless looks like the above?
Edit: If I additionally assume that $R^if_*\mathbb{C}$ are local systems for all $i$, can one conclude that $f$ is smooth? I understand that the limit mixed Hodge structure is pure as there is no monodromy around the singular fibers.
One can find examples where $f$ is not smooth, even with the extra condition in the edit, e.g., this can happen when there are nonreduced fibres. For a specific example, take $X$ to be a hyperelliptic surface and $f$ the natural map to $\mathbb{P}^1$.
Thank you very much for the comment. Unfortunately I am not sure how to see that the decomposition theorem for your example satisfies all the properties in my question. If you could please expand a bit, I will really appreciate it.
OK, here are some details. To make this short, although it is not necessary, I will assume the result in Donu Arapura's answer. It is obvious that $R^0f_\mathbb{C}$ and $R^2f_\mathbb{C}$ are constant sheaves, so it suffices to consider $R^1f_\mathbb{C}$. All the fibres of the map are elliptic curves, but some are not reduced. Using proper base change, the stalks of $L :=R^1f_\mathbb{C}$ are all isomorphic. To see that it is a local system it suffices to show that $L$ has no summand supported on any point corresponding to a nonreduced fibre.
cont. In the simplest kind of hyperelliptic surface, there are 4 singular fibres and the local behaviour around each of them is the same. If there are a skyscraper sheaf corresponding to these points, using the Leray spectral sequence to compute $H^1(X,\mathbb{C})$, you get that this is at least 4 dimensional . However, it is easy to see from the description of $X$ as a quotient of a product of elliptic curves that this cohomology group is actually 2 dimensional. (There should, of course, be a purely local proof, but I have not worked that out.)
Dear @ulrich, thank you so much for the beautiful explanation. This seems to be an answer and I would happily accept it if you could please post it as an answer instead of a comment.
Perhaps I am being slow, but the only point I still do not understand is why $R^1f_*\mathbb{C}$ could not be of the form $IC(\mathbb{L})$ for some non-trivial rank 2 local system $\mathbb{L}$ outside the 4 points. I agree with all the other statements you made.
I think I see your point now. Since you ruled out skyscraper sumands, if $L=IC(\mathbb{L})$ for some local system $\mathbb{L}$ on the punctured curve, we could use the fact that $IC(\mathbb{L} = j_\mathbb{L}$ on curves, where $j$ is the inclusion of the punctured curve. Since the stalk of $j_\mathbb{L}$ at the puncture is given by $H^0(\Delta^*, \mathbb{L})$ for small enough punctured discs, and we already knew that the stalks of $L$ are all equidimensional we can conclude that $\mathbb{L}$ has no monodromy around the puncture and hence extends over the puncture as a local system.
@ulrich I am not sure how to turn your comment into an answer. Would you mind posting it as an answer?
OK, I have posted a simplified version of the argument as an answer.
Here is an example where $f$ is not smooth but $Rf_* \mathbb{C}$ behaves as if it were:
Let $X$ be a hyperelliptic surface and $f$ the natural morphism to $Y \cong\mathbb{P}^1$. All reduced fibres of $f$ are elliptic curves, but there is a nonzero number of nonreduced fibres, the number dependending on $X$.
The singular cohomology of $X$ is given by $H^0(X, \mathbb{C}) \cong H^4(X,\mathbb{C}) \cong \mathbb{C}$ and $H^1(X, \mathbb{C}) \cong H^3(X, \mathbb{C}) \cong \mathbb{C}^2$. Furthermore, the restriction map $H^1(X, \mathbb{C}) \to H^1(F, \mathbb{C})$ is an isomorphism for any fibre $F$ of $f$.
It is clear that $R^0 f_* \mathbb{C}_X \cong R^2f_* \mathbb{C}_X \cong \mathbb{C}_Y$, so let us consider $R^1f_* \mathbb{C}_X$. Since $H^1(X, \mathbb{C}) \cong \mathbb{C}^2$, we get a natural map $\mathbb{C}^2_Y \to R^1f_* \mathbb{C}$. By evaluating this on stalks and using the proper base change theorem, we see that this is an isomorphism.
Finally, since we know exactly what each sheaf $R^i f_* \mathbb{C}_X$ is, the same proof as in the case $f$ smooth can be used to show that $Rf_*\mathbb{C}_X$ decomposes as a direct sum of its (shifted) cohomology sheaves.
One may ask if a similar statement holds whenever all reduced fibres are smooth (and say $f$ is flat); I did not think about this. It would also be interesting to know if there are examples with non-smooth reduced fibres. Also, note that in the example $R^1 f_* \mathbb{Z}_X$ is not a local system.
Oh, it's you...
@DonuArapura: Yes, Hi Donu!
I don't know how to characterize such morphisms, which I think is your first question. However, this can certainly happen, even if $f$ is not smooth. (By the way, your comment about absence of local monodromy, and purity of limit MHS isn't correct.)
Prop (Zucker). If $Y$ is a curve, then
$$R f_*\mathbb{C} = \bigoplus_i R^if_* \mathbb{C}[-i]$$
Since Zucker in section of 15 of his 1979 Annals paper proves a
slightly weaker statement. Let me sketch a proof using stuff that appeared since. I can flesh it out if needed.
Sketch. Let $D\subset Y$ be the discriminant, $j:U\to Y$ the complement. By the decomposition theorem of BBDG, the object above decomposes as a sum $\bigoplus L_i$, where $L_i$ are translates of pure perverse sheaves. We can assume the $L_j$ are translates of minimal extensions. By restricting to $Y-D$ and applying Deligne (Théoremes de Lefschetz...), we can identify $L_i|_U=R^if_*\mathbb{C}|_{U}[-i]$, after reindexing. It follows that $L_i=j_*R^if_*\mathbb{C}|_{U}[-i]$ for sheaves supported on $Y$. There may be other summands supported on $D$ which need to accounted for. Use the local invariant cycle theorem to get a surjection $R^if_*\mathbb{C}\to L_i$. By purity (in the sense of Hodge modules, say) we can split this. So that we can absorb all $L_k$ with support on $D$ into some $R^if_*\mathbb{C}$
Added Comment: Regarding the latest question, I think I was too hasty in my comment. The example I had in mind doesn't satisfy all your requirements, but it may still be interesting to describe. One has a pencil of genus 2 curves degenerating to a union of 2 elliptic curves at each singular fibre. Contracting one of the elliptic curves from each pair results in singular surface mapping to a curve such that the higher direct images are constant. This is probably similar to what Ulrich Naf was suggesting. (Rmk, Oct 11/20: in fact it's different.)
Dear Prof. Arapura, thanks so much for the beautiful answer. And of course, I meant to also add that $R^if_*\mathbb{C}$ are all local systems (if I understand correctly the comment about local monodromy is true in this case.). Under this additional assumption, can one conclude smoothness of $f$?
Since the answer is really helpful and explicit, I will leave the question as it is and add the part about the local system assumption at the end.
You're welcome. OK, with added assumptions about direct images, I agree that the local monodromy is trivial. Surprisingly, this is not enough to guarantee smoothness. I'll expand my answer later.
Thanks! I will look forward to it. Somehow I have the feeling that such $f$ (additional assumption inclusive) cannot have divisorial discriminant locus. But I don't have an argument or a counter-example.
Isn't the discriminant always of pure codimension one (if non-empty)?
@Chris That is unfortunately not correct. Blow-ups already give you examples where discriminant is not of pure codim 1.
|
2025-03-21T14:48:31.920743
| 2020-09-01T12:12:28 |
370596
|
{
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370596"
}
|
Stack Exchange
|
If a joint density factorizes on a square, does this imply that the marginal random variables are locally independent?
Let $Z=(X,Y) : \Omega\rightarrow\mathbb{R}^2$ be a Borel-measurable random vector and $U\subset\mathbb{R}$ be open. Suppose that $Z$ is absolutely continuous with Lebesgue density $\zeta$.
I was wondering about the following question: If we assume that $\zeta$ factorizes on the square $U\times U$, i.e. that the restriction
$$\tag{1}q:=\left.\zeta\right|_{U\times U} \quad\text{is such that}\quad q\equiv q(x,y)=q_1(x)q_2(y) \quad\text{for some } \ q_1, q_2 : U\rightarrow\mathbb{R},$$
does this imply that $X$ and $Y$ are independent on $U$ in the sense that
$$\tag{2}\mathbb{P}(X^{-1}(A)\cap Y^{-1}(B)) = \mathbb{P}_X(A)\cdot\mathbb{P}_Y(B)\quad\forall\, A, B\subseteq U \ (\mathrm{Borel})?$$
Clearly this holds for both the special cases $U=\mathbb{R}$ and $q_i = \zeta_i$ ($i=1,2$, with $\zeta_i$ the marginal densities of $\zeta$), however I couldn't show it in the above generality; do you have a counterexample?
The answer is no. E.g., let $U=(0,1)$ and
$$\zeta=\frac12\times 1_{(0,1)^2}+\frac12\times 1_{(1,2)^2};$$
that is, the joint distribution of $(X,Y)$ is the half-and-half mixture of the uniform distributions on the the squares $(0,1)^2$ and $(1,2)^2$. Then
$$P_{X,Y}(U\times U)=\frac12\ne\frac12\times\frac12=P_X(U)P_Y(U).$$
No. For example suppose $Z=(X,Y)$ distributes uniformly in the triangle $\{(x,y):x,y\geq 0, x+y\leq 1\}$, and $U=(0,\frac 1 2)$. Let $A=(0,\frac 1 2)$ and $B=(0,\frac 1 {10})$. Then $P(Z\in A\times B)=\frac 1 {10}$, and $P(X\in A)P(Y\in B)=\frac 3 4 \times \frac{19}{100}$.
|
2025-03-21T14:48:31.920886
| 2020-09-01T12:57:48 |
370597
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"DSblizzard",
"Per Alexandersson",
"Random",
"Wlod AA",
"https://mathoverflow.net/users/1056",
"https://mathoverflow.net/users/110389",
"https://mathoverflow.net/users/50088",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/370597"
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|
Stack Exchange
|
Maximal length of trajectories in billiard
Consider discrete rectangular billard on lattice with integer dimensions a*b and n balls with radius $\frac{\sqrt 2}{2}$ and equal mass. In one time step ball runs from one lattice point to neighboring point in horizontal or vertical direction. When two balls meet on neighboring diagonal points they reflect according to laws of physics; when ball reaches border, it reflects back. Other types of collisions are forbidden (we halt this run and choose other ball positions), in particular collisions with more than 2 balls. This billiard is reversible and all allowable paths are cyclic. What is the maximal number of steps before all balls will be in position in which they started depending on a, b and n? Is it exponential or polynomial when n grows linearly and ratio of a and b to n is constant?
EDIT: Now I use modification of this billiard: balls are squares with diagonal 2 and diagonals parallel to axes. This allows new type of collision - straight (angle to angle). Manhattan distance between centers of squares should always be even.
Do all balls have the same mass? Also, I am not quite sure what "reflects according to the laws of physics" means, because unless I am mistaken a ball that has experienced a collision will not move on lattice points anymore (after 1 time step the ball will not be on a lattice point).
Yes, mass is the same. Some collisions can reflect balls perpendicularly to their previous movement. You can test this here: https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html ("advanced" page)
Perhaps consider the case with square board first, and do some computer simulations, and look for hits in the oeis?
Already tried this without success. And computer can compute only very small boards with irregular sequence of path lengths, so I cannot discover pattern.
Would you initiate the system?! What balls are present, and what are their moving directions (if any) at the moment ZERO?
The formulation of the question is only a sketch to me. Fully developed q. is missing. #### I would also suggest an equivalent mathematization of the question with balls being replaced by moving points, and the rules of "reflections" being defined explicitly without relying on physics. This way, this would open an easy way for natural and precise generalizations.
I computed lengths of cycles and counts of different cycle lengths for small values of square balls and obtained results which suggest that both of these numbers grow exponentially. When ball count raises by 1 power of polynomial for cycle lengths or counts also raises approximately by 1.
Results for cycle lengths for 6 balls for different lattice sizes:
6: 142
7: 740
8: 1214
9: 2836
10: 4978
11: 6260
12: 18974
13: 54248
14: 40902
15: 69684
16: 116994
17: 224644
18: 247622+
19: 152844+
20: 371832+
21: 921420+
22: 527196+
23: 260288+
25: 4574916+
29: 6751260+
33: 17788732+
Cycle counts for 2 balls:
even lattice size: 5
odd lattice size: 6
Cycle counts for 3 balls:
4: 3
5: 5
6: 13
7: 17
8: 31
9: 44
10: 58
11: 91
12: 121
13: 150
14: 182
15: 243
16: 275
17: 345
18: 375
19: 461
20: 533
21: 615
22: 688
23: 806
24: 876
25: 989
26: 1072
27: 1230
28: 1328
29: 1470
30: 1600
31: 1806
32: 1898
Cycle counts for 4 balls:
4: 5
5: 11
6: 27
7: 46
8: 84
9: 130
10: 174
11: 271
12: 371
13: 486
14: 636
15: 860
16: 1065
17: 1364
18: 1677+
Cycle counts for 6 balls:
6: 22
7: 34
8: 93
9: 134
10: 177
11: 298
12: 386
13: 534+
14: 568+
C program with which I obtained these results (Mersenne twister from https://github.com/ESultanik/mtwister):
#include <inttypes.h>
#include <iso646.h>
#include <math.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include "mtwister.h"
#define ball_len 4
MTRand randomize() {
srand((unsigned int) time(NULL));
MTRand mtrand = seedRand(rand());
return mtrand;
}
int rnd(int r_bound, MTRand* p_mtrand) {
return genRandLong(p_mtrand) % r_bound;
}
int* make_new_balls(int ball_count) {
return (int*) malloc(ball_count * ball_len * sizeof(int));
}
int* get_ball(int* balls, int ball_num) {
return balls + ball_len * ball_num; // x, y, v_x, v_y
}
int get_ball_prop(int* ball, int coord_or_v, int x_or_y) {
return ball[coord_or_v * 2 + x_or_y];
}
void set_ball_prop(int* ball, int coord_or_v, int x_or_y, int val) {
ball[coord_or_v * 2 + x_or_y] = val;
}
void inc_ball_prop(int* ball, int coord_or_v, int x_or_y, int val) {
ball[coord_or_v * 2 + x_or_y] += val;
}
void mul_ball_prop(int* ball, int coord_or_v, int x_or_y, int val) {
ball[coord_or_v * 2 + x_or_y] *= val;
}
void print_ball(int* ball) {
printf("(%d, %d), (%d, %d)\n", ball[0], ball[1], ball[2], ball[3]);
}
void print_balls(int ball_count, int* balls) {
printf("\n");
for (int ball_num = 0; ball_num < ball_count; ++ball_num) {
int* ball = get_ball(balls, ball_num);
print_ball(ball);
}
}
int get_is_new_state_long(int ball_count, int* balls, int* first_balls) {
for (int i = 0; i < ball_count * ball_len; ++i) {
if (balls[i] != first_balls[i]) {
return 1;
}
}
return 0;
}
int run_two_dim_balls_lattice_with_square_balls(int field_0, int field_1, int ball_count, int* balls, int* first_balls, int* new_balls, int* collision_counts, int* close_counts, int* vs_history, int* coll_history, int* orig_balls) {
int is_running = 1;
int vs_history_pos = -ball_count * 2;
int bit_history_pos = -field_0;
int coll_num = 0;
int* parity_ball;
int iter_num = -1;
while (is_running) {
iter_num += 1;
vs_history_pos += ball_count * 2;
bit_history_pos += field_0;
for (int ball0_num = 0; ball0_num < ball_count; ++ball0_num) {
int* ball0 = get_ball(balls, ball0_num);
if (get_ball_prop(ball0, 0, 0) < 0 or get_ball_prop(ball0, 0, 0) >= field_0 or get_ball_prop(ball0, 0, 1) < 0 or get_ball_prop(ball0, 0, 1) >= field_1) { // out of field
return -1;
}
for (int ball1_num = ball0_num + 1; ball1_num < ball_count; ++ball1_num) {
int* ball1 = get_ball(balls, ball1_num);
if (abs(get_ball_prop(ball0, 0, 0) - get_ball_prop(ball1, 0, 0)) + abs(get_ball_prop(ball0, 0, 1) - get_ball_prop(ball1, 0, 1)) < 2) { // balls are too close
return -1;
}
}
}
for (int ball_num = 0; ball_num < ball_count; ++ball_num) {
close_counts[ball_num] = 0;
}
for (int ball0_num = 0; ball0_num < ball_count; ++ball0_num) {
int* ball0 = get_ball(balls, ball0_num);
for (int ball1_num = ball0_num + 1; ball1_num < ball_count; ++ball1_num) {
int* ball1 = get_ball(balls, ball1_num);
if (abs(get_ball_prop(ball0, 0, 0) - get_ball_prop(ball1, 0, 0)) == 1 and abs(get_ball_prop(ball0, 0, 1) - get_ball_prop(ball1, 0, 1)) == 1) {
close_counts[ball0_num] += 1;
close_counts[ball1_num] += 1;
}
}
}
for (int ball_num = 0; ball_num < ball_count; ++ball_num) {
if (close_counts[ball_num] > 1) { // more than 2 balls are near each other
return -1;
}
}
for (int i = 0; i < ball_count * ball_len; ++i) {
new_balls[i] = balls[i];
}
// handle collisions
int is_all_collisions_resolved = 0;
int coll_iter_num = -1;
while (not is_all_collisions_resolved) {
coll_iter_num += 1;
if (coll_iter_num == 1000) {
return -1;
}
for (int ball_num = 0; ball_num < ball_count; ++ball_num) {
collision_counts[ball_num] = 0;
}
for (int ball0_num = 0; ball0_num < ball_count; ++ball0_num) {
int* ball0 = get_ball(balls, ball0_num);
int* new_ball0 = get_ball(new_balls, ball0_num);
// collisions with borders
if (get_ball_prop(ball0, 0, 0) == 0 and get_ball_prop(ball0, 1, 0) < 0) {
collision_counts[ball0_num] += 1;
mul_ball_prop(new_ball0, 1, 0, -1);
} else if (get_ball_prop(ball0, 0, 0) == field_0 - 1 and get_ball_prop(ball0, 1, 0) > 0) {
collision_counts[ball0_num] += 1;
mul_ball_prop(new_ball0, 1, 0, -1);
} else if (get_ball_prop(ball0, 0, 1) == 0 and get_ball_prop(ball0, 1, 1) < 0) {
collision_counts[ball0_num] += 1;
mul_ball_prop(new_ball0, 1, 1, -1);
} else if (get_ball_prop(ball0, 0, 1) == field_1 - 1 and get_ball_prop(ball0, 1, 1) > 0) {
collision_counts[ball0_num] += 1;
mul_ball_prop(new_ball0, 1, 1, -1);
}
// collisions with other balls
for (int ball1_num = ball0_num + 1; ball1_num < ball_count; ++ball1_num) {
int* ball1 = get_ball(balls, ball1_num);
int* new_ball1 = get_ball(new_balls, ball1_num);
if (abs(get_ball_prop(ball0, 0, 0) - get_ball_prop(ball1, 0, 0)) == 1 and abs(get_ball_prop(ball0, 0, 1) - get_ball_prop(ball1, 0, 1)) == 1) {
// parallel collisions
if (get_ball_prop(ball0, 1, 0) == -get_ball_prop(ball1, 1, 0) and (get_ball_prop(ball0, 0, 0) - get_ball_prop(ball1, 0, 0)) * get_ball_prop(ball0, 1, 0) < 0) {
collision_counts[ball0_num] += 1;
collision_counts[ball1_num] += 1;
if (get_ball_prop(ball0, 0, 1) > get_ball_prop(ball1, 0, 1)) {
set_ball_prop(new_ball0, 1, 1, 1);
set_ball_prop(new_ball1, 1, 1, -1);
} else {
set_ball_prop(new_ball0, 1, 1, -1);
set_ball_prop(new_ball1, 1, 1, 1);
}
set_ball_prop(new_ball0, 1, 0, 0);
set_ball_prop(new_ball1, 1, 0, 0);
} else if (get_ball_prop(ball0, 1, 1) == -get_ball_prop(ball1, 1, 1) and (get_ball_prop(ball0, 0, 1) - get_ball_prop(ball1, 0, 1)) * get_ball_prop(ball0, 1, 1) < 0) {
collision_counts[ball0_num] += 1;
collision_counts[ball1_num] += 1;
if (get_ball_prop(ball0, 0, 0) > get_ball_prop(ball1, 0, 0)) {
set_ball_prop(new_ball0, 1, 0, 1);
set_ball_prop(new_ball1, 1, 0, -1);
} else {
set_ball_prop(new_ball0, 1, 0, -1);
set_ball_prop(new_ball1, 1, 0, 1);
}
set_ball_prop(new_ball0, 1, 1, 0);
set_ball_prop(new_ball1, 1, 1, 0);
// perpendicular collisions
} else if (get_ball_prop(ball0, 0, 0) + get_ball_prop(ball0, 1, 0) == get_ball_prop(ball1, 0, 0) + get_ball_prop(ball1, 1, 0) and get_ball_prop(ball0, 0, 1) + get_ball_prop(ball0, 1, 1) == get_ball_prop(ball1, 0, 1) + get_ball_prop(ball1, 1, 1)) {
collision_counts[ball0_num] += 1;
collision_counts[ball1_num] += 1;
int temp0 = get_ball_prop(ball1, 1, 0);
int temp1 = get_ball_prop(ball1, 1, 1);
int temp2 = get_ball_prop(ball0, 1, 0);
int temp3 = get_ball_prop(ball0, 1, 1);
set_ball_prop(new_ball0, 1, 0, temp0);
set_ball_prop(new_ball0, 1, 1, temp1);
set_ball_prop(new_ball1, 1, 0, temp2);
set_ball_prop(new_ball1, 1, 1, temp3);
}
// direct collisions
} else if ( (abs(get_ball_prop(ball0, 0, 0) - get_ball_prop(ball1, 0, 0)) == 2 and get_ball_prop(ball0, 0, 1) == get_ball_prop(ball1, 0, 1)) or (abs(get_ball_prop(ball0, 0, 1) - get_ball_prop(ball1, 0, 1)) == 2 and get_ball_prop(ball0, 0, 0) == get_ball_prop(ball1, 0, 0)) ) {
if ( (get_ball_prop(ball0, 0, 0) + get_ball_prop(ball0, 1, 0) == get_ball_prop(ball1, 0, 0) + get_ball_prop(ball1, 1, 0)) and (get_ball_prop(ball0, 0, 1) + get_ball_prop(ball0, 1, 1) == get_ball_prop(ball1, 0, 1) + get_ball_prop(ball1, 1, 1)) ) {
collision_counts[ball0_num] += 1;
collision_counts[ball1_num] += 1;
mul_ball_prop(new_ball0, 1, 0, -1);
mul_ball_prop(new_ball0, 1, 1, -1);
mul_ball_prop(new_ball1, 1, 0, -1);
mul_ball_prop(new_ball1, 1, 1, -1);
}
}
} // for ball1_num
} // for ball0_num
is_all_collisions_resolved = 1;
for (int ball_num = 0; ball_num < ball_count; ++ball_num) {
if (collision_counts[ball_num] > 0) {
is_all_collisions_resolved = 0;
if (collision_counts[ball_num] > 1) {
return -1;
}
}
}
for (int i = 0; i < ball_count * ball_len; ++i) {
balls[i] = new_balls[i];
}
} // while (not is_all_collisions_resolved)
if (iter_num == 0) {
for (int i = 0; i < ball_count * ball_len; ++i) {
first_balls[i] = balls[i];
}
} else {
is_running = get_is_new_state_long(ball_count, balls, first_balls);
}
for (int ball_num = 0; ball_num < ball_count; ++ball_num) {
int* ball = get_ball(balls, ball_num);
inc_ball_prop(ball, 0, 0, get_ball_prop(ball, 1, 0));
inc_ball_prop(ball, 0, 1, get_ball_prop(ball, 1, 1));
}
} // while (is_running)
return iter_num;
}
int main() {
MTRand mtrand = randomize();
int ball_count = 6;
int field_0 = 8;
int field_1 = field_0;
printf("ball_count, field_0 = %d, %d\n", ball_count, field_0);
int* balls = make_new_balls(ball_count);
int* first_balls = make_new_balls(ball_count);
int* orig_balls = make_new_balls(ball_count);
int* new_balls = make_new_balls(ball_count);
int v_count = 4;
int* vs = (int*) malloc(v_count * 2 * sizeof(int));
int history_len = 1000000;
int* vs_history = (int*) malloc(history_len * sizeof(int));
int* bit_history = (int*) malloc(history_len * sizeof(int));
int* coll_history = (int*) malloc(history_len * sizeof(int));
vs[0] = -1; vs[1] = 0; vs[2] = 0; vs[3] = -1; vs[4] = 0; vs[5] = 1; vs[6] = 1; vs[7] = 0;
int vs_history_pos;
int bit_history_pos;
int* close_counts = (int*) malloc(ball_count * sizeof(int));
int* collision_counts = (int*) malloc(ball_count * sizeof(int));
int v;
int bit;
int iter_count;
int run_count = 0;
int max_iter_count = -1;
int horiz_ball_count;
int more_count = 0;
int max_coord_pair_count = field_0 * field_1;
int* coord_pairs = (int*) malloc(max_coord_pair_count * 2 * sizeof(int));
int coord_pair_count = 0;
for (int i_x = 0; i_x < field_0; ++i_x) {
for (int i_y = 0; i_y < field_1; ++i_y) {
if ((i_x + i_y) % 2 == 1) {
continue;
}
coord_pairs[coord_pair_count*2] = i_x;
coord_pairs[coord_pair_count*2 + 1] = i_y;
++coord_pair_count;
}
}
printf("coord_pair_count = %d\n\n", coord_pair_count);
int rnd0, rnd1;
int zero_bit;
int* ball;
int pair_num;
int iter_counts_len = 100000000;
int* iter_counts = (int*) malloc(iter_counts_len * sizeof(int));
int iter_counts_pos = 0;
for (int i = 0; i < iter_counts_len; ++i) {
iter_counts[i] = 0;
}
int is_iter_count_found;
int max_wait = 0;
int prev_change_num = 0;
while (1) {
run_count += 1;
for (int ball_num = 0; ball_num < ball_count; ++ball_num) {
ball = get_ball(balls, ball_num);
pair_num = rnd(coord_pair_count, &mtrand);
ball[0] = coord_pairs[pair_num*2];
ball[1] = coord_pairs[pair_num*2 + 1];
zero_bit = rnd(2, &mtrand);
ball[3 - zero_bit] = rnd(2, &mtrand)*2 - 1;
ball[2 + zero_bit] = 0;
}
iter_count = run_two_dim_balls_lattice_with_square_balls(field_0, field_1, ball_count, balls, first_balls, new_balls, collision_counts, close_counts, vs_history, coll_history, orig_balls);
if (iter_count > 0) {
is_iter_count_found = 0;
for (int i = 0; i < iter_counts_pos; ++i) {
if (iter_counts[i] == iter_count) {
is_iter_count_found = 1;
break;
}
}
if (not is_iter_count_found) {
iter_counts[iter_counts_pos] = iter_count;
iter_counts_pos += 1;
if (run_count - prev_change_num > max_wait) {
max_wait = run_count - prev_change_num;
}
prev_change_num = run_count;
}
}
if (iter_count > max_iter_count) {
max_iter_count = iter_count;
printf("run_count, max_iter_count = %d, %d\n", run_count, max_iter_count);
}
if (run_count % 1000000 == 0) {
printf("\n");
printf("ball_count, field_0, run_count, max_iter_count, max_wait, iter_counts_pos = \n%d, %d, %d, %d, %d, %d\n", ball_count, field_0, run_count, max_iter_count, max_wait, iter_counts_pos);
}
}
printf("ball_count, field_0, max_iter_count = %d, %d, %d\n", ball_count, field_0, max_iter_count);
return 0;
}
|
2025-03-21T14:48:31.921704
| 2020-09-01T13:27:21 |
370601
|
{
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"sort": "votes",
"url": "https://mathoverflow.net/questions/370601"
}
|
Stack Exchange
|
Is unirationality decidable?
When the answer to the Lüroth problem is affirmative, the genus (for curves) or Castelnuovo's criteria (for separably unirational surfaces) give computable invariants which decide if a given variety is unirational.
I am interested in the simplest cases where non-rational examples are known: namely surfaces over $\overline{\mathbb{F}_p}$ and 3-folds over $\mathbb{C}$.
Is it known whether there exists an algorithm to determine if a given variety of the above type is unirational?
A naive hope would be to produce a bound on the degree of a rational map $\mathbb{P}^n \to X$ however this is clearly impossible since rational maps $\mathbb{P}^n \to \mathbb{P}^n$ have unbounded degree.
The next best thing would be to put a computable upper bound on the minimal degree of such a rational map. Are such bounds known for the two cases I outlined above? Are such bounds sufficient, in principle, to provide an algorithm?
This is a widely studied problem, and I think with the current technology it looks unlikely this will be answered. Let me stick to characteristic $0$ for simplicity; the story gets much richer in positive characteristic (even if you add the adjective 'separably' everywhere).
Unirational varieties are rationally connected, which in turn implies $H^0(X,(\Omega_X^1)^{\otimes m}) = 0$ for all $m > 0$. Mumford conjectured that this vanishing conversely implies that $X$ is rationally connected, but I think little is known about this conjecture. At least if you believe Mumford (and maybe with a bound on which $m$ you have to try, analogous to Castelnuovo's criterion), then rational connectedness should be somewhat decidable.
However, a well-known open problem is whether every rationally connected variety is unirational (I think people expect this to be false ― I certainly do). The reason we can't answer this is that we don't have any obstructions that can distinguish between the two. Therefore, I think it's highly unlikely that we can give a characterisation that decides whether a variety is unirational.
I'm not sure what the status is on unirationality in families, but (stable) rationality is not a deformation invariant of smooth projective varieties by Hassett–Pirutka–Tschinkel. So discrete [i.e. locally constant in families] invariants like cohomology vanishing are not enough to detect (stable) rationality, and very likely the same is true for unirationality.
As for surfaces over $\bar{\mathbf F}_p$, some K3 surfaces are unirational, but most are not, so once again it is not a deformation invariant. Shioda conjectured that if $X$ is a surface over an algebraically closed field $k$ of characteristic $p > 0$ with $\pi_1^{\operatorname{\acute et}}(X) = 0$, then $X$ is unirational if and only if $H^2_{\operatorname{cris}}(X/K)$ is supersingular, i.e. all slopes of Frobenius are $1$. (This is clearly a necessary condition, since $H^2_{\operatorname{cris}}(X/K) \hookrightarrow H^2_{\operatorname{cris}}(Y/K)$ is injective and preserves Frobenius actions if $Y \twoheadrightarrow X$ is a dominant map of smooth projective varieties; see e.g. Prop. 1.2.4 in Kleiman's paper Algebraic cycles and the Weil conjecutres in Dix Exposés.)
But this doesn't say anything about what happens if $\pi_1^{\operatorname{\acute et}}(X) \neq 0$, although it has to be finite and its order indivisible by $p$ (see e.g. this note by Chambert-Loir). So you're first supposed to compute the universal cover and then compute slopes ― I'm not sure if there is a more direct way to do this. Assuming Shioda's conjecture this gives a criterion.
I realise that my answer is mostly about the general situation, and doesn't really address the low dimensional cases of the OP.
Thank you Remy! As you say in your answer, the general case seems pretty hopeless. For now, I will leave the question open because I'm interested in any progress made on the low-dimensional case.
There is an obstruction to unirationality proposed by Koll'ar. The obstruction has not been computed.
@wnx You can read more about this in Section 7 of Koll'ar's article, "Which are the simplest algebraic vareities", available here: https://www.ams.org/journals/bull/2001-38-04/S0273-0979-01-00917-X/S0273-0979-01-00917-X.pdf
@JasonStarr I don't see where Kollar proposes an obstruction. Could you perhaps point me to it? Thank you.
Koll'ar's "obstruction" is existence / non-existence of a rational surface containing a very general point of the variety. The parameter spaces of rational curves on many classes of rationally connected complex manifolds are understand in some detail. The obstruction is equivalent to proving that every uniruled subvariety of the parameter space does not parameterize rational curves that sweep out the manifold. This has been proved for parameter spaces of low degree curves.
|
2025-03-21T14:48:31.922091
| 2020-09-01T14:35:26 |
370605
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370605"
}
|
Stack Exchange
|
Reading material for an analytical aspect of Kähler Geometry
This question was originally posted on MSE.
But I would like to post it here to see whether anyone could recommend some reference for me.
I am currently reading the paper "Three-circle theorem and dimension estimate for holomorphic functions on Kähler manifolds" by Gang Liu and would like to see if anyone could recommend some books on the analytical aspect of Kähler geometry. More specifically, are there any books in analysis on Kähler geometry/ Kähler-Ricci flow that is written in a way like Peter Li's "Geometric Analysis" or Chow-Lu-Ni's "Hamilton's Ricci Flow"?
Your help is very much appreciated.
In the original post, I found Wells' and Ballmann's books to be useful, but not exactly what I am looking for.
@HassanJolany Sorry for the late reply, I did not check the forum for a day. Your answers are all amazing, thank you so much for the recommendation! As for the last comment, you mean any book that concerns "Entire and holomorphic functions"?
I'll give an answer that is specifically tailored towards the Kahler-Ricci flow. Hopefully other answers can give some good materials for geometric analysis on Kahler manifolds more generally. For KR flow, I found the following manuscripts to be very informative. All of them can be found for free online as well, which is a plus.
Boucksom, S. et al (ed.) (2013). An introduction to the Kähler-Ricci flow (Vol. 2086). Cham: Springer.
This is really a collection of chapters, but it has a good overview of various different aspects of KR flow analysis.
Song, J., & Weinkove, B. (2012). Lecture notes on the K" ahler-Ricci flow. arXiv preprint arXiv:1212.3653.
This is a good overview with a focus towards the analytic minimal model program.
Tosatti, V. (2018). KAWA lecture notes on the Kähler–Ricci flow. In Annales de la Faculté des sciences de Toulouse: Mathématiques (Vol. 27, No. 2, pp. 285-376).
This is another good introduction to the flow, with a lot of relevant background.
Lecture notes by Otis Chodosh and Christos Mantoulidis based off a class by Richard Bamler.
These aren't specifically about Kahler-Ricci flow, but I like them because they give details about Uhlenbeck's trick and the intuition for Hamilton-Li-Yau inequalities (Hamilton's original paper on the subject also has good insights, but I suppose that's not really an introduction per se).
I am so grateful that you provided so many resources for me! It is difficult to find a book in Kähler geometry that is not algebraic, at least on my own. I am glad I asked on overflow!
Ben Weinkove 5 lectures
The Kähler–Ricci flow on compact Kähler manifolds which has been collected in
Bray, Hubert L. (ed.); Galloway, Greg (ed.); Mazzeo, Rafe (ed.); Sesum, Natasa (ed.), Geometric analysis. Lecture notes from the graduate minicourse of the 2013 IAS/Park City Mathematics Institute session on geometric analysis, Park City, UT, USA, 2013, IAS/Park City Mathematics Series 22. Providence, RI: American Mathematical Society (AMS); Princeton, NJ: Institute for Advanced Study (IAS) (ISBN 978-1-4704-2313-1/hbk; 978-1-4704-2881-5/ebook). xvi, 438 p. (2016). ZBL1343.53002.
and the second chapter of famous book of Bennett Chow et all:
Chow, Bennett; Chu, Sun-Chin; Glickenstein, David; Guenther, Christine; Isenberg, James; Ivey, Tom; Knopf, Dan; Lu, Peng; Luo, Feng; Ni, Lei, The Ricci flow: techniques and applications. Part I: Geometric aspects, Mathematical Surveys and Monographs 135. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-3946-1/hbk). xxiii, 536 p. (2007). ZBL1157.53034.
Thank you too! Weinkove's notes seem great! I have the Chow's Ricci Flow "series", but it only has one chapter about the KRF, I am not sure whether it is enough or not. Anyhow, I still use it as a reference for Ricci flow in general.
|
2025-03-21T14:48:31.922314
| 2020-09-01T16:09:19 |
370612
|
{
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"authors": [
"Daniel Loughran",
"Dr. Pi",
"GH from MO",
"https://mathoverflow.net/users/11919",
"https://mathoverflow.net/users/5101",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/370612"
}
|
Stack Exchange
|
Equidistribution of $\{\alpha p\}$ for $p$ in an arithmetic progression
Let $\alpha$ be irrational. A famous theorem of Vinogradov says that $\{ \alpha p\}$ is equidistributed in $[0,1]$ as $p$ runs over all primes.
Let $a,q$ be natural numbers with $\gcd(a,q) = 1$. Then is the sequence $\{ \alpha p\}$ equidistributed in $[0,1]$, as $p$ runs over primes with $p \equiv a \bmod q$?
Almost certainly this must be known. So I'm looking for a precise reference in the literature as I need it in a paper. Ideally, it would be nice to have an effective version which makes explicit the speed of convergence (via the Erdős-Turán inequality, say).
I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero integer $k$, and for every Dirichlet character $\chi$ modulo $q$, we have
$$\sum_{p<x}\chi(p)e(k\alpha p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$
The function $n\mapsto\chi(n)$ can be written as a linear combination of additive characters $n\mapsto e((a/q)n)$ with $a\in\mathbb{Z}$, hence it suffices to show that
$$\sum_{p<x}e((a/q+k\alpha)p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$
This in turn follows from the equidistribution of $\{(a/q+k\alpha)p\}$, because $a/q+k\alpha$ is an irrational number. QED
The error terms will be similar as in Vinogradov's theorem, but with additional constants depending on $q$. The rate of convergence will depend on how well $\alpha$ can be approximated by rational numbers. See Chapter XI and the subsequent notes in Vinogradov's "The method of trigonometrical sums in the theory of numbers". I am sure there is a modern reference, but I am no expert in this subject.
Great answer, thanks.
@DanielLoughran: Thank you. Actually, we don't even need to use Dirichlet characters. The condition $p\equiv a\pmod{q}$ can be detected directly by additive characters modulo $q$, so one arrives at the last display (with different $a$'s) directly.
Yes I had realised this as well. It's a nice trick.
A more interesting problem seems to be the following. Let $K/\mathbb{Q}$ be a number field. Then is ${\alpha p}$ equidistributed as $p$ runs over all primes completely splti in $K$? (Or more general Chebotarev sets.) This doesn't seem to be immediately reducible to Vinogradov's result.
It is not difficult to extend Vinogradov's approach to $K/\mathbb Q$. Vaughan's identity is is a decomposition of $-\zeta'/\zeta$ and here you only have to replace zeta by the Dedekind zeta.
@DanielLoughran: The previous comment (by Captain Darling) was meant for you.
|
2025-03-21T14:48:31.922471
| 2020-09-01T17:13:18 |
370616
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370616"
}
|
Stack Exchange
|
Rational cohomology of p-adic general linear groups
I wanted to compute the cohomology ring $H^*(GL_n(\mathbb{Z}_p); \mathbb{Q}_p)$ (with $p$ fixed prime as usual). I found some incomplete notes stating that the computation should go as follows.
First we use a theorem by Lazard stating that for $G$ a compact p-adic analytic group we have
\begin{equation} H^*(G; \mathbb{Q}_p)\cong H^*(\mathfrak{g};\mathbb{Q}_p)^G \end{equation}
where $\mathfrak{g}$ is the rationalization of the logarithm algebra associated to an open subset of $G$ which is a uniformly powerful pro-p group. The most modern reference I could find is "Cohomology of p-adics Analytic Groups" of P. Symonds and T. Weigel, theorem 5.2.4. This algebra $\mathfrak{g}$ can also be identified with the tangent space of $G$ with the usual Lie bracket.
If $G=GL_n(\mathbb{Z}_p)$ we have $\mathfrak{g}\cong p.M_n(\mathbb{Z}_p)\otimes_{\mathbb{Z}_p}\mathbb{Q}_p$ where $p.M_n(\mathbb{Z}_p)$ is the module of matrices which are multiples of $p$: it has a Lie bracket induced by its algebra structure.
First of all we have to show there is a bijection between the following sets.
Let $P^d$ be the set of functions $f\colon p.M_n(\mathbb{Z}_p) \rightarrow \mathbb{Q}_p$ which are homogeneous polynomial of degree $d$ in the entries of the argument matrix. Let $S^d$ be the set of symmetric multilinear maps $g \colon (p.M_n(\mathbb{Z}_p))^d \rightarrow \mathbb{Q}_p$. The map $S^d \rightarrow P^d$ is given just by $g\mapsto f$ with $f(m)=g(m,m,\dots,m)$ while the inverse is given by polarization
\begin{equation} g(m_1,\dots,m_d)= \frac{(-1)^d}{d!}\sum_I(-1)^{|I|}f\bigg( \sum_{i \in I}m_i\bigg) \end{equation}
where the indexing set $I$ ranges over all the subsets of $\{1,2,\dots,d \}$.
This bijection is nothing particular to the situation: it should hold if we were to replace $\mathbb{Q}_p$ with a a field $K$ of characteristic zero and $p.M_n(\mathbb{Z}_p)$ with $\text{End}(M)$ for any $M$ finite dimensional $K$-vector space.
We now define $f_k\colon p.M_n(\mathbb{Z}_p)\rightarrow \mathbb{Q}_p$ to be the element of $P^k$ sending $u$ to the coefficient of $t^{n-k}$ in the characteristic polynomial $\det(u-It)$. Using the bijection above we obtain $g_k \in S^k$ and extending by $\mathbb{Q}_p$-linearity we get a symmetric multilinear map $g_k\colon \mathfrak{g}^k \rightarrow \mathbb{Q}_p$.
Using this we define a map $\mathfrak{g}^{\otimes (2k-1)}\rightarrow \mathbb{Q}_p$ by
\begin{equation} u_1\otimes \dots \otimes u_{2k-1}\mapsto g_k([u_1,u_2], [u_3, u_4], \dots, u_{2k-1}) \end{equation}
and by antisymmetrizing we get an element $a_{2k-1}\in \Lambda^{2k-1}\mathfrak{g}^{\ast}$.
The mentioned notes claim without proof that these elements are cocycles of the Chevalley complex computing $H^*(\mathfrak{g}; \mathbb{Q}_p)$ and in fact $H^*(\mathfrak{g}; \mathbb{Q}_p)\cong \Lambda[a_1,\dots, a_{2n-1}]$.
I managed to show that these $a_{2k-1}$ are in fact cocyles, from the invariance of $f_k$ under conjugation it is clear that if we prove $H^*(\mathfrak{g}; \mathbb{Q}_p)$ is the exterior algebra generated by these elements then it coincides with $H^*(GL_n(\mathbb{Z}_p); \mathbb{Q}_p)$.
But I could not prove that these $a_{2k-1}$'s are actually generators and my question would be if you know how to do this. I thought that maybe I have to employ a spectral sequence coming from a filtration of the Chevalley complex but I could not find a nice multiplicative filtration doing the job.
I computed explicitly the Chevalley complex for $n=2$ to see if the claim holds and indeed it was confirmed true for this case, even if I do not see why it should be true in general.
Since this computation is considered folklore it should be written somewhere, but I did not have any luck in finding a source. Do you know a reference for this? Also is there a good reference for computations of cohomology (rational or not) groups of pro-p groups and analytic p-adic groups? I mentioned before the paper by Symonds and Weigel but it deals more with the homological foundations of continuous cohomology rather than concrete examples.
|
2025-03-21T14:48:31.922678
| 2020-09-01T17:46:13 |
370617
|
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|
Stack Exchange
|
Does a fully faithful functor always preserve limits and colimits?
I read on this n-lab page that a fully faithful functor $F: C\to D$ reflects all limits and colimits by the universal property.
On the other hand, I think a fully faithful functor does not always preserve limits and colimits since a "testing object" in $D$ is not necessarily in the image of $F$.
Are there any easy counter examples?
What your concern points at is that limits and colimits are not necessarily preserved by a fully faithful functor.
I provide an explicit counterexample to a fully faithful functor preserving limits in my answer
In fact, I'd say that it's relatively rare for a fully faithful functor $F: \mathcal C \to \mathcal D$ to preserve both limits and colimits. After all, if things are nice enough so that adjoint functor theorems can be applied everywhere, and if $F$ preserves both limits and colimits, then $F$ admits both a left and a right adjoint, $\mathcal C$ is simultaneously a reflective and coreflective subcategory of $\mathcal D$.
The forgetful functor from abelian groups to groups is fully faithful, and does not preserve coproducts. For example, in abelian groups, $\mathbb Z\coprod \mathbb Z=\mathbb Z\times \mathbb Z$, but in groups $\mathbb Z\coprod \mathbb Z$ is the free group on two generators.
The point is that the cone starts out in $C$, then you test it in $D$.
Being a limit cone in $D$ is more than you need by virtue of what you mentioned about some test objects in $D$ not being visible to $F$.
Suppose you had a cone $\ell$ in $C$ whose image under $F$ is a limit cone in $D$. Let $c$ be another cone in $C$ for the same diagram, then $Fc$ will be a cone in $D$ and so by the universal property (of $F\ell$ in $D$) this admits a unique morphism $Fc\to F\ell$.
Now, since $F$ is fully faithful, this morphism arises as a unique morphism $c\to\ell$ in $C$, proving that $\ell$ has the universal property of being a limit cone in $C$ as well.
I suspect you might be confusing this with the property of preserving limits. If a fully faithful functor cannot see all the objects of $D$ (i.e., is not essentially surjective) then the fact that there are test objects in $D$ that $F$ cannot see will make it possible that a limit cone in $C$ will no longer be a limit cone in $D$.
For an explicit example, let $C=\{0\}$ be a one object category, and $D=\{0\to1\}$ the walking arrow category. Take $F:C\to D$ to be the inclusion sending $0\mapsto0$, then $F$ is fully faithful. However, $0$ is a limit cone for the empty diagram in $C$ (since it is the terminal object), but $F(0)=0$ is not a limit cone for the empty diagram in $D$.
The problem is that not all cones in $D$ are of the form $Fc$.
When you're talking about reflecting limits, you only care about cones that come from $C$. What you know is that the image of $\ell$ is a limit cone in $D$, so fully faithfulness allows you to move the problem to $D$ (where you know by assumption that there is a factoring morphism) and then bring it back to $C$.
Reflecting limits means that "cones in $C$ that are limits in $D$ become limits in $C$"
Reflecting limits means that "cones in $C$ that are limits in $C$ become limits in $D$".
Let $l$ be a limit cone in $C$. Although for any cone $c$ in $C$, there is a unique morphism $Fc\to F l$, this does not mean that for any cone $d$ in $D$ there is a unique morphism $d\to F l$ since $d$ is not necessarily of the form $Fc$.
In the definition of a reflected limit, we have that $F$ reflects limits if $\ell$ is a limit in $C$ whenever $F\ell$ is a limit in $D$. In particular, this means that when testing if $F$ reflects limits, you are already assuming that any cone $d$ in $D$ admits a unique factoring morphism $d\to F\ell$. We then only care about those cones of the form $Fc$ for some cone $c$ in $C$.
What you are doing is the opposite, which is regarding the property of preserving a limit. You're correct that a limit cone $\ell$ in $C$ is not necessarily a limit cone $F\ell$ in $D$ even if $F$ is fully faithful, but I even provide an example of this in my answer.
It seems that I misunderstand the term "preserve". I will modify my question. Thank you.
|
2025-03-21T14:48:31.922922
| 2020-09-01T18:15:27 |
370619
|
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|
Stack Exchange
|
Pro-étale locally simply connected schemes
In topology, topological manifolds are locally simply connected. However, in the étale topology of schemes, the analogue statement is not true: If $k$ is a field then finite separable field extensions $k^{sep}/l/k$ are cofinite in the étale coverings, and we have
$$\pi_1^{ét}(\text{Spec}(l),k^{sep})=\text{Gal}(k^{sep}/l)$$
and thus unless $[k^{sep}:k]$ is finite, $\text{Spec}(k)$ is not étale locally simply connected. However, the field extension $k^{sep}/k$ is pro-étale, and thus in the pro-étale topology, $\text{Spec}(k)$ is simply connected. Hence I'm wondering if every scheme of finite type over a field $k$ is pro-étale locally simply connected.
There is some relevant discussion here: https://golem.ph.utexas.edu/category/2020/03/pyknoticity_versus_cohesivenes.html#c057620
|
2025-03-21T14:48:31.923001
| 2020-09-01T18:16:58 |
370621
|
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|
Stack Exchange
|
Factorization in formal power series versus in convergent power series over the complexes
Let $R=\mathbb C\{x_1,...,x_n\}\subset S=\mathbb C [[x_1,...,x_n]]$ denote the ring of convergent, respectively formal, power series over $\mathbb C$.
Suppose $f\in R$ is irreducible in $R$. Does it remain irreducible in $S$?
When crossposting, it's important to link all versions together to avoid duplication of work. Here is the post on MSE.
What partial results do you know of? I think Nagata proved this for prime instead of irreducible.
The ring of germs of holomorphic functions is a UFD (see section 2.2 of http://people.math.harvard.edu/~yifei/Weierstrass_theorems.pdf . I think I remember this being in Gunning and Rossi, but I can't find my copy right now.) So prime = irreducible. Does this finish the question?
Right, but Arno Fehm then says that Nagata proved this with prime in place of irreducible.
Okay, I communicated badly somewhere. What someone should do now is to track down the Nagata paper and track down the UFD fact to canonical references, and check that I am not missing some other logic. I haven't posted an answer because I haven't done that work. I agree that someone should do it.
I do find the comments of the OP quite rude. That said, I think that prime elements stay prime because Nagata claims that prime ideals stay prime in the introduction to his 1953 paper "Some remarks on local rings, II", but I haven't actually looked into the proofs.
@ArnoFehm Thank you very much for the reference!
Thank you @David for going through the details!
@DavidESpeyer: You were right about Gunning and Rossi: https://books.google.co.vi/books?id=wsqFAwAAQBAJ&lpg=PP1&pg=PA65#v=onepage&q&f=false.
As Arno Fehm points out, this follows from results in Nagata's Some Remarks on Local Rings II. Both $R$ and $S$ are UFD's, so $f$ is irreducible, in $R$ or $S$ respectively, if and only if the ideal it generates, in $R$ or $S$ respectively, is prime. At the bottom of page 1 of Nagata's paper, he states that, if $\mathfrak{p}$ is a prime ideal of $R$, then $\mathfrak{p}S$ is prime as well.
I found it hard to absorb all of Nagata's vocabulary; here is a route to get the desired claim from his results while skipping some of the sophisticated language.
Let $R'$ and $S'$ denote the versions of $R$ and $S$ with $n-1$ variables. Suppose that $f=gh$ for $f \in R$ and $g$, $h \in S$ nonunits. Use the Weierstrass preparation theorem to factor $f = pu$, $g=qv$ and $h = rw$ where $p \in R'[x_n]$, $q \in S'[x_n]$ and $r \in S'[x_n]$ are Weierstrass polynomials and $u \in R^{\times}$, $v \in S^{\times}$ and $w \in S^{\times}$ are units. Then $qr$ is a Weierstrass polynomial of $S$, and $vw \in S^{\times}$, so $f = (qr) (vw)$ and $f = pu$ are both Weierstrass factorizations in $S$. Since such factorizations are unique, we have $p = qr$ and $u = vw$.
Write $p(x_n) = x_n^a + \sum_{i=0}^{a-1} p_i x_n^i$, $q(x_n) = x_n^b + \sum_{i=0}^{a-1} q_i x_n^i$, $r(x_n) = x_n^c + \sum_{i=0}^{c-1} r_i x_n^i$. Then the $q_i$ and $r_i$ are polynomial combinations of the roots of $p$, so the $q_i$ and $r_i$ are integral over $R'$. But Nagata, in his proof of Theorem 5, shows that $R'$ is integrally closed in $S'$, and this proof is extremely concrete. So that shows that the $q_i$ and $r_i$ land in $R'$. Thus $q$ and $r \in R$, and we deduce that $f$ factors in $R$ as well.
Thanks a lot for your remarkable work, David. Your extracting an answer to my question so quickly from Nagata's hard to read aricle (and idiosyncratic terminology) is a real *tour de force" .
|
2025-03-21T14:48:31.923521
| 2020-09-01T20:08:27 |
370631
|
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|
Stack Exchange
|
Smallest $\mathrm{D}(Q\|P)$ given fixed marginals $\mathrm{D}(Q_X\|P_X)$ and $\mathrm{D}(Q_Y\|P_Y)$
Let $P$ be a distribution on a set $U\times V$ with marginal distributions $P_X$ and $P_Y$.
Suppose we have two values $d_x, d_y\in\mathbb R$, and we want to find the distribution $Q$ absolutely continuous with respect to $P$ that minimizes
$$\mathrm{D}(Q\|P)$$
given $\mathrm{D}(Q_X\|P_X) = d_x$ and $\mathrm{D}(Q_Y\|P_Y) = d_y$.
What can we say about $Q$? Do we even know that such a distribution exists? Or can there be an infinite sequence of $Q'$ with respectively smaller $\mathrm{D}(Q'\|P)$?
I'm particularly interested in what the marginals of $Q$ look like.
If instead we had fixed $E_P[(X,Y)]$ (assuming now $X,Y\in\mathbb R$), we know from e.g. Corollary 12.1. in Yury Polyanskiy and Yihong Wu that the distribution minimizing $\mathrm{D}(Q\|P)$ given $E_Q[(X,Y)]=E_P[(X,Y)]$ is a titled version of $P$.
That is $dQ = \frac{e^{\phi}}{Ee^{\phi}}dP$ where $\phi = \lambda_1 X+\lambda_2 Y$.
I wonder if there is a similar simple family of distributions one might restrict to given my requirement on $Q$?
Below is a visualization belonging to said Corollary 12.1.
Here is a related problem: Fix $Q_X$ and $Q_Y$ and ask what coupling $Q$ minimizes $D(Q||P)$. In this case and under mild assumptions, the minimum exists since the set of couplings $\Pi(Q_X,Q_Y)$ is compact in the weak topology and relative entropy is weakly lower semicontinuous. This problem is known as the Schrödinger problem. See this survey by Christian Léonard. As you can see, in this case you can't generally expect a simple family of solutions like the one quoted.
Addressing your setting where you now optimize over marginals $Q_X, Q_Y$ satisfying the relative entropy constraints, I don't expect things become any easier. Unlike the situation you quote, these are nonconvex sets of marginals that you are constrained to. Unlike above, it isn't even clear that the desired extremizer exists since these sets are also not closed in general.
Even if there is no simple form, at least it's a tractable convex optimization problem. Minimizing $D(Q|P)-a D(Q_1|P_1)-b D(Q_2|P_2)$ I think is equivalent to finding the hypercontractive norm of a positive matrix, by Lemma 1.1 in https://thomasahle.com/papers/supermajority.pdf . But that problem appears to possibly be NP hard, though there is no proof afaik.
Or maybe minimizing $\frac{D(Q|P)}{a D(Q_1|P_1)+b D(Q_2|P_2)}$ actually, over all $Q$, so the scaling doesn't matter.
Regarding your first comment on hypercontractivity, you are mostly correct. This equivalence holds in a general sense. $P$ being $(1/a,1/b)$-hypercontractive is equivalent to the infimum of $(||)−(1||1)−(2||2)$ over all $Q$ being zero. More generally, the infimum of this problem will be nonzero, which by duality then corresponds to something like hypercontractivity, but with a prefactor in the norm inequality. See, e.g., page 22 here: https://arxiv.org/pdf/1702.06260.pdf, and references therein (in particular, Carlen and Cordero-Erausquin, 2009).
to add to above, this equivalence does not actually make your problem any easier. It just converts from an entropy problem to an equivalent functional one. Except in special cases (e.g., where $P$ is product of rho-correlated Gaussians or Rademachers), a characterization of the extremizers is likely non-explicit. However, if $P$ is a product measure, you can use tensorization of Brascamp--Lieb inequalities (and their entropic equivalents) to conclude that the extremizers will also have product structure. You should also be able to see this directly.
At this point I would be excited for just computability. Even an exponential time algorithm for the case where $P$ has finite support.
|
2025-03-21T14:48:31.923902
| 2020-09-01T21:30:43 |
370637
|
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|
Stack Exchange
|
Mittag-Leffler for non-compact Riemann surfaces
Quote from Grauert & Remmert's Theory of Stein spaces: 'Behnke and Stein showed in 1948 that the Mittag-Leffier Partial Fraction Theorem and the Weierstrass
Product Theorem (i.e. the Cousin Theorems) are valid on non-compact Riemann
surfaces. The following lemma appears at the end of their paper:
Hilfssatz C: Let $D$ be a discrete set in a non-compact Riemann surface $X$. For
every $p\in D$ let $z_p$ be a local coordinate at p. Suppose that at all $p \in D$ there is
prescribed a finite Laurent-Taylor series $h_p = \sum_{\nu=-m_p}^{n_p}a_\nu z_p^{\nu}$, $0\leq m_p,n_p<\infty$. Then
there exists a function $H$ which is meromorphic on $X$, holomorphic on $X\setminus D$, and
whose Laurent development at $p$ with respect to $z_p$ agrees with $h_p$ up to the $n_p$-th term.'
Does $H$ still exist if we ask it to have no zeroes in $X\setminus D$? That is, I want to $H$ to realize exactly a prescribed divisor. References are welcomed.
I can't see any reason why not, maybe check Coherent Analytic Sheaves by GR, or Forster's book on Riemann surfaces.
Considering the reciprocal function, it is sufficient to construct a holomorphic function with prescribed zeros, and prescribed finite portions of Taylor series at those zeros. For the plane and the unit disk this is an interpolation problem whose positive solution follows from Weierstrass's theorem (and its generalization to the unit disk).
I do not understand how you get a function on your original Riemann surface from a function on its universal cover.
|
2025-03-21T14:48:31.924050
| 2020-09-01T23:29:56 |
370641
|
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|
Stack Exchange
|
What are immediate applications of the classification of connected reductive groups?
After years of putting it off, I finally sat down, read, and understood the classification of connected reductive groups via root data.
That's a non-trivial theory! I'm hoping that now that I am done I can reap some benefits. What are some immediate applications/corollaries of this theory? Now that I understand this classification, what can I do with it? At the moment, I don't really feel like I can do anything I couldn't do before...
Number theoretic and algebro-geometric applications are great! But I'm open to anything.
Worth noting is that much of the machinery used to classify reductive groups is also extremely useful in classifying/studying representations of these groups.
Since this seems not to be attracting much of a big-list, maybe it should be retagged gr.group theory?
There are at least two different ways to interpret this question. One is, what are some applications of the machinery (e.g., root systems) that was developed to classify reductive groups? A second is, what corollaries are there of the classification theorem itself? The latter includes theorems proved by checking each group case by case. As for the former, I'm tempted to say that it'd be easier to list major results that don't make use of the root system. I'm not an expert but it feels like every paper in this area begins, "Let $\mathfrak g$ be a Lie algebra and $\Phi$ its root system..."
An immediate application is the existence of the Langlands complex dual group. If $G$ is a connected reductive group over a field $F$ and $\overline{F}/F$ is an algebraic closure, then $G^\vee$ is the connected reductive group over $\mathbb{C}$ whose root datum is dual to that of $G_{\overline{F}}$, i.e. is obtained by interchanging roots with coroots and simple roots with simple coroots.
In the case when $F$ is a local or global field, Langlands's discovery of $G^\vee$ and the related $L$-group ${}^LG$ in 1966 lead him to, and these groups figure prominently in: the local and global Langlands correspondences, the general definition of local and automorphic $L$-functions, and the principle of functoriality.
Come that far, one can begin to exploit the subgroups, turn one's attention to symmetric spaces, of which there are the compact and noncompact ones, the riemannian and the hermitian ones. And one can begin to exploit the arithmetic subgroups and so delve into locally symmetric varieties with various geometric and arithmetic properties. Come that far, whole universes open up: examples, counterexamples, theorems in Differential Geometry, and the whole garden of Number Theory and Arithmetic Geometry with Shimura varieties, compactificaions of locally symmetric varieties of all sorts relating to moduli spaces, Langlands conjectures...
Needless to say, quite useless to give references - such a list would be endless and beyond. So it's up to you to get lost your own way.
Given a prime $p$ and a connected reductive algebraic group $G$ over $\mathbb{F}_p^{\mathrm{alg}}$ with a Frobenius map $F$, the fixed points $G^F$ are a finite group 'of Lie type'. The finite groups of Lie type are the main case in the Classification Theorem of Finite Simple Groups. They can all be obtained in a uniform way by this construction, except for the Suzuki and Ree groups. (Roughly speaking, these are also obtained using the data from root systems and Dynkin diagrams, but they require further automorphisms that do not descend from the algebraic group.) Various structural properties of finite groups of Lie type follow easily from their analogues in the algebraic group, for example that they have $BN$ pairs.
I'm surprised this wasn't already an answer. It seems close enough to an 'application' to me to be worth mentioning. I admit it is far from an 'immediate application'.
Is this really an application of the classification? In the following sense: does the proof of CSFG actually use the fact that we have an exhaustive list of reductive groups, or does it only use the constructions of the reductive groups without the fact that the classification is exhaustive?
The latter, according to my understanding. (And I don't think this will be changed by any of the programmes to revise the proof of CFSG.) But I would defend my answer on the grounds that many useful structural properties of the finite groups do follow fairly immediately from the analogues in the algebraic group. E.g. existence of tori (and conjugation, although split/non-split tori in the finite case complicate things), $BN$ pairs, and the (related) Bruhat decomposition.
check work of David Vogan, and a book of Trappa Vogan ...
"Work of David Vogan" is only marginally more specific than "the papers of George Lusztig". It would probably be more helpful to mention some good entrées.
By "a book of Trappa Vogan" do you mean Representations of Reductive Groups: In Honor of the 60th Birthday of David A. Vogan, Jr., edited by Monica Nevins and Peter E. Trapa?
Sorry i confused you, i mean the book The Langlands classification and irreducible characters (D. Barbasch, J. Adams and D. Vogan), Birkhauser (1992), Boston-Basel-Berlin.
|
2025-03-21T14:48:31.924423
| 2020-09-02T00:33:37 |
370642
|
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|
Stack Exchange
|
Limit law of eigenvalue of random matrix with mean different to 0
If $X$ denotes a $m \times n$ random matrix whose entries are independent identically distributed random variables with mean $\mu$ and $\sigma^2 < \infty$, let
$$Y = X X^T$$
with $X^T$ the transpose of $X$. Let $\lambda_1 , \lambda_2 ,\ldots, \lambda_m$ be the eigenvalues of $Y$ (viewed as random variables).
If $\mu = 0$, it is known that the law of $\lambda$ converges to Marchenko–Pastur distribution:
https://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution
My question is that in the case $\mu \neq 0$ what is the limit distribution of $\lambda$?
The addition of the same constant $\mu$ to all elements of $X$ (so that their mean becomes $\mu$) is a rank-one perturbation of the matrix, which has no effect on the distribution of the eigenvalues of $XX^T$ in the limit $n,m\rightarrow\infty$ at fixed $n/m$ --- this limiting distribution remains the Marchenko-Pastur distribution. The rank-one perturbation introduces a single outlier, but the bulk of the spectrum is unchanged. See for example The singular values and vectors of low rank perturbations of large rectangular random matrices (2011).
|
2025-03-21T14:48:31.924540
| 2020-09-02T01:29:17 |
370643
|
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|
Stack Exchange
|
Genus 2 Heegaard decompositions of rational homology 3-spheres
I'm trying to build an example of a rational homology 3-sphere $M$ (that is not an integral homology 3-sphere) with an irreducible genus 2 Heegaard splitting so that $M$ is not a lens space or connect sum of lens spaces.
I haven't been successful trying to draw a Heegaard diagram for such a 3-manifold, so I suspect the problem is overdetermined. Is there an obvious reason for why this is?
Also, I suspect that some small Seifert fibered spaces might be a good place to look, but I haven't yet understood how to generally construct a Heegaard diagram for a general Serifert fibered space.
Manifolds with Heegaard genus 2 are double branched covers of 3-bridge links. Once one has a bridge diagram, one can in principle derive a Heegaard diagram for the double branched cover. See also: https://doi.org/10.1016/S0166-8641(98)00063-7
Most Heegaard splittings of small Seifert fibered spaces are "vertical". See Moriah or Boileau, Collins, Zieschang for the full classification.
Suppose that $K$ is the figure eight knot in the three-sphere $S^3$. Let $n(K)$ be a small open neighbourhood of $K$. Let $X = S^3 - n(K)$. You can find a Heegaard diagram for $X$ as follows.
Let $N = N(K)$ be a closed neighbourhood of $K$, slightly larger than $n(K)$. Draw both in a knot diagram of $K$. Add to $N$ a small neighbourhood $P$ of an "unknotting arc" (also called a "tunnel"). So $V = N \cup P$ is a handlebody, and $V' = N \cup P - n(K)$ is a compression body. Note that $W = S^3 - \mathrm{interior}(V)$ is a handlebody as well.
So $(V', W)$ is a genus two Heegaard splitting of $X$. It is an exercise to give a diagram for this splitting. (Hint: The diagram has only three curves, not four. To find it, start by drawing a knot diagram for $K$. Now add the unknotting arc $\alpha$ along the core of $A$.) Any non-longitudinal Dehn filling of $X$ gives a rational homology three-sphere. Most of these are not integral homology spheres. The filling slope gives the fourth curve in the desired Heegaard diagram.
Note that something similar works for all two-bridge knots and links (and tunnel number one knots, in general). Doing this for torus knots will yield many Seifert fibered examples - the examples above are mostly hyperbolic.
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