User david stewart - MathOverflow

Answer by David Stewart for Irreducibility of fundamental Weyl modules

2012-10-04T14:22:05Z

You might also find the data on Frank Luebeck's website useful:

http://www.math.rwth-aachen.de/~Frank.Luebeck/chev/WMSmall/index.html

Answer by David Stewart for Kostant partition function: asymptotics and specifics

2012-07-25T08:59:05Z

I have some answers.

I couldn't find a computer program giving values of $\mathfrak P$ so I wrote one. Assuming it's giving the correct numbers, $\mathfrak P_{A_n}(2\rho)$ is $$1,3,26,898,128826,82462230$$ with natural logs $$0,1.10,3.26,6.80,11.77,18.22.$$ The latter looks roughly quadratic to me. Note that the sequence does not come up on the OEIS.

As for proving some bounds...

First of all, [Cline, Parshall, Scott; Reduced standard modules and cohomology] has a statement of a bound on page 5258. They state that it's easy to prove $\mathfrak P(2(h-1)\rho)\leq N!P(N(h-1))$ where $N=|\Phi^+|$ is the number of positive roots and $P$ is the usual partition function. Wikipedia tells me that $$P(n)\sim \frac{1}{4n\sqrt 3}\exp\left(\pi\sqrt{\frac{2n}{3}}\right)$$ so this gives $\log \mathfrak P(2(h-1)\rho)$ is $O(n^{3/2})$.

I think I can improve on this (just for type $A_n$ but it should be representative):

Observation 1: Looking up in the tables in Bourbaki, one gets $2\rho=n\cdot1\alpha_1+(n-1)\cdot2\alpha_2+\dots$. So the lowest coefficient is $n$ and the highest coefficient is roughly $(\frac{n+1}{2})^2$.

Observation 2: Once the coefficients of the non-simple roots in a sum of positive roots has been chosen, one completes the expression by adding in a determined number of simple roots to make up the difference. So $\mathfrak P(2r\rho)$ can be obtained by counting sets of positive non-simple roots whose sum is less than or equal to $2r\rho$ (in the dominance ordering).

Lower bounds: just use roots of height 2. Choose $\lfloor n/2\rfloor$ coefficients $b_i$ so that $$b_1(\alpha_1+\alpha_2)+b_2(\alpha_3+\alpha_4)+\dots\leq 2r\rho.$$ By Observation 2, each $b_i$ can be anything up to $rn$. Get $$\mathfrak P(2r\rho)\geq (rn)^{n/2}.$$ Probably not very good, but at least it shows it's exponential.

Upper bounds: There are $n(n-1)/2$ non-simple roots and by Obs 2 each can occur with any coefficient which at most $r(n+1)^2$. So get $$\mathfrak P(2r\rho)\leq (r(n+1))^{n(n-1)}.$$

Combining these two gives that $\log \mathfrak P(2\rho)$ having growth rate strictly more than linear and strictly less than cubic. The upper bound says that $\log \mathfrak P(2\rho)$ is $O(n^2\log n)$. The lower bound says that $\log \mathfrak P(2\rho)$ grows at least as fast as $n\log n$.

I think one can probably do better, particularly on the lower bound. I may have another look at it at some later stage. But basically I'm content with the statement that $\log \mathfrak P(2\rho)$ is roughly quadratic with the rank.

Kostant partition function: asymptotics and specifics

2012-07-19T11:36:05Z

Let $\Phi$ denote a root system and let $\mathfrak P$ denote the associated Kostant partition function. Thus $\mathfrak P(\lambda)$ is the number of ways of writing $\lambda$ as a sum of elements of $\Phi^+$. For example, if $\Phi=A_3$ then the longest root $\alpha_0$ can be written as $\alpha_0=(\alpha_1+\alpha_2+\alpha_3)$, $(\alpha_1)+(\alpha_2)+(\alpha_3)$, $(\alpha_1+\alpha_2)+\alpha_3$ or $\alpha_1 + (\alpha_2+\alpha_3)$. Thus $\mathfrak P(\alpha_0)=4$. There is a recursion $$\mathfrak P(\mu)=-\sum_{1\neq w\in W} (-1)^{l(w)}\mathfrak P(\mu+w(\rho)-\rho)$$ due to Kostant.

I can't find very many general results on $\mathfrak P$.

One knows that $\mathfrak P((p-1)\rho)$ is the dimension of the $0$ weight space of the Steinberg module (see Jantzen, RAGS, II.10.12). I conjecture that one also has $\mathfrak P(r\rho)$ is the dimension of the $0$ weight space in $L_{\mathbb C}(r\rho)$, where $L_{\mathbb C}(r\rho)$ is the irreducible representation of high weight $r\rho$ for a complex Lie algebra with root system $\Phi$ (which may be zero if $r$ is odd). A reference even for this basic statement would be good!

[RE-EDIT: This paragraph, including the conjecture, is completely wrong. See Chuck Hague's and Jim Humphreys' remarks below. Of course one must have $\dim H^0(\lambda)_0\leq \mathfrak P(\lambda)$ for all $\lambda$, so this gives a lower bound.]

I'd also like some data on asymptotics. I have used the Maple `Coxeter/Weyl' package to compute $\mathfrak P(2\rho)$ for $A_n$ up to $n=5$. The numbers are $1,3,15,219,7834\dots$ and correspond to sequence A007081 on the OEIS.

[EDIT: These calculated values are in fact the dimensions of the $0$ weight space of $H^0(2\rho)$. Since the conjecture that $\mathfrak P(2\rho)=\dim H^0(2\rho)_0$ is incorrect these do not coincide with the values of $\mathfrak P(2\rho)$ as claimed.]

I guess $\mathfrak P(2\rho) < n^n$. [EDIT: This looks very optimistic---suspect it's wrong for $n=4$.] Any ideas? What about $\mathfrak P (2r\rho)<(nr)^{nr}$ or something better? [EDIT: Similarly.] Lower bounds?

Connected extensions of finite by connected algebraic groups

2011-08-26T10:44:25Z

Let $1\to H\to E\to G\to 1$ be a short exact sequence of algebraic groups defined over an algebraically closed field $k$ of characteristic $p$. Suppose $H$ is a finite group, and $G$ and $E$ are connected. Does it follow that $G\cong E$?

[Edit: of course not in general, since, as Max points out, E=SL_n, H=Z(E) and G=PSL_n yield counterexamples]

More specifically for my purposes, if $G$ is a vector group, (i.e. $G$ is isomorphic to the direct product of $n$ copies of the additive group $G_a(k)$ of the field $k$) must $E$ be one as well? (Let $E$ also be unipotent, if it helps.)

I note that this follows when $G=G_a(k)$, the additive group of the field, since there are only two connected 1-dimensional groups and $G_m(k)$ does not surject onto $G_a(k)$.

Differentiating, one sees that $L(E)\cong L(G)$, for what it's worth.

Comment by David Stewart

2012-07-20T07:44:36Z

Many thanks for the answer. In defence of the weight space conjecture, I think we know that $\mathfrak P(r\rho)$ coincides with $\dim L_\mathbb C(r\rho)_0$ whenever $r=p^s−1$, since then $L(r\rho)=St_s$ is the $s$th Steinberg module and equal to $\mathrm{Ind}^{G_s}_{B_s}(r\rho)=k[U_s]⊗r\rho$. So I suspect the identification with the weight space follows in this case---see the Jantzen reference. So I'm somewhat doubtful about finding a counterexample as $r$ would have to be some number not of the form $p^s−1$ for any $p$ and $s$\dots

Comment by David Stewart

2012-07-20T06:33:24Z

Yes, why not? Call it Stewart's conjecture, if you like. I have no idea why it would be true. At least if that connection were proved, one could use the upper bounds in the McKay paper referenced in the OEIS to give upper bounds for $\mathfrak P(2\rho)$, though I doubt it would be as tight as $n^n$.

Comment by David Stewart

2011-08-26T13:54:53Z

A vector group is an algebraic group isomorphic to a direct product of n copies of $G_a(k)$. I suppose it's distinct from a vector space in the sense that there's no assumed action of $k^*$ on it.

Comment by David Stewart

2011-08-26T13:37:50Z

I suppose it is still possible that if $G$ is a vector group, then $E$ will be also.