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?

  • $\begingroup$ Are you conjecturing that $\mathfrak P(2\rho)$ for $A_n$ is the number of labeled Eulerian oriented graphs with n nodes? or is there an explanation? $\endgroup$ – Bruce Westbury Jul 19 '12 at 22:07
  • $\begingroup$ 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 $n^n$. $\endgroup$ – David Stewart Jul 20 '12 at 6:33
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    $\begingroup$ I have done a hand calculation for $\mathfrak{P}(2ρ)$ for $A_3=SL(4)$ and got more than 15. Did I make a mistake? For $A_n$ with $n<3$ I got agreement. $\endgroup$ – Bruce Westbury Jul 20 '12 at 13:51
  • $\begingroup$ Indeed, here's a counterexample. Take $G$ of type $A_3$ and $p=3$. Then the dimension of the $0$-weight space of $H^0((p-1)\rho) = H^0(2\rho)$ is 15 (as follows from, say, a LiE computation), but $\mathfrak P(2\rho) = 26$. $\endgroup$ – Chuck Hague Jul 20 '12 at 17:29
  • $\begingroup$ Regarding your edit -- actually, the conjecture does not become true for $r = p^s-1$ and $s$ sufficiently big; in fact I believe it fails for every $s$ as long as the rank of $G$ is $>2$. Jantzen's statement requires you to first fix $\lambda = p^t - 1$ for whatever chosen value of $t$ you want; and then the statement is true for $s \gg 0$. But $t$ and $s$ definitely need not be equal -- Jantzen's statement tells you that for sufficiently large $s$, the $(p^s-1)\rho - (p^t-1)\rho$-weight space (which is not the zero weight space!) of $St_s$ is given by $\mathfrak P((p^t-1)\rho)$. $\endgroup$ – Chuck Hague Jul 22 '12 at 4:44

[EDIT] Concerning your specific question, my earlier answer was too offhand. After trying this with pencil and paper, I'm very doubtful that the multiplicity of the zero weight in the finite dimensional module with highest weight $r\rho$ for $r>0$ is always given by applying the partition function to $r\rho$. Sorry to have confused the issue here. Maybe I can identify a specific small counterexample, though a computer program makes such things easier.

In more detail, $\rho$ has two basic characterizations: as the sum of fundamental dominant weights (showing that it is dominant) and as the half-sum of all positive roots (showing at least that $2\rho$ lies in the root lattice, though $\rho$ itself might or might not). From representation theory it follows that the value of the partition function at $r\rho$ is the dimension of the zero weight space of the corresponding infinite dimensional Verma module.

Kostant's basic insight was that the finite dimensional simple quotient of any Verma module with dominant highest weight has formal character given by an alternating sum of naturally related Verma module characters (which later got encoded in the BGG resolution). So your underlying representation theory problem is to show that the alternating sum in this special case will not reduce the size of the zero weight space. But I don't see this following from Kostant's weight multiplicity formula.

In general, Kostant explained in the introduction to his 1959 paper here how his intuition developed toward the rigorous formula. Later work by Bernstein-Gelfand-Gelfand streamlined the derivation of this formula (and Cartier pointed out earlier the logical equivalence of the formula with Weyl's), but it may be useful to take a look back at the original paper. [Note that the symbol $g$ is used there for what later became standardized as $\rho$.]

Concerning asymptotics of the partition function, I'm not sure how much has been written down. But for example an appendix in Verma's 1966 Yale thesis makes some observations along this line; as I recall he developed a polynomial bound. Much later Lusztig (and then R.K. Brylinski) studied a $q$-version of the partition function as well. Such work is done in a purely theoretical mode, of course, but there may be other computational approaches of interest for your question.

[ADDED] The discussion has been getting more convoluted, starting with David's original attempt to quote Jantzen's II.10.12 (which turns out to be only marginally relevant to the question about Kostant's partition function). Now that I've gone back over that material in RAGS more carefully, I can see that it's one of those places where Jantzen's notation gets devious. In 10.11 he defines $\mathrm{St'}_r$ to be the module for $G_r T$ obtained from the usual Steinberg module (for $G$) of highest weight $(p^r-1)\rho$ by tensoring with that weight. So for example the zero weight of the Steinberg module becomes the $(p^r-1)\rho$ weight of the new module. This weight of course varies with increasing $r>0$, whereas Jantzen asks you to fix a weight $\lambda$ of the new module and then find large enough $r$, etc. As I say, it's devious.

Some of the comments indicate that the conjecture involving $2\rho$ and such will have counterexamples. But the basic question about asymptotic behavior of Kostant's function may be worth further study. As I mentioned above, this did come up in Verma's thesis where he proved existence of a polynomial bound when the partition function is expressed in terms of the $\ell$ coefficients of an element in the root lattice ($\ell$=rank). But his reason for doing all this got bypassed in later literature such as Dixmier's book.

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  • $\begingroup$ 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 $\endgroup$ – David Stewart Jul 20 '12 at 7:44
  • $\begingroup$ @David: At first I assumed the conjecture would follow directly from Kostant's multiplicity formula, but that doesn't seem to be so obvious. When I have enough time I'll look at Jantzen's method more closely, to see if it carries over to quantum groups at a root of unity; that would be the quickest way to get a uniform result of similar flavor. In any case, type $A_n$ is highly symmetric and therefore likely to behave better than other types. Similarly for asymptotics of Kostant's function. $\endgroup$ – Jim Humphreys Jul 20 '12 at 15:06

To expand on Jim's answer, it's a result of Kostant that for any weight $\mu$ and any dominant weight $\lambda$ the dimension of the $\mu$-weight space of the induced module $H^0(\lambda)$ with highest weight $\lambda$ is given by $$ \sum_{w \in W}(-1)^{l(w)} \mathfrak P( w(\lambda + \rho) - \mu - \rho ) .$$ (Remark that equivalently this gives the dimension of the $\mu$-weight space of the Weyl module with highest weight $\lambda$, and in characteristic 0 the induced and Weyl modules of highest weight $\lambda$ are isomorphic to each other and simple so this gives weight multiplicities for simple modules in characteristic 0 too). As Jim mentions, this formula should imply the result you want for the 0-weight space of $H^0(r\rho)$. [EDIT: Or perhaps not -- see Jim's updated answer] [SECOND EDIT: It's not true in general that the dimension of the $0$-weight space of $St_r$ is $\mathfrak P( (p^r-1)\rho )$.].

Furthermore, there are a lot of interesting facts about the Kostant partition function and its $q$-analog; in particular, the Kostant partition function is equal to the evaluation at 1 of certain Kazhdan-Lusztig polynomials, and it's also connected to facts about nilpotent orbits. If you want to know more about this story, the paper "Limits of weight spaces, Lusztig's $q$-analogs, and fiberings of adjoint orbits" by R.K. Brylinski is a good paper to look at.

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  • $\begingroup$ @Chuck: I think your newer statement of Jantzen's result in the comments is still out of focus, since he isn't actually looking at the weights of a Steinberg module but instead tensors it with its highest weight. (I tried to explain this in my own added remarks.) $\endgroup$ – Jim Humphreys Jul 21 '12 at 14:47
  • $\begingroup$ I made a mistake in my initial comment, which I hopefully fixed in my later comment (unless it's still wrong) -- oh for an edit window on comments. I think the point is that for any weight $\lambda \geq 0$ and dominant $\mu \gg 0$ the dimension of the $\mu-\lambda$-weight space of $H^0(\mu)$ is given by $\mathfrak P(\lambda)$, which isn't actually a statement about Steinberg modules but about weight spaces of the zero Verma module. $\endgroup$ – Chuck Hague Jul 21 '12 at 17:02

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.

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  • $\begingroup$ It has been added as A214808. $\endgroup$ – Charles Jul 30 '12 at 16:51

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