2

Let $L$ be a semi simple Lie algebra over an algebraically closed field $F$ with

  1. Cartan decomposition $L = h \oplus n_+ \oplus n_- $,

  2. Root system $\Phi$,

  3. Set of positive roots $\Phi_+$,

  4. Simple roots $\Delta$.

Consider the universal enveloping algebra $U(n_+)$ of $n_+$ which can be consider as a $n_+$ - module.

What is the dimension of the weight space $U(n_+)_\beta$ for a root $\beta$ ?

I am trying to prove that, this dimension is equal to $K(\beta,q)$ where $K(.;q)$ is the q-kostant partition function.

$K(\beta,q)$ = co-efficient of $e^{-\beta}$ in the product $\prod_{\alpha \in \Phi_+}(1-qe^{-\alpha})^{-mult \alpha}$

Thanks in Advance.

Hanno
  • 19,510
GA316
  • 4,324
  • 27
  • 50

1 Answers1

3

Denoting ${\mathscr S}(V)$ the symmetric algebra over a vector space $V$, realized as the subalgebra of symmetric tensors within the tensor algebra ${\mathscr T}(V)$ (as opposed to a quotient), the PBW-Theorem implies that $$\bigotimes\limits_{\alpha\in\Phi_+} {\mathscr S}({\mathfrak g}_{\alpha})\to\bigotimes\limits_{\alpha\in\Phi_+}{\mathscr U}({\mathfrak n}_+)\xrightarrow{\text{mult}}{\mathscr U}({\mathfrak n}_+)$$ is an isomorphism of vector spaces, which moreover is quickly seen to be ${\mathfrak h}$-linear. Hence, $$\chi\left({\mathscr U}({\mathfrak n}_+)\right) = \prod\limits_{\alpha\in\Phi_+}\chi\left({\mathscr S}({\mathfrak g}_{\alpha})\right).\quad\quad(\ddagger)$$ Similarly, denoting ${\mathbb C}_\alpha$ the simple ${\mathfrak h}$-module with weight $\alpha$, we have ${\mathscr S}({\mathfrak g}_\alpha)\cong {\mathscr S}({\mathbb C}_\alpha)^{\otimes \dim({\mathfrak g}_\alpha)}$, so $$\chi\left({\mathscr S}({\mathfrak g}_\alpha)\right)=\chi\left({\mathscr S}({\mathbb C}_\alpha)\right)^{\dim({\mathfrak g}_\alpha)}=(1+e^{\alpha} + e^{2\alpha} + ...)^{\dim({\mathfrak g}_\alpha)}=(1-e^{-\alpha})^{-\dim({\mathfrak g}_\alpha)},$$ which together with $(\ddagger)$ gives your claim, up to the $q$ of which I also don't know what it's supposed to be.

Hanno
  • 19,510
  • In the first expression, you mean just $U(n_+)$ I think, instead of tensoring over it. am I correct or not. Thanks. – GA316 Feb 01 '15 at 11:29
  • First, each ${\mathscr S}({\mathfrak g}\alpha)$ naturally maps to ${\mathscr U}({\mathfrak n}+)$, and then I apply the multiplication map. – Hanno Feb 01 '15 at 11:30
  • Please explain what is a simple $h$ module $ \mathbb{C}_\alpha$ with weight $\alpha$. Thanks. – GA316 Feb 01 '15 at 11:31
  • It's the $1$-dimensional ${\mathfrak h}$-module on which ${\mathfrak h}$ acts by $\alpha$. – Hanno Feb 01 '15 at 11:33
  • thanks. nice. I have understand. please clarify this, in the line in which you have written the isomorphism, it is symmetric algebra of tensor product of $\mathbb{C}\alpha$ or tensor product of symmetric algebra of $\mathbb{C}\alpha$? thanks. – GA316 Feb 01 '15 at 11:39
  • Tensor product of copies of the symmetric algebra over ${\mathbb C}_\alpha$. – Hanno Feb 01 '15 at 11:41
  • thanks. very nice answer. can you suggest me any reference please? – GA316 Feb 01 '15 at 11:45
  • The statement itself is mentioned and used in Humphrey's book on semisimple Lie algebras, but I don't think much is said about the proof. – Hanno Feb 01 '15 at 11:54
  • Thanks a lot for your answer and clarifications. – GA316 Feb 01 '15 at 12:08
  • 1
    @GA316 Note that this also directly gives the alternative description of the Kostant partition function, which is the number of ways to write $\beta$ as a sum of positive roots. – Tobias Kildetoft Feb 02 '15 at 08:59