Let $G$ be a complex simple reductive group. Then the set of isomorphy classes $Irr G$ is isomorphic to the set of dominant weights $\Lambda_+$ in the weight lattice of the maximal torus of a Borel subgroup of $G$. Further, $\Lambda_+$ is isomorphic to $\mathbb{N}_0^\ell$ for some integer $\ell$. Thus we can write $\rho = n_1\lambda_1 + ... + n_{\ell}\lambda_{\ell} = \underline{n}$ for any irreducible representation $\rho\colon G \to End(V_{\rho})$.

Now let $R$ be a ring with $G$-action, $M$ an $R$-$G$-module with isotypic decomposition $M \cong \bigoplus_{\underline{n} \in \mathbb{N}_0^{\ell}} M_{\underline{n}}\otimes_{\mathbb{C}} V_{\underline{n}}$. Then one defines the Hilbert function of $M$ as $h(\underline{n}) := rk (M_{\underline{n}})$. Further, consider the function $P(M,\underline{z}) := \sum_{\underline{n} \in D} h(\underline{n})z_1^{n_1}...z_{\ell}^{n_\ell}$ for some finite subset $D \subset \mathbb{N}_0^{\ell}$.

What is the behaviour of $P(M,\underline{z})$ if $D$ is sufficiently large? Is it a rational function?

If $M'$ is a submodule of $M$, how are $P(M',\underline{z})$ and $P(M,\underline{z})$ or $h'$ and $h$ related? Does $\frac{h'(\underline{n})}{h(\underline{n})}$ converge for $\underline{n} \not\in D$ as $D$ becomes large? Which assumptions on $R$ and $M$ are necessary? What is the limit?

I am especially interested in the case $G = Sl_2$ (i.e. $\ell = 1$) but I would also like to know the answer in the general case.

  • $\begingroup$ I was wondering, do you have any representative examples of these objects you're interested in? For instance, my suspicion is that at least some finiteness assumption on M is missing (eg if R=C with trivial action, then M is just a rep (right?) of G, so we probably want it ifinite dimensional). $\endgroup$ – Peter McNamara Mar 31 '11 at 5:26
  • $\begingroup$ You are right, I want M (and M') to be finitely generated as an R-algebra, but it need not be finite as an R-module. An example is to take G = Sl_2, R = C, M the regular representation of Sl_2, so that h(n) = n+1. How does h' look like for subrepresentations M' of M? $\endgroup$ – Tanja Becker Mar 31 '11 at 11:18
  • $\begingroup$ I do not understand the question. What is the point of the finite subset $D$? By truncating a power series like that you make it a polynomial. Then why do you ask if it is a rational function? $\endgroup$ – Wilberd van der Kallen Aug 25 '11 at 8:12

I am not sure if I completely understand all of the assumptions (especially the role of $R$), but let me give it a try.

Let me assume that $R$ is finitely generated over the complex numbers. Since you want to assume that $M$ is also a finitely generated $R$-algebra, let's just throw out $R$ for now and consider $M$, which is a finitely generated C-algebra. I will also assume that the multiplication in $M$ is compatible with the $G$-action.

It seems you've already chosen a set of positive roots, so let $B$ be a Borel subgroup corresponding to these positive roots, and let $U$ be its unipotent radical. The $U$-invariants of $M$ are just the highest weight vectors and we get a decomposition

$\displaystyle M^U = \bigoplus_{\underline{n} \in \Lambda_+} M_{\underline{n}}$

where the dimension of $M_{\underline{n}}$ is the same as the one you are considering. Since the multiplication in $M$ is compatible with $G$, $M^U$ is actually multigraded by $\Lambda_+$. So the function $P(M,\underline{z})$ is a rational function in $\underline{z}$ if we know that $M^U$ is finitely generated (as an algebra).

A general fact: if reductive $G$ acts on a finitely generated algebra $A$, then the $G$-invariants $A^G$ is also finitely generated. Let ${\bf C}[G/U]$ be the coordinate ring of the coset space $G/U$ (which is a quasi-projective variety). Now take $A = M \otimes {\bf C}[G/U]$ where $G$ acts diagonally. As a $G$-module, ${\bf C}[G/U]$ is a direct sum of all irreducible representations of $G$, each appearing with multiplicity 1. Hence the $G$-invariants of $A$ can be identified with the $U$-invariants of $M$, so $M^U$ is finitely generated.

(Warning: in general, if the action of $U$ did not come from an action of $G$, the invariant subalgebra need not be finitely generated)

I can't comment on the second part about submodules. I've never seen a situation where one takes a quotient of the Hilbert series.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.