6
$\begingroup$

Let $G$ be a compact connected Lie group and $T\subset G$ a maximal torus. Let $V$ be a representation of $G$ and $U=\{v\in V: tv=v\textrm{ for all }t\in T\}$. For any $g\in N(T)$ we have for all $t\in T$ and $v\in U$ that $g^{-1}tgv=v \Rightarrow t(gv)=gv$. This shows that for all $v\in U$ we have $gv\in U$ as well. From this we can define a representation of the Weyl group $W$ on $U$. I have the following two questions:

  1. Does this $W$-module structure on $U$ depend on the choice of the maximal torus $T$?

  2. Assume that $V$ is irreducible and $U$ nontrivial. Is there a way to understand when $U$ is irreducible as $W$-module? Is it always the case?

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ For (1). We consider $U=U(G,T)$ and $W=W(G,T)$. If $T'\subset G$ is another maximal torus, then there exists $g\in G$ such that $T'=gTg^{-1}$. The isomorphism $g\colon (G,T)\to (G,T')$ induces compatible isomorphisms $U(G,T)\overset{\sim}{\to} U(G,T')$ and $W(G,T)\overset{\sim}{\to} W(G,T')$. Therefore, the answer to Question (1) is No. $\endgroup$
    – Mikhail Borovoi
    Commented Aug 7, 2020 at 9:54

1 Answer 1

Reset to default
7
$\begingroup$

This paper of Humphreys addresses your second question (the first is answered in the comments - the $W$-module structure is independent of the choice of torus): https://people.math.umass.edu/~jeh/pub/zero.pdf

Here is a quote from the paper (section 1.4):

Indeed, it is usually unclear how to determine directly whether or not the W-module $L_\lambda(0)$ [i.e. $U$] is simple, even if its dimension is compatible with simplicity.

The $W$-module $U$ is not always irreducible. Indeed in type $A_2$ (i.e. $G=SU(3)$), there is a formula for the dimension of $U$ in Section 2.2, which shows that the dimension may be unbounded.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .