0
$\begingroup$

For $G$ a semisimple simply connected linear algebraic group, $R(G)$ its representation ring, there is a canonical embedding of algebras $$ R(G) \hookrightarrow \mathcal O(G)^G = \{\text{regular class functions on } G\} $$ obtained by mapping a representation to its character. Here, I understand a regular class function to mean a class function $ G \to \mathbb C$ that is regular in the sense of a morphism of varieties. It is a 'well-known fact' that if $G$ is reductive then the induced map given by complexification of the left-hand side $$ \mathbb C \otimes_{\mathbb Z} R(G) \to \mathcal O(G)^G$$ is an isomorphism.

Now any algebra homomorphism $R(G) \to \mathbb C$ can be identified with evaluation of a character $z \in R(G)$ at a semisimple element $a \in G$, where here I take semisimple to mean a diagonalisable element, i.e. up to conjugation lies in (some) maximal torus $T \subset G$. Why is this bijection true?

$\endgroup$
0

1 Answer 1

0
$\begingroup$

It is because for $f$ a class function on $G$ and for $x \in G$, $f(x)=f(x_s)$, where $x=x_sx_u$ is the semisimple-unipotent (Jordan-Chevalley multiplicative) decomposition. Thus really we identify maps $R(G) \to \mathbb C$ with evaluation at any element of $G$, but we may as well take its semisimple part (since there is a nice bijection of semisimple classes with elements of $T//W$).

$\endgroup$

You must log in to answer this question.

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