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I'm trying to prove as much as possible of classical results of finite groups representation theory without using the notion of character, especially orthogonal relations, since I want to work with representations over algebraic closures of finite fields.

I managed to prove that the regular representation is the sum of irreducible representations with multiplicities matching their dimensions using basically Schur's Lemma and Mashke's theorem. I wonder if I can prove that the number of representations of finite group $G$ over an algebraically closed field $k$ with $\mathrm{char}(k)\nmid |G|$ is equal to the number of conjugacy classes in G.

Can one proceed attempting to find the dimension of $Z(k[G])$ (center of group algebra)?

Matthew Willow
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1 Answers1

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Yes, computing the dimension of the center works. As you say, using Maschke's theorem and Schur's lemma we can get the Artin-Wedderburn decomposition

$$K[G] \cong \prod_i M_{d_i}(K)$$

where $d_i$ are the dimensions of the irreducibles, without using character theory. It follows that the number of irreducibles is $\dim Z(K[G])$. The key observation from here is that an element $z$ of the group algebra is central iff $gz = zg$ for all $g \in G$, hence iff

$$gzg^{-1} = z.$$

So $Z(K[G])$ is the subspace of fixed points of the conjugation action of $G$ on $K[G]$.

General fact: Let $G$ be a group acting on a set $X$. The subspace of fixed points of the action of $G$ on the free vector space $K[X]$ over $X$ has a basis given by sums $\sum_{x \in O} x$ where $O$ runs over the finite orbits of the action of $G$ on $X$.

This is a nice exercise. It follows that $Z(K[G])$ has a basis given by sums over conjugacy classes, hence $\dim Z(K[G])$ is the number of conjugacy classes, as desired.

Qiaochu Yuan
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