Let $G$ be a finite group and let $\chi_1, \dotsc, \chi_t$ be a complete collection of irreducible complex characters of $G$ with corresponding representations $V_1, \dotsc, V_t$ (i.e. every irreducible complex representation of $G$ is isomorphic to precisely one of the $V_i$).
I am searching for some intuition behind the following statement:
The characters $\chi_1, \dotsc, \chi_t$ span $C(G)$, the space of complex-valued class functions on $G$.
(One could also take all characters of $G$ instead of only the irreducible ones without changing the question.)
There seem to be two standard proofs for this statement, but none of them helped me get a better understanding of the problem so far.(I have included both of them below.)
Is also suspect that my lack of intuition may come from a lack of understanding of what class functions actually are.While I tend to think about them as central elements of the group algebra $\mathbb{C}[G]$, I’m not sure how well this works when it comes to characters.
Any help would be appreciated.
The two proofs
- One can identify $C(G)$ with the center $\operatorname{Z}(\mathbb{C}[G])$ as vector spaces. (For some people these are already the same.)From $\mathbb{C}[G] \cong \prod_{i=1}^t M_{n_i}(\mathbb{C})$ (with $n_i = \dim V_i$) it then follows that$$ \operatorname{Z}(\mathbb{C}[G]) \cong \operatorname{Z}\left( \prod_{i=1}^t M_{n_i}(\mathbb{C}) \right) = \prod_{i=1}^t \operatorname{Z}( M_{n_i}(\mathbb{C}) ) \cong \prod_{i=1}^t \mathbb{C},$$so $\dim C(G) = t$.Since $\chi_1, \dotsc, \chi_t$ are linearly independent (because they are orthonormal) they already span $C(G)$.
The gist of this proof is that dimensions match up, but it doesn’t really seem to explain why they should do so in the first place.
- Since $\chi_1, \dotsc, \chi_t$ are orthonormal it sufficies to show that every $\alpha \in C(G)$ which is orthogonal to every $\chi_i$ vanishes.This is equivalent to $\beta \in C(G)$ with $\beta(g) = \alpha(g^{-1})$ vanishing, which in turn is equivalent to the central element $\hat{\beta} := \sum_{g \in G} \beta(g) g \in \operatorname{Z}(\mathbb{C}[G])$ being zero.This holds iff the map $\mathbb{C}[G] \to \mathbb{C}[G]$, $x \mapsto \hat{\beta} x$ vanishes.This map is $G$-equivariant because $\hat{\beta}$ is central, and $\mathbb{C}[G]$ decomposes into a direct sum of irreducible subrepresentations, each of which is invariant under $\hat{\beta}$;it therefore sufficies to show that for every irreducible representation $V_i$ the $G$-equivariant map$$ f_i \colon V_i \to V_i, \quad v \mapsto \hat{\beta}{.}v = \sum_{g \in G} \beta(g) g{.}v$$vanishes.By Schur’s Lemma the map $f_i$ is given by multiplication with $\lambda_i \in \mathbb{C}$, and since$$ n_i \lambda_i = \operatorname{tr} f_i = |G| \langle \alpha, \chi_i \rangle = 0,$$it follows that $\lambda_i = 0$, and therefore that $f_i = 0$.
This proof sems to suggest that one should think about class functions as acting on representations, and that the inner product $\langle \alpha, \chi_V \rangle$ tells us something about how $\alpha$ (or more precisely $\beta$ with $\beta(g) = \alpha(g^{-1})$ ) acts on a representation $V$ (or maybe how $\alpha$ acts on $V^*$).But so far I have not been able to make this idea more precise.