## § Burnside lemma by representation theory.

Recall that burnside asks us to show that given a group $G$ acting on a set $S$, we have that the average of the local fixed points $1/|G|(\sum_{g \in G} |\texttt{Fix}(g)|)$ is equal to the number of orbits (global fixed points) of $S$, $|S/G|$. Let us write elements of $g$ as acting on the vector space $V_S$, which is a complex vector space spanned by basis vector $\{ v_s : s \in S \}$. Let this representation of $G$ be called $\rho$. Now see that the right hand side is equal to
\begin{aligned} &1/|G| (\sum_g \in G Tr(\rho(g))) \\ &= 1/|G| (\sum_g \in G \chi_\rho(g) ) \\ &\chi \rho \cdot \chi_1 \end{aligned}
Where we have:
• $\chi_1$ is the charcter of the trivial representation $g \mapsto 1$
• The inner product $\langle \cdot , \cdot \rangle$ is the $G$-average innerproduct over $G$-functions $G \rightarrow \mathbb C$:
$\langle f , f' \rangle \equiv \sum_{g \in G} f(g) \overline{f'(g)}$
So, we need to show that the number of orbits $|S/G|$ is equal to the multiplicity of the trivial representation $1$ in the current representation $\rho$, given by the inner product of their characters $\chi_1 \cdot \chi_\rho$. let $s* in S$ whose orbit we wish to inspect. Build the subspace spanned by the vector $v[s*] \equiv \sum_{g \in G} \rho(g) v[s]$. This is invariant under $G$ and is 1-dimensional. Hence, it corresponds to a 1D subrepresentation for all the elements in the orbit of $s*$. (TODO: why is it the trivial representation?)