§ Burnside lemma by representation theory.


Recall that burnside asks us to show that given a group GG acting on a set SS, we have that the average of the local fixed points 1/G(gGFix(g))1/|G|(\sum_{g \in G} |\texttt{Fix}(g)|) is equal to the number of orbits (global fixed points) of SS, S/G|S/G|.
Let us write elements of gg as acting on the vector space VSV_S, which is a complex vector space spanned by basis vector {vs:sS}\{ v_s : s \in S \}. Let this representation of GG be called ρ\rho.
Now see that the right hand side is equal to
1/G(gGTr(ρ(g)))=1/G(gGχρ(g))χρχ1 \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:
f,fgGf(g)f(g) \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|S/G| is equal to the multiplicity of the trivial representation 11 in the current representation ρ\rho, given by the inner product of their characters χ1χρ\chi_1 \cdot \chi_\rho.
let sinSs* in S whose orbit we wish to inspect. Build the subspace spanned by the vector v[s]gGρ(g)v[s]v[s*] \equiv \sum_{g \in G} \rho(g) v[s]. This is invariant under GG and is 1-dimensional. Hence, it corresponds to a 1D subrepresentation for all the elements in the orbit of ss*. (TODO: why is it the trivial representation?)