Let \(G\) be a finite group and work with finite-dimensional complex representations.
- Define a representation and an irreducible representation of \(G\).
- Prove that every finite-dimensional complex representation of \(G\) is a direct sum of irreducible representations.
- Construct a \(G\)-invariant Hermitian inner product.
Solution
1. A representation of \(G\) is a homomorphism\[ \rho:G\to\operatorname{GL}(V) \]for a finite-dimensional complex vector space \(V\). It is irreducible if its only \(G\)-stable subspaces are \(0\) and \(V\).
A subspace \(W\subseteq V\) is \(G\)-stable when \(\rho(g)W\subseteq W\) for every \(g\in G\).
2. Every representation admits a \(G\)-invariant positive-definite Hermitian inner product.
Start with any positive-definite Hermitian inner product \(\langle-,-\rangle_0\) on \(V\) and average it over \(G\): \[ \langle v,w\rangle_G = \frac1{|G|}\sum_{g\in G} \langle\rho(g)v,\rho(g)w\rangle_0. \] This remains Hermitian and positive definite. For \(h\in G\), \[ \langle\rho(h)v,\rho(h)w\rangle_G = \frac1{|G|}\sum_{g\in G} \langle\rho(gh)v,\rho(gh)w\rangle_0 = \langle v,w\rangle_G, \] because \(g\mapsto gh\) permutes \(G\).
3. If \(W\subseteq V\) is \(G\)-stable, then its orthogonal complement \(W^\perp\) is also \(G\)-stable.
For \(v\in W^\perp\), \(w\in W\), and \(g\in G\), invariance of the inner product gives \[ \langle\rho(g)v,w\rangle_G = \langle v,\rho(g^{-1})w\rangle_G=0, \] since \(\rho(g^{-1})w\in W\). Hence \(\rho(g)v\in W^\perp\).
4. Every finite-dimensional complex representation of \(G\) is completely reducible.
Induct on \(\dim V\). If \(V\) is irreducible, there is nothing to prove. Otherwise choose a nonzero proper invariant subspace \(W\). By 3, \[ V=W\oplus W^\perp \] as \(G\)-representations. Apply induction to the two lower-dimensional summands.
Step 4 is Maschke’s theorem over \(\CC\).