How is a $G$-invariant inner product constructed from an arbitrary inner product using the group average?

Conditions

• $G$ is a finite group.
• $\rho: G \to GL(V)$ is a representation.
• $\langle \cdot, \cdot \rangle$ is an arbitrary initial inner product on $V$.
• $|G|$ is the order (cardinality) of the finite group $G$.

Reasoning, step by step

1. Start with any inner product $\langle \cdot, \cdot \rangle$ on $V$.
2. Define a new form $(v, w)$ by summing $\langle \rho(g)v, \rho(g)w \rangle$ over all $g \in G$.
3. Normalize the sum by dividing by $|G|$.
4. Verify that this new form is $G$-invariant by showing $(\rho(h)v, \rho(h)w) = (v, w)$ using the reindexing trick $g \mapsto gh$.
5. Confirm that the new form satisfies inner product axioms (bilinearity is immediate, positivity is an exercise).

Example

The video defines the averaged inner product as $(v,w):=\frac{1}{|G|}\sum_{g\in G}\langle \rho(g)v,\rho(g)w\rangle$ and proves its invariance by reindexing the sum with $gh$.

Common misconceptions

• Assuming the average is over a continuous parameter; it is a finite sum over the discrete group elements.
• Forgetting the normalization factor $\frac{1}{|G|}$, which is necessary for the form to behave like a standard inner product average.

Watch the explanation

Explore next

• Why can we replace $\sum_{g\in G} f(g)$ by $\sum_{g\in G} f(gh)$?
• What remains to be checked to prove that the averaged form is truly an inner product?