Weyl's Unitarity Trick

# Weyl's Unitarity Trick

Theorem 1 (Weyl's Unitarity Trick): Let $G$ be a finite group and let $(V, \rho)$ be a representation of $G$. Then there exists an inner product on $V$ for which $(V, \rho)$ is unitary. |

**Proof:**Let $(V, \rho)$ be a representation of $G$. Let $\langle -, - \rangle$ be an inner product on $V$. Define a new inner product $(-, -)$ on $V$ defined for all $v, w \in V$ by:

\begin{align} \quad (v, w) = \sum_{g \in G} \langle [\rho(g)](v), [\rho(g)](w) \rangle \end{align}

- Since $G$ is a finite group, the above sum has no convergence issues. Moreover, $(-, -)$ is indeed an inner product since $\langle -, - \rangle$ is an inner product.

- Lastly, $(V, \rho)$ with respect to the inner product $(-, -)$ is unitary since for all $h \in G$ and for all $v, w \in V$ we have that:

\begin{align} \quad ([\rho(h)](v), [\rho(h)](w)) = \sum_{g \in G} \langle [\rho(g)]([\rho(h)](v)), [\rho(g)]([\rho(h)](w)) \rangle = \sum_{g \in G} \langle [\rho(gh)](v), [\rho(gh)](w) \rangle \overset{(*)} = \sum_{g \in G} \langle [\rho(g)](v), [\rho(g)](w) \rangle = (v, w) \end{align}

- (Where the equality at $(*)$ comes from the fact that for each fixed $h \in G$, the map $g \to gh$ is an automorphism of $G$, and so the sums are just rearrangements of one another). $\blacksquare$