Theorem (Weyl's Unitary Trick): Let $G$ be a compact Lie group and $\rho: G \to GL(V)$ a continuous finite-dimensional representation. Then there exists a $G$-invariant inner product on $V$.

Proof:

Step 1: Choose an arbitrary inner product

Start with any inner product $\langle \cdot, \cdot\rangle_0$ on $V$. This exists since $V$ is finite-dimensional over $\mathbb{R}$ or $\mathbb{C}$.

Step 2: Average over the group

Define a new inner product by averaging $\langle \cdot, \cdot\rangle_0$ over $G$ using Haar measure: $\langle v, w\rangle_G = \int_G \langle \rho(g)v, \rho(g)w\rangle_0 \, d\mu(g)$

We must verify this is well-defined and has the required properties.

Step 3: Verify positive definiteness

For $v \neq 0$: $\langle v, v\rangle_G = \int_G \langle \rho(g)v, \rho(g)v\rangle_0 \, d\mu(g) > 0$

The integrand is non-negative everywhere and positive on a set of positive measure (where $\rho(g)v \neq 0$), so the integral is positive.

Step 4: Verify $G$-invariance

For any $h \in G$: $\langle \rho(h)v, \rho(h)w\rangle_G = \int_G \langle \rho(g)\rho(h)v, \rho(g)\rho(h)w\rangle_0 \, d\mu(g)$

Substituting $g' = gh$ and using left-invariance of Haar measure $d\mu(gh) = d\mu(g)$: $= \int_G \langle \rho(g')v, \rho(g')w\rangle_0 \, d\mu(g') = \langle v, w\rangle_G$

Therefore $\langle \cdot, \cdot\rangle_G$ is $G$-invariant.

Step 5: Conclude unitarity

With respect to $\langle \cdot, \cdot\rangle_G$, every $\rho(g)$ preserves the inner product: $\langle \rho(g)v, \rho(g)w\rangle_G = \langle v, w\rangle_G$

Thus $\rho(g)$ is a unitary (or orthogonal) operator for all $g \in G$. □

Corollary (Complete Reducibility): Every representation of compact $G$ is completely reducible.

Proof: Use induction on dimension. If $V$ is irreducible, done. Otherwise, there exists proper invariant subspace $W \subset V$. The invariant inner product gives $V = W \oplus W^\perp$ with $W^\perp$ also invariant. By induction, both $W$ and $W^\perp$ decompose into irreducibles, giving a decomposition of $V$. □