Complete reducibility of finite-dimensional compact-group representations

Base case. If $\dim V=0$, then $V=\{0\}$ by [F5], and $V$ is the empty direct sum of irreducible subrepresentations, which [F7] allows; hence the zero representation is completely reducible.

Induction step setup. Fix a natural number $n\ge 1$ and assume the induction hypothesis: every continuous finite-dimensional complex representation of $K$ on a complex vector space of dimension $m<n$ is completely reducible. Let $\rho :K\to \mathrm{GL}(V)$ be a continuous finite-dimensional complex representation with $\dim V=n$; then $V\mathrm{\ne }\{0\}$ by [F5].

By [F11], fix a normalized Haar probability measure $\mu$ on $K$. By [F5] fix an ordered basis $(b_1,\dots ,b_n)$ of $V$ and let $h_0$ be the Hermitian inner product in these coordinates, $h_0({\sum }_i{a}_i{b}_i,{\sum }_i{c}_i{b}_i):={\sum }_i{a}_i\overline{c_i}$. By [F1] the averaged form $h$ is a positive-definite $K$-invariant Hermitian form on $V$, so $h$ is an inner product on $V$ and every $\rho (g)$ is a unitary operator for $h$. Since $V$ has the ordered basis $(b_1,\dots ,b_n)$ of finite length, [F8] makes $(V,\mathrm{\parallel }\cdot {\mathrm{\parallel }}_h)$ a Banach space for the norm induced by $h$, so $(V,h)$ is a complex Hilbert space.

The representation $\rho$ is strongly continuous for the norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_h$: for fixed $v\in V$ and $g_0\in K$, the operator norm inequality of [F9] gives $\mathrm{\parallel }\rho (g)v-\rho (g_0)v{\mathrm{\parallel }}_h\le \mathrm{\parallel }\rho (g)-\rho (g_0){\mathrm{\parallel }}_h\mathrm{\parallel }v{\mathrm{\parallel }}_h$, and $g\mapsto \rho (g)$ is continuous at $g_0$ for the operator norm because it is continuous for the topology of some norm on the finite-dimensional complex space $\mathrm{End}(V)$ and all such norms are equivalent by [F9]. Hence $g\mapsto \rho (g)v$ is continuous, and $\rho :K\to U(V,h)$ is a strongly continuous unitary representation of $K$ on the complex Hilbert space $(V,h)$.

Irreducible case. If $\rho$ is irreducible, then by [F7] its only subrepresentations are $0$ and $V$; since $V\mathrm{\ne }\{0\}$ by step 1.2, the space $V$ is itself an irreducible subrepresentation and $V=V$ is a direct sum with the single summand $V$, so $\rho$ is completely reducible.

Non-irreducible case. If $\rho$ is not irreducible, then, since $V\mathrm{\ne }\{0\}$, [F7] provides a subrepresentation $M\subseteq V$ with $M\mathrm{\ne }\{0\}$ and $M\mathrm{\ne }V$.

In the situation of step 2.3, $M$ is a finite-dimensional subspace of $V$, hence closed in $(V,\mathrm{\parallel }\cdot {\mathrm{\parallel }}_h)$ by [F4], and it is invariant by definition; applying the complement lemma [F2] to the strongly continuous unitary representation $\rho$ on the Hilbert space $(V,h)$ from step 2.1, the orthogonal complement $M^{\perp }$ is a closed invariant subspace. By the orthogonal decomposition [F3], $V=M\oplus M^{\perp }$; by the dimension formula [F10] and $\dim M\ge 1$, one has $\dim M<n$ by [F5], and $\dim M^{\perp }=\dim V-\dim M<n$.

Apply the induction hypothesis of step 1.2 to the restrictions ${\rho }_M(g):=\rho (g){\mathrm{\mid }}_M$ and ${\rho }_{M^{\perp }}(g):=\rho (g){\mathrm{\mid }}_{M^{\perp }}$. These are continuous finite-dimensional complex representations: invariance makes $\rho (g){\mathrm{\mid }}_M$ a linear self-map of $M$ and injectivity of $\rho (g)$ makes it invertible, the homomorphism property is inherited, and continuity follows from $\mathrm{\parallel }\rho (g){\mathrm{\mid }}_M-\rho (g_0){\mathrm{\mid }}_M\mathrm{\parallel }\le \mathrm{\parallel }\rho (g)-\rho (g_0)\mathrm{\parallel }$, with the same argument for $M^{\perp }$. Since $\dim M<n$ and $\dim M^{\perp }<n$ by step 3.1, the induction hypothesis gives irreducible subrepresentations with $M=M_1\oplus \cdots \oplus M_r$ and $M^{\perp }=M_{r+1}\oplus \cdots \oplus M_s$; these $M_i$ are also irreducible subrepresentations of $\rho$, and concatenating with $V=M\oplus M^{\perp }$ gives $V=M_1\oplus \cdots \oplus M_s$, so $\rho$ is completely reducible in this case as well.

Discharge. Every continuous finite-dimensional complex representation of $K$ of dimension $n\ge 1$ is completely reducible, by the two cases of steps 2.2 and 4.1 together with the induction hypothesis of step 1.2, and the case $n=0$ is step 1.1; strong induction [F6] therefore proves that every continuous finite-dimensional complex representation of $K$ is completely reducible.