representations of compact groups are equivalent to unitary representations

representationrepresentationsequivalent

Theorem - Let $G$ be a compact topological group. If $(\pi ,V)$ is a finite-dimensional representation (http://planetmath.org/TopologicalGroupRepresentation) of $G$ in a normed vector space $V$, then $\pi$ is equivalent (http://planetmath.org/TopologicalGroupRepresentation) to a unitary representation.

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Proof: Let $⟨\cdot ,\cdot ⟩$ denote an inner product in $V$. Define a new inner product in the vector space $V$ by

${⟨v,w⟩}_1:={\displaystyle {\int }_G}⟨\pi (s)v,\pi (s)w⟩𝑑\mu (s)$

where $\mu$ is a Haar measure in $G$. It is easy to see that ${⟨\cdot ,\cdot ⟩}_1$ defines indeed an inner product, noting that $⟨\pi (\cdot )v,\pi (\cdot )w⟩$ is a continuous function in $G$.

Now we claim that, for every $s\in G$, $\pi (s)$ is a unitary operator for this new inner product. This is true since

${⟨\pi (t)v,\pi (t)w⟩}_1$$=$${\displaystyle {\int }_G}⟨\pi (s)\pi (t)v,\pi (s)\pi (t)w⟩𝑑\mu (s)$

$=$${\displaystyle {\int }_G}⟨\pi (st)v,\pi (st)w⟩𝑑\mu (s)$

$=$${\displaystyle {\int }_G}⟨\pi (s)v,\pi (s)w⟩𝑑\mu (s)$

$=$${⟨v,w⟩}_1$

Denote by $V^{\prime }$ the space $V$ endowed the inner product ${⟨\cdot ,\cdot ⟩}_1$. As we have seen, $(\pi ,V^{\prime })$ is a unitary representation of $G$ in $V^{\prime }$. Of course, $(\pi ,V^{\prime })$ and $(\pi ,V)$ are equivalent representations, since

$\pi ={\mathrm{id}}^{-1}\pi \mathrm{id}$

where $\mathrm{id}:V⟶V^{\prime }$ is the identity mapping. Thus, $(\pi ,V)$ is equivalent to a unitary representation. $\mathrm{\square }$