Unitary representation

Remarks

Clause 2 says that $\pi$ is continuous for the strong operator topology on $U(H)$. It is equivalent to weak continuity, the continuity of $g \mapsto \ip{\pi(g)u}{v}$ for all $u, v \in H$.

Example: $G = (\R, +)$ on $H = L^2(\R)$ by $(\pi(t)f)(x) = f(x - t)$. Translation preserves $\int \abs{f}^2$ and has inverse $\pi(-t)$, so each $\pi(t)$ is unitary, and $\pi(s + t) = \pi(s)\pi(t)$. For continuity, $\norm{\pi(t)f - f}_2 \to 0$ as $t \to 0$ holds for compactly supported continuous $f$ by uniform continuity, and then for all $f \in L^2(\R)$ since these are dense and each $\pi(t)$ has norm $1$.

Non-example: let $\ell^2(\R)$ be the Hilbert space of square-summable functions on the set $\R$, with orthonormal basis $(\delta_s)_{s \in \R}$, and let $\pi(t)\delta_s = \delta_{s + t}$. Each $\pi(t)$ is unitary and $\pi$ is a homomorphism, but clause 2 fails for the usual topology of $\R$. Indeed $\norm{\pi(t)\delta_0 - \delta_0} = \sqrt{2}$ for every $t \ne 0$, so $\pi(t)\delta_0$ does not tend to $\delta_0$ as $t \to 0$.

Two unitary representations $\pi_1$ on $H_1$ and $\pi_2$ on $H_2$ are unitarily equivalent if there is a unitary $T \colon H_1 \to H_2$ with $T\pi_1(g) = \pi_2(g)T$ for all $g \in G$. A unitary representation is irreducible if $H \ne 0$ and the only closed subspaces invariant under all $\pi(g)$ are $0$ and $H$.