Let $G$ be a compact group and $u: G \to B(H)$ be a strongly continuous unitary representation on the Hilbert space $H$. Then is $u: G \to B(H)$ strictly continuous?

That is, give $B(H)$ the topology induced by the $*$-isomorphism $M(B_0(H))\cong B(H)$. Explicitely, a net $(x_i)$ in $B(H)$ converges strictly to $x$ iff $\|x_i y -x y\|\to 0$ and $\|yx_i \to yx\| \to 0$ for all compact operators $y \in B_0(H)$. On bounded subsets this agrees with the $*$-strong topology.


1 Answer 1


As you note, on bounded sets, the strict topology and the strong-$\ast$ topology agree on bounded sets. As the set of unitary operators is bounded, we can just work with the strong-$\ast$ topology. If $(u_i)$ is a net of unitary operators converging strongly to $u$ a unitary, then for $\xi\in H$,

$$ \| u_i^\ast(\xi) - u^\ast(\xi)\|^2 = \|u_i^\ast(\xi)\|^2 - (u_i^\ast(\xi)|u^\ast(\xi)) - (u^\ast(\xi)|u_i^\ast(\xi)) + \|u^\ast(\xi)\|^2 $$

Here I write $(\cdot|\cdot)$ for the inner product. As $u_i,u$ are unitary, this is equal to

$$ 2\|\xi\|^2 - (\xi|u_iu^\ast(\xi)) - (u_iu^\ast(\xi)|\xi). $$

As $u_i(\eta)\rightarrow u(\eta)$ for any $\eta$, this converges to

$$ 2\|\xi\|^2 - (\xi|u u^\ast(\xi)) - (u u^\ast(\xi)|\xi) = 0. $$

Thus $u_i\rightarrow u$ strong-$\ast$. This is just a proof that the strong and strong-$\ast$ topologies agree on the set of unitaries; this is surely in standard textbooks, if you look hard.


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