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Let $G$ be a compact group and let $\lambda: G \rightarrow \mathcal{U}(L^2(G))$ be the left regular representation, i.e. $\lambda_sf(t)=f(s^{-1}t)$. Why is the induced group action $\overline{\lambda}:G \rightarrow \mathrm{Aut}(\mathcal{K}(L^2(G)))$, $\overline{\lambda}_s(T)= \lambda_s T\lambda_{s^{-1}}$ strongly continuous?

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Let $g_j\to g$ a convergent net in $G$ and let $T\in\mathcal K$. We need to show that $\lambda_{g_j}T\lambda_{g_j^{-1}}$ converges to $\lambda_g T\lambda_{g^{-1}}$ in the strong topology. Now $\lambda_{g_j}$ converges to $\lambda_g$ strongly, i.e., $\lambda_{g_j}f$ tends to $\lambda_gf$ in the $L^2$ norm. This is seen for continuous $f$ by uniform convergence and for general $f$ by $L^2$-approximation with continuous functions. Now write $$ \lambda_{g_j}T\lambda_{g_j^{-1}}f-\lambda_{g}T\lambda_{g^{-1}}f =(\lambda_{g_j}T\lambda_{g_j^{-1}}f-\lambda_{g_j}T\lambda_{g^{-1}}f) +(\lambda_{g_j}T\lambda_{g^{-1}}f-\lambda_{g}T\lambda_{g^{-1}}f) $$ and use the triangle inequality as well as the fact that $\lambda_g$ preserves norms.

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  • $\begingroup$ Am I right in thinking that compactness of $T$ is not used? $\endgroup$
    – Yemon Choi
    Commented Mar 17, 2017 at 16:13
  • $\begingroup$ It would appear so. $\endgroup$
    – user1688
    Commented Mar 17, 2017 at 17:00

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