Strongly Continuous Group Actions on the $C^{\ast}$-Algebra of Compact Operators on a Hilbert Space

Let $$\mathcal{H}$$ be a not-necessarily-separable Hilbert space. Let $$G$$ be a locally compact Hausdorff group. It is easy to see that if $$U: G \to \mathbb{U}(\mathcal{H})$$ is a norm-continuous homomorphism from $$G$$ to the group of unitary operators on $$\mathcal{H}$$, then we can define a strongly continuous action $$\alpha$$ of $$G$$ on the $$C^{\ast}$$-algebra $$\mathbb{K}(\mathcal{H})$$ of compact operators on $$\mathcal{H}$$ by $$\forall g \in G: \qquad \alpha_{g} \stackrel{\text{df}}{=} \left\{ \begin{matrix} \mathbb{K}(\mathcal{H}) & \to & \mathbb{K}(\mathcal{H}) \\ T & \mapsto & U(g) \circ T \circ U(g^{-1}) \end{matrix} \right\}.$$

Question. If $$\alpha$$ is a strongly continuous action of $$G$$ on $$\mathbb{K}(\mathcal{H})$$, does it necessarily arise from a norm-continuous homomorphism from $$G$$ to $$\mathbb{U}(\mathcal{H})$$ as in the manner described above?

I am pretty sure that the answer to this is known already, but I am frustrated by my inability to locate a reference.

Thank you so much for your assistance!

Clarification

To avoid any confusion, I wish to clarify that a group acts on a $$C^{\ast}$$-algebra by $$\ast$$-automorphisms.

• Obviously not, if the action of $G$ on $K(H)$ does not preserve the multiplication (i.e., is not by non-unital algebra automorphisms). Also it should commute with the $\ast$-involution. Are these missing hypotheses? The answer should be clear for $G=\mathbf{Z}$, to start with. – YCor Oct 25 '18 at 18:50
• I have deleted my comments as they are seemingly not helpful – Matthew Daws Oct 25 '18 at 19:51
• I am coming late to this, but as someone who is merely a practising functional analyst, it seems clear to me from the context of the question that the OP is asking about homomorphisms $\alpha: G \to {\rm Aut}{\mathbb K}({\mathcal H})$ where the automorphisms are in the category of ${\rm C}^*$-algebras -- which makes them isometric, $*$-preserving, non-degenerate, etc etc. – Yemon Choi Oct 25 '18 at 19:58
• @YCor: that is correct, $U(H)$ is the automorphisms group of $K(H)$. ($K(H)$ has only one irrep up to unitary equivalence.) – Nik Weaver Oct 26 '18 at 0:45
• @NikWeaver: Shouldn’t it be $\mathbb{U}(\mathcal{H})$ modulo the circle group instead? – Transcendental Oct 26 '18 at 0:51

The answer is no. Let $$G$$ be the Cartesian product of a sequence of copies of the unit circle, identified with the functions in $$l^\infty$$ whose modulus is constantly $$1$$. For $$f \in G$$ let $$U_f \in B(l^2)$$ be multiplication by $$f$$. Let $$G$$ act on $$K(l^2)$$ by conjugation by these unitaries.
The action is strongly continuous because convergence in $$G$$ is pointwise, so if $$f_\alpha \to f$$ and $$T \in K(l^2)$$ then $$U_{f_\alpha}TU_{f_\alpha}^* \to U_fTU_f^*$$ in norm, since $$T$$ approximately lives on finitely many coordinates. However, if we let $$f_n$$ be the function which is $$-1$$ in the $$n$$th entry and $$1$$ everywhere else, then $$f_n \to 1$$ in $$G$$ but $$U_{f_n}$$ does not go to $$I$$ in norm.
• Thanks for your counterexample, Nik! Actually, I just found out that my question has a negative answer by way of what’s called the “Mackey obstruction”. It’s an element of ${H^{2}}(G,\mathbf{T})$ associated to every strongly continuous action $\alpha$ of $G$ on $\mathbb{K}(\mathcal{H})$ with the property that if it isn’t trivial, then $\alpha$ can’t be implemented by even an algebraic homomorphism from $G$ to $\mathbb{U}(\mathcal{H})$, much less a norm-continuous one. – Transcendental Oct 26 '18 at 0:44
• I wonder, if I may, I could make a comment that as I see it, there are two separate things going on here. One is the issue of projective representations. The other is an issue of topology. If we give $U(H)$ the SOT topology, then in Nik's example, we do have that $U_{f_n}\rightarrow 1$. I believe that if you equip $U(H)$ with the SOT then only the cohomological obstruction is left. I don't know how to characterise the norm-continuous case (perhaps look at actions on all of $B(H)$?) – Matthew Daws Oct 26 '18 at 7:59
• @NikWeaver: This may sound like a silly question, Nik. Do you happen to know if every strongly continuous action of $G$ on $\mathbb{K}(\mathcal{H})$ extends to a strongly continuous action of $G$ on $\mathbb{B}(\mathcal{H})$? I’m assuming $G$ to be an arbitrary locally compact Hausdorff group. – Transcendental Nov 1 '18 at 0:16