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!


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

  • 1
    $\begingroup$ 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. $\endgroup$
    – YCor
    Oct 25 '18 at 18:50
  • $\begingroup$ I have deleted my comments as they are seemingly not helpful $\endgroup$ Oct 25 '18 at 19:51
  • $\begingroup$ 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. $\endgroup$
    – Yemon Choi
    Oct 25 '18 at 19:58
  • 1
    $\begingroup$ @YCor: that is correct, $U(H)$ is the automorphisms group of $K(H)$. ($K(H)$ has only one irrep up to unitary equivalence.) $\endgroup$
    – Nik Weaver
    Oct 26 '18 at 0:45
  • 1
    $\begingroup$ @NikWeaver: Shouldn’t it be $ \mathbb{U}(\mathcal{H}) $ modulo the circle group instead? $\endgroup$ 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.

  • 2
    $\begingroup$ 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. $\endgroup$ Oct 26 '18 at 0:44
  • 1
    $\begingroup$ Right, I remember that now. You always get a projective unitary representation though, don't you? $\endgroup$
    – Nik Weaver
    Oct 26 '18 at 3:03
  • $\begingroup$ Yes, we always get a projective representation. $\endgroup$ Oct 26 '18 at 6:22
  • 1
    $\begingroup$ 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 $B(H)$?) $\endgroup$ Oct 26 '18 at 7:59
  • $\begingroup$ @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. $\endgroup$ Nov 1 '18 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.