2
$\begingroup$

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.

Improve this question
$\endgroup$
13
  • 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
    Commented Oct 25, 2018 at 18:50
  • $\begingroup$ I have deleted my comments as they are seemingly not helpful $\endgroup$
    – Matthew Daws
    Commented Oct 25, 2018 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
    Commented Oct 25, 2018 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
    Commented Oct 26, 2018 at 0:45
  • 1
    $\begingroup$ @NikWeaver: Shouldn’t it be $ \mathbb{U}(\mathcal{H}) $ modulo the circle group instead? $\endgroup$
    – Transcendental
    Commented Oct 26, 2018 at 0:51

1 Answer 1

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • 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. $\endgroup$
    – Transcendental
    Commented Oct 26, 2018 at 0:44
  • 1
    $\begingroup$ Right, I remember that now. You always get a projective unitary representation though, don't you? $\endgroup$
    – Nik Weaver
    Commented Oct 26, 2018 at 3:03
  • $\begingroup$ Yes, we always get a projective representation. $\endgroup$
    – Transcendental
    Commented Oct 26, 2018 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 $B(H)$?) $\endgroup$
    – Matthew Daws
    Commented Oct 26, 2018 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. $\endgroup$
    – Transcendental
    Commented Nov 1, 2018 at 0:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .