# Covariant representations and crossed products of von Neumann algebras

Let $$(M,G,\alpha)$$ be a $$W^\ast$$-dynamical system with $$G$$ locally compact abelian (I am mostly interested in the case $$G=\mathbb{R})$$. A covariant representation of $$(M,G,\alpha)$$ is a pair $$(\pi,u)$$ consisting of a normal representation $$\pi$$ of $$M$$ on a Hilbert space $$H$$ and a strongly continuous unitary representation $$u$$ of $$G$$ on $$H$$ such that $$\pi(\alpha_g(x))=u_g\pi(x)u_g^\ast$$ for all $$x\in M$$ and $$g\in G$$.

In several places (for example Kostecki - $$W^*$$-algebras and noncommutative integration) I have seen the claim that covariant representations of $$(M,G,\alpha)$$ are in one-to-one correspondence with normal representations of the crossed product $$M\rtimes_\alpha G$$. However, I have been unable to find a proof.

Since $$G$$ is abelian, the full and reduced $$C^\ast$$-algebraic crossed product coincide. Hence every covariant representation induces a representation $$\pi\rtimes u$$ of $$M^c\rtimes_{r,\alpha}G$$, where $$M^c=\{x\in M\mid \text{g\mapsto \alpha_g(x) continuous}\}$$. Thus the question can be rephrased as "Is it true that $$\pi\rtimes u$$ is $$\sigma$$-weakly continuous if we view $$M^c\rtimes_{r,\alpha}G$$ as a subalgebra of $$B(G;L^2(M))$$?"

Even more than a proof I would welcome a reference for this fact (if true). It does not seem to be contained in Takesaki's books or the original articles by Haagerup and Takesaki on the subject.

• I'm not familiar with the operator-algebra lingo. Does "a normal representation" just mean "a representation in the normal sense", i.e., an algebra homomorphism $M \to \operatorname{End}_\text{Hilbert}(H)$, or is it such a representation whose image lands in the normal operators? Commented May 4, 2021 at 13:09
• @LSpice Neither of these two. A normal representation is one that is continuous with respect to the $\sigma$-weak operator topologies. The different meanings of "normal" in this field can be somehwat unfortunate. Commented May 4, 2021 at 13:11
• @MaoWao I am a little late to the party, but the statement you are after is true for compact (quantum) groups $G$. Commented Aug 26 at 20:48

No, such a one-to-one correspondence does not hold. For instance, if $$G$$ is a countable infinite group and $$G \curvearrowright (X,\mu)$$ is an essentially free, ergodic, probability measure preserving action, the crossed product $$M = L^\infty(X) \rtimes G$$ is a II$$_1$$ factor. At the same time, the representation $$\pi : L^\infty(X,\mu) \to B(L^2(X,\mu))$$ as multiplication operators and the unitary representation $$(u_g \xi)(x) = \xi(g^{-1} \cdot x)$$ form a covariant pair. By ergodicity, the von Neumann algebra generated by $$\pi(L^\infty(X,\mu))$$ and the unitaries $$(u_g)_{g \in G}$$ equals $$B(L^2(X,\mu))$$. So, $$(\pi,u)$$ does not give a normal $$*$$-homomorphism $$M \to B(L^2(X,\mu))$$.