Covariant representations and crossed products of von Neumann algebras

Question

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.

Answer 1

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 $M \to B(L^2(X,\mu))$.