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](https://www.fuw.edu.pl/%7Ekostecki/wstarint.pdf)) 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.



Covariant representations and crossed products of von Neumann algebras
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 $M \to B(L^2(X,\mu))$.


