# For a separable v-N algebra M, how to see $M \rtimes \mathbb{R}$ as a subalgebra of $M \otimes B(L^2)$?

For a v-N algebra $M$ acting as bounded operators on a separable Hilbert space $H$, how to see $M \rtimes \mathbb{R}$ as a subalgebra of $M \otimes B(L^2(\mathbb{R})$?

Why I am confused is because $x \in M$ is already naturally sitting inside $M \otimes B(L^2(\mathbb{R}))$ as $x \otimes id$, where as it is sitting inside $M \rtimes \mathbb{R}$ as a different object.

In fact action of $x$ on $H \otimes L^2(\mathbb{R})$ is given by the following equation:

$xf(s) = \alpha_{-s}(x) f(s)$ for all $s \in \mathbb{R}, f \in H \otimes L^2(\mathbb{R})$

where $\alpha$ is the action of $\mathbb{R}$ on $M$.

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Is this not in Kadison & Ringrose volume 2, or Takesaki? (I may have misremembered) – Yemon Choi May 12 '11 at 18:53
I think Yemon is right to suggest Takesaki. If I recall there are some delicate points regarding actions of $\mathbb{R}$. If the group were discrete, then it may be instructive for you to look at how things look in the case of, e.g., an action of the integers mod 2. In this case the algebra $M$ acts "diagonally" twisted in each diagonal entry by a different group element, and the group elements act essentially as permutation matrices. I'm sure my latter suggestion is too simple for what you want, so good luck with $\mathbb{R}$! – Jon Bannon May 12 '11 at 21:42
Thanks. Looks like it is about $(M \rtimes \mathbb{R}) \rtimes \mathbb{R})$ being isomorphic to $M \otimes B(L^2(\mathbb{R}))$. Started reading now. – Madhushree May 13 '11 at 5:59
I think the precise statement is Takesaki's duality theorem. If $\mathbb{R}$ acts on $M,$ then there is a dual action'' of $\mathbb{R}$ (or more precisely $\widehat{\mathbb{R}}$) and Takesaki's duality theorem shows that $(M\rtimes \mathbb{R})\rtimes \mathbb{R}\cong M\overline{\otimes} B(L^{2}(\mathbb{R}).$ This is done in Takesaki II Theorem $X.2.3$ (iii) (and actually the more general case of crossed products by locally compact abelian groups is considered). If I recall, the proof is just playing around a lot with Fourier transforms. – Benjamin Hayes Jun 13 '11 at 6:11