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I was reading J. Renault's paper "The Fourier Algebra of a Measured Groupoid" and I am confused about his approach to the predual of a von Neumann Algebra.

Let $M:= VN(\mathcal{G})$ be the von Neumann algebra of a measured groupoid; I think the definition is irrelevant, we just need to know that $M$ is a von Neumann Algebra.

For example in Theorem 2.3, when he wants to refer to an element in the predual $u \in M_*$, he views it as a normal linear map $u:M \rightarrow M_n(\mathbb C)$, for $n \in \mathbb N$. I am not familiar with this caracterization of the predual of a von Neumann algebra and I am wondering if someone has a reference or can tell me more details about it.

An hypothesis I put is that $u:M \rightarrow M_n(\mathbb C)$ is just a diagonal operator made of a normal functional $u_0:M \rightarrow \mathbb C$, but that doesn't seem the case (I might be wrong).

So is there a caracterization of the predual of a von Neumann algebra that I am not aware of? Or does this have to specifically with the precise definition of the von Neumann algebra of a groupoid, or am I just not getting it and is just a diagonal operator?

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    $\begingroup$ This seems a bit weird. The predual is just the space of normal linear functionals. Not sure where the matrix algebra comes from. $\endgroup$
    – David Gao
    Mar 31 at 19:52
  • $\begingroup$ Unless it’s not the predual of $M$ but of $M_n(M)$? $\endgroup$
    – David Gao
    Mar 31 at 21:11
  • $\begingroup$ @DavidGao Thank you so much for your time, I spent the afternoon trying to understand this and by now I am quite sure $u$ is just a diagonal operator. I was trying to compute $u$ without sucess but I think this way of viewing $u$ was to prove that $u_0$ is a completely bounded operator. Should I delete the post? $\endgroup$ Mar 31 at 21:50

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For example in Theorem 2.3, when he wants to refer to an element in the predual $u \in M_*$, he views it as a normal linear map $u:M \rightarrow M_n(\mathbb C)$, for $n \in \mathbb N$.

That's not what happens there. Jean is proving that the map $a^*\odot\varphi\odot b\longmapsto u$, already stated to be linear and completely contractive, is completely isometric and onto. For this he shows that the $n$-amplification of this map is onto and increases norm. It is because of that that he takes $u\in M_n(\text{VN}(G)^*)=\text{Lin}(\text{VN}(G),M_n)$, in order to find an element of $M_n(L^2(X)^*\otimes_{hX}A(G)\otimes_{hX}L^2(X))$ that maps to $u$. .

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  • $\begingroup$ Thank you for your reply. However, you can see in the 1st paragraph of the proof that he regards an element $u \in M_*$ as a linear map $VN(G) \rightarrow M_n$. By now I am convinced he did so to prove complete boundness at the same time. $\endgroup$ Apr 6 at 21:09

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