# Characterization of the Predual of a von Neumann Algebra

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?

• This seems a bit weird. The predual is just the space of normal linear functionals. Not sure where the matrix algebra comes from. Mar 31 at 19:52
• Unless it’s not the predual of $M$ but of $M_n(M)$? Mar 31 at 21:11
• @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? Mar 31 at 21:50

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 $$u$$. .
• 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. Apr 6 at 21:09