# Finite von-Neumann algebra

Let $G\subset B(H)$ be a group of unitary operators on separable Hilbert space $H$, and $\xi\in H$ such that $\{g\xi\}_{g\in G}$ is a basis for $H$. Is $W^*(G)$(von-Neumann algebra generated by $G$) a finite von Neumann algebra?

This question is a proposition in a paper, and the following is its proof:

By defining a unitary operator $W : H → l2(G) \otimes \Bbb C \xi$ such that $Wg\xi = l_gχ_e \otimes \xi, ∀i, ∀g ∈ G,$ we can assume that $G$ has the form of $\{l_g \otimes I : g ∈ G\}$, where I is the identity operator on $\Bbb C\xi$. Since the commutant of $\{l_g \otimes I : g ∈ G\}$ is $\{l_g : g ∈ G\}'\otimes \Bbb C\xi$, and $\{l_g : g ∈ G\}'$ and $W^*(\{l_g : g ∈ G\})$ are finite von Neumann algebras by R. Kadison and J. Ringrose, “Fundamentals of the Theory of Operator Algebras,” vols. I and II, the lemma follows.

I could not understand this proof.

We may assume that $\|\xi\|=1$. Note also that the fact that $H$ is separable guarantees that $G$ is countable.
First, the hypothesis that $\{g\xi:\ g\in G\}$ is an orthonormal basis is what makes $W$ a unitary: given $\eta\in H$, we have $\eta=\sum_g x_g\,g\xi$. Then, in the case where $x_g\ne0$ only for a finite $F\subset G$, \begin{align} \|W\eta\|^2&=\|\sum_g x_g\,(\chi_g\otimes \xi)\|^2=\sum_{g,h}x_g\overline{x_h}\,\langle \chi_g\otimes\xi,\chi_h\otimes\xi\rangle =\sum_{g,h}x_g\overline{x_h}\langle\chi_g,\chi_h\rangle\,\|\xi\|^2\\ \ \\ &=\sum_g|x_g|^2=\|\eta\|^2. \end{align} So $W$ is isometric on a dense subset of $H$, and thus it extends to an isometry. As all the sums $\sum_{g\in F}x_g\,\chi_g\otimes\xi$ appear in the range of $W$, it follows that its range is dense; being an isometry, it is surjective and so a unitary.
So now one can consider $WGW^*\subset B(\ell^2\otimes\mathbb C\xi)$ (the choice of $W$ is kind of unfortunate here, since now we want to consider the von Neumann algebra $W^*(WGW^*)$, with two different meanings for $W$ and for $*$ in the expression). To avoid this mess, we simply work with the image of $G$ (since unitary conjugation will preserve anything), which is the group $\{l_g\otimes I:\ g\in G\}$. Write $L(G)\subset B(\ell^2(G))$ for the von Neumann algebra generated by $\{l_g:\ g\in G\}$. It is not hard to see that $L(G)\otimes I$ is the von Neumann algebra generated by $\{l_g\otimes I:\ g\in G\}$, since linear combinations, products, and limits of elements $l_g\otimes I$ are of the form $x\otimes I$ with $x\in L(G)$.
As mentioned in the proof you quote, it is well-known that $L(G)$ is finite, since it has the trace $a\longmapsto \langle a\,\chi_e,\chi_e\rangle$.
• No, sorry. This is a place to answer questions, not to develop theory. That $L(G)$ is a finite von Neumann algebra when $G$ is discrete is something that appears in every single text on von Neumann algebras. Besides, the fact that the map above is a trace is an easy (if annoying) exercise. – Martin Argerami Dec 14 '17 at 15:04