2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $a\longmapsto \langle a\,\chi_e,\chi_e\rangle$.

$\endgroup$
  • $\begingroup$ Please give more explanation about the last line. In fact, it is not well known for me. $\endgroup$ – saeed Dec 14 '17 at 14:48
  • $\begingroup$ 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. $\endgroup$ – Martin Argerami Dec 14 '17 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.