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Let $\mathcal{T}$ be the trivalent directed tree, with two parents and one child for each vertex (see below). Let $\mathcal{V}$ be the set of vertices of $\mathcal{T}$ and $H$ be the Hilbert space $\ell^2(\mathcal{V})$. Let $S$ be the shift operator on $H$ defined so that $S(v)$ is the child of the vertex $v$. Note that $S$ is bounded and its adjoint $S^{\star}$ is given by reversing the direction of each edge.

Question: What is the von Neumann algebra generated by $S \in B(H)$ and denoted $W^{\star}(S)$?
Is it hyperfinite? Is it a factor? of each type?

Note that $W^{\star}(S)$ is not abelian because the operator $S$ is not normal, because for any vertex $v$, $$(SS^{\star}-S^{\star}S) \cdot v = v-v'$$ with $v'$ the other parent of the child of $v$.

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    $\begingroup$ Since $S$ is (up to a scalar) a coisometry, the $C^\ast$-algebra generated by $S$ will be isomorphic to the Toeplitz algebra, which is a type I $C^\ast$-algebra. So $W^\ast(S)$ must be type I. $\endgroup$
    – user85913
    Commented Dec 21, 2017 at 9:08
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    $\begingroup$ Moreover, it seems to me that no $C^\ast(S)$-invariant subspace of $H$ is annihilated by $SS^\ast -S^\ast S$, since a recursive argument using the formula that you found shows that any element of such a subspace would have to be constant on larger and larger subsets of $\mathcal V$. Now the Wold decomposition implies that $H$ is isomorphic to a multiple of the standard representation of the Toeplitz algebra, and so it seems to me that $W^\ast(S)$ is the factor of type $I_\infty$. $\endgroup$
    – user85913
    Commented Dec 21, 2017 at 9:52
  • $\begingroup$ Yes, $\frac{\sqrt{2}}{2}S^{\star}$ is a proper isometry, so by Coburn's theorem, $C^{\star}(S)$ is the Toeplitz algebra. $\endgroup$ Commented Dec 21, 2017 at 15:05

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