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Mar 11 at 19:38 comment added David Gao Note that in the second link you provided, the left shift operator is on $l^2(\mathbb{Z})$, in which case it is the bilateral shift, a unitary that is actually contained in a tracial vNa, namely the group algebra of $\mathbb{Z}$. In contrast, as Mikael pointed out, the unilateral shift on $l^2(\mathbb{N})$ cannot be contained in a tracial vNa, rendering this question meaningless unless you can supply a definition of Brown measure which can work in the nontracial setting.
Mar 11 at 14:48 comment added Mikael de la Salle At least in the link that you provide, the Brown measure is defined only for operators in a von Neumann with a tracial normal state. Such a von Neumann algebra cannot contain the shift (otherwise the trace of $1-SS^*$ would be nonzero, because $S S^* \neq 1$, and equal the trace of $1-S^*S=0$). So what is the precise meaning of your question?
Mar 11 at 10:31 history asked Ma Joad CC BY-SA 4.0