0
$\begingroup$

This looks an extremely simple question - I am just trying to give an example of Brown measure, https://en.wikipedia.org/wiki/Brown_measure, so I try to compute it for the left/right-shift operator on $\ell^2 (\mathbb N).$ However, I quickly get stuck, and I haven't found a good book with such examples.

I tried to use the formula here: Brown measure of left shift operator. While I can work out the spectral measure of $(A-\lambda)^\ast (A-\lambda),$ which is supported on $[1+|\lambda|^2-2|\lambda|,1+|\lambda|^2+2|\lambda|],$ the integral becomes really hard to evaluate because we have $\log$.

I don't know if it is possible to find an explicit formula for the density of the Brown measure. If not possible, how can I understand the shape of the Brown measure of the shift? [Is there a reference for a plot of that or something?]

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Commented Mar 11 at 14:48
  • $\begingroup$ 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. $\endgroup$
    – David Gao
    Commented Mar 11 at 19:38

0

You must log in to answer this question.