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Let $L$ be the left shift operator on $\ell^2(\mathbb{Z})$ with trace $\tau(T) := \langle T \delta_0, \delta_0 \rangle$.

How can I show that the Brown measure of $L$ is the uniform measure on the unit circle?

The Brown measure is defined as follows: For each $z \in \mathbb{C}$ define $\nu_{z}$ to be the spectral measure of the (self-adjoint) operator $(L - z)^*(L-z)$. Then let $$f(z) = \frac{1}{2} \int_0^\infty \log x \,d\nu_z(x).$$ Then the Brown measure is defined as $$\mu_L := \frac{1}{2\pi} \Delta f,$$ where the Laplacian is taken in the sense of distributions. Motivation for the definition can be found at Terence Tao's notes on the circular law for random matrices.

Edit: posted at math stackexchange https://math.stackexchange.com/questions/2258115/brown-measure-of-left-shift-operator

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  • $\begingroup$ What's the relevance of $\tau (T)$ for this question? $\endgroup$
    – Christian Remling
    Commented Apr 29, 2017 at 17:15
  • $\begingroup$ To address my own comment, I guess what you're saying is that $\nu_z$ is defined as the spectral measure $d\|E(x)\delta_0\|^2$ of $\delta_0$ ("the spectral measure," meaning a positive measure on $\mathbb C$, is not unambiguously defined)? $\endgroup$
    – Christian Remling
    Commented Apr 29, 2017 at 17:17
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    $\begingroup$ The spectral measure of $L$ wrt $\delta_0$ is Lebesgue measure on $S^1$. Using this as your starting point, it should be possible to work everything out explicitly. I believe your question would have been better suited for math.stackexchange.com $\endgroup$
    – Christian Remling
    Commented Apr 29, 2017 at 17:26
  • $\begingroup$ I was having trouble computing the spectral measure of $(L-z)^*(L-z)$, however. Is that still easy? And fair enough, I posted here because the only other references to Brown measure I could find were in research papers. I just posted in stackexchange. $\endgroup$
    – keej
    Commented Apr 29, 2017 at 19:51
  • $\begingroup$ Actually, it might be better to do it differently. I'll answer at MSE. $\endgroup$
    – Christian Remling
    Commented Apr 30, 2017 at 2:37

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Since the two-sided shift is a normal operator this is not really a question about the Brown measure, but about the spectral measure, so it suffices to compute the $*$-moments $\tau(L^nL^{*m})$. For an introduction to Brown measure you might also see Chapter 11 of the book https://www.math.uni-sb.de/ag/speicher/publikationen/Mingo-Speicher.pdf

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