Support conjecture for Brown measures of multiplicative Brownian products

Let $(\mathscr{A},\tau)$ be a $W^*$-probability space and let $b_0$ be an invertible operator in it. For $z\in\mathbb{C}$, define

$$\tilde{p}_0(z)=\tau[|b_0-z|^{-2}],\qquad \tilde{p}_2(z)=\tau[b_0^*b_0|b_0-z|^{-2}].$$

For any operator $a$, write $|a|^2=a^*a$. Define $T(b_0,z)$ by

$$T(b_0,z)=\frac{1}{|z|^2\tilde{p}_2(z)-\tilde{p}_0(z)}\log\left(\frac{|z|^2\tilde{p}_0(z)}{\tilde{p}_2(z)}\right),$$

and, for $t>0$, define

$$\Sigma(b_0,t)=\{z\in\mathbb{C}:T(b_0,z)<t\}.$$

Let $b(t,\zeta)$ be a free multiplicative Brownian motion freely independent of $b_0$, and let $\Psi^{b_0}_{t,\zeta}$ be the map defined in the source. Support conjecture for multiplicative Brownian products. The support of the Brown measure of $b_0b(t,\zeta)$ equals $\overline{\Sigma(b_0,t,\zeta)}$, the image of $\overline{\Sigma(b_0,t)}$ under $\Psi^{b_0}_{t,\zeta}$. This conjecture extends the known positive-definite case to arbitrary invertible initial operators; the supplied evidence states that for positive definite $b_0$ the support was proved to be contained in the conjectured region, apart from an exception at $0$, while equality remains open in the stated generality.