# gr.group theory – Covergence of Brown measures For each $$nin mathbb N$$, let $$mathcal M_n$$ be a finite von Neumann algebra with a faithful trace $$tau_n$$. Fix a non-principal ultrafilter $$omega$$ on $$mathbb N$$. Let $$mathcal M^omega$$ be the tracial ultraproduct of $$(mathcal M, tau_n)$$ (see the definition in https://arxiv.org/abs/1212.5457).

Assume first that $$A=(A_i)in mathcal M^omega$$ and $$A_i$$ are self-adjoint (or normal). Denote by $$mu_{A_i}$$ and $$mu_A$$ the spectral measures corresponding to the operators $$A_i$$ and $$A$$ respectively. It is known that the weak ultralimit of $$mu_{A_i}$$ (with respect to $$omega$$) is $$mu_A$$.

If $$A=(A_i)$$ is not self-adjoint, one can define the Brown measure $$mu_{A_i}$$ and $$mu_A$$ (see the definition in https://arxiv.org/abs/math/0605251). We can ask whether the ultralimit of $$mu_{A_i}$$ is $$mu_A$$. The following trivial example gives a negative answer. Assume $$mathcal M_i=M_i(mathbf C)$$ and $$A_iin mathcal M_i$$ is the matrix having 1 under the diagonal and 0 in the rest of entires.

I want to modify slightly the previous question. Let $$R$$ be a $$*$$-algebra and $$phi_i:Rto mathcal M_i$$ a $$*$$-homomorphism such that for every $$xin R$$, $$sup |phi_i(x)|. Then we can define $$phi=(phi_i):Rto mathcal M^omega$$. Assume, moreover that $$phi$$ is injective.

Let $$xin R$$. Is it true that the ultralimit of $$mu_{phi_i(x)}$$ is
$$mu_{phi(x)}$$? Posted on Categories Articles