4

It's not true. For simplicity, suppose $A$ is unital, so that its spectrum is compact Hausdorff. If $X$ is a compact Hausdorff space and $C(X)$ is separable, then you can show that $X$ is second countable. (Let $\{f_k\}$ be a countable dense subset of $C(X)$, and $\{U_n\}$ a countable basis of open sets in $\mathbb{C}$; then the countable collection of ...


2

The problem is still ill-posed, because the set of nilpotent bounded operators is not a vector space. (The sum of two nilpotents need not be nilpotent.) Let's interpret the question as: which algebras admit faithful representations as algebras of nilpotent bounded operators? Note that "nilpotent" usually means $T^n = 0$ for some $n$, not $T^2 = 0$. But ...


1

For separable von Neumann algebras the Haagerup property for $N$ is equivalent to the existence of a real closable derivation $\delta$ such that $\delta^*\delta$ has compact resolvents in $\mathcal B(L^2(N))$. This is true even in the non-tracial case. See Theorem 7.7 in: Martijn Caspers, Adam Skalski, The Haagerup Approximation Property for von Neumann ...


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