Do you know of any (necessary and) sufficient conditions for a C* algebra to be separable? Reference to bibliography is welcome.


I don't think there is a condition that is interesting. But it is easy to show that $A$ is separable if and only if it is generated by a finite or a countable set.

So as long as you have finite or countably many generators, your algebra will be separable. The vast majority of the commonly used C$^*$-algebras are like that. Exceptions would be the infinite-dimensional von Neumann algebras (and AW$^*$-algebras, but these are not as sexy).

The proof is extremely simple: if $A$ is separable, take $X\subset A$ countable and dense; then $A=C^*(X)$. Conversely, if $X$ is finite or countable and $A=C^*(X)$, then $$Y=\left\{\sum_{j=1}^m (a_j+ib_j)x_{1,j}\cdots x_{n,j}:\ n,m\in\mathbb N,\ a_j,b_j\in\mathbb Q,\ x_{k,j}\in X\cup X^*\right\}$$ is countable and dense.

  • $\begingroup$ Your answer interests me a lot:how can it be shown that an algebra is separable iff it is generated by a finite or a countable set of elements? Please, just provide a reference if available. I am interested in infinite von Neumann algebras, especially the type III algebras found in Algebraic Quantum Field Theory, which by the aforementioned argument are non-separable. Aren't they? $\endgroup$ – val 72 Sep 25 '18 at 6:10
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    $\begingroup$ It's a two-liner, so I don't think there's a reference; I have included the argument. And yes, any type III von Neumann algebra is not separable as a C$^*$-algebra. They are usually separable as von Neumann algebras, though, when one uses the sot or wot topologies. $\endgroup$ – Martin Argerami Sep 25 '18 at 14:22

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