In many theories about $C^*$-algebras, the $C^*$-algebras are always assumed to be separable. I have a question: Why few people discuss the inseparble $C^*$-algebras? Are they more difficult to handle?

  • $\begingroup$ Is this really a common assumption? I do not think it is. Sometimes a von Neumann algebra is called seperable if it can be faithfully represented on $\ell^2(\Bbb N)$, but this does not mean the algebra is seperable in the "usual" sense. $\endgroup$
    – s.harp
    May 21 '20 at 16:57

It really depends on what you are doing with your $C^*$-algebras. I wouldn't say that few people study nonseparable $C^*$-algebras, but a lot of modern areas of research (in particular, classification) are currently dealing with separable algebras, because, in short, they are easier to deal with (see my last paragraph). There isn't one all-encompassing reason why one would consider separable algebras over nonseparable algebras, but there are several reasons that come to mind. Here's a short list of possible reasons:

  1. For a separable $C^*$-algebra, one can guarantee the existence of a sequential approximate unit (of course, if this is the only reason, you can just assume your algebra is $\sigma$-unital).
  2. For separable $C^*$-algebras, there is always a faithful representation on a separable Hilbert space, and the space of bounded operators on separable Hilbert spaces is more well-behaved than the space of operators on nonseparable Hilbert spaces.
  3. Any $C^*$-algebra is an inductive limit of its separable $C^*$-subalgebras. So if a given property passes to inductive limits, it suffices to check it on separable subalgebras.
  4. Many naturally occurring classes of $C^*$-algebras are separable (examples in no particular order:$C(X)$ for $X$ compact and metrizable, group $C^*$-algebras of countable groups, AF-algebras, AF-embeddable algebras, AT-algebras, rotation algebras, Cuntz algebras, $\ldots$).

In regards to your last question, generally speaking, anytime someone adds an adjective to the class of objects they're studying, it's to make the problem simpler/ more accessible. You remove adjectives (or put "non" in front of them) to make the problem harder.

  • $\begingroup$ In the 3rd point, you mentioned that any $C^*$-algebra is the inductive of its separable subalgebras.( Wolud you mind telling the thought of the proof?) So the non-separable $C^*$-algebra is the inductive of its separable subalgebras? $\endgroup$ May 21 '20 at 21:30
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    $\begingroup$ Depending on how much detail you want, this could be a separate question. The vague idea is as follows: Given $A$, inclusions of separable subalgebras into the entire algebra gives a map from the inductive limit to $A$ by universality. Since all maps are injective (inclusions), the universal map is injective. But every $a\in A$ is contained in a separable $C^*$-subalgebra, so the universal map is surjective, and thus an isomorphism. $\endgroup$
    – Aweygan
    May 22 '20 at 1:37
  • $\begingroup$ I saw an open question: Is every exact $C^*$-algebra a subalgebra of a nuclear $C^*$-algebra? The above question is true when the exact $C^*$-algebra is separable, but it is unknown in the non-separable case. According to your statement, a non-separable $C^*$-algebra is the inductive limit of its separable subalgebras, can we conclude that the above conclusion also holds in the non-separable case? $\endgroup$ May 25 '20 at 0:25
  • $\begingroup$ The problem is that the property "is a subalgebra of a nuclear $C^*$-algebra" may not pass to inductive limits. $\endgroup$
    – Aweygan
    May 25 '20 at 1:31
  • $\begingroup$ ,does there exist an example to show that "is a subalgebra of nuclear c*-algebra may not pass to inductive limits? $\endgroup$ May 25 '20 at 4:35

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