# When are two C*-algebras isomorphic as Banach spaces?

We may consider each $C^*$-algebra as a Banach space (by forgetting the multiplication and adjoint). I wonder how drastic this step is, i.e., which properties of the $C^*$-algebra are reflected by its Banach space structure.

Clearly, finite-dimensionality and other cardinality properties can be detected. Thus, to avoid these cases, let us restrict to separable, infinite-dimensional $C^*$-algebras.

For commutative $C^*$-algebras the problem has been solved by Milutin. In particular, $C(X)$ and $C(Y)$ are isomorphic as Banach spaces for any uncountable, compact, metric spaces $X$ and $Y$.

In the paper
Hamana. On linear topological properties of some $C^*$-algebras, Tohoku Math. J., II. Ser. 29, 157-163 (1977).
it is shown that the Banach space structure also reflects if all irreducible representations of the $C^*$-algebra are finite-dimensional.

Let us ask what happens at the 'opposite' end of the scale:

Question: Are all simple, separable, infinite-dimensional $C^*$-algebras isomorphic as Banach spaces?

• Is an isomorphism of Banach spaces a bounded linear map with bounded linear inverse or an isometric isomorphism? – Qiaochu Yuan Feb 3 '14 at 23:45
• I assume "isomorphic as Banach spaces" means linear homeomorphism. It's well known that there are non-isomorphic C*-algberas that are isometric as Banach spaces. (Google "C*-algebra not isomorphic to its opposite algebra.) The correct statement is: A and B are isomorphic as C*-algebras if and only if they are completely isometric as operator spaces. – Nik Weaver Feb 4 '14 at 3:14
• Hi Hannes, Kirchberg showed in his subalgebras of CAR algebra paper that all separable nuclear, non Type I C*-algebras are isomorphic as operator spaces (much stronger than just isomorphic as Banach spaces). I think Christensen and others also did some things in this direction for von Neumann algebras: those results are in Pisier's operator space book, but my copy isn't with me. – Caleb Eckhardt Feb 4 '14 at 3:52
• A complement to Caleb's remark: Being type I is invariant under a Banach space isomorphism (at least in the separable case). This follows from Haagerup--Rosenthal--Sukochev's classification of noncommutative $L^1$-spaces up to Banach isomorphism. – Narutaka OZAWA Feb 4 '14 at 8:49
• By Szankowski (Acta Math 1981), there is a separable (simple, unital, etc.) $C^*$-algebra which does not have the approximation property. Such a $C^*$-algebra cannot be Banach isomorphic to a nuclear one. – Narutaka OZAWA Feb 4 '14 at 8:57