By an operator space structure on a Banach space $X$ I mean a sequence of norms on spaces $M\_n \otimes X$ that satisfies Ruan's axioms.


Among such admissible sequences there is always the smallest one (if we impose some normalisation, say $\|e\_{11} \otimes x\|=\|x\|$), the one obtained by embedding $X$ in some $C(K)$-space; it is sometimes referred to as "commutative" operator space structure. It is then reasonable to ask: can a non-commutative $C^{\ast}$-algebra (which has a privileged operator space structure induced by $\ast$-homomorphic embedding into $B(\mathcal{H})$) be completely isomorphic (or isometric) to a minimal operator space?


I suspect that the answer is no (at least in the completely isometric case) but I can't come up with a proof.



Can a non-commutative C*-algebra be a minimal operator space?
I don't think so. If A is a noncommutative C\*-algebra with the min operator space structure then it's double dual $A^{\*\*}$ will also have the min operator space structure (this fact can be found in any operator space text) and be noncommutative. Then $A^{\*\*}$ will have two noncommuting projections. One can then use ideas from Takesaki Vol 1, Chapter V.1 (the last topic in the section) to build a 2x2 matrix algebra inside of $A^{\*\*}.$ This would force the natural operator space structure on 2x2 matrices to be minimal, which it isn't. 


