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.


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.

  • $\begingroup$ Thank you very much; this settles the completely isometric case. The completely bounded case, as it stands, is obviously false since $M_2 \oplus L_{\infty}$ (just to have it infinite dimensional) provides a (trivial) counterexample. $\endgroup$ Jan 24 '14 at 16:31
  • 2
    $\begingroup$ @Mateusz Wasilewski: $A$ is cb isomorphic to min iff it is subhomogeneous. A $\mathrm{C}^*$-algebra $A$ is said to be subhomogeneous if it has only finite-dimensional irreducible representations (which is also equivalent to that it is isomorphic to a subalgebra of $M_n(C(X))$). Indeed, if $A$ is not subhomogeneous, then $A^{**}$ has a direct summand isomorphic to $B(H)$ with infinite-dimensional $H$, which is not cb isomorphic to a min space. $\endgroup$ Jan 25 '14 at 8:06
  • $\begingroup$ @NarutakaOZAWA: Thank you very much. $\endgroup$ Jan 25 '14 at 10:24

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