TITLE: Which C*-algebras are complemented in their bidual?
QUESTION [6 upvotes]: Every von Neumann algebra is 1-complemented in its bidual, and so is every injective C*-algebra. Also, if $C_0(X)$ is infinite-dimensional and separable then it is not complemented in its bidual, and $\mathcal{K}(\mathcal{H})$ is not complemented in its bidual when $\mathcal{H}$ is infinite-dimensional.
How much more than this is known? Is every infinite-dimensional separable C*-algebra not complemented in its bidual? Is anything known about the nonseparable case?

REPLY [5 votes]: It seems that only for C*-algebras that are 1-complemented in the second dual we may say something definite (they are AW*-algebras as observed by Hannes). 
A separable C*-algebra $A$ that is complemented in $A^{**}$ must be finite-dimensional, as by a result of Pfitzner, it is a Grothendieck space. It is an easy exercise on the Eberlein–Šmulian theorem that a separable Grothendieck space is reflexive. Reflexive C*-algebras are finite-dimensional.
Beyond this case we cannot say much more even in the commutative case. For example, kernels of non-weak*-continuous characters on $\ell_\infty$ are complemented in their biduals (being Banach-space isomorphic to $\ell_\infty$). The question of for which (locally) compact space $K$ the space $C_0(K)$ is isomorphic to a dual space (such spaces are complemented in their second duals) is discussed in the forthcoming book by Dales, Dashiell, Lau and Strauss. Even in this case, the current state of affairs it far from a complete characterisation. Indeed, it is still open whether injective Banach spaces are isomorphic to 1-injectives ones (that is, to commutative AW*-algebras) and a commutative C*-algebra complemented in the second dual is injective.