# Which C*-algebras are complemented in their bidual?

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?

• Cameron, what does 1-complemented mean? – Sergei Akbarov Nov 26 '16 at 8:52
• It means complemented with a projection of norm 1. – Cameron Zwarich Nov 26 '16 at 9:07
• Every dual Banach space is 1-complemented in its bidual. – Ruy Nov 26 '16 at 11:31

(1) If $A$ is $1$-complemented in its bidual, then $A$ is an AW*-algebra.

Indeed, assume that $A$ is $1$-complemented in its bidual, via a contractive projection $p\colon A^{**}\to A$. By a theorem of Tomiyama, $p$ is a conditional expectation. This implies that $A$ is monotone closed, that is, every upward directed family of self-adjoint elements in $A$ has a supremum. In particular, $A$ is an AW*-algebra. (A C*-algebra is an AW*-algebra if every maximal abelian $*$-subalgebra is monotone complete, see [1].)

(2) Every separable AW*-algebra is finite-dimensional.

Indeed, assume $A$ is a separable AW*-algebra. Consider a maximal abelian $*$-subalgebra (=masa) $B$ of $A$. Then $B\cong C(X)$ for some compact, Hausdorff, extremally disconnected space $X$ (such spaces are also called Stonean). Every metrizable, extremally disconnected space is discrete. Therefore, $X$ is finite. Thus, every masa in $A$ is finite-dimensional. This implies that $A$ is finite-dimensional. (Every infinite-dimensional C*-algebra contains a positive element with infinite spectrum, and hence an infinite-dimensional masa.)

Combining (1) and (2) we obtain that a separable, infinite-dimensional C*-algebra is never $1$-complemented in its bidual.

[1] arxiv.org/abs/1501.02434

• That separable, infinite-dimensional C*-algebras are never complemented in their biduals (with any constant) follows also from an earlier result of Pfitzner that von Neumann algebras are Grothendieck spaces. – Tomek Kania Nov 26 '16 at 17:33
• I see, that is a much more direct and more general argument. Very nice! – Hannes Thiel Nov 26 '16 at 18:25

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.

• Hannes only said that a C*-algebra that is 1-complemented in its bidual must be monotone complete, not that every monotone complete algebra is 1-complemented in its bidual. I couldn't find any evidence of the converse by searching online, but maybe it's known. And if we knew that every AW*-algebra was 1-complemented in its bidual then we would know that every AW*-algebra is monotone complete. – Cameron Zwarich Nov 27 '16 at 21:16
• Every injective C*-algebra is monotone complete (for the reason noted by Hannes), but not every monotone complete C*-algebra is injective, e.g. any non-injective von Neumann algebra. – Cameron Zwarich Nov 28 '16 at 1:03