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Is every von Neumann algebra complemented in its bidual? It is certainly true for commutative von Neumann algebras as their spectrum is hyperstonian. Is it 1-complemented?

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Every dual Banach space $X^{\ast}$ is complemented in its bidual $X^{\ast\ast\ast}$ with a projection of norm $1$: Indeed, the adjoint $(\iota_X)^{\ast}:X^{\ast\ast\ast} \to X^{\ast}$ of the canonical inclusion $\iota_{X}: X \to X^{\ast\ast}$ is left inverse to the canonical inclusion $\iota_{X^{\ast}}: X^{\ast} \to X^{\ast\ast\ast}$, thus $p = \iota_{X^{\ast}} (\iota_X)^{\ast}$ is a norm one projection onto $\iota_{X^{\ast}}(X^{\ast})$. Since by Sakai's theorem a von Neumann algebra is precisely a $C^{\ast}$-algebra whose underlying Banach space is a dual space, a positive answer to both your questions follows immediately.

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    @Yemon: I'm pretty sure that the abelian case was established (or at least nearly proved) by Grothendieck in *Sur une caracterization vectorielle-métrique des espaces $L^1$* where he proved the abelian version of Sakai's theorem. [See also my answer here](http://math.stackexchange.com/questions/55386/), where you'll find the bibliographic details and a link. Thanks for pointing it out, though.2011-11-23