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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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Maybe there are easier proofs, but I'm no operator theorist and my intuition says that it must essentially boil down to what I'm saying here. If I remember the usual proof of Sakai's theorem correctly, at some point it is even shown that the predual is complemented in its bidual, but I may be wrong here. –  t.b. Jun 21 '11 at 11:30
    
Just chanced on this while wasting time at office. It is certainly known that the predual of a vN algebra M is complemented in $M^*$ via an $L$-projection -- in the abelian case this theorem is older and has a name, which for now I forget. However, I don't know if this fact follows from the proof(s) of Sakai's theorem –  user16299 Nov 23 '11 at 3:47
    
@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, where you'll find the bibliographic details and a link. Thanks for pointing it out, though. –  t.b. Nov 23 '11 at 3:54
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