Is a C*-algebra with an isomorphic predual a von Neumann algebra? It is well-known that a C*-algebra $A$ is a von Neumann algebra if and only if it has an isometric predual, that is, if and only if there exists a Banach space $X$ such that $A$ is isometrically isomorphic to $X^*$. Does the same hold for isomorphic preduals?

Let $A$ be a C*-algebra such that $A$ is (not necessarily isometrically) isomorphic to $X^*$ for some Banach space $X$. Is $A$ a von Neumann algebra?

 A: Via my colleague Garth Dales, some observations which answer your question in the negative, even in the abelian case:$\newcommand{\N}{{\mathbb N}}$

We know that $K$ is hyper-Stonean iff $C(K)$ is isometrically dual. So you are asking for locally compact spaces $K$ such that $C_0(K)$  is isomorphically dual, but not isometrically a dual space.
The easiest example is to look at $\beta\N$  and choose a point $p\in \beta\N \setminus\N$, and consider the maximal ideal
  $M_p$ of functions that vanish at $p$.   This is isomorphic to $\ell^\infty$, but $M_p= C_0(\beta\N \setminus \{p\})$ and $\beta\N \setminus \{p\}$  is not even compact.
Probably you want a compact space $K$ with this property. In our book we give a compact $K$ such that $C(K)$ is isomorphically dual, but $K$ is not even Stonean. (It is totally disconnected.)
... The standard example is $K=G_I$, the Gleason cover of the unit interval. $K$ is an inﬁnite, separable Stonean space without isolated points and $C(K)$ is  isomorphically a bidual space because $C(K)$ is isomorphic to $\ell^\infty$. But $K$ is not hyper-Stonean.

The book he refers to, co-authored with Dashiell and Lau and Strauss, is this one, which should appear later in 2016.
A: Now that Yemon has given a counterexample, I thought it worth giving a stronger hypthoesis under which we have a positive answer.
I looked at something similar here: http://arxiv.org/abs/math/0604372 see Section 4.

Theorem: Let $M$ be a commutative von Neumann algebra, and let $N$ be a dual Banach algebra.  Any bounded algebra isomorphism between $M$ and $N$ is automatically weak$^*$-continuous.

With Le Pham and White we remove the commutative condition in http://arxiv.org/abs/0804.3764
With ten years' of hindsight, I find that I can now improve this result and give an easy proof...  A dual Banach algebra is a Banach algebra which is a dual space, in such a way that the multiplication is separately weak$^*$-continuous.  Thus the above result says that a von Neumann algebra has a unique weak$^*$-topology making the multiplication map weak$^*$-continuous.
If $E$ is a Banach space, then whenever $E$ is isomorphic to $F^*$ for some Banach space $F$, say under $\theta: E \rightarrow F^*$ we can look at the image of $F$ under the adjoint $\theta^*$.  This gives a closed subspace of $E^*$, say $X$, such that $E$ is canonically isomorphic to $X^* = E^{**} / X^\perp$, where $X^\perp$ is the annihilator of $X$ in $E^{**}$.  This means that:


*

*For each $0\not=x\in E$ there is $\mu\in X$ with $\mu(x)\not=0$;

*For each $\Phi\in E^{**}$ there is $x\in E$ with $\mu(x) = \Phi(\mu)$ for all $\mu\in X$.


I call $X$ a "concrete predual" of $E$.  The $x\in E$ given by (2) is unique, because of condition (1).  Let $\alpha : E^{**} \rightarrow E$ be the map so given.  It is a projection of $E^{**}$ onto $E$, once we canonically identify $E$ with its image in $E^{**}$.

Theorem: Let $A$ be a C$^*$-algebra which is isomorphic to a dual space $F^*$, with the additional assumption that $F$ is a Banach $A$-bimodule, and the isomorphism $A \cong F^*$ is a bimodule map.  Then $A$ is a von Neumann algebra (and the weak$^*$-topology is the canonical one).

To prove this, argue as above to reduce to a concrete predual $X\subseteq A^*$.  The bimodule condition becomes that $X$ is $A$-invariant for the usual action of $A$ on $A^*$.  One can check that $X$ being $A$-invariant is actually equivalent to the map $\alpha:A^{**}\rightarrow A$ being an algebra homomorphism.  Thus $\ker\alpha$ is a closed 2-sided ideal in $A^{**}$ and so is $*$-closed.  This in turn is equivalent to $X$ being $*$-closed, which in turn implies that $\alpha$ is actually a $*$-homomorphism.  Then $\alpha$ is a contractive projection which implies that $A$ is isometrically isomorphic $X^*$, hence a von Neumann algebra.
(This proof is obviously influenced by Tomiyama's proof that W$^*$-algebras and von Neumann algebras are the same thing.)
