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?



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> 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?
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Is a C*-algebra with an isomorphic predual a von Neumann algebra?
Via my colleague Garth Dales, some observations which answer your question in the negative, even in the abelian case:$\newcommand{\N}{{\mathbb N}}$



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> 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.
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> 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.
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> 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.)
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> ... 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.
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The book he refers to, co-authored with Dashiell and Lau and Strauss, is [this one](http://www.springer.com/gb/book/9783319323473), which should appear later in 2016.


