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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?

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  • $\begingroup$ Is your definition of "isomorphism" "linear homeomorphism" (see mathoverflow.net/questions/80567/…) ? $\endgroup$
    – jjcale
    Commented Jul 17, 2016 at 6:18
  • $\begingroup$ Just change the norm on $X$ according to the duality with the right norm on $A$, and $A$ will be the isometric dual. $\endgroup$
    – Andreas Thom
    Commented Jul 17, 2016 at 6:25
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    $\begingroup$ @AndreasThom At least for general Banch spaces this is not always possible: There are Banach spaces that have a predual but no isometric predual. Thus, there exists a Banach space $Y$ such that $Y$ is isomorphic to some $X^*$, but not isometrically isomorphic to $Z^*$ for any $Z$. My question is whether such a behaviour is possible for C*-algebras. $\endgroup$
    – Hannes Thiel
    Commented Jul 17, 2016 at 8:07
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    $\begingroup$ Ok, you are right. $\endgroup$
    – Andreas Thom
    Commented Jul 17, 2016 at 11:12
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    $\begingroup$ To my knowledge, this was first shown in this article: Davis, Johnson: A renorming of nonreflexive Banach spaces, Proc. Amer. Math. Soc. 37 (1973) The result of Davis and Johnson was later generalized, for instance here: Godun: Equivalent norms on nonreflexive Banach spaces, Dokl. Akad. Nauk SSSR 265 (1982) $\endgroup$
    – Hannes Thiel
    Commented Jul 17, 2016 at 15:04

2 Answers 2

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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.

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  • $\begingroup$ Thank you very much. Do you know why the Gleason cover $K=G_I$ of the unit interval is not hyper-Stonean, and why $C(G_I)$ is isomorphic to $\ell^\infty$? The book of Dales et al. will appear in August, so maybe I just have to be patient. $\endgroup$
    – Hannes Thiel
    Commented Jul 20, 2016 at 7:34
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    $\begingroup$ @HannesThiel: I think the Gleason cover is another name for the max ideal space of the injective envelope. Looking in arxiv.org/abs/0706.2995v1, pages 10-11, one sees that $C(K)$ is isomorphic to a complemented subspace of $L^\infty(I)$, hence isomorphic to a complemented subspace of $\ell^\infty$, hence isomorphic to $\ell^\infty$ by the Pelczynski method. It's not clear to me why $C(K)$ is not a von Neumann algebra but perhaps this follows from general properties of injective envelopes? $\endgroup$
    – Yemon Choi
    Commented Jul 20, 2016 at 11:10
  • $\begingroup$ It seems to me that this might even show that every injective, commutative C*-algebra (i.e., every commutative AW*-algebra) is isomorphic to $\ell^\infty(S)$ for some set $S$. And since there are commutative AW*-algebras that are not von Neumann algebras, this answers the question. $\endgroup$
    – Hannes Thiel
    Commented Jul 20, 2016 at 16:27
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    $\begingroup$ @YemonChoi The proof that the "Dixmier algebra" is not a W*-algebra is that it cannot be separated by normal states, or even by states preserving countable directed suprema. This is because such states correspond to countably additive Borel probability measures on [0,1] vanishing on meagre sets, and such measures do not exist by a variation of the usual argument for producing meagre set of measure 1. $\endgroup$
    – Robert Furber
    Commented Jul 21, 2016 at 20:17
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    $\begingroup$ @HannesThiel, let $\lambda$ be an uncountable set. Then $L_\infty(\{0,1\}^\lambda)$ is not Banach-space isomorphic to $\ell_\infty(S)$ for any $S$. $\endgroup$
    – Tomasz Kania
    Commented Aug 4, 2016 at 9:48
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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:

  1. For each $0\not=x\in E$ there is $\mu\in X$ with $\mu(x)\not=0$;
  2. 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.)

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