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Suppose that $\tilde{X}$ is a compact space. If $C(\tilde{X})$ is isometrically isomorphic to the second dual of a Banach space, does there exist a locally compact space $X$ such that $C(\tilde{X})=C_0(X)^{**}$?

P.S. Note that the converse is always true, namely if $X$ is a locally compact space, there is a compact space $\tilde{X}$ such that $C_0(X)^{**}=C(\tilde{X})$. The Banach space $C_0(X)^{**}$ turns into a Banach algebra with the first (second) Arens product. This is a non-trivial product on $C_0(X)^{**}$.

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    $\begingroup$ I don't see why that's true. Here is an example. If $X$ is a locally compact space, then $C_0(X)^{**}$ is a unital commutative $C^*-$algebra and therefore it is in the form of $C(\tilde{X})$ for a compact space $\tilde{X}$. This ($C(\tilde{X})$) gives an example of a $C(X)$ that is a dual of a Banach space. $\endgroup$
    – Bob
    Commented Feb 27, 2014 at 3:51
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    $\begingroup$ @PaulSiegel just take an abelian von Neumann algebra. Diffuse would get you a dual space, atomic would get you a bidual space. $\endgroup$
    – Yemon Choi
    Commented Feb 27, 2014 at 4:21
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    $\begingroup$ If $K$ is a compact space and $C(K)$ is isomorphic to a second dual, then it is injective and, by a result of Haydon, it is isomorphic to $\ell_\infty(\Gamma)$ for some set $\Gamma$. Hence $C(K)$ is isomorphic to $c_0(\Gamma)^{**}$. $\endgroup$
    – M.González
    Commented Feb 27, 2014 at 7:39
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    $\begingroup$ @Yemon Choi: Yes. He only proves isomorphic. This is the reason I did not put it as an answer. I will put a more detailed version as an answer. $\endgroup$
    – M.González
    Commented Feb 27, 2014 at 15:26
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    $\begingroup$ Continuing: Maharam's theorem gives a characterization of $L_1(\mu)$ spaces, but which ones of these are isometric to dual spaces? The answer could be easy, but is unknown to me. $\endgroup$
    – Bill Johnson
    Commented Feb 28, 2014 at 1:18

2 Answers 2

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Only a partial answer, but too long for comments.

This was question 3 considered at a BIRS meeting in 2012 — I wonder if your question is motivated by seeing it here or raised elsewhere by Dales or his collaborators?

In any case, according to the final report (see the sidebar) the answer is yes when $C(\widetilde{X})$ is assumed to be isometric to the bidual of a separable Banach space, by old results of H. E. Lacey:

H. E. Lacey. A note concerning $A^{\ast} =L_1(\mu )$. Proc. Amer. Math. Soc. 29 (1971) 525—528. Link to paper.

If I understand Lacey's result correctly, it says that if $C(\widetilde{X})=A^{**}$ for $A$ a separable Banach space then $C(\widetilde{X})$ is the bidual of $c_0$ or of $C[0,1]$.

It appears that at the time that report was written, the general case was still open. Perhaps if Fred Dashiell is still on MathOverflow he might be able to inform us of any further progress.

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    $\begingroup$ I have a feeling Lacey's paper has been mentioned elsewhere on MO before, but I could be misremembering. $\endgroup$
    – Yemon Choi
    Commented Feb 27, 2014 at 4:32
  • $\begingroup$ Thank you very much Yemon! The question is motivated by the work of Dales, Lau and Strauss, but I didn't know about the current state of the art. $\endgroup$
    – Bob
    Commented Feb 27, 2014 at 4:57
  • $\begingroup$ @Bob the question rang a bell because Dales mentioned it to me in 2007 and I was unable to say anything intelligent then :) $\endgroup$
    – Yemon Choi
    Commented Feb 27, 2014 at 4:59
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    $\begingroup$ To the best of my knowledge Dales et al. are going to publish (rather soon) a survey article on this and related problems. I will keep you posted. $\endgroup$
    – Tomasz Kania
    Commented Mar 15, 2014 at 20:18
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This is a detailed version of a previous comment that gives a partial answer. If $K$ is a compact space and $C(K)$ is isometrically isomorphic to $X^{**}$, then $X$ is a $\mathcal{L}_{\infty}$-space. Hence $X^{**}$ is injective.

It was proved by R. Haydon [Israel J. Math. 31 (1978), 142-152] that an injective bidual space is isomorphic to $\ell_\infty(\Gamma)$ for some set $\Gamma$. Hence $C(K)$ is isomorphic to $c_0(\Gamma)^{**}$.

I do not know if $C(K)$ is isometrically isomorphic to $c_0(\Gamma)^{**}$.

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    $\begingroup$ Perhaps I am misunderstanding something: but I think that if $C[0,1]^{**}$ were isometrically isomorphic to $c_0(\Gamma)^{**}$ then uniqueness of isometric preduals would imply that $M[0,1]=C[0,1]^*$ is isometrically isomorphic to $\ell^1(\Gamma)$. But this is impossible since $M[0,1]$ contains a complemented copy of $L^1[0,1]$, while all complemented subspaces of $\ell^1(\Gamma)$ are $\ell^1$ of something. $\endgroup$
    – Yemon Choi
    Commented Feb 27, 2014 at 16:20
  • $\begingroup$ Yemon, Haydon in his proof uses Pełczyński's decomposition method hence his result is only isomorphic not isometric, so the trick with uniqueness of preduals doesn't work. This situation is analogous to the fact that there exists a Banach-space between apparently very different Banach algebras $\ell_\infty$ and $L_\infty$. I am not sure if the OP is happy with Banach-space isomorphisms, though. $\endgroup$
    – Tomasz Kania
    Commented Mar 15, 2014 at 20:15
  • $\begingroup$ @TomekKania I was responding to the last part of this post, giving a reason why $C(K)$ cannot be isometric to any $c_0(\Gamma)^{**}$ when $K$ is the max ideal space of $C([0,1])^{**}$. I had noticed the point you raise when looking at Haydon's proof. $\endgroup$
    – Yemon Choi
    Commented Mar 16, 2014 at 3:03

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