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For a Hilbert space $H$ it is well known that the algebra $B(H)$ has a unique predual; the Banach space of trace class operators.

If $E$ is a Banach space then is it known whether

  1. $B(E)$ is always a dual Banach algebra?

  2. The predual is always unique?

I'm aware that 2 can fail in the case of a general `dual Banach algebra', so if the answer is "No!" can we place appropriate conditions on $E$ to ensure that 1 and 2 hold? If this is well-known then appropriate references would be useful.

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From Section 4 of Daws's paper arxiv.org/abs/math/0604372 - if E is reflexive then B(E) is a dual Banach algebra with respect to the predual $E\hat{\otimes} E^*$, and this is unique as an isometric DBA predual; if E also has the AP then I think Daws's results imply this predual is unique as a DBA predual, but I may be misreading. –  Yemon Choi Jul 3 '11 at 4:04
    
Thanks Yemon, that looks promising! –  Ollie Margetts Jul 3 '11 at 12:00
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3 Answers 3

If a Banach space X has the a.p. then the nuclear operators on X, $N(X)$, equipped with the 'nuclear norm' is an isometric predual of $B(X^*)$.

Apply this for $X^*=\ell_1$. This is overkill since $\ell_1$ has continuum many non-isomorphic preduals (even totally incomparable). It seems to me that if $X$ and $Y$ are non-isomorphic then $N(X)$ and $N(Y)$ cannot be isomorphic, however, if both have duals isomorphic to $\ell_1$, $B(\ell_1)$ would be isomorphic to both $B(X)$ and $B(Y)$. Of course unique up to isomorphism and isometric isomorphism are different things.

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You claim that there are $2^\omega$ many preduals of $\ell^1$. I know "only" $\omega_1$ many, i.e. all the spaces $C[0,\omega^{\omega^\alpha}], \alpha<\omega_1$ and some (finitely many) other preduals of $\ell^1$ including the Argyros-Haydon example. What are the missing ones? –  Tomek Kania Jul 3 '11 at 10:31
    
Tomasz, Richard Haydon showed about a dozen years ago that the original somewhat-reflexive spaces of the Bourgain-Delbaen construction are in fact $\ell_p$-saturated for $p$ depending on real parameters $a$ and $b$ chosen at the beginning of the construction; there are continuum many such possible $p$. –  Philip Brooker Jul 3 '11 at 11:34
    
The aforementioned paper of Haydon can currently be found at matwbn.icm.edu.pl/ksiazki/sm/sm139/sm13935.pdf –  Philip Brooker Jul 3 '11 at 11:39
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Yeah, I should have been more precise. I am not sure that it is known whether there are a continuum of non isomorphic spaces all of whose duals are ISOMETRIC to $\ell_1$. –  Bill Johnson Jul 3 '11 at 15:08
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Tomasz: It good to hear from you again. As Philip said on the isomorphic preduals. In fact in the Acta paper of Bourgain-Delbaen from 1981 they show that there are continuum many non-isomorphic spaces with dual isomorphic to $\ell_1$ by just considering different parameters. As for ISOMETRIC preduals of $\ell_1$: they must be $c_0$ saturated. Fonf showed that every infinite-dimensional Banach space whose dual closed unit ball contains but countably many extreme points is $c_0$ saturated. –  Kevin Beanland Jul 5 '11 at 15:28
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  1. No, not even isomorphically. Take an $E$ that is not complemented in its bidual (and hence not complemented in any dual space).

1.1. What Kevin said in his first sentence.

$2$. Again, what Kevin said. Take preduals $X$ and $Y$ of $\ell_1$, so that duals of both $N(X)$ and $N(Y)$ are isometric to $B(\ell_1)$. $X$ and $Y$ need not be isomorphic. Probably you can prove e.g. if $X=C(A)$ and $Y=C(B)$ are non isomorphic $C(K)$ spaces with $A$, $B$ countable ordinals, then $N(X)$ and $N(Y)$ are not isomorphic. The proof would use the Szlenk index of the spaces. But just to see an example, all you need to do is fix $A$ and let $B$ vary through the countable ordinals, so that the sup over such $Y=C(B)$ of the Szlenk indices of $Y$ and hence of $N(Y)$ is the first uncountable ordinal, whence not all such $Y$ can embed into $N(X)$.

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What the heck?!? In my typing last paragraph begins with 2., but it comes out 1. How can that happen? –  Bill Johnson Jul 2 '11 at 17:45
    
I could get it to display as 2 only by placing it within dollar signs. Go figure. –  Bill Johnson Jul 2 '11 at 17:47
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@Bill: The software thinks it knows better than you. In this connection, Yuri Gurevich has on several occasions mentioned what is apparently a common saying in Russia: "What's worse than an idiot? An idiot with initiative." –  Andreas Blass Jul 2 '11 at 18:25
    
Christian Samuel has published a proof that for countable, compact $A$ and $B$, the spaces $\mathcal{N}(C(A))$ and $\mathcal{N}(C(B))$ are isomorphic only if $C(A)$ is isomorphic to $C(B)$; see Section 4 of On spaces of operators on $C(Q)$ spaces ($Q$ countable metric space) Proc. Amer. Math. Soc. 137 (2009), 965-970. This paper is currently available for free from Samuel's web page: samuel.u-3mrs.fr/op_CQ.pdf –  Philip Brooker Jul 3 '11 at 13:33
    
I presume that following Bill's suggested path, one wants to use the original definition of the Szlenk index (i.e., from Szlenk's paper) rather than the modified definition in terms of the fragmentation index with respect to the dual norm and the $w^\ast$-topology. –  Philip Brooker Jul 3 '11 at 13:41
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As Yemon mentioned ages ago (sorry!) I explored this a bit in http://arxiv.org/abs/math.FA/0604372

We say that a Banach algebra $A$ is a dual Banach algebra if $A$ is isomorphic to $E^*$ for some Banach space $E$, such that the multiplication in $A$ becomes separately weak$^*$-continuous. If $X$ is a dual space, then $B(X)$ is the dual of $N(X_*)$ but a little calculation shows that the multiplication is only weak$^*$-continuous on one side. To get a dual Banach algebra, you need $X$ to be reflexive.

My little result is that if $X$ is reflexive, and also has the approximation property, then $N(X)$ is the unique dual Banach algebra predual of $B(X)$. To be precise, if $B$ is another dual Banach algebra, and $\theta:B(X)\rightarrow B$ is a linear bijection, is bounded, and is an algebra homomorphism, then $\theta$ is necessarily weak$^*$-continuous (so I don't need to assume that $\theta$ is an isometry).

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