Preduals of B(E) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T01:51:39Z http://mathoverflow.net/feeds/question/69343 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69343/preduals-of-be Preduals of B(E) Ollie Margetts 2011-07-02T16:53:38Z 2011-08-06T12:56:46Z <p>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.</p> <p>If $E$ is a Banach space then is it known whether</p> <ol> <li><p>$B(E)$ is always a dual Banach algebra? </p></li> <li><p>The predual is always unique? </p></li> </ol> <p>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.</p> http://mathoverflow.net/questions/69343/preduals-of-be/69347#69347 Answer by Kevin Beanland for Preduals of B(E) Kevin Beanland 2011-07-02T17:25:34Z 2011-07-02T17:25:34Z <p>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 $B(X^*)$. </p> <p>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.</p> http://mathoverflow.net/questions/69343/preduals-of-be/69349#69349 Answer by Bill Johnson for Preduals of B(E) Bill Johnson 2011-07-02T17:43:24Z 2011-07-02T17:43:24Z <ol> <li>No, not even isomorphically. Take an $E$ that is not complemented in its bidual (and hence not complemented in any dual space).</li> </ol> <p>1.1. What Kevin said in his first sentence.</p> <p>$2$. 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 $N(X)$. </p> http://mathoverflow.net/questions/69343/preduals-of-be/72237#72237 Answer by Matthew Daws for Preduals of B(E) Matthew Daws 2011-08-06T12:56:46Z 2011-08-06T12:56:46Z <p>As Yemon mentioned ages ago (sorry!) I explored this a bit in <a href="http://arxiv.org/abs/math.FA/0604372" rel="nofollow">http://arxiv.org/abs/math.FA/0604372</a></p> <p>We say that a Banach algebra $A$ is a dual Banach algebra if $A$ is isomorphic to <code>$E^*$</code> for some Banach space $E$, such that the multiplication in $A$ becomes separately weak<code>$^*$</code>-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<code>$^*$</code>-continuous on one side. To get a dual Banach algebra, you need $X$ to be reflexive.</p> <p>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).</p>