Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How does one define "predual" and the surrounding notions? More specifically:

Why must there be only one predual of $X$ when $X$ is a Banach space? What is the correct notion of similarity here that gives this uniqueness? Is it isomorphic homeomorphism of Banach spaces? I'm also interested in the corresponding algebraic statement. Is it true that if $V$ is a vector space, then it has at most one predual?

I have noticed from looking online that the predual of $B(H)$ is the trace class operators, and the predual of that is the compact operators, which strangely enough means that taking preduals doesn't always reduce the "size" of the space (I'm not able to be precise since I don't know the true meaning of the uniqueness of predual), even though in the algebraic setting, one always has the usual injection of a vector space into its dual. I suppose that this discrepancy is because in the analytic definition of dual, we require continuity, so that the dual vector of a vector $x$ in $X$ when $X$ is a Banach space need not actually be in the continuous dual $X^*$ of $X$?

share|cite|improve this question
What is true is that the predual of a von Neumann algebra is unique up to isometric isomorphism (that's part of Sakai's theorem characterizing von Neumann algebras as $C^\ast$-algebras that are dual spaces as Banach spaces). Generally, a Banach space has many preduals: $\ell^1$ has uncountably many pairwise non-isometrically isomorphic preduals (among which the function spaces $C(K)$ with $K$ a countable and compact Hausdorff space). – t.b. Jun 4 '12 at 18:22
Thanks, that helps. Also, it just occurred to me that although the trace class operators may be a subset of the compact operators, there can still be an injection, even a nonsurjective one, from the compacts to the trace class operators. – Jeff Jun 4 '12 at 18:27
After doing some internet searching, I have found a conspicuous level of avoidance of Sakai's theorem, as well as that people never define Von Neumann Algebras abstractly. Is this because the theorem is not considered pedagogically important, or because its proof is extremely difficult and not suitable for graduate students, or both? It seems I'd have to get a copy of Sakai's paper, which isn't available on Arxiv, not to mention I'm not even sure I have the prereqs to read it. – Jeff Jun 4 '12 at 18:42
A paper from the fifties on ArXiV? :) Here's the original. Sakai's theorem is proved e.g. around Cor. III.3.9, p.135 in Takesaki, vol. I. I'm not an operator theorist but my suspicion is that (apart from historical reasons) many vNa's people look at "in practice" arise concretely, so that's a natural framework in some sense. The theory can easily and elegantly developed from the abstract point of view, that's done e.g. in Sakai's book. – t.b. Jun 4 '12 at 19:00

As pointed out in comments, preduals of general Banach spaces are not unique, even up to (non-isometric) isomorphism. For example, see the paper by Benyamini and Lindenstrauss, A predual of $\ell_1$ which is not isomorphic to a $C(K)$ space (Israel J. Math. 13 (1972), 246-254) or related MathOverflow threads Preduals of B(E) and separable $L^1$ predual.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.