14
$\begingroup$

Are there some good questions on functional analysis whose solution depends on tools in set theory? My major is mathematical logic, I think tools in set theory, especially infinity combinatorics and forcing, should be used to solve some questions in functional analysis. For functional analysis, I just have read the main part of Conway's textbook. In this book, I have not found such questions.

$\endgroup$
5
  • 2
    $\begingroup$ "should" or "could" be used? $\endgroup$
    – Yemon Choi
    Mar 22, 2010 at 17:08
  • 1
    $\begingroup$ Tom, would you object less to the question if it asked about applying "advanced tools of set theory", or something along those lines? $\endgroup$ Mar 22, 2010 at 23:19
  • 1
    $\begingroup$ I kind of disagree with the premise. Almost all of functional analysis, as customarily phrased, depends on set theory. A Banach space is defined as a set equipped with certain structure. An operator is a function with certain properties. When you take the direct sum of vector spaces, you use the set-theoretic theorem/axiom that cartesian products of sets exist. When you define "invariant subspace", you use the notion of the image of a subset under a function. I know that's not what you're asking, but I think it's a disservice to set theory to speak as if it consists only of "exotic stuff". $\endgroup$ Mar 22, 2010 at 23:20
  • 1
    $\begingroup$ Mark, yes, I would object less, or not at all. But I think there's a bit more than that going on in the background, to do with social attitudes towards set theory and the extent to which it's viewed as esoteric. A MO comment isn't the place for an explanation, though. (By the way, I slightly edited my comment, i.e. deleted and replaced it, at the same time as you were writing your comment, which is why they appear out of order.) $\endgroup$ Mar 22, 2010 at 23:24
  • $\begingroup$ It's slightly strange (for me) to read you say, Tom, that "a Banach space is defined as a set equipped with certain structure", since I've had a vague but long-standing feeling that this POV has hindered functional analysts from fully profiting from categori(c)al ideas. But this is connected to my feeling that the forgetful functor from Ban to Set is not the most profitable one to lean on... $\endgroup$
    – Yemon Choi
    Mar 23, 2010 at 16:04

6 Answers 6

17
$\begingroup$

One of the most elementary examples is the use of the infinite version of Ramsey's theorem to prove Rosenthals $\ell_1$ theorem. See Chapter 10 in Albiac & Kalton, Topics in Banach Space Theory. You'll find ultraproducts in Chapter 11. More model theory than set theory, but still logic. Deeper set theory (large cardinals, for example) have also been used. Check out, e.g., Todorcevic, S* on MathSci Net for recent things.

$\endgroup$
3
  • $\begingroup$ This book should be what I was searching for. Thank you. $\endgroup$ Mar 22, 2010 at 14:44
  • 5
    $\begingroup$ Also, Ilijas Farah has several related results. $\endgroup$ Mar 22, 2010 at 17:12
  • 2
    $\begingroup$ Was just writing a long answer, but you beat me to a mention of Farah $\endgroup$
    – Yemon Choi
    Mar 22, 2010 at 17:16
16
$\begingroup$

The survey article ``Set theory and C*-algebras'' by Nik Weaver might have some things along the lines you are looking for ( Bull. Symbolic Logic Volume 13, Issue 1 (2007), 1-20; see also math/0604198 on the arXiv). For a particular example of Weaver's recent work in this area, see Akemann & Weaver's paper

Consistency of a counterexample to Naimark's problem

We construct a C*-algebra that has only one irreducible representation up to unitary equivalence but is not isomorphic to the algebra of compact operators on any Hilbert space. This answers an old question of Naimark. Our construction uses a combinatorial statement called the diamond principle, which is known to be consistent with but not provable from the standard axioms of set theory (assuming those axioms are consistent). We prove that the statement ``there exists a counterexample to Naimark's problem which is generated by $\aleph_1$ elements'' is undecidable in standard set theory.

There has also been some work by I. Farah on applying set-theoretical techniques to operator-algebraic problems.

$\endgroup$
2
  • 1
    $\begingroup$ Thanks for adding this Yemon. I think Farah's work is fascinating and many of the functional analysts that I know have told me similar things. $\endgroup$ Mar 22, 2010 at 20:25
  • $\begingroup$ Ah, I was going to mention Weaver, but you beat me to it! $\endgroup$ May 7, 2010 at 20:57
9
$\begingroup$

One interesting example is A discontinuous homomorphism from C(X) without CH, by W. Hugh Woodin, which begins with the following introduction.

Suppose that X is an infinite compact Hausdorff space and let C(X) be the algebra of continuous real-valued functions on X. Then C(X) is a commutative Banach algebra relative to the supnorm: || f || = sup{ f(p) | p ∈ X }. A well-known question of I. Kaplansky posed around 1947 asks whether every algebra homomorphism of C{X) into a Banach algebra B is necessarily continuous.

There is a discontinuous homomorphism of C(X) if and only if there is a discontinuous homomorphism of C(X, C), the C*-algebra of continuous complexvalued functions on X. We prefer to deal with the real case; some of the references adopt the complex view. The question is now known to be independent of the axioms of set theory, ZFC. H. G. Dales [1] and J. Esterle [3] independently constructed discontinuous homomorphisms of C{X) for any infinite space X assuming the Continuum Hypothesis, CH. About the same time R. Solovay [7] proved that it is relatively consistent with ZFC that every homomorphism of C(X) for any space X is necessarily continuous. Solovay's result was improved [8] fairly soon thereafter, to obtain the relative consistency with ZFC + Martin's Axiom (ZFC + MA) that every homomorphism of C(X) for any space X is continuous. We refer the reader to [2] for an exposition of the latter result concerning MA, historical points and related results. After these results several questions remained. This paper is concerned with the question of whether the existence of a discontinuous homomorphism of C{X) is possible given the failure of the Continuum Hypothesis.

A standard method to obtain the consistency of a proposition with the negation of the Continuum Hypothesis (¬CH) when a proof of the proposition assuming CH is known is to attempt to use MA + ¬CH in place of CH and prove the proposition. However by the result indicated above, this approach will not succeed. The main theorem of this paper, formulated using the terminology of forcing, is the following.

THEOREM. Assume CH. Let P be the Cohen partial order for adding ω2 Cohen reals. Then in VP there exists a discontinuous homomorphism of C(X)for every infinite compact Hausdorff space X.

$\endgroup$
3
  • $\begingroup$ en. This is a good example. Also, do you know any questions as this style which is still open now? $\endgroup$ Mar 22, 2010 at 14:32
  • 3
    $\begingroup$ Woodin mentions the principal open question in this area on the second page of that article, which is whether ZFC+MA+not-CH is consistent with a discontinuous homomorphism. But I would expect that any such question is hard, given that Solovay and Woodin have both worked on this topic. $\endgroup$ Mar 22, 2010 at 14:37
  • 3
    $\begingroup$ There are many questions in this field worth working on. To get a good idea, you may want to consult Dales's monograph on Automatic continuity and his survey papers; for closely related material, se ethe book by Dales and Woodin on Super-real fields. $\endgroup$ Jan 20, 2011 at 5:51
8
$\begingroup$

Chris Phillips and Nik Weaver wrote a paper called The Calkin Algebra has Outer Automorphisms, where they showed that the Continuum Hypothesis implies that the Calkin algebra $\mathcal{B(H)/K(H)}$ has outer automorphisms. See also this paper of Farah, McKenney, and Schimmerling, and references therein; they show that it is relatively consistent with ZFC that the Calkin algebra has only inner automorphisms, and hence the question of existence of outer automorphisms is independent of ZFC.

$\endgroup$
2
  • 1
    $\begingroup$ That paper by Farah et al. dealt with generalized Calkin algebras. The consistency of all automorphisms being inner was shown in an earlier paper by Farah alone, All automorphisms of the Calkin algebra are inner (arXiv:0705.3085). $\endgroup$
    – Nik Weaver
    Apr 19, 2013 at 1:46
  • $\begingroup$ Sure, thanks for mentioning my paper. Incidentally, Ilijas gave another proof of our consistency result in his paper, and I think his proof is better. Our overall strategy seems more natural but making it work was really hard. His proof goes in an unexpected direction but it ends up being easier. $\endgroup$
    – Nik Weaver
    Apr 19, 2013 at 18:10
7
$\begingroup$

Although the construction of Tsirelson's space doesn't use set theory per se, in this short essay Tsirelson recounts (among other things) how his construction was inspired by forcing.

$\endgroup$
1
  • 2
    $\begingroup$ Thanks for the link, Mark. In a follow up paper to Tsirelson's, Figiel and I wrote "The 'forcing method' employed by Tsirelson..." because it reminded us of forcing as used in set theory, but we did not know that Tsirelson really had been exposed to forcing. $\endgroup$ Mar 22, 2010 at 18:02
5
$\begingroup$

Though this is a little more advanced, there is actually some very exciting research right now at the intersection of descriptive set theory, ergodic theory, and von Neumann algebras. It is quite striking that the three areas have powerful tools for looking at similar problems, and yet tend to be applicable in different cases. For a nice introduction to some of these ideas from a more set-theoretical point of view I would say check out "Topics in Orbit Equivalence" by Kechris and Miller.

doi:10.1007/b99421

Here is a link where you can download it (though you might need a subscription but many universities will have it so it should work on a department computer.) It is actually quite elementary, you need some basic descriptive set theory and measure theory, but arrives at quite deep theorems.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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