7
$\begingroup$

I have posed this question to some experts at my university who would probably know the answer if there were a complete one, so my expectations are limited. It's possible that the question deserves the "open problems" tag, but I'm not that certain.

Here is the setup. By "K-homology" I mean the generalized homology theory dual to topological K-theory. The model of K-homology most relevant to this question, due to Kasparov, comes from functional analysis. Given a locally compact Hausdorff space $X$, one defines a Fredholm module to be a triple $(\rho,H,F)$ where $H$ is a Hilbert space, $\rho$ is a representation of the C*-algebra $C_0(X)$ on $H$, and $F$ is a bounded operator on $H$ which is compatible with the representation in a suitable sense. In Kasparov's model, the K-homology groups of $X$ are generated by equivalence classes of Fredholm modules where the equivalence relation identifies Fredholm modules which are operator homotopic, unitarily equivalent, or compact perturbations of one another.

Baaj and Julg proved that the K-homology groups of a space (in fact, the KK groups of a pair of C*-algebras) are generated by "unbounded Fredholm modules" in which the operator above is a densely defined unbounded operator (and some additional axioms are satisfied). Specifically, they associated to any unbounded Fredholm module an ordinary (bounded) Fredholm module and showed that all K-homology classes have an unbounded representative. This implicitly yields an unbounded model for K-homology: just say two unbounded Fredholm modules are equivalent if the associated bounded Fredholm modules are.

My question is: can anyone express the equivalence relation on unbounded Fredholm modules which determines K-homology more explicitly (i.e. without referring to the bounded theory)? There are some serious technical challenges; for example two Fredholm modules could be constructed from the same representation on the same Hilbert space but with operators whose domains are disjoint.

Searching the literature, I can find a number of papers written about how to define the Kasparov product in the unbounded model, but as far as I can tell nobody has sorted out the relations. Is there any progress on this?

$\endgroup$
2
  • 2
    $\begingroup$ Hmmm... knowing that you are at PSU, you have the best possible experts at hand... I think you can safely re-tag the question as "open problem"! $\endgroup$ Jan 26 '12 at 21:51
  • 1
    $\begingroup$ You picked a strange tag - "functinoal-analysis" -:)) Some weeks ago, I asked myself the same question, especially since Blackadar ''leaves'' this ''to the reader'' in his K-theory book, 2nd edition, bottom of page 165. Searching his references does not yield any result. $\endgroup$ Jan 26 '12 at 22:15
1
$\begingroup$

Some recent progress in addressing this open question can be found in https://arxiv.org/pdf/1503.07398.pdf. They also explicitely cite Blackadar's question in their introduction

“We leave to the reader the task of appropriately formulating the equivalence relations on unbounded cycles corresponding to the standard relations on bounded cycles.”

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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