MathOverflow is a question and answer site for professional mathematicians. 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

There is a simple, intuitive "construction" of twisted K-theory if we are allowed to ignore that many things only hold up to homotopy. We know that maps to $K(Z,2)$ give line bundles on a space and that $K(Z,2)$ forms a group corresponding to the tensor product of line bundles. Line bundles also act as endomorphisms of K-theory given by the tensor product. Thus, there is an action of $K(Z,2)$ on $F$ (where $F$ is the classifying space for $K^0$). $K(Z,2)$ principal bundles are classified by maps to $BK(Z,2) \cong K(Z,3)$, ie, elements of $H^3$. Choosing such a map, we get a principal $K(Z,2)$ bundle, $E$, and we can form the associated bundle $E \times_{K(Z,2)} F$. Twisted K-theory is then the homotopy classes of sections of this bundle.

The usual constructions of twisted K-theory that I have seen make the above precise by choosing representatives of the relevant objects so that all the needed relations hold on the nose. My question is whether you can avoid doing that. In other words, can you define all the various notions up to homotopy and obtain a definition of twisted K-theory that way?

share|cite|improve this question
up vote 11 down vote accepted

The answer is yes if you're working on the level of $\infty$-categories (and I'm pretty sure no if you're working on the level of homotopy categories). In other words, in the $\infty$-world there's no problem talking about a principal bundle for K(Z,2)=BBZ on any space, and they're indeed classified by maps (in the $\infty$-category of spaces) to BBBZ, as are elements of $H^3$. Moreover for any spectrum $E$ there's a well defined group object $GL_1(E)$, and we have a map $BBZ\to GL_1(E)$ in the case of E=K-theory. For any principal $GL_1(E)$ bundle (eg one induced from a BBZ bundle) we have an associated "E line bundle", and its global sections are the twisted K-theory of your space (i.e. a spectrum whose homotopy groups are the usual twisted K-groups).

Or again in short, everything works intuitively as you think without chosing representatives or strictifying if you work in the wonderful world of $\infty$-categories.. for example, the idea of sheaves of spectra (hence twisted cohomology theories) are as simple to work with formally as ordinary sheaves. When it comes to calculating things.. well that's a whole other story.

share|cite|improve this answer
Thanx. I think I even remember you telling me this once :). Is there a reference (besides, I suppose, reading DAG I-DXVIII)?. – Aaron Bergman Nov 29 '09 at 23:04
The closest reference I can think of is by Ando, Blumberg, Gepner, Hopkins and Rezk (maybe Charles Rezk will weigh in more authoritatively). There are of course lots of treatments of related topics in model categorical language, in particular the monograph by May and Sigurdsson (0716 on the K-theory archive) (and I assume you've looked at the Freed-Hopkins-Teleman oeuvre).. but I do think the answer closest to what you ask for is an oo-answer.. – David Ben-Zvi Nov 29 '09 at 23:41

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.