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

I read in a paper by Christopher Douglas that third cohomology twists of $K$-theory may be interpreted as TMF-classes via a map $K(\mathbb{Z},3) \to TMF$, which is related to String orientations. How exactly is this map constructed? Could it be that there is an extension to higher twists, i.e. is there an extension to $BBU_{\otimes} \to TMF$?

EDIT: I know that $BBU_{\otimes}$ splits off $K(\mathbb{Z},3)$ as a factor. Therefore there is of course a map $BBU_{\otimes} \to TMF$, which factors over $K(\mathbb{Z},3)$. The corresponding classes of TMF, however, only see the ordinary third-cohomology twists. So, I reshould restate the second question as something like: Is there an extension $BBU_{\otimes} \to TMF$, which "sees" higher twists?

share|cite|improve this question
I can't answer your question about $BBU_\otimes$. All I can say is that one of the most natural places that possess those kinds of orientations is $K(ku)$, the algebraic K-theory of the complex K-theory spectrum $ku$. Ausoni-Rognes showed that $K(ku)$ possesses chromatic features similar to elliptic cohomology, and Baas-Dundas-Richter-Rognes showed that $K(ku)$ can be interpreted in terms of 2-vector bundles. So far as I know there isn't a direct comparison to elliptic cohomology theories. – Tyler Lawson Nov 14 '11 at 17:59
In case it was not clear -- the relation to $K(ku)$ comes from the maps $BBU_\otimes \to BGL_1(ku) \to BG_\infty(ku) \to K(ku)$. I would be happy just to see a ring spectrum map from $K(ku)$ to $E_2$, not necessarily factoring through $EO_2$ or $TMF$. – John Rognes Nov 16 '11 at 17:47

Inspired by the title of your question, you should look at Twists of K-theory and TMF by Ando-Blumberg-Gepner. For twists of $TMF$, there is a map $K(\mathbb{Z},4) \to BGL_1(TMF)$, and the latter classifies the most general twists that can occur. The discussion of what you're interested in starts on page 21, and twists of K-theory are indeed related to $BBU_\otimes$.

It depends on the existence of an orientation constructed by Ando, Hopkins, and Rezk in "Multiplicative orientatons of KO-theory and the spectrum of topological modular forms".

share|cite|improve this answer
It's quite likely that either Neil or Charles will be able to answer your question in more depth. – Tyler Lawson Nov 13 '11 at 22:15
I don't know what I could say that's not in the ABG paper. – Charles Rezk Nov 13 '11 at 23:30
I would just mention that ring maps $\Sigma^\infty_+K(\mathbb{Z},3)\to E$ (for suitable even periodic $E$) correspond to Weil pairings on the formal group for $E$. For any elliptic curve $C$ there is a canonical Weil pairing on $C$ and therefore on the infinitesimal part of $C$, which is a formal group. This provides a canonical map from $\Sigma^\infty_+K(\mathbb{Z},3)$ to any elliptic spectrum and thus (modulo handwaving) to $TMF$. This is the same as the map coming from the string orientation. I don't know if there is a similarly rich algebraic story for $BBU_{\otimes}$. – Neil Strickland Nov 14 '11 at 9:25

You can construct the map $K(\mathbb{Z},3) \to tmf$ as follows: first there is the String orientation of tmf, which you already mention. This is a map $$ MString \to tmf$$ Then String is by definition a $K(\mathbb{Z},2)$-fibration over Spin. This yields in particular a map $$ K(\mathbb{Z},3) \to MString $$ Then you can construct the map $K(\mathbb{Z},3) \to tmf$ as the composition of the above two maps. In order to extend this constuction you had to find a map $BBU_\otimes \to MString$. I think such a map does not exist apart from the one you describe, but I am not entirely sure.

share|cite|improve this answer
Actually there is a map from the fibre of $BString \to BO(n)$ to $MSting$. This is slightly biggher then $K(\mathbb{Z},3)$. I think this fibre is maybe the 3-type of the twists of K-Theory. – Thomas Nikolaus Feb 4 '12 at 16:12
Sorry, BO and not BO(n) in the comment above. – Thomas Nikolaus Feb 4 '12 at 16:12

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.