12
$\begingroup$

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?

Improve this question
$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Tyler Lawson
    Commented Nov 14, 2011 at 17:59
  • 2
    $\begingroup$ 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 $TMF$. $\endgroup$
    – John Rognes
    Commented Nov 16, 2011 at 17:47

2 Answers 2

Reset to default
10
$\begingroup$

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".

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ It's quite likely that either Neil or Charles will be able to answer your question in more depth. $\endgroup$
    – Tyler Lawson
    Commented Nov 13, 2011 at 22:15
  • 1
    $\begingroup$ I don't know what I could say that's not in the ABG paper. $\endgroup$
    – Charles Rezk
    Commented Nov 13, 2011 at 23:30
  • 6
    $\begingroup$ 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 $BBU_{\otimes}$. $\endgroup$
    – Neil Strickland
    Commented Nov 14, 2011 at 9:25
5
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Thomas Nikolaus
    Commented Feb 4, 2012 at 16:12
  • $\begingroup$ Sorry, BO and not BO(n) in the comment above. $\endgroup$
    – Thomas Nikolaus
    Commented Feb 4, 2012 at 16:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .