most recent 30 from http://mathoverflow.net 2013-06-19T22:52:18Z http://mathoverflow.net/feeds/question/80844 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80844/twists-of-k-theory-and-tmf Ulrich Pennig 2011-11-13T21:08:04Z 2012-02-04T15:54:12Z

<p>I read in a <a href="http://arxiv.org/abs/math/0402082" rel="nofollow">paper</a> 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 $BBU_{\otimes} \to TMF$?</p> <p>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?</p>

http://mathoverflow.net/questions/80844/twists-of-k-theory-and-tmf/80848#80848 Tyler Lawson 2011-11-13T22:09:15Z 2011-11-13T22:15:40Z

<p>Inspired by the title of your question, you should look at <a href="http://arxiv.org/abs/1002.3004" rel="nofollow">Twists of K-theory and TMF</a> by Ando-Blumberg-Gepner. For twists of $TMF$, there is a map <code>$K(\mathbb{Z},4) \to BGL_1(TMF)$</code>, 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 <code>$BBU_\otimes$</code>.</p> <p>It depends on the existence of an orientation constructed by Ando, Hopkins, and Rezk in "<a href="http://www.math.uiuc.edu/~mando/papers/koandtmf.pdf" rel="nofollow">Multiplicative orientatons of KO-theory and the spectrum of topological modular forms</a>".</p>

http://mathoverflow.net/questions/80844/twists-of-k-theory-and-tmf/87533#87533 Thomas Nikolaus 2012-02-04T15:48:25Z 2012-02-04T15:54:12Z

<p>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.</p>