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May 22, 2014 at 0:16 comment added David Roberts @AndréHenriques - it may just be that the answer is "no", or better, "no, the derived space is this: ..."
May 21, 2014 at 11:54 comment added David Roberts Also, I gave the definition as in the Baez-Schreiber paper: your edit (replacing the dual of the wedge of the cotangent bundle with the wedge of the tangent bundle) makes this one step removed from what they actually wrote.
May 21, 2014 at 11:52 comment added David Roberts @AndréHenriques - it's not my groupoid. A physicist told me that higher bundles over this thing should be interesting, and my guess as outlined in the post (that it is somehow the derived fixed-point locus) is the only reason I could think why.
May 21, 2014 at 8:17 history edited André Henriques CC BY-SA 3.0
removed an unnecessary double dualization.
May 21, 2014 at 8:15 comment added André Henriques I'm not convinced that your groupoid is very interesting... Maybe a better question is: what is the derived fixed-point locus of the free loop space? -- is there a down-to-earth (i.e. finite dimensional) description of that derived space?
May 21, 2014 at 7:59 comment added André Henriques I thought that the derived loop space $map(S^1,M)=map((\text{cohomology of $S^1$}),M)=map((\text{odd line}),M)=\Pi TM$ is the odd tangent bundle of $M$ (the space such that functions on it are differential forms on $M$). Is that right? That doesn't look like the gerbe you describe in your first paragraph.
May 21, 2014 at 7:04 history edited David Roberts CC BY-SA 3.0
added 242 characters in body
May 21, 2014 at 7:02 comment added David Roberts @AndréHenriques- well, it's in the title, but I guess I can make it more clear.<br> -Done!
May 21, 2014 at 7:01 comment added André Henriques Hi David. It's not quite clear what you're asking exactly. Can you be more precise?
May 21, 2014 at 6:58 history asked David Roberts CC BY-SA 3.0