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Sep 27, 2015 at 20:05 vote accept user51223
Jul 30, 2015 at 11:28 comment added Dmitri Pavlov @user51223: I am not sure what your definition of “moduli space” is in the first place, but Madsen and Weiss's M(F) corresponds to my F(pt) (not F[pt]).
Jul 30, 2015 at 8:27 comment added user51223 Actually, as I understand it, the bordism group that you mentioned is not what we get as a moduli space just from plane definition; compare to the definition on Second page of Madsen and Weiss. In fact, I think, in the particular case of surfaces, the discussions on the first pages of Madsen-Weiss answers my question. But, I am not sure that in general, at least in the case of higher dimensional manifolds, $F[pt]$ becomes the moduli space.
Jul 29, 2015 at 17:05 comment added Dmitri Pavlov @user51223: The general framework in the paper of Galatius and Randal-Williams is the same as in the GMTW paper, which I described above.
Jul 29, 2015 at 17:04 comment added Dmitri Pavlov @user51223: In particular, this gives you the homotopy equivalence (1) that you asked for. As for (2), F[pt] in this case is simply the bordism group of (d−1)-manifolds, which I presume is what you mean by M_C.
Jul 29, 2015 at 16:47 comment added Dmitri Pavlov @user51223: Galatius—Madsen—Tillmann—Weiss apply this construction to several different stacks, see the very first paragraph in §4 of their paper. They show that the concordance classifying space of D_d gives the Madsen—Tillmann spectrum and all three maps induce weak equivalences of concordance classifying spaces, which proves their theorem.
Jul 28, 2015 at 16:02 comment added user51223 +as you may know, a work of Galatius and Randal-Williams has a ``weaker'' analogue of Mumford's conjecture where they replace $\Gamma_\infty^+$ with the diffeomorphism group of $g$-fold connected sum of $D^{n}\times D^n$ if I am not mistaken. Of course, I understand that one difficulty is that we don't know about the classification of higher dimensional manifolds, so talking about the moduli space of these objects maynot give the right feeling/construction, whereas in the case of dimension $2$ one knows that oriented surfaces are classified by genus. Shouldn't they consider the above approach?
Jul 28, 2015 at 15:56 comment added user51223 So, for $X=pt$, $F[X]$ has to be the moduli space that one has to look for. Working backwards, given a category $\mathbf{C}$ related to manifolds at least, a possible strategy to get a positive answer to the above question is that one may look for a sheaf or stack $F$ so that 1) there is a homotopy equivalence $CF(pt)\to B\mathbf{C}$; 2) there is a homotopy equivalence $F[pt]\to \mathcal{M}_\mathbf{C}$. Of course, I am still thinking about the work of Randal-Williams and Ebert who showed that higher dimensional analogue of Mumford's conjecture does not hold. How would you comment on that?
Jul 28, 2015 at 12:32 comment added Dmitri Pavlov @user51223: I added a detailed description of the map. Note that for the case of sets (i.e., discrete spaces) this is the same map that Galatius—Madsen—Tillmann—Weiss use in the formula (2.7) in their paper.
Jul 28, 2015 at 12:20 history edited Dmitri Pavlov CC BY-SA 3.0
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Jul 28, 2015 at 11:20 comment added user51223 Thank you very much for the answer. However, I still couldn't follow how you get a map from the classifying space to the moduli space/stack or the way around?!? Even, in the case of Madsen-Tillmann map, this does not happen, unless we accept that the moduli space of Riemannian surfaces is that same as $B\Gamma_\infty^+$?!
Jul 27, 2015 at 20:07 history answered Dmitri Pavlov CC BY-SA 3.0