most recent 30 from http://mathoverflow.net 2013-05-25T01:40:15Z http://mathoverflow.net/feeds/question/115528 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115528/cobordism-of-orbifolds unknown (google) 2012-12-05T18:48:07Z 2012-12-05T23:24:18Z

<p>Is it possible to setup classical cobordism theory in the context of orbifolds? For example, let's consider the free abelian group generated by oriented smooth orbifolds and quotient by those which are null cobordant (that is, they are the boundary of a smooth oriented orbifold with boundary). Is the resulting ring expressible as the homotopy groups of some Thom spectrum? Can we prove this just by smoothly embedding an orbifold in $\mathbb R^n/S_n$ and following the classical proof?</p> <p>My motivation is that in Gromov--Witten theory, the moduli space of stable maps $\bar M_{g,n}(X,A)$ is a smooth oriented orbifold (assuming it is cut out transversally, and assuming we have smooth charts for gluings) and it is defined up to cobordism. Thus instead of taking its fundamental class and pushing forward to $H_\ast(\bar M_{g,n}\times X^n)$ to get Gromov--Witten invariants, we could consider the class it represents in the generalized cohomology theory which we might call "oriented orbifold cobordism" of $\bar M_{g,n}\times X^n$, and get a slightly more refined invariant.</p>

http://mathoverflow.net/questions/115528/cobordism-of-orbifolds/115533#115533 David C 2012-12-05T19:08:46Z 2012-12-05T19:08:46Z

<p>Andres Angel has written some papers on this subject, what I know is a preprint entitled "Orbifold cobordism" where he computes some orbifold cobordism rings in terms of bordism groups of classifying spaces of groups.</p> <p>Papers available here: <a href="http://www.math.uni-bonn.de/people/aangel79/papers.html" rel="nofollow">http://www.math.uni-bonn.de/people/aangel79/papers.html</a></p>