What is the "right" deifinition of the homology(cohomology) of an orbifold? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:13:57Z http://mathoverflow.net/feeds/question/77246 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77246/what-is-the-right-deifinition-of-the-homologycohomology-of-an-orbifold What is the "right" deifinition of the homology(cohomology) of an orbifold? Lin Jianfeng 2011-10-05T15:35:13Z 2011-10-05T16:37:47Z <p>What is the "right" analog in the orbifold case of a singular homology of a topological space? We can not just take the homology of the underlying space, because it does not contain much imformation. For example, is there any kind homology of orbifold such that the first homology is abelien of the fundamental group of the orbifold? And a n dim orbifold will have all homology group higher than n dim equal to zero? And if there is such a homology, will there be Poincare duality in the orbifold case???</p> <p>Thanks very much! </p> http://mathoverflow.net/questions/77246/what-is-the-right-deifinition-of-the-homologycohomology-of-an-orbifold/77252#77252 Answer by Angelo for What is the "right" deifinition of the homology(cohomology) of an orbifold? Angelo 2011-10-05T16:04:26Z 2011-10-05T16:04:26Z <p>One can define singular homology directly for an orbifold, mimicking the standard constructions, but I don't know if this has been written up. The alternative is to take a space that is homotopy equivalent to your orbifold, and take the homology of that. This has been worked out by Behrang Noohi in <a href="http://arxiv.org/abs/0808.3799" rel="nofollow">http://arxiv.org/abs/0808.3799</a>.</p> <p>If $G$ is a group, then the homology of the classifying orbifold of $G$ (the stack quotient of a point by the trivial action of $G$) is the group homology of $G$; so you see that it does not vanish above the dimension of the orbifold, which is 0, except in the trivial case.</p> http://mathoverflow.net/questions/77246/what-is-the-right-deifinition-of-the-homologycohomology-of-an-orbifold/77254#77254 Answer by Michael Joyce for What is the "right" deifinition of the homology(cohomology) of an orbifold? Michael Joyce 2011-10-05T16:08:44Z 2011-10-05T16:08:44Z <p>I do not know the current state of the art, but I can point you to the paper of <a href="http://arxiv.org/abs/math/0004129" rel="nofollow">Chen and Ruan</a> which should have everything you are asking for.</p> http://mathoverflow.net/questions/77246/what-is-the-right-deifinition-of-the-homologycohomology-of-an-orbifold/77257#77257 Answer by John Mangual for What is the "right" deifinition of the homology(cohomology) of an orbifold? John Mangual 2011-10-05T16:21:17Z 2011-10-05T16:21:17Z <p>A math definition of orbifold Euler characteristic appears in <a href="http://www.springerlink.com/content/r4t436n20942t2x9/" rel="nofollow">On the Euler Number of an Orbifold</a>. If there is a group acting on a manifold $X / G$ we could try to write $\chi(X/G) = \frac{1}{|G|} \sum_{g \in G} \chi(X^g)$ where $X^g$ is the fixed-point set of $g \in G$. Instead it seems to be better to sum over conjugacy classes $[g]$ of $G$. $\chi(X,G) = \sum_{[g]} \chi(X/C(g))$ where $C(g) = \{ h: hgh^{-1}=g\}$ is the <em><a href="http://en.wikipedia.org/wiki/Centralizer_and_normalizer" rel="nofollow">centralizer</a></em> of $g \in G$. For any group action $|[g]|\cdot |C(g)| = G$.</p> <p>This definition was motivated by some physicists in the last 1980's and by the mid 90's this idea was extended to homology. See <a href="http://arxiv.org/abs/hep-th/9408074" rel="nofollow">A Strong Coupling Test of S-Duality</a> $H^*(X/G) = \bigoplus_g H^*(X^g)^{C(g)}$</p> <p>The direct sum of the centralizer-invariant part $[\cdot] ^{C(g)}$ of the cohomology classes $H^*(\cdot)$ of fixed point sets $X^g$. As $g$ runs over the conjugacy classes of $G$.</p> <p>This was used to find the Euler characteristic of the Hilbert scheme of points on a non-singular space (originally found by Lothar G&#246;ttsche) </p> <p>$\sum_{n=0}^\infty q^n \chi(X^{[n]}) = \frac{1}{\prod_{i=1}^\infty (1-q^n)^{\chi(X)}}$</p> <p>So there's connection to the Dedekind <a href="http://en.wikipedia.org/wiki/Dedekind_eta_function" rel="nofollow">&eta;-function</a>. This generating function might be different if $X$ itself is singular.</p> http://mathoverflow.net/questions/77246/what-is-the-right-deifinition-of-the-homologycohomology-of-an-orbifold/77260#77260 Answer by David Roberts for What is the "right" deifinition of the homology(cohomology) of an orbifold? David Roberts 2011-10-05T16:37:47Z 2011-10-05T16:37:47Z <p>If your orbifold is given to you as a proper etale Lie/topological groupoid, you can:</p> <ul> <li>take the nerve of this groupoid, to get a simplicial manifold/space, </li> <li>take the levelwise singular set to get a bisimplicial set, </li> <li>take the diagonal to get a simplicial set, </li> <li>take geometric realisation to get a topological space, </li> <li>and then take (co)homology. </li> </ul> <p>This recipe at the very least gives you the homotopy type of the orbifold as a space (to which you can apply the (co)homology functor you want), but it perhaps too big and unwieldy for your purposes. </p>