# Cobordism of orbifolds?

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?

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.

• Joyce writes on various kinds of bordism and cobordism for orbifolds in his book available at people.maths.ox.ac.uk/joyce/dmbook.pdf, Section 13.3. He also introduces bordism and cobordism for what he calls d-orbifolds. Using those, you should be able to treat the spaces M_g,n (X,A) even if they are not smooth. Dec 5, 2012 at 20:32
• In the 1960's, Conner and Floyd used cobordism theory to study actions of finite groups on compact manifolds. The theory is explained in Differentiable Periodic Maps, LNM 738, 1979. In my understanding, the main problem in the equivariant setting (and therefore for orbifolds) is that transversality does not always hold, so homotopy alone does not capture the full story. Dec 5, 2012 at 23:25