Does the Deligne-Mumford space module $S_{n}$ action have a fundamental chain?

Question

Does the Deligne-Mumford space (without ordering for marked points) $\bar M_{g,n}/S_{n}$ has fundamental chain in signular simplicial chains? (because I read Costello's paper GW potential to TCFT, as a orbifold,what's the fundamental chain?)

Answer 1

A compact orbifold without boundary will have a fundamental chain in rational singular homology if and only if it is orientable. The fundamental chain will satisfy the same sorts of properties as for a manifold. In particular, collapsing the complement of a small disc will send the fundamental chain to the generator of $H_n(D^n,\partial D^n)$. The reason that closed oriented orbifolds have fundamental chains in rational homology is essentially because, to the eyes of rational homology, they look like manifolds.

So the first part of what you are asking amounts to the question of whether $\overline{\mathcal{M}}_{g,n}/S_n$ is orientable or not.

Just so we're clear, $\overline{\mathcal{M}}_{g,n}$ is the moduli space of stable curves of genus $g$ with $n$ marked points labelled $1 \ldots n$. The action of the symmetric group $S_n$ is by permuting the labels of the marked points. Before taking the quotient by $S_n$, the space $\overline{\mathcal{M}}_{g,n}$ is a complex orbifold, so the complex structure induces a well-defined orientation just as for complex manifolds.

So the question is now whether the $S_n$ action preserves the orientation or not. The answer is that it does indeed. The symmetric group action is in fact algebraic, so it preserves the complex structure and hence the orientation.

As an aside: There is a homotopy equivalence between the moduli space $M^{rib}_{g,n}$ of metric ribbon graphs (that thicken to a genus $g$ surface with $n$ labelled boundary components) and the uncompactified moduli space $\mathcal{M}_{g,n}$. The action of $S_n$ on $M^{rib}_{g,n}$ is not orientation preserving. This is because, via Strebel differentials (or Penner's Lambda lengths) one can show that $M^{rib}_{g,n}$ is homeomorphic to $\mathcal{M}_{g,n} \times \mathbb{R}_+^n$ and under this homeomorphism the symmetric group action corresponds to the usual action on the first factor and the permutation action on the coordinates of the second factor. Thus it doesn't act in an orientation preserving way.

Answer 2

A pseudomanifold is a finite-dimensional topological space $X$ (say Hausdorff and locally compact) that admits a closed subspace $Y$ of dimension $\dim X-2$ such that $X-Y$ is a manifold (see e.g. Goresky-MacPherson, Intersection homology 1, Topology 19, 135-162). If in addition to that $X-Y$ is connected and orientable, then there is a fundamental (cellular) chain: the homology long exact sequence inplies that the $H_{\dim X}(X,\mathbf{Z})=\mathbf{Z}$; take any representative of any of the two generators and this will be a fundamental chain. If $X$ is a CW-complex and $Y$ is a subcomplex, then a cellular fundamental chain can be constructed as follows: take the sum $[X]$ of all cells of the highest dimension with the orientation induced by some orientation of $X-Y$. This is a cycle since any cell of dimension one less will occur twice in $\partial [X]$, once with a plus and once with a minus.

The above conditions are satisfied if $X$ is an irreducible compact complex algebraic variety and $Y$ is a closed subvariety, since by Lojasiewicz's theorem one can triangulate $X$ so that $Y$ is a subpolyhedron. If $X$ is complex algebraic but not compact and $Y$ is still closed, then the above construction still works but the number of simplices will no longer be finite, so the fundamental class will live in the Borel-Moore homology.

If you have a finite group acting biregularly on an irreducible complex algebraic variety, then the quotient is not necessarily algebraic: it may happen that some orbits do not lie inside an affine open subset. But the fundamental class exists nonetheless: take the union of the singular locus and all points with nontrivial stabilizers. This is a subvariety whose real codimension is at least 2. Using the homology long exact sequence again one can see that the highest homology group is $\mathbf{Z}$, so any representative will ba a fundamental chain

Remark: the argument in the above paragraph is not really necessary in the case you are interested in. The quotient of the Deligne-Mumford compactifications of the moduli spaces of curves by the symmetric groups are algebraic and are coarse moduli spaces for the functor ``a family of smooth or nodal curves plus a set-valued section that does not intersect the nodes''.