Non-orientable manifolds and mod 2 homology

Question

I was reading the wonderful book "The wild world of 4-manifolds" by Alexandru Scorpan and I found the following sentence:

"We are able to orient $\mathfrak{M}$ (else we only get modulo 2 invariants)."

Here $\mathfrak{M}$ is the moduli space of instantons on a four-manifold. The thing is that I have seen similar statements before in the literature. These statements seem to associate singular homology $H^{\ast}(M,\mathbb{Z}_{2})$ with $\mathbb{Z}_{2}$ coefficients to non-orientable manifolds and singular homology $H^{\ast}(M,\mathbb{Z})$ with $\mathbb{Z}$ coefficients to orientable manifolds. However I don't understand this. Certainly one can define $H^{\ast}(M,\mathbb{Z}_{2})$ for any manifold $M$, oriented or not, and one can define $H^{\ast}(M,\mathbb{Z})$ again for any manifold, oriented or not. So why are the $\mathbb{Z}_{2}$ coefficients related to non-orientable manifolds?

Thanks.

Answer 1

I think the basic source of the statement comes from Poincare-duality. This builds on the fact that a closed oriented manifold has a fundamental holomology class in in $H_n(M,\mathbb Z)$, where $n=\dim(M)$. This class is obtained from local fundamental classes which are determined up to sign and the condition that these classes can be chosen consistently to fit together to a global class exactly is orientability. If one drops the orientability requirement, then there is no fundamental class, but the construction still can be pushed through for coefficients in $\mathbb Z_2$. Therefore any compact manifold admits a fundamental homology class in $H_n(M,\mathbb Z_2)$ and Poincare-duality works in general with $\mathbb Z_2$-coefficients.

Starting from this basic ingredient, it turns out that several constructions for oriented closed manifolds, which involve integral cohomology, admit an a non-orientable analog with $\mathbb Z_2$-coefficients.

Answer 2

I'm not sure I understand your quesiton. The Donaldson invariants are actually defined by taking $\mathcal M$ and intersecting it with submanifolds defined by homology classes $[\Sigma_i]$, and when you eventually get something of dimension 0, you count the number of points. (This defines a polynomial in $H^2(X;\Bbb Z)$, more or less.) This last bit - counting the points - is where we needed the orientation of $\mathcal M$, for then we get a signed count of points. If you cannot count signed points, you get mod 2 invariants, just like in degree theory/intersection theory. (I don't know if Scorpan talks about this, but the orientation of $\mathcal M$ depends on a little more than just the orientation of the underlying manifold; it depends also on something called a homology orientation. Not super important, though.)

I have never thought about the non-orientable theory, but the point is that when you cut down by surfaces in $2H^2(X;\Bbb Z)$, you end up with an even number of points, and hence your mod 2 invariants are zero. So they're really only $\Bbb Z/2$-valued polynomials in $H^2(X;\Bbb Z/2)$; there's no value whatsoever in defining it as a polynomial in $H^2(X;\Bbb Z)$ - we're just going to be writing down lots of extra zeroes.