# Intersection pairing in cohomology and Poincare duality

This question is similar to but not exactly the same as the old question (the issue of torsion).

In Exercise 11.8 of Dan Freed's notes he claims that the following pairing being nondegenerate is equivalent to Poincare duality $$\bar{I}_M: \text{Free}\,H^{2k}(M;\mathbb{Z}) \times \text{Free}\,H^{2k}(M;\mathbb{Z}) \longrightarrow \mathbb{Z}$$

My question is, how should we interpret the following possibly degenerate (when $H^{2k}(M;\mathbb{Z})$ has torsion) pairing using Poincare duality? $$I_M: H^{2k}(M;\mathbb{Z}) \times H^{2k}(M;\mathbb{Z}) \longrightarrow \mathbb{Z}$$ Does it invalidate the duality (of course not)?

• I wouldn't call that pairing degenerate. I would call a pairing of a finitely generated abelian group with itself nondegenerate if it's nondegenerate on the free quotient (when you mod out by torsion). Your pairing kills torsion, so I'm not sure what the question is.
– user98602
Commented Aug 2, 2016 at 23:09
• @MikeMiller: by degenerate what I mean is if you put in a class $c_1$ belonging to the torsion subgroup then $I_M(c_1,c_2) = 0$ for any cohomology class $c_2$, then we cannot identify it as a linear functional of the cohomology group and in turn a homology class (the goal is to write it in the standard form of Poincare duality $H_{2k}(M;\mathbb{Z}) \cong H^{2k}(M;\mathbb{Z})$) Commented Aug 2, 2016 at 23:49
• Yeah, I don't really understand Freed's point. But if you can show that the pairing is nondegenerate with coefficients in any field $\Bbb F$, you've certainly proven Poincare duality.
– user98602
Commented Aug 4, 2016 at 16:55
• @MikeMiller: thanks Mike! Maybe torsion is the obstruction to extending the nondegeneracy to arbitrary field and that is the reason why Freed chooses the pairing on the free part not the entire cohomology group. Commented Aug 5, 2016 at 0:21
• That's true, but he also just says that - the pairing necessarily kills torsion, so he quotients out by torsion.
– user98602
Commented Aug 5, 2016 at 0:22