# Intersection of two homology classes

Studying the first pages of Gompf-Stipsicz's 4-Manifolds & Kirby Calculus forced me to worry about the geometric meaning of homology and cohomology classes; in particular page 7 contains the following sentence, which I feel quite obscure:

Let $X$ be a compact, oriented, topological 4-manifold. When $X$ is oriented, it admits a fundamental class $[X]\in H_4(X,\partial X; \mathbb Z)$ [which is, as explained before, a top-degree homology class generating $H^4(X,\partial X; \mathbb Z)$].

Definition. The symmetric bilinear form $$Q_X\colon H^2(X,\partial X;\mathbb Z)\times H^2(X,\partial X;\mathbb Z) \to \mathbb Z$$ defined by $Q_X(a,b)=\langle a\cup b,[X]\rangle$ is called the intersection form of $X$.

My first question is: how's that pairing $\langle-,-\rangle$ defined?

I think I have a slight confidence with the idea of homology-cohomology classes as geometric objects (immersed submanifold, cycles as sub-simplices...).

I'm also able to forsee that all the different pairing definable in various (co)homology theories can someway can someway be reconduced to a single idea.

So my second question is in fact a modified version of the first one: how can I link the former "intersection" pairing to the similar one $$\langle -,-\rangle_{dR}\colon H^k_{dR}(M)\times H^{n-k}_{dR}(M)\to \mathbb R\colon (a,b)\mapsto \int_M a\land b$$ in de Rham cohomology?

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You may want to look at cap product. Wedging ($k$-form) with a $(n-k)$ form and integrate over $M$ and capping a $k$-dimensional cohomology class with the fundamental class are the "same" thing, only that the latter one is the general construction for compact orientable (topological) manifolds. –  Soarer Jul 26 '11 at 19:41

## 1 Answer

Recall that $H_k(X)$ is a subquotient of $C_k(X)$, the dimension-$k$ singular simplices in $X$, and that $H^k(X)$ is a subquotient of $C^k(X)=\mbox{Hom}(C_k(X),\mathbb{Z})$. So a $k$-dimensional cohomology class is represented by a cocycle (up to coboundaries), and its evaluation on a homology class is easily seen to be well-defined: if $[\alpha]\in H^k(X)$ and $[c]\in H_k(X)$, then $\langle \alpha+\delta \beta, c + \partial d \rangle = \langle \alpha, c \rangle$ since $\delta \alpha=0$ and $\partial c=0$. (Recall the adjunction $\langle \delta \gamma , e \rangle = \langle \gamma, \partial e \rangle$ -- this is in fact nothing more than the definition of $\delta$.)

As for your second question, the wedge product of differential forms is the de Rham analogue of the cup product, and under Poincare duality, cup product in cohomology corresponds to intersection product in homology. In the de Rham setting, this is proved relatively early on in Bott & Tu's "Differential Forms in Algebraic Topology".

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And it should be pointed out that if cycles are represented by submanifolds, then generically they intersect in points, for dimension reasons, and these points can be given a sign, based on how the two orientations of the intersecting manifolds compare with the ambient orientation. Then the intersection pairing really does count these intersections, albeit with a sign. Greenberg and Harper's book has a detailed account, although I'm sure there are other good references. –  Grumpy Parsnip Aug 18 '11 at 12:52
even more is true. if two cohomology classes $x,y$ are such that their poicaré duals are represented by embedded submanifolds $A,B$, then the cup product $x \cup y$ may be computed by intersecting A and B transversally and then forming the poincaré dual of this intersection. –  mland Apr 29 '12 at 9:00
This extends the notion of intersection product to all homology classes (instead of just those that can be represented by submanifolds). –  Aaron Mazel-Gee Apr 29 '12 at 21:48