# Intersection paring on homology and cup product cohomology?

One advantage of working with cohomology rather than homology is that cohomology is naturally endowed with a ring structure via cup product. However the cup product is often understood via intersection pairing on homology classes via Poincare duality. In this sense, I think homology is also naturally equipped with a ring structure.

My question is why people say that only cohomology is naturally a ring but not homology. Of course if the space is singular or non-compact, the intersection paring won't work well. Is this the only reason? If the space is good enough, say smooth compact manifold, then the ntersection paring on homology and the cup product cohomology are the same?

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There are several reasons for this.

One reason is, as you already said, that for a general space, what is the intersection product supposed to be?

For smooth manifolds usually the idea is as follows. Given to smooth submanifolds, $N_1$ and $N_2$ of a manifold $M$, say of dimension $d_i = \dim N_i$ and $n=\dim M$, then they should represent an element in $H_*(M)$ the homology on $M$. So how does this work? Well, if both $N_i$ are compact and orientable, they have a fundamental class $\lbrack N_i \rbrack \in H_{d_i}(N_i)$ which gives us an element $i_*(\lbrack N_i \rbrack) \in H_{d_i}(M)$ via the inclusion of the submanifolds.

Now if also $M$ is closed and oriented we have Poincar\'e duality and hence get classes $\alpha_i \in H^{n-d_i}(M)$ which are Poincar\'e dual to these homology classes $i_*(\lbrack N_i \rbrack)$.

The natural question to ask is whether the cup product $\alpha_1 \cup \alpha_2 \in H^{2n-d_1-d_2}(M)$ allows any geometric interpretation. Now how can one make precise what "geometric interpretation" shall be for cohomology classes?

Well, again it is natural to ask that its Poincar\'e dual $PD(\alpha_1\cup\alpha_2) \in H_{d_1+d_2-n}(M)$ is represented by a submanifold (in the same manner as before, by pushing forward the fundamental class along the inclusion).

So note that there are two steps going on here, firstly you want that this homology class IS represented by SOME submanifold, and now you want to know, if this is the case, can it be represented by a submanifold which also has something to do with the manifolds you started out to begin with (the $N_i$).

It turns out, that given the situation I have described, it is indeed true that $PD(\alpha_1\cup\alpha_2)$ is represented by a nice submanifold, namely by a transverse intersection of the two submanifolds $N_1$ and $N_2$.

This means, that you morally want to take the intersection of the two submanifolds (which should then ideally have dimension $d_1+d_2 -n$). But in general this is not a submanifold, so you have to change this intersection to make them intersect transversally, then the intersection indeed is a smooth submanifold.

So let me put together what we have used, and what we were able to say in certain special cases.

In order to get something like an intersection product, you need a relation between (suitably nice) subspaces of your space and homology classes in the big space that are represented by these subspaces. (This is possible for smooth manifolds).

Moreover you would want that every homology class can be obtained in this way (THIS for example is in general NOT possible even for smooth manifolds, indeed Peter Teichner constructed a concrete 6 manifold, and a cohomology class whose poincar\'e dual is not represented by any manifold).

So indeed, via Poincar\'e duality, there is a ring structure on homology, but in general you cannot describe it in terms of the geometry of the manifold itself, you really have to use Poincar\'e duality and the cup product.

So I think, once one understands what it really means, that the intersection product in homology is "dual" to the cup product in cohomology, you see why in general you should not expect a ring structure on (integral singular) homology of an arbitrary space.

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Thank you for the detailed answer. Yes, this is exactly what I was vaguely thinking about but not really convinced. So I really need cohomology to think of the ring structure. –  M. K. Dec 10 '12 at 4:07