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Suppose $M$ is an oriented compact $n$-dimensional manifold, then Poincaré duality is an isomorphism \begin{equation} \alpha \in H^i(M,\mathbb{Z}) \mapsto [M] \cap \alpha \in H_{n-i}(M,\mathbb{Z}),~ \alpha \end{equation} where $[M]$ is the homology class associated to the manifold $M$. For details, see e.g. Algebraic Topology by Allen Hatcher.

In Griffiths and Harris' book Principles of Algebraic Geometry, the intersection of two homology classes $A$ and $B$ is defined, and if we let $A^\vee$ be the Poincaré dual of $A$, etc, then there is \begin{equation} A^\vee \cup B^\vee =(A\cap B)^{\vee} \end{equation}

I am wondering whether there are interesting explanations of this equality or interesting examples which could show the essence of it?

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This gives a geometric interpretation of the cup product. If you think about it, based just on the definition, it's hard to say what the cup product really "means". It looks like some kind of algebraic mumbo-jumbo with no easy geometric interpretation. Well, this formulation tells you that the cup product is dual to intersection of submanifolds. Now that's a nice geometric interpretation!

Here is an example. Consider the torus, $T = S^1 \times S^1$. You have two obvious generators for the homology, $a$ and $b$, which go exactly once around either the first $S^1$ or the second $S^1$. These two homology classes have Poincaré dual cohomology classes $\alpha$ and $\beta$.

Now how do you compute the cup products? For $\alpha \smile \beta$, your formula tells you that you can consider the intersection $a \frown b$. This is just a point. The Poincaré dual of a point is the top cohomology class $\omega \in H^2(T)$, hence $\alpha \smile \beta = \omega$. That was easy!

And let's say you also want to compute $\alpha \smile \alpha$ (nevermind that we know it's zero for algebraic reasons: $\alpha$ has odd degree...). Your formula tells you that you need to consider the intersection $a \frown a$. Of course you can't take the intersection randomly, it has to be transverse intersection. So you move $a$ a little bit to the right, it becomes $a'$, the same homology class, but now the intersection is empty (hence transverse): $a \frown a' = \varnothing$, the zero homology class. So $\alpha \smile \alpha = 0$, because the Poincaré dual of $0$ is $0$.

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