Geometrical interpretation of cup product.

Question

I know geometrically the meaning of cohomology groups of topological spaces. Is their any geometrical interpretation of cup product?

Answer 1

If you are willing to restrict your attention to a compact, connected, oriented manifold $M$, rather than a general topological space, then yes, there is a geometric interpretation. It's called the Poincare Duality Theorem.

In brief, letting $m = \text{dimension}(M)$, the Poincare Duality isomorphism $f : H^i(M) \to H_{m-i}(M)$ (I'll use $\mathbb{R}$-coefficients) produces for each $c \in H^i(M;\mathbb{R})$ an $m-i$ dimensional chain representing the homology class $f(c)$ which I'll denote $c^\perp$. This has the property that given $c \in H^i(M)$ and $d \in H^j(M)$, and assuming that the chains $c^\perp$ and $d^\perp$ are chosen to be transverse to each other in an appropriate sense, $f(c \cup d) \in H_{m-(i+j)}(M)$ is defined by the $m-(i+j)$ dimensional "intersection chain" $c^\perp \cap d^\perp$.

Just as an example, if $i+j=m$ then $c \cup d$ is equal to a scalar multiple of the "fundamental" cohomology class. In this case, to say that $c^\perp$ and $d^\perp$ are transverse to each other implies that $c^\perp \cap d^\perp$ is a set of points with signed numbers on them, i.e. a 0-chain, the scalar is equal to the sum of these numbers.