Cup Product and Ring Structure

The cup product endows cohomology with the structure of a graded ring, providing strictly finer invariants than the additive structure of cohomology groups alone.

Definition of the Cup Product

Definition

Let $\varphi \in C^p(X; R)$ and $\psi \in C^q(X; R)$ be cochains with coefficients in a commutative ring $R$ . The cup product $\varphi \smile \psi \in C^{p+q}(X; R)$ is defined on a singular $(p+q)$ -simplex $\sigma : \Delta^{p+q} \to X$ by $(\varphi \smile \psi)(\sigma) = \varphi(\sigma|_{[v_0, \ldots, v_p]}) \cdot \psi(\sigma|_{[v_p, \ldots, v_{p+q}]})$ where $\sigma|_{[v_0, \ldots, v_p]}$ denotes the front $p$ -face and $\sigma|_{[v_p, \ldots, v_{p+q}]}$ denotes the back $q$ -face of $\sigma$ .

The cup product satisfies the Leibniz rule $\delta(\varphi \smile \psi) = (\delta \varphi) \smile \psi + (-1)^p \varphi \smile (\delta \psi)$ , which ensures it descends to cohomology.

Definition

The cohomology ring of $X$ with coefficients in $R$ is $H^*(X; R) = \bigoplus_{n \geq 0} H^n(X; R)$ equipped with the cup product $\smile : H^p(X; R) \otimes H^q(X; R) \to H^{p+q}(X; R)$ defined by $[\varphi] \smile [\psi] = [\varphi \smile \psi]$ . This makes $H^*(X; R)$ a graded-commutative $R$ -algebra.

Graded Commutativity

Theorem7.3Graded Commutativity

For cohomology classes $\alpha \in H^p(X; R)$ and $\beta \in H^q(X; R)$ , we have $\alpha \smile \beta = (-1)^{pq}\, \beta \smile \alpha$ In particular, if $\alpha \in H^{2k+1}(X; R)$ and $R$ has no $2$ -torsion, then $\alpha \smile \alpha = 0$ .

ExampleDistinguishing spaces via cup product

The spaces $\mathbb{CP}^2$ and $S^2 \vee S^4$ have isomorphic cohomology groups: $H^0 \cong H^2 \cong H^4 \cong \mathbb{Z}$ and all others zero. However, if $\alpha$ generates $H^2(\mathbb{CP}^2)$ , then $\alpha \smile \alpha$ generates $H^4(\mathbb{CP}^2)$ . In contrast, $H^2(S^2 \vee S^4) \smile H^2(S^2 \vee S^4) = 0$ . Thus the cup product structure distinguishes these spaces.

Naturality and Applications

RemarkRing homomorphisms from continuous maps

A continuous map $f : X \to Y$ induces a ring homomorphism $f^* : H^*(Y; R) \to H^*(X; R)$ , since $f^*(\alpha \smile \beta) = f^*(\alpha) \smile f^*(\beta)$ . This multiplicative compatibility gives the cup product its power: it converts topological maps into algebra homomorphisms, where the algebraic constraints are often very rigid.

The cohomology ring is the primary invariant distinguishing spaces that homology alone cannot tell apart. Its computation is central to obstruction theory, characteristic classes, and the classification of fiber bundles.