Cup Product & the Cohomology Ring

Multiplying cohomology classes: the cup product on cochains, the graded commutative ring it builds, and the spaces it separates that homology cannot.

The idea

The cup product makes the cohomology of a space into a ring, and the ring is a strictly finer invariant than the groups. By the universal coefficient theorem, every $H^{n}(X;R)$ is determined by the homology groups, so two spaces with matching homology have matching cohomology in every degree. The product is what cohomology adds.

Cochains can be multiplied because a cochain is a function on simplices with values in a ring $R$. Given $\varphi$ of degree $k$ and $\psi$ of degree $\ell$, evaluate $\varphi$ on the front $k$-face of a $(k+\ell)$-simplex, evaluate $\psi$ on the back $\ell$-face, and multiply the two results in $R$. The result is the cup product $\varphi \smile \psi$, a cochain of degree $k + \ell$.

Theorem (The cohomology ring).

Let $X$ be a space and $R$ a commutative ring. The cup product of two cocycles is a cocycle whose class depends only on the classes of the factors, so it defines a product $H^{k}(X;R) \times H^{\ell}(X;R) \longrightarrow H^{k+\ell}(X;R)$ that makes $H^{}(X;R) = \bigoplus_{n} H^{n}(X;R)$ an associative graded ring with identity, and this ring is graded commutative: $\alpha \smile \beta = (-1)^{k\ell}\, \beta \smile \alpha$ for $\alpha \in H^{k}(X;R)$ and $\beta \in H^{\ell}(X;R)$. A continuous map $f \colon X \to Y$ induces a ring homomorphism $f^{} \colon H^{}(Y;R) \to H^{}(X;R)$.

Cutting a simplex into a front face and a back face is not a symmetric operation, and the sign in graded commutativity is the one trace of the asymmetry that survives into cohomology.

Because $f^{*}$ is a ring homomorphism, homotopy equivalent spaces have isomorphic cohomology rings — and there exist spaces whose cohomology groups agree in every degree while their rings do not. The ring separates such spaces; no list of groups can.

Ways to work on it

• Walkthrough. The cup product formula, why it descends to cohomology, the sign in graded commutativity, and the rings of the torus, real and complex projective space.
• Proof. Two spaces with identical cohomology groups in every degree, told apart by their cohomology rings.
• Practice. Compute cup products from the cochain formula and do arithmetic in cohomology rings.
• Hardest. Rule out maps between real projective spaces using the cohomology ring.

Not sure where to start? Take the ten-question placement test.