The Cup Product — Mathematics Knowledge System

Overview

The cup product is a bilinear operation on singular cohomology that turns H*(X; R) into a graded ring. While homology groups only carry additive information, the cup product structure encodes multiplicative information that can distinguish spaces with the same homology groups. It is graded-commutative: α ∪ β = (-1)^{pq} β ∪ α for α ∈ H^p, β ∈ H^q.

Intuition

Cohomology classes can be thought of as 'measuring devices' for cycles. The cup product of two measuring devices α (for p-cycles) and β (for q-cycles) produces a new device α∪β that measures (p+q)-cycles by first applying α to the front p vertices and β to the back q vertices. On a torus, the two fundamental circle classes cup-product to give the surface class -- this is a nontrivial product that does not occur on the wedge sum S^1 ∨ S^1 ∨ S^2.

Formal Definition

Definition

For singular cochains C^*(X; R), the cup product ∪: C^p × C^q → C^{p+q} is defined by (φ ∪ ψ)(σ) = φ(σ|_{[v₀,...,v_p]}) · ψ(σ|_{[v_p,...,v_{p+q}]}) for a singular (p+q)-simplex σ. This descends to cohomology: H^p(X;R) ⊗ H^q(X;R) → H^{p+q}(X;R). The graded cohomology ring H*(X;R) = ⊕_n H^n(X;R) with cup product is a graded-commutative R-algebra.

(\varphi \cup \psi)(\sigma) = \varphi(\sigma|_{[v_0,\ldots,v_p]}) \cdot \psi(\sigma|_{[v_p,\ldots,v_{p+q}]})

Cup product on cochains

\alpha \cup \beta = (-1)^{pq}\, \beta \cup \alpha \quad \text{for } \alpha \in H^p,\, \beta \in H^q

Graded commutativity

H^*(X;R) = \bigoplus_{n \ge 0} H^n(X;R)

Cohomology ring

\delta(\varphi \cup \psi) = \delta\varphi \cup \psi + (-1)^p \varphi \cup \delta\psi

Leibniz rule for cup product

Notation

Notation	Meaning
$\alpha \cup \beta$	Cup product of cohomology classes α and β
$H^*(X; R)$	Cohomology ring of X with coefficients in R
$H^p(X; R)$	p-th cohomology group with R-coefficients
$\varphi\|_{[v_0,\ldots,v_k]}$	Restriction of cochain to front k-face

Theorems

Theorem 1: Theorem 1

For α ∈ H^p(X;R) and β ∈ H^q(X;R), we have α ∪ β = (-1)^{pq} β ∪ α in H^{p+q}(X;R). In particular, if p is odd, then α ∪ α = -α ∪ α so 2(α ∪ α) = 0.

Theorem 2: Theorem 2

For spaces X and Y with cohomology free over a PID R, there is a ring isomorphism H*(X×Y; R) ≅ H*(X; R) ⊗_R H*(Y; R) where (α⊗β)∪(α'⊗β') = (-1)^{|β||α'|}(α∪α')⊗(β∪β').

Theorem 3: Naturality of Cup Product

For a continuous map f: X to Y, the induced map f*: H*(Y;R) to H*(X;R) is a ring homomorphism satisfying f*(alpha cup beta) = f*(alpha) cup f*(beta).

Worked Examples

1
H^0(T^2)=Z, H^1(T^2)=Z², H^2(T^2)=Z with generators 1, α, β, and γ.
$H^*(T^2;\mathbb{Z}) \cong \mathbb{Z}[\alpha,\beta]/(\alpha^2,\beta^2,\alpha\beta + \beta\alpha)$
2
The cup product: α∪β = γ (the fundamental class), and α∪α = β∪β = 0 by graded commutativity (odd degree).
$\alpha \cup \beta = \gamma \ne 0$
3
For S^1 ∨ S^1 ∨ S^2: H^1 = Z², H^2 = Z, same as T^2.
4
However on S^1 ∨ S^1 ∨ S^2, the cup product of any two H^1 classes is zero (by the wedge sum formula), while on T^2, α∪β ≠ 0. So the rings are not isomorphic.

✓ Answer

H*(T^2;Z) ≅ Z[α,β]/(α²,β²) as a graded ring with α∪β≠0, distinguishing T^2 from S^1∨S^1∨S^2.

Practice Problems

Mediumproof writing

Prove that if α ∈ H^{2k+1}(X;Z) then 2(α ∪ α) = 0.

Mediumfree response

Why cannot CP^2 be homeomorphic to S^4?

Common Mistakes

Common Mistake

The cup product is commutative.

The cup product is graded-commutative: α∪β = (-1)^{|α||β|} β∪α. For classes in odd degree, this means α∪α has 2-torsion. Full commutativity only holds when working over fields of characteristic 2 or when all degrees are even.

Common Mistake

Two spaces with the same cohomology groups must have the same cohomology ring.

The ring structure (cup products) carries additional information beyond the groups. The torus T^2 and S^1∨S^1∨S^2 have the same cohomology groups but different cup product structures, distinguishing them.

Quiz

Historical Background

The cup product was introduced by Alexander and Whitney in the 1930s as a way to define a product structure on cohomology. Lefschetz had earlier identified intersection forms on manifolds, which the cup product generalizes via Poincaré duality. The ring structure on cohomology proved powerful for distinguishing spaces: the torus and the wedge S^1 ∨ S^1 ∨ S^2 have the same homology groups but different cup product structures.

1935
Alexander and Whitney independently define the cup product on simplicial cochains
Alexander, Whitney
1941
Eilenberg shows the cup product is natural and defines cohomology rings
Eilenberg
1952
Steenrod operations generalize the cup product to power operations
Steenrod

Summary

The cup product ∪: H^p × H^q → H^{p+q} makes the cohomology H*(X;R) into a graded-commutative ring.
Graded commutativity: α∪β = (-1)^{pq} β∪α, so the ring is not commutative in general.
The cup product is natural: continuous maps induce ring homomorphisms on cohomology.
The Künneth formula computes H*(X×Y) as the graded tensor product of H*(X) and H*(Y).
Key examples: H*(CP^n;Z) ≅ Z[α]/(α^{n+1}) with |α|=2; H*(T^2;Z) has nontrivial cup product α∪β≠0.

References

BookHatcher, A. Algebraic Topology. Cambridge University Press, 2002. Section 3.2.
BookMay, J.P. A Concise Course in Algebraic Topology. University of Chicago Press, 1999.
WebsiteWikipedia: Cup product