Cup product and cohomology ring - Imported Knowlpedia Content

Definition

Let $X$ be a topological space and $R$ a commutative ring with identity. The cup product is the natural bilinear operation

\smile:H^p(X;R)\times H^q(X;R)\longrightarrow H^{p+q}(X;R)

on singular cohomology. It is induced by multiplying cochains after restricting a singular $(p+q)$ -simplex to its front $p$ -face and back $q$ -face. With this multiplication and the class $1\in H^0(X;R)$ , the graded module

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

is the cohomology ring of $X$ .

Cochain construction

For $\alpha\in C^p(X;R)$ , $\beta\in C^q(X;R)$ , and a singular simplex $\sigma:\Delta^{p+q}\to X$ , the Alexander–Whitney convention gives

(\alpha\smile\beta)(\sigma) = \alpha(\sigma|[v_0,\ldots,v_p])\, \beta(\sigma|[v_p,\ldots,v_{p+q}]).

The identity

\delta(\alpha\smile\beta) = \delta\alpha\smile\beta+(-1)^p\alpha\smile\delta\beta

shows that the product descends to cohomology. Different standard cochain models induce the same product on cohomology; see Hatcher, §3.2.

Algebraic properties

The cup product is associative, unital, natural under pullback, and graded-commutative:

a\smile b=(-1)^{pq}b\smile a

for $a\in H^p(X;R)$ and $b\in H^q(X;R)$ . Consequently $H^*(X;R)$ is a graded-commutative graded ring. The ring structure can distinguish spaces whose cohomology groups are additively isomorphic.

Examples and scope

If $x\in H^2(\mathbb{CP}^n;\mathbb Z)$ is the standard generator, then

H^*(\mathbb{CP}^n;\mathbb Z)\cong\mathbb Z[x]/(x^{n+1}).

By contrast, all positive-degree products vanish in the reduced cohomology ring of a suspension.

References

Allen Hatcher, Algebraic Topology, Cambridge University Press, 2002. Author-hosted book record. Relevant: §3.2, cup product and the cohomology ring.