Cup product

Definition

Let $X$ be a topological space and $R$ a commutative ring. The cup product can be viewed as a bilinear map:

$H^{i} (X; R) \times H^{j} (X; R) \to H^{i + j} (X; R)$

or equivalently as a linear map:

$H^{i} (X; R) \otimes H^{j} (X; R) \to H^{i + j} (X; R)$

defined as follows. Let $a \in H^{i} (X; R), b \in H^{j} (X; R)$ . Pick representing cocycles $α$ for $a$ and $β$ for $b$ . We will now produce an $(i + j)$ -cocycle.

To do this, let $s$ be any $(i + j)$ -simplex in $X$ . Then via the diagonal embedding of $X$ in $X \times X$ , $s$ becomes an $(i + j)$ -simplex in $X \times X$ , and the Alexander-Whitney map then sends $s$ to an element of $S i n g_{.} (X) \otimes S i n g_{.} (X)$ . Look at the component for $S i n g_{i} (X) \otimes S i n g_{j} (X)$ , and evaluate $α \otimes β$ on this element. This gives a scalar (element of $R$ ). This scalar is the value on the simplex $s$ .

It needs to be checked that the cochain defined in this manner is indeed a cocycle, and that its cohomology class is independent of the choices for representatives $α$ and $β$ .

Importance

Further information: Cohomology ring of a topological space

The cup product does not depend specifically on the Alexander-Whitney map, but rather on the Alexander-Whitney map upto chain homotopy, and by the theory of acyclic models, there is only one such map upto chain homotopy. Thus, it yields a natural multiplication on the direct sum of all the cohomology groups.

It turns out that this multiplication is associative on the nose for the usual choice of Alexander-Whitney map (for other choices, it is associative only upto homotopy). Also, multiplication is graded-commutative (sometimes called supercommutative) if the ground ring is commutative.