Most algebraic topology books (for instance, Hatcher) contain a recipe for computing cup products in singular or simplicial homology. In other words, given two explicit singular or simplicial cocycles, they contain a recipe for computing an explicit cocycle representing the cup product of the cocycles in question.
Is there a similar recipe in cellular cohomology? In other words, if I have a very explicit CW complex and two explicit cellular cocycles, then is there a recipe for computing a cellular cocycle representing their cup product?
Of course, one answer is to subdivide everything up into a simplicial complex, but that is messy (and not always possible). Is there a better way?
I'm especially interested in the special case of 2-dimensional CW complexes, where the only interesting cup products are between elements of $H^1$.
Best Answer
Cup products $\cup: H^1(X,\mathbb{Z})\times H^1(X,\mathbb{Z}) \to H^2(X,\mathbb{Z})$ are easy to visualize. Given a cellular 2-complex $X$, the 1-skeleton of $X$ is a graph $\Gamma$, and the 2-skeleton is a collection of disks with boundaries attached to the graph $\Gamma$, which (up to homotopy) are encoded by closed paths in $\Gamma$. The 1-cycles $\alpha_1,\alpha_2 \in H^1(X,\mathbb{Z})$ are encoded by maps $\alpha_i: X\to S^1=K(\mathbb{Z},1)$, and the 2-cycle $\alpha_1\cup\alpha_2$ is represented by the map $(\alpha_1\times\alpha_2)^{\ast}: H^2(S^1\times S^1)\to H^2(X)$. Choose the standard cell structure on $S^1 =c_0\cup c_1$. Subdivide $\Gamma$ by putting a vertex in the middle of each edge $e\subset \Gamma$ into intervals $e_1\cup e_2$. Then up to homotopy, we may assume that $\alpha_i$ sends $e_i$ to $c_0$ (and is extended over the 2-cells in any fashion). Then $$\alpha_1\times \alpha_2: \Gamma \to (c_0\times c_1\cup c_1\times c_0)= S^1 \vee S^1\subset S^1\times S^1,$$ in such a way that each edge $e_1$ is sent to a horizontal edge, and each edge $e_2$ is sent to a vertical edge. To figure out the degree of this map, for each 2-cell $f\subset X$, we lift $\alpha_1\times \alpha_2:\partial f\to \mathbb{R}^2$, the universal cover of $S^1\times S^1$. Then we compute the winding number of the path represented by $f$ with respect to the midpoints of the lattice $(\widetilde{S^1\vee S^1} \subset \mathbb{R}^2$ giving $\alpha_1\cup\alpha_2(f)$. For example, the closed path in this picture has winding number 7.