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Cup product

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We start on the cochain level: if $\varphi ^p$ and $\psi ^q$ are two cochains of degrees $p$ and $q$ respectively, then their cup product is a certain $(p+q)$-cochain $\varphi ^p \smile \psi ^q$. The definition is then extended to cohomology classes.

For simplicial cochains this cochain is defined by $$(\varphi ^p \smile \psi ^q)(\sigma) = \varphi ^p(\sigma _{0,1, ... p}) \cdot \psi ^q(\sigma _{p, p+1 ,..., p + q}),$$ where $\sigma$ is a $(p+q)$-simplex in the simplicial complex $X$ and $\sigma _{0,1, ..., p}$ is the $p$-th "front face" and $\sigma_{p, p+1, ..., p + q}$ is the $q$-th "back face", respectively, of $\sigma$.

For a cubical complex in the $n$-dimensional space, cochains are defined on the cubes: $$Q=\Pi _{k=1}^{n}A _k,$$ where each $A_k$ is either a vertex or an edge in the $k$th component of the space. If we omit the vertices, a $(p+q)$-cube can be rewritten as $$Q=\Pi _{i=1}^{p+q}I _i,$$ where $I_i$ is its $i$th edge. Now, the cochain of the cup product is given by $$(\varphi ^p \smile \psi ^q)(I)=\Sigma _s (-1)^{\pi (s)}\varphi ^p(\Pi _{i=1}^{p}I _{s(i)}) \cdot \psi ^q(\Pi _{i=p+1}^{p+q}I _{s(i)}),$$ with summation taken over all permutations of $\{1,2,...,p+q\}$ and $\pi (s)$ stands for the parity of permutation $s$.

The cup product corresponds to the wedge product under the de Rham map.

The coboundary of the cup product is given by $$\delta(\varphi ^p \smile \psi ^q) = \delta{\varphi ^p} \smile \psi ^q + (-1)^p(\varphi ^p \smile \delta{\psi ^q}).$$ The cup product of two cocycles is again a cocycle, and the product of a coboundary with a cocycle is a coboundary. Therefore the cup product is well defined on cohomology as a bilinear operator: $$\smile :H^p(X) \times H^q(X) \to H^{p+q}(X).$$

Exercise. Define cup product for singular cohomology.

The cup product as an operator on the cohomology of $X$ can also be defined as the composition: $$\displaystyle C^*(X) \times C^*(X) \overset{K}{\to} C^*(X \times X) \overset{\Delta^*}{\to} C^*(X),$$ where $K$ is the Kunneth map and $\Delta ^*$ is induced by the diagonal map $\Delta \colon X \to X \times X$.

The cup product in cohomology is skew-commutative: $$\alpha^p \smile \beta^q = (-1)^{pq}(\beta^q \smile \alpha^p).$$

The cup product is natural (or functorial): $$f^*(\alpha \smile \beta) =f^*(\alpha) \smile f^*(\beta),$$ where $f^*\colon H^*(Y)\to H^*(X)$ is the cohomology operator induced by a map $f\colon X\to Y$. This makes $f^*$ a graded ring homomorphism.

Compare to cap product.