Coboundary Formula

Question

When $K$ is a simplicial complex, the dual complex $C^*(K)$ to the chain complex $C_*(K)$ has a concrete interpretation: an element in $C^n(K)$ (a cochain) is given by assigning an integer to every oriented $n$-simplex in $K$.

My question: I am am aware of an explicit formula for the boundary map on a chain complex, but I am not sure about how to derive an explicit formula for the coboundary of a cochain as described above. Will an explicit formula involve the explicit formula for the boundary of a chain at all? Any help would be appreciated, whether answers, comments, or directions to textual/online references.

Answer 1

The correct way to define the coboundary $\delta_i:C^i(K)\to C^{i+1}(K)$ is by using the boundary map. You should be able to get an explicit formula for the coboundary by using the boundary. Specifically, if $\theta$ is a cochain, define $(\delta_i\theta)(\sigma) = \theta(\partial_i\sigma)$ for every $(i+1)$-simplex $\sigma$, then extending to $C_{i+1}(K)$ by linearity.

Concretely, if $K$ is a (finite) simplicial complex with $i$-simplices $\sigma^i$ and $(i+1)$-simplices $\sigma^{i+1}$, the boundary map $\partial_i:C_{i+1}(K)\to C_i(K)$ is given by a matrix $A_i$. Since our chain complex is based, we may give the cochain complex the dual basis, $\theta^i_s(\sigma^i_r)=\delta_{sr}$. In the dual basis, we can explicitly write down a matrix for the coboundary map: $$(D_i)_{jk} = (\delta_i\theta^i_k)(\sigma^{i+1}_j) = \theta^i_k(\partial_i\sigma^{i+1}_j) = A_{kj}$$

Which is what you should expect. The coboundary is dual to the boundary, so its matrix should just the transpose of the boundary matrix.