Exact forms are orthogonal to co-closed forms

Recall that a Riemannian manifold is equipped with an inner product (aka Riemannian metric or metric tensor). As a result we also have an inner product of differential forms. Now, suppose $A=da$ is exact and $B$ is co-closed. Then $$<A,B> = <da, B> = <a,d^*B> = <a,0> = 0.$$ So, they are orthogonal, in this sense.

The computation isn't limited to smooth manifolds but also applies to combinatorial manifolds equipped with a Riemannian metric.