With indices and without indices.

Question

Let $S = (S^{\mu\nu})$ be a two form on a four space and $\tilde S$ is the (Hodge) dual of $S$.

On what condition can we have

$$\partial_{\mu}(S^{\mu\nu}+\tilde{S}^{\mu\nu}) = 0\Leftrightarrow d(S+ \tilde S) = 0?$$

Answer 1

Using that $ \tilde S$ is the Hodge dual of $S$, we have

$$*(S + \tilde S) = (S+ \tilde S),$$

where $*$ is the Hodge star operator (Acting on two forms).

Then

$$\partial_{\mu}(S^{\mu\nu}+\tilde{S}^{\mu\nu}) = 0$$

if and only if $d^* (S+ \tilde S) = 0$, where $d^*$ is the adjoint of $d$. But $d^* = \pm *d*$, so the equality is the same as

$$0=d (*(S+ \tilde S)) = d(S+ \tilde S).$$