# Hodge star operator in Yang-Mills theory and derivation of the YM equations

I'm confused about application of the Hodge star operation within Yang Mills theory. Using differential forms, and given the connection $A$, the curvature in Yang-Mills theory is $F=dA+A \wedge A$. The Bianchi identities and the field equations are then $d_{A}F=0$ and $*d_{A}*F=0$, where $d_{A}$ is the covariant derivative $d_{A}F=dF+[A,F]$. Now the commutator of a $P$-form and $Q$-form is $[P,Q]=P\wedge Q-Q\wedge P$ if either or both of $P$, $Q$ are even, and $[P,Q]=P\wedge Q+Q\wedge P$ if both $P$ and $Q$ are odd. I want to express the Bianci identity and the YM equations in full in terms of $A$ and the wedge product. In full, the Bianchi identity in terms of connections and wedges is $$d_{A}F=dA\wedge A-A\wedge dA + A\wedge dA + A\wedge A\wedge A - dA\wedge A - A\wedge A \wedge A= 0.$$ But what is the corresponding full expression in terms of $A$ for the YM equations $*d_{A}*F$? And/or how is it derived directly from the action $\int *F \wedge F$? It is the correct use of the Hodge star operation that I'm messing up on. (Can't find any notes/texts that answers this in detail.)

The Yang-Mills action is $$S = \int F \wedge \ast F$$ Its variation is \begin{align} \delta S &= 2 \int \delta F \wedge \ast F = 2 \int ( d \delta A + \delta A \wedge A + A \wedge \delta A ) \wedge \ast F \\ &= 2 \int \delta A \wedge ( d \ast F + A \wedge \ast F - \ast F \wedge A ) \\&= 0 \end{align} The equations are then $$d \ast F + A \wedge \ast F - \ast F \wedge A = d_A \ast F = 0 ~.$$