Let $(X, g)$ be a smooth, oriented, Riemannian 4-diemnsional manifold. Let $\Lambda^2$ denote the bundle of 2-forms on $X$. Then the Hodge-star operator decomposes $\Lambda^2$ into the space of self-dual and anti-self-dual 2-forms$$\Lambda^2 = \Lambda^2_+ \oplus \Lambda^2_- .$$Both $\Lambda^2_+$ and $\Lambda^2_-$ are $3$-dimensional bundles. We then have the projection$$P_+: \Lambda^2\to \Lambda^2_+$$and the operator $d_+: \Lambda^1\to \Lambda^2_+$ given by$$d_+=P_+\circ d$$where $d: \Lambda^1\to \Lambda^2$ is the de Rham operator.
The Riemannian curvature tensor could be considered as a map $\mathcal{R}: \Lambda^2\to \Lambda^2$ and under the decomposition $\Lambda^2 = \Lambda^2_+ \oplus \Lambda^2_- $, $\mathcal{R}$ could be decomposed as$$\mathcal{R}=\begin{bmatrix}A & B \\ B^* & C\end{bmatrix}.$$
The Riemannian manifold $(X, g)$ is called self-dual if we have$$C-\frac{1}{3}tr(C)=0.$$
Now let $\omega$ and $\theta$ be $1$-forms on $X$. Then $\omega\wedge \theta$ is a $2$-form and we can consider $P_+(\omega\wedge \theta)$ and the $3$-form $d(P_+(\omega\wedge \theta))$. On the other hand we have $3$-forms $(d_+\omega)\wedge \theta$ and $\omega\wedge (d_+\theta)$.
My question is: do we have$$d(P_+(\omega\wedge \theta))=(d_+\omega)\wedge \theta-\omega\wedge (d_+\theta)$$if $(X,g)$ is a self-dual manifold?
