Do we have $d(P_{+}(\omega\wedge \theta))=d_{+}\omega\wedge \theta-\omega\wedge d_{+}\theta$ on a self-dual manifold?

Question

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?

Answer 1

No. If this formula were true, then we would have $$ \mathrm{d}\bigl(P_+(\mathrm{d}f\wedge\mathrm{d}g)\bigr) = 0 $$ for all smooth functions $f$ and $g$, since $\mathrm{d}_+(\mathrm{d}f) =P_+\bigl(\mathrm{d}(\mathrm{d}f)\bigr) = 0$.

Now, consider $\mathbb{R}^4$ with its standard flat metric $g = (\mathrm{d}x_1)^2+(\mathrm{d}x_2)^2+(\mathrm{d}x_3)^2+(\mathrm{d}x_4)^2$ and orientation $\mathrm{d}x_1\wedge\mathrm{d}x_2\wedge\mathrm{d}x_3\wedge\mathrm{d}x_4>0$, which is clearly self-dual.

Let $f=f(x_1)$ and $g = g(x_2)$, then $$ P_+(\mathrm{d}f\wedge\mathrm{d}g) = \tfrac12\,f'(x_1)g'(x_2)\bigl(\mathrm{d}x_1\wedge\mathrm{d}x_2 + \mathrm{d}x_3\wedge\mathrm{d}x_4\bigr), $$ so $$ \mathrm{d}\bigl(P_+(\mathrm{d}f\wedge\mathrm{d}g)\bigr) = (f''(x_1)g'(x_2)\,\mathrm{d}x_1 + f'(x_1)g''(x_2)\,\mathrm{d}x_2)\wedge\mathrm{d}x_3\wedge\mathrm{d}x_4\, $$ which will not usually be zero.