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?
> > 
> > 
> > 
> 
> 
> 



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


