I'm reading Besse's book on Einstein manifolds.
Theorem 13.14 (due to Atiyah, Hitchin, and Singer) states that a Riemannian 4-manifold is Einstein iff the Levi-Civita connection on $\Lambda^+$ is self-dual.
Here $\Lambda^2=\Lambda^+\oplus\Lambda^-$ is the space of two forms, which decomposes due to the fact that the Hodge star operator $\ast\in \mathrm{End}(\Lambda^2)$ is an involution ($\ast^2=1$) in dimension 4.
Question: Does this theorem imply that an Einstein 4-manifold is self-dual (i.e. $W^-=0$, where $W=W^++W^-$ is the Weyl tensor of the manifold)? If not, when is an Einstein 4-manifold self-dual?
Note that the curvature operator $R:\Lambda^2\to \Lambda^2$ has the decomposition $$ R=\begin{pmatrix} A & B \\ B^* & D \end{pmatrix} $$ Note that this decomposition is based on $\Lambda^2=\Lambda^+\oplus {\Lambda}^-$. Hence the curvature $F$ of $\Lambda^+$ is given by the first column in the block, i.e. $F=A+B^*$. Now $\lambda^+$ is an $SO(3)$ bundle with adjoint bundle $\mathfrak{g}\simeq \Lambda^+$. Hence $F\in \Lambda^+\otimes \Lambda^2$. Again from the decomposition, we know $A\in End(\Lambda^+)\simeq \Lambda^+\otimes {\Lambda^+}^*$ and $B^*\in Hom(\Lambda^+,\Lambda^-)\simeq \Lambda^+\otimes {\Lambda^-}^*$. So clearly $B^*$ corresponds to the anti-selfdual parts of the curvature, i.e. $F^-$. Hence $F^-=0$ iff $B^*=0$ iff $M$ is Einstein.
However, $B^*=0$ has nothing to do with the Weyl tensor.