I'm studying the paper https://arxiv.org/abs/dg-ga/9506002 from LeBrun and I'm stuck on the the last part of the proof of Theorem 2. The author claims that on a Kähler surface $(M,g,J)$ with constant scalar curvature $s$, the Ricci and the Kähler form are related by $$ \rho^+ = \frac{s}{4}\omega $$ where $\rho^+$ is the self-dual component of the Ricci form.
I suppose it could be a consequence of some property of cscK surfaces but I haven't found any reference for this fact. Can someone gently help to enlight on this?
Thank you for any advice or just for your time
I think to have found an answer to my question: on a 4-dimensional Kähler surface $(M,g)$ the Lefschetz decomposition writes as $$ \mathcal{A}^{1,1}_{\mathbb{C}} = \mathbb{C}\omega \oplus P^{1,1} $$ where $\omega$ is the Kähler 2-form and $P^2$ is the set of primitive $(1,1)$-forms on $M$. So the Ricci form, which is of course of type $(1,1)$, can be uniquely written as $$ \rho = a \omega + \eta \tag{1} $$
As proved in Proposition 6.29 of the book "Hodge Theory and Complex Algebraic Geometry 1" from Claire Voisin, on a Kähler manifold of complex dimension $n$ the Hodge-dual of a primitive $k$-form of type $(p,q)$ can be expressed as $$ \star \eta = (-1)^{\frac{k(k+1)}{2}} \hspace{2ex} i^{p-q} \hspace{1ex} \frac{L^{n-k}}{(n-k)!}\eta $$ In the previous particular case this formula implies $$ \star \eta = -\eta $$ id est, on a Kähler 4-manifold all primitive forms of type $(1,1)$ are anti self-dual. Therefore, since the Kähler form is always self-dual, by projecting $(1)$ on the space of self-dual $(1,1)$ forms one gets $$ \rho^+ = a\omega $$ Finally, as 4-forms are multiple of the volume element $\frac{\omega^2}{2}$, an easy computation on the components of the Ricci tensor proves $$ \rho \wedge \omega = \frac{s}{4} \omega \wedge \omega $$ thus $a = \frac{s}{4}$ and the required expression follows.
It should be possible to prove this fact also by observing that on a Kähler 4-manifold the space of complex-valued self-dual forms can be decomposed as $$ \Lambda^2_+(M) \otimes \mathbb{C} = \mathbb{C}\omega \oplus \mathcal{A}^{2,0} \oplus \mathcal{A}^{0,2} $$ thus, as the Ricci form is a real-valued $(1,1)$-type form, its self-dual component must be nothing but a real scalar multiple of the Kähler form.