Kähler–Einstein condition Let $(M,\omega)$ be a Kähler manifold, and $g$ the Riemannian metric such that $\omega(X,JY)=g(X,Y)$. If there is a function $f$ such that $\operatorname{Ric}=f\omega$ does this mean that $\operatorname{Ric}=fg$? I'm confused on what the Einstein condition is. Is there a difference between the Ricci curvature being proportional to the metric and the Ricci curvature being proportional to the Kähler form?
 A: This depends on what definition of the Ricci curvature you are working with. If you view it as a tensor, writing in index notation, the condition for a Kahler manifold to be Einstein is that $$R_{i \bar{j} } = \alpha g_{i\bar{j}} ,$$ for some constant $\alpha$. But one can also define the curvature form (also known as the Ricci form) to be $$ R = iR_{i\bar{j}} dz^i \wedge d\bar{z}^{\bar{j}},$$ so in that notation, its proportional to the Kahler form, which is defined as $$\omega = i g_{i\bar{j}} dz^i \wedge d\bar{z}^{\bar j}.$$ So one can either say that the Ricci tensor is proportional to the metric tensor, or that the curvature two-form is proportional to the Kahler two-form, and both statements mean the same thing.
A: We need the following lemma 
Lemma1: Let $X$ be a connected complex manifold, and $h$ be a Kahler metric. Let $ω$
be the associated Kahler form. Show that if $\dim X ≥ 2$, the wedge product
with $ω$ is injective on $1$-forms.
Let we have a Kahler-Einstein condition, then $Ric(\omega)=f\omega$ where $f$ is a function. Now since $Ric(\omega)$ is closed, hence $dRic(\omega)=0$, so $$df\wedge \omega+d\omega\wedge f=0$$
So, since $\omega$ is Kahler, $d\omega=0$, but from the Lemma1, we get $df=0$, hence $f$ is constant. 
Now the constant $f$ can be computed as follows
$$f=\frac{\int_Xc_1(TX)\wedge \omega^{n-1}}{\int_X\omega^n}$$
