For a (compact) Kahler manifold $M$, the **Ricci tensor** is the symmetric $2$-form
$$
r(u,v) = \text{tr}\big( w \mapsto (D\_wD\_u - D\_uD\_w - D\_{[u,w]})v\big). 
$$
The **Ricci curvature** is the $2$-form 
$$
r(u,v) := r(I(u),v).
$$


So the question! The first Chern class of the Levi--Civita connection for $\Omega^{(0,1)}$ is a $(1,1)$-form. The Ricci curvature of $M$ is also a $(1,1)$-form. Are these two forms related in any way? For example, might one be a scalar multiple of the other? 



Ricci Curvature and the Chern Class of the Levi-Civita
Yes, this is a classic result of Chern's. If $\rho = \sum R\_{a\overline b} dz^a \wedge dz^{\overline b}$ is the Ricci form, then $\frac{i}{2\pi} \rho$ represents the first Chern Class of $M$.


Among the results of Yau's classic paper [Calabi’s conjecture and some new results in algebraic geometry](http://www.pnas.org/content/74/5/1798.abstract), one finds a stronger statement. Any $(1,1)$-form on a compact Kähler manifold $(M,g)$ which represents the first Chern class of $M$ is the Ricci form of a unique Kähler metric in the same Kähler class: that is, one whose Ricci form is in the same class as the Ricci form of the original metric.


