# Ricci Curvature and the Chern Class of the Levi-Civita

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?

• Up to a constant multiple, they are the same. Look in any decent textbook on Kähler geometry. – Robert Bryant Apr 4 '15 at 11:49

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, 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.