A question on Ricci curvature and Ricci form.

Question

It seems to me that there are two definition of Ricci flatness; real and complex ones. The real Ricci flatness claims vanishing of Ricci curvature. To define complex Ricci flatness, one needs to introduce so called the Ricci form for a Kahler manifold $(X,g)$ defined by $$ Ric(\omega)=-i\partial \overline{\partial}\log\det(g_{i,\overline{j}}), $$ where $g=(g_{i,\overline{j}})$ is a Kahler metric. A Kahler manifold$(X,g)$ is called Ricci flat if $Ric(\omega)$ vanishes.

I wonder if the complex one coincides with real one for a Kahler manifold $(X,g)$. How Ricci curvature and Ricci form are related?

Answer 1

The whole point is that if the metric is Kähler, then the Chern connection coincides with the complexification of Levi-Civita connection of the riemanian metric on the underlying real manifold given by the real part of the Kähler hermitian metric.

Moreover, this happens if and only if the starting metric is Kähler (and is indeed my favorite definition of Kähler metric).

So, let me give you a hint.

If $\omega$ is any hermitian metric on $X$, then $\det\omega$ is the induced metric on $\det T_X$ (this line bundle is usually referred as to the anticanonical bundle $K_X^{*}$). The Chern curvature $\Theta$ of this metric is given by $$ i\Theta(K_X^*,\det\omega)=-i\partial\bar\partial\log\det\omega. $$ Now, in general, the Chern curvature (with respect to the induced hermitian metric) of the determinant $\det E$ of a hermitian holomorphic vector bundle $(E,h)$ is given by the trace of the Chern curvature of $E$: $$ i\Theta(\det E,\det h)=i\operatorname{Tr}\Theta(E,h). $$ Therefore, $$ -i\partial\bar\partial\log\det\omega=i\operatorname{Tr}\Theta(T_X,\omega). $$ But $\Theta(T_X,\omega)$ is the complexification of the curvature $R^{LC}$ of the Levi-Civita connection of $X$ $\nabla^{LC}_{X,\Re\omega}$ with respect to $\Re\omega$, where $\Re\omega$ is the induced riemannian metric on the underlying real manifold.

Finally, by definition, the Ricci curvature of $(X,\Re\omega)$ is the trace of $R^{LC}$.

There is a small missing point in my sketchy argument: the riemannian Ricci curvature is a symmetric $2$-controvariant tensor, while the Ricci form is a closed real $(1,1)$-form. You have to show that in fact the riemannian Ricci curvature of (the real part of) a Kähler metric is invariant with respect to the complex structure $J$, then you apply the $1-1$ correspondence between real $(1,1)$-forms and symmetric $J$-invariant $2$-controvariant tensors.

Answer 2

If you only have an almost complex manifold with a compatible metric i.e. an almost Hermitian manifold $(M,g,J)$ then you can define the Ricci form globally by $$\rho(X,Y):=Ric(JX,Y).$$ And vanishing of $Ric$, $\rho$ are equivalent. You don't need your manifold to be complex or Kähler.