Fano manifolds and positive Ricci curvature.

Question

I already know that if a Kahler manifold, $M$, is Fano, then $M$ has Kahler metrics with positive Ricci curvature. What about the converse? If $Ric(M)$ is positive, does this mean that $-K_M$ is ample?

Answer 1

I'm sorry, I misunderstood and thought you were asking about Kahler-Einstein metrics. As an apology, here's a real answer.

If you just want Kahler metrics instead of Kahler-Einstein ones the situation is relatively simple:

Let $M$ be a Kahler manifold of dimension $n$ (no compactness needed). It is equivalent to give a smooth Hermitian metric $h$ on $K_M$ and to give a smooth volume form $dV$ on $M$. This holds because given a volume form $dV$ and local sections $\sigma,\tau$ of $K_M$ we can define $h$ by $$ i^{n^2} \sigma \wedge \overline \tau = h(\sigma,\overline\tau)\, dV. $$ Conversely, given $h$ the same equation defines a volume form $dV$.

Suppose now that we have a Hermitian metric $\omega$ on $M$. This defines a volume form $dV$ and thus a metric $h$ on $K_X$. A standard computation shows that one of the three Ricci curvatures of $\omega$ is $$ \operatorname{Ric} \omega = - \frac{i}{2\pi} \Theta_{K_M,h} = \frac{i}{2\pi} \Theta_{-K_M,h^{-1}}. $$ Thus, if $M$ admits a Kahler metric (in which case all the three Ricci forms are equal) with positive Ricci curvature, then $-K_M$ is ample.

The converse, as pointed out in Ballmann's book, is true but a little trickier (and needs compactness of $M$). Given a positive Hermitian metric on $-K_M$ we pass to a volume form and then have to find a Kahler metric which has that volume form (the problem can be formulated in different ways; we can also look for a Kahler metric with a given Ricci form). There's no problem to find such a Hermitian metric, but to get a Kahler one we have to solve some hard PDEs.

A more delicate question is whether we can take the Kahler metric $\omega$ to be a Kahler-Einstein metric. Up to scaling and normalization, this means that we want a Kahler metric $\omega$ such that $\operatorname{Ric} \omega = \omega$. The simplest example of such a thing is the Fubini-Study metric on $\mathbb P^n$. Now, if we have such a metric, then clearly $-K_M$ is ample because the Ricci curvature is positive. However, there are compact Kahler manifolds with $-K_M$ ample that do not admit Kahler-Einstein metrics (the simplest one is maybe the projective plane blown up in 9 well chosen points). The work of Tian and Chen, Donaldson and Sun (and others) is aimed at finding the necessary and sufficient condition for a manifold with $-K_M$ ample to admit a Kahler-Einstein metric. If the preprints in the links are correct, this happens when the tangent bundle $T_M$ is stable in some sense.