Hermitian semipositive and Ricci semipositive.

Question

In the paper Compact Kähler manifolds with Hermitian semipositive anticanonical bundle, the authors defined Hermitian positivity of a line bundle:

Let $X$ be a compact Kähler manifold, a line bundle $L$ is said to be Hermitian semipositive if it can be equipped with a smooth Hermitian metric of semipositive curvature form.

And my question is : wether the condition anticanonical bundle $-K_X$ is Hermitian semipositive is equivalent to the Ricci form of $X$ is semipositive? If I'm right, semipositive is equivalent to non-negative, so if Hermitian semipositivity of $-K_X$ is equivalent to Ricci semipositivity, then $X$ can be divided into 2 cases: Ricci flat manifolds and Fano(Ricci positive) manifolds, is that right? Any comments are welcome, thanks in advance.

Answer 1

From your definition, the $-K_X$ is semipositive if you can give $-K_X$ a Hermitian metric $h$ so that the curvature two form of $h$ is semipositive. This $h$ might have nothing to do with the original Kähler metric $g$.

On the other hand, we have the following

Theorem (Calabi-Yau): Let $(M, \omega)$ be a compact Kähler manifold. Then for any two form $\Omega$ representing the class $\pi c_1(-K_X)$, there is an unique Kähler metric $\bar \omega \in [\omega]$ so that $\operatorname{Ric}(\bar \omega) = \Omega$.

Thus if $-K_X$ is semipositive, one can find a Kähler metric $\bar \omega$ on $M$ with semipositive Ricci form.

It is not true that there is only the Ricci-flat and Fano case. Indeed given a non-negative matrix, you cannot say that either all eigenvalues are zero or all are positive.

For example, $\mathbb P^1 \times T^2$ with the standard Fubini-Study metric on $\mathbb P^1$ and Euclidean metric on $\mathbb T^2$ is not Ricci flat, not Fano, but has a semipositive Ricci curvature.