Let $(E,h)\to X$ be a holomorphic Hermitian vector bundle over a compact Kähler manifold. Denote by $F_h$ the curvature of its Chern connection. Can we know a priori the sign of the quantity
$$\int_X\operatorname{Tr}(F_h^2)\wedge \omega^{n2}$$
or if $X$ is a surface the sign of
$$\int_X\operatorname{Tr}(F_h^2)?$$
1 Answer
The short answer: No.
The longer answer: The $(2,2)$form $\operatorname{Tr}(F_h^2)$ represents the second Chern class of $E$, or $c_2(E)$. In general, the integral of that class over the manifold doesn't have a sign independent of the bundle. What we do have in general is the pointwise identity $$ \left \frac{i}{2\pi} F_h \right^2 \omega^{n}/n! = (2 c_2(E,h)  c_1(E,h)^2) \wedge \omega^{n2} / (n2)! + \left \operatorname{Tr}_{\omega} \frac{i}{2\pi} F_h \right^2 \omega^n / n!, $$ where $c_k(E,h)$ is the representative of the $k$th Chern class of $E$ defined by the curvature form of the metric. We don't need $\omega$ to be Kahler for this to hold, nor $X$ to be compact; this holds on general Hermitian manifolds. (This is known, and usually implicit somewhere in any differentialgeometric proof of the KobayashiLubke inequality. See here for one proof.)
To be able to say anything more I think we have to assume that $\omega$ is Kahler and $E = T_X$, in which case $\operatorname{Tr}_\omega \frac{i}{2\pi} F_h = c_1(T_X,\omega)$. The above then reduces to $$ \left \frac{i}{2\pi} F_h \right^2 \omega^{n}/n! = 2 c_2(T_X,\omega) \wedge \omega^{n2} / (n2)! + s(\omega)^2 \omega^n / n!, $$ where $s(\omega)$ is the scalar curvature of $\omega$. This still doesn't force $c_2(T_X,\omega) \wedge \omega^{n2}$ to have a definite sign unless we assume the scalar curvature of $\omega$ is constantly zero, in which case $\omega$ will have to be a Ricciflat KahlerEinstein metric. In that case, yes, $c_2(T_X,\omega) \wedge \omega^{n2} \geq 0$ with strict inequality unless the metric is flat.

$\begingroup$ Shouldn't it be $Tr(F^2_h)=c_1^2(E)c_2(E)$? $\endgroup$ Commented Dec 29, 2021 at 5:14