Sign of $\int_X\operatorname{Tr}(F_h^2)$ 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
$$\int_X\operatorname{Tr}(F_h^2)\wedge \omega^{n-2}$$
or if $X$ is a surface the sign of
$$\int_X\operatorname{Tr}(F_h^2)?$$
 A: 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
$$
\left|
\frac{i}{2\pi} F_h
\right|^2 \omega^{n}/n!
= (2 c_2(E,h) - c_1(E,h)^2) \wedge \omega^{n-2} / (n-2)!
+ \left|
\operatorname{Tr}_{\omega} \frac{i}{2\pi} F_h
\right|^2 \omega^n / n!,
$$
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 differential-geometric proof of the Kobayashi-Lubke 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
$$
\left|
\frac{i}{2\pi} F_h
\right|^2 \omega^{n}/n!
= 2 c_2(T_X,\omega) \wedge \omega^{n-2} / (n-2)!
+ s(\omega)^2 \omega^n / n!,
$$
where $s(\omega)$ is the scalar curvature of $\omega$. This still doesn't force $c_2(T_X,\omega) \wedge \omega^{n-2}$ 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 Ricci-flat Kahler-Einstein metric. In that case, yes, $c_2(T_X,\omega) \wedge \omega^{n-2} \geq 0$ with strict inequality unless the metric is flat.
