How to bound the curvature tensor? If a manifold is Kahler, and its Ricci curvature is bounded two side. How to bound the curvature tensor in L2 sense by a topology invariant which only depend on the first and second chern class?
 A: Let's write $X$ for the underlying complex manifold, $\omega$ for the
$(1,1)$-form of the Kahler metric and set $\dim_{\mathbb C} = n$. We also
write $\frac{i}{2\pi} \Theta_\omega$ for the curvature tensor of
$\omega$ and $\mathrm{Ric}_\omega$ for the Ricci-form of $\omega$. Then we
have
$$
\Bigl(\Bigl|\frac{i}{2\pi} \Theta_\omega\Bigr|^2 
- |\mathrm{Ric}_\omega|^2\Bigr) \, \omega^{[n]}
= (2c_2 - c_1^2) \wedge \omega^{[n-2]}
$$
at all points of $X$, where $c_k$ is the $k$-th Chern form defined by $R$
and $\omega^{[k]} := \omega^k / k!$. This actually holds for any
Hermitian vector bundle $(E,h) \to X$; the correct equation there is
$$
\Bigl(\Bigl|\frac{i}{2\pi} \Theta_h\Bigr|^2 
- \Bigl|\Lambda_\omega \frac{i}{2\pi} \Theta_h \Bigr|^2\Bigr) \, \omega^{[n]}
= (2c_2 - c_1^2) \wedge \omega^{[n-2]},
$$
where $\Lambda_\omega$ is the adjoint of the Lefschetz operator
defined by $\omega$. In particular, we have this equality on plain
Hermitian manifolds, but there I'm not sure exactly which of the three
Ricci-forms is involved. When $X$ is s compact we can integrate this
pointwise equality over the manifold, so when $X$ is also Kahler the
right-hand side is entirely cohomological.
This equality is implicit in all papers that prove the Kobayashi-Lubke
inequality for Hermite-Einstein bundles by differential-geometric
methods, in particular in the ones by Chen and Oguie, Lubke and
Demailly, since that inequality is proved by calculating the norm
of the curvature tensor and then using Cauchy-Schwarz to estimate the
adjoint Lefschetz term. I've never seen the equality stated
explicitly, but it shouldn't be difficult to wade through the local
coordinate calculations of the standard proofs and
see where it pops up. I have a little note where I obtain it by mostly
formal calculations with the Lefschetz operator and its adjoint on the
exterior algebra of the total space of $E$ (it's really just the usual
local coordinate proof, written in a coordinate-invariant way), if you
want I can send it to you.
