## Curvature as a bilinear form

Here’s something I thought of when I couldn’t sleep last night.

The curvature tensor of a Kähler metric can be viewed as a Hermitian form on $$\bigwedge^{1,1} T_X^*$$ by mapping $$\operatorname{End} T_X \to \bigwedge^{1,1} T_X^*$$ via the metric.

If we’re on a compact Kähler manifold with zero first Chern class, then for each Kähler class $$\omega$$ and $$(1,1)$$-classes $$u, v$$, we can pick the Ricci-flat metric in $$\omega$$ and the harmonic representatives of $$u, v$$. If $$R$$ is the curvature tensor of the metric, viewed as a Hermitian form, we can then set $b(u,v)(\omega) := \int_X R(u, v) \, dV_{\omega}.$ This defines a smooth bilinear form $$b$$ on the tangent space of the Kähler cone of $$X$$.

Besides being fun times, can we say anything interesting about $$b$$? For example, what is its norm with respect to the Riemannian metric on the Kähler cone, or its trace with respect to that metric? Can we integrate it over some subset of the cone?