In Huybrechts' Complex Geometry- an introduction, the author defined a contraction of curvature form:
The contraction of the curvature $F_{\nabla} \in \mathcal{A}^{1,1}\left(\operatorname{End}\left(T^{1,0} X\right)\right)$ with the Kähler form $\omega$ yields an element $\Lambda_\omega F \in \mathcal{A}^0\left(X, \operatorname{End}\left(T^{1,0} X\right)\right)$ or, equivalently, an endomorphism $T^{1,0} X \rightarrow T^{1,0} X$. Its composition with the isomorphism $T^{1,0} X \rightarrow$ $\wedge^{0,1} X$ induced by the Kähler form will be denoted
$$
\tilde{\omega}\left(\Lambda_\omega F\right): T^{1,0} X \stackrel{\Lambda_\omega F_{\nabla}}{\longrightarrow} T^{1,0} X \stackrel{\omega}{\longrightarrow} \Lambda^{0,1} X .
$$
He claimed that $\tilde{\omega}\left(\Lambda_\omega F\right) \in \mathcal{A}^{1,1}(X)$. But I don't know how to prove it.
Also, he tried to compute $\tilde{\omega}\left(\Lambda_\omega F\right)(u,v)$ for $u,v\in TM$ as follows:
$$
\begin{aligned}
2 i \cdot \tilde{\omega}\left(\Lambda_\omega F\right)(u, v) & =i \cdot \omega\left(\Lambda_\omega F(\xi(u)), \overline{\xi(v)}\right)=\omega\left(\Lambda_\omega(F(\xi(I(u)), \overline{\xi(v)})\right. \\
& =\omega\left(\sum F\left(x_i, y_i\right) \xi(I(u)), \overline{\xi(v)}\right) \\
& =\omega\left(\xi\left(\sum R\left(x_i, y_i\right) I(u)\right), \overline{\xi(v)}\right) \\
& =2 g\left(I\left(\sum R\left(x_i, y_i\right) I(u)\right), v\right)=-2 \sum g\left(R\left(x_i, y_i\right) u, v\right)
\end{aligned}
$$
where $\xi:TM\to T^{1,0}M$ defined by $\xi(u)=\frac{1}{2}(u-iJu)$ and $\omega$ is the Kahler form.
I don't know why the first equal sign and the penultimate equal sign work. In fact, I think $2\omega(u,v)\neq \omega(\xi(u),\overline{\xi(v)})$. Can anyone help me with these two problems? Thanks in advance.
Best Answer
I prefer to do things in a more local coordinate fashion, so let $(X, \omega_g)$ be a Kähler manifold with underlying metric $g$. Let $R$ denote the (Chern or Riemannian) curvature tensor of $g$, whose components in a local frame are denoted by $R_{i \bar{j} k}{}^{\ell}$. The curvature is skew-symmetric in the first two indices (i.e., the $i,j$ indices) and therefore, by tracing over $k,\ell$, we see that the (first Chern) Ricci curvature $\text{Ric}_{i \bar{j}}^{(1)} : = g^{k \overline{q}} g_{\ell \bar{q}} R_{i \bar{j} k}{}^{\ell}$ defines a $(1,1)$-form. The third and fourth indices are the endomorphism part of the curvature, and therefore, tracing over the first two indices yields the (second Chern) Ricci curvature $\text{Ric}^{(2)}{}_k{}^{\ell} : = g^{i \bar{j}} R_{i\bar{j} k}{}^{\ell}$, which is an endormorphism. This second Chern Ricci curvature is what you are denoting by $\Lambda_{\omega} F_{\nabla}$.
The isomorphism $T^{1,0} X \to \Lambda^{1,0}X$ induced by the Kähler form is just the musical isomorphism. Hence, with $\Lambda_{\omega}(F_{\nabla}) = \text{Ric}^{(2)}{}_k^{\ell}$, we can use the metric to lower the index, we see that $$\widetilde{\omega} \Lambda_{\omega}(F_{\nabla})_{k \bar{q}} = \text{Ric}^{(2)}{}_k^{\ell} g_{\ell \bar{q}} = g^{i \bar{j}} R_{i \bar{j}k}{}^{\ell} g_{\ell \bar{q}}.$$
The claimed $(1,1)$-form is then locally given by $\widetilde{\omega} \Lambda_{\omega}(F_{\nabla})_{k \bar{q}} dz^k \wedge d\overline{z}^q = \sqrt{-1} g^{i \bar{j}} R_{i \bar{j} k}{}^{\ell} g_{\ell \bar{q}} dz^k \wedge d\bar{z}^q.$
The Liu--Yang paper here has some calculations that you may find beneficial in learning this stuff.