At P211 in 'Complex Geometry, An Introduction' by Huybrechts:
The contraction of the curvature $F_{\nabla}\in\mathscr{A}^{1,1}(\mathrm{End}(T^{1,0}X))$ with the Kahler form $\omega$ yields an element $\Lambda_{\omega}F\in\mathscr{A}^0(X,\mathrm{End}(T^{1,0}X))$.
I know the definition of 'contraction' of tensors and 'contraction' of differential form and vector fields, but I don't know what is the definition of 'contraction' here.
Thank you for your help!!
Best Answer
The symmetries of the curvature tensor $R$ of the Levi-Civita connection imply that $R$ is an endomorphism-valued $(1,1)$-form. Write $R_{i \overline{j} k\overline{\ell}}$ for the components of $R$. The Kähler metric $g$ can be used to contract the back indices to produce a $(1,1)$--form $$\text{Ric}_{i \overline{j}} = g^{k \overline{\ell}} R_{i \overline{j} k \overline{\ell}},$$ which is the same as a metric (tensor) contraction. This $(1,1)$--form is called the Ricci form, and is a closed $(1,1)$--form representing the first Chern class of the anti-canonical bundle.
If you contract the first two indices, you get an endomorphism $$g^{i \overline{j}} R_{i \overline{j} k \overline{\ell}}.$$
If the metric is Kähler, these contractions coincide, but in the general context of Hermitian geometry, they differ.
The contraction over the third and fourth indices is the first Levi-Civita Ricci curvature, the contraction over the first two indices is the second Levi-Civita Ricci curvature, contraction over the second and third gives the third Levi-Civita Ricci curvature, and the contraction of the first and fourth indices gives the fourth Levi-Civita Ricci curvature.
Moreover, since the Chern connection fails to coincide with the Levi-Civita connection in the non-Kähler case, you get four more Ricci curvatures -- the so called Chern--Ricci curvatures.
The following paper of Liu--Yang has more information concerning Ricci curvature in Hermitian geometry.