In Huybrechts Complex Geometry, on page 217, the definition of a Hermitian-Einstein structure is given as:
$$i\Lambda_{\omega}F_{\nabla} = \lambda \, \mathrm{Id}_E,$$
where $\Lambda_{\omega}$ denotes the contraction by $\omega$.
What is the precise definition of contraction by a form?
Specifically, does $\Lambda_{\omega}F_{\nabla}$ mean $\sum_{i,j} F_{\nabla}(z_i, \bar{z_j})$?
*Edit:
I have come across 3 ways to define the connection on the vector bundle (all of them are defined by the Leibniz rule, so I am focusing on how they are defined as maps), following my comment I would like to make some comments on that. Let $E\to M$ be a complex vector bundle.
-
As a linear map $\nabla:\Gamma(E)\to \Gamma(E\otimes T^*M)$
$$\nabla(fs)=df\otimes s + f\nabla(s)$$ -
As a $\mathbb{C}$-linear sheaf homomorphism $\nabla:A^0\otimes \Gamma(E)\to A^1(E)$, (Huybrechts).
$$\nabla(f\otimes s)=df\otimes s + f\nabla(s)$$ -
As a map $\nabla: \mathfrak{X}(M)\times \Gamma(E)\to \Gamma(E)$, (Tu).
$$ \nabla_X(fs)=(df)(X)s+f\nabla s$$
Here $\Gamma(E)$ is the space of sections on $E$, $\mathfrak{X}(M)$ is the vector fields on $M$ and $f\in C^{\infty}(M)$. Also
$A^{p,q}(E):=\Gamma(X,A^{p,q}\otimes_{\mathscr O_X} E)$, and
$A^k(E):=\bigoplus_{p+q=k}A^{p,q}(E)$
$1$ and $2$ are the same since $A^0\otimes \Gamma(E)\cong \Gamma(E)$ via $f\otimes s\to f\cdot s$ is again a section. And $\Gamma(E\otimes T^*M)\cong \Gamma(E)\otimes A^1.$
And as far I understand $1,2$ are the same as $3$, but we are suppressing the vector field. So an evaluation on a vector field at $p$ would look like $(\nabla(fs)\big)_p(X)=df(X)s(p)+f(p)(\nabla s)_p(X)$.
Now for the curvature Huybrechts defines it as map
$$ F_{\nabla }:=\nabla^2:A^0(E)\to A^2(E)$$
For this definition to make sense, we need a natural extension, let $a$ be a $k$-form,
$$ \nabla: A^k(E)\to A^{k+1}(E)$$
$$ \nabla(a\otimes s)=d(a)\otimes s+(-1)^k a\wedge \nabla(s) \qquad(*)$$
and using the fact that any connection is of the form $\nabla=d+A$ where $A$ is a matrix of one forms, we obtain $F_{\nabla }=d(A) +A\wedge A$.
But I think again we are suppressing the vector fields/sections.
According to Tu
$$F_{\nabla}:\mathfrak{X}(M)\times \mathfrak{X}(M)\times \Gamma(E)\to \Gamma(E)$$ defined as
$$F_{\nabla}(X,Y,s):=\nabla_{X}\nabla_{Y}s-\nabla_{Y}\nabla_{X}s-\nabla_{[X,Y]}s$$
Tu proves that locally they are the same.
So locally if we pick a frame $e_1,\ldots,e_n$, we have
$$F_{\nabla}(X,Y,e_i)=\nabla(\nabla (e_i))(X,Y)=\nabla(\sum_{j\leq n}A^j_i\otimes e_i)(X,Y)=\sum_{j\leq n}\nabla(A^j_i\otimes e_i)(X,Y)\overset{*}{=}$$
$$\overset{*}{=}\sum_{j\leq n}dA^j_i\otimes e_j-A^j_i\wedge \nabla (e_j)(X,Y)= \sum_{j\leq n}(dA^j_i-\sum_{r\leq n}A^r_i\wedge A^j_r)\otimes e_j)(X,Y)$$
In this context, what is $\Lambda_{\omega}F_{\nabla}$ ?
I found a post in mathoverflow Dual Lefschetz Operator and Contraction with the Fundamental Form.
Gunnar Þór Magnússon comment suggest that $\Lambda_{\omega}F_{\nabla}=\text{tr}(\omega^{-1} F_{\nabla})$.
Is this what we mean here ? Then $\Lambda_{\omega}$ is just the dual of the Lefschetz operator $\Lambda=*^{-1}\circ L\circ *$ ?
Can we get a local expression for $\Lambda_{\omega}F_{\nabla}$?
Again, the question remains what we mean by contraction.
Best Answer
Presumably you are considering $F_\nabla$ as the curvature of the Levi-Civita connection or equivalently the Chern connection? If not, just consider a local frame for $\operatorname{End}(E) = E^* \otimes E$, express $F_\nabla$ using this and follow the incoming considerations. Note that $F_\nabla \in \Omega^2(M, \operatorname{End}(T^{1,0}M))$, and that $\omega = \sqrt{-1}g_{j\bar{k}}dz^j \wedge d\bar{z}^k$. The Lefschetz operator $L : \Omega^{p,q}(M) \to \Omega^{p+1,q+1}(M)$ is given by $$\alpha \mapsto \alpha \wedge \omega.$$ Let $e_k:\Omega^{p,q}(M) \to \Omega^{p+1,q}(M)$ be the operator given by $\alpha \mapsto dz^k\wedge \alpha$ and define $\bar{e}_k:\Omega^{p,q}(M) \to \Omega^{p,q+1}(M)$ similarly. Recall now also that the wedge product has the interior product as its adjoint. Denote these by $\iota_k$ and $\bar{\iota}_k$ respectively. Now $$ \begin{align*} L(\alpha) &= \alpha \wedge \omega \\ &= \sqrt{-1}g_{j\bar{k}}\alpha\wedge dz^j\wedge d\bar{z}^k \\ &= \sum_{j,k} \sqrt{-1}g_{j\bar{k}}e_j\bar{e}_k(\alpha). \end{align*} $$
The adjoint $\Lambda$ is thus given by $$ \Lambda = -\sum_{j,k}\sqrt{-1}g^{j\bar{k}}\bar{\iota}_k\iota_j. $$
Expressing $F_\nabla$ now locally, one has
\begin{align*} F_\nabla &= \sum \Omega^i_j \otimes dz^j \otimes \frac{\partial}{\partial z^i} \\ &= \sum \left(\sum R^i_{j\alpha\bar{\beta}} dz^\alpha \wedge d\bar{z}^\beta\right) \otimes dz^j \otimes \frac{\partial}{\partial z^i}. \end{align*}
Then
$$ \begin{align*} \Lambda F_\nabla &= \Lambda\left(\sum \Omega^i_j \otimes dz^j \otimes \frac{\partial}{\partial z^i}\right) \\ &= \Lambda\left(\sum \left(\sum R^i_{j\alpha\bar{\beta}} dz^\alpha \wedge d\bar{z}^\beta\right) \otimes dz^j \otimes \frac{\partial}{\partial z^i}\right) \\ &= \sum \left(\sum R^i_{j\alpha\bar{\beta}} \Lambda(dz^\alpha \wedge d\bar{z}^\beta)\right) \otimes dz^j \otimes \frac{\partial}{\partial z^i} \\ &= -\sqrt{-1}\sum \left(\sum g^{\alpha\bar{\beta}}R^i_{j\alpha\bar{\beta}} \right) \otimes dz^j \otimes \frac{\partial}{\partial z^i} \end{align*} $$ If you are working with an orthonormal frame the term $g^{\alpha\bar{\beta}}R^i_{j\alpha\bar{\beta}}$ simplifies to $R^i_{j\alpha\bar{\alpha}}$ and one obtains the expression you wrote, modulo the factor of $-\sqrt{-1}$.