# Positive curvature on holomorphic vector bundles

There must be a mistake in my understanding the definition of positivity for the curvature. Let me summarize: Let $(L,\nabla,h) \rightarrow M$ be a hermitian hol line bundle with Chern connection. Then one can show (e.g. Huybrechts, complex geometry Prop. 4.3.8) that the curvature $F = \nabla^2 \in \mathcal{A}^{1,1}(M,\mathbb{C})$ is a (1,1)-Form with the property that

(i) $h(F_{X,Y}\sigma,\tau)=-h(\sigma,F_{X,Y}\tau)$.

Now, writing $F$ in local coordinates $(z_i)$ of $M$, we see $F=\sum_{i,j}F_{Z_i,\bar Z_j}dz_i \wedge d\bar z_j$ where

$a_{ij}:=F_{Z_i,\bar Z_j}$

is a hermitian symmetric matrix. Now my question is:

From the first equality (i) we deduce, that $F_{X,Y}$ is a purely imaginary complex number. Why isn't this a contradiction to $a_{ij}$ being hermitian symmetric? The matrix entries of $a_{ij}$ are special $F_{X,Y}$, so we get a matrix with purely imaginary entries which cannot be positive definite.

Where is my mistake?!

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In the expression $F_{X,Y}$, $X$ and $Y$ are vectors living in the real tangent space $TM \subset TM \otimes \mathbb C$. But of course $Z_i$ and $\bar Z_i$ do not lie in the real tangent space.
For example, consider the two-form $\omega = idx \wedge dy$. Then using $dx = \frac{1}{2}(dz + d\bar z), dy = \frac{1}{2i}(dz - d\bar z)$ you see that $\omega = \frac{1}{2} d\bar z \wedge dz$. So writing it in terms of real differential forms gives a purely imaginary coefficient but in terms of the complex coordinates you get a real coefficient.