# Local expression of hermitian metric

I have really hard times reading Zheng's Complex Differential Geometry and I find the following sentence especially baffling (sec. 7.4, page 170):

"Let $M^n$ be a complex manifold. A Hermitian metric on $M^n$ is just a Hermitian metric on the holomorphic tangent bundle $T_M$ [...], i.e. $$h = \sum_{i,j=1}^n h_{i \overline{j}} dz_{i} \otimes d\overline{z_j},$$ where $(z_1, \ldots, z_n)$ is a local holomorphic coordinate, $h = (h_{i\overline{j}})$ is an $n \times n$ matrix of smooth functions which is Hermitian symmetric and positive definite."

What I absolutely DONT understand is: how precisely should I think about this formula and what are these $dz_{i}$'s and $d\overline{z_j}$'s? I mean, if the formula is somehow analogous to the usual formula for the Riemannian metric, say, $$g = \sum_{i,j=1}^n g_{ij} dx_{i} \otimes dx_{j},$$ then I would expect it to eat two holomorphic vector fields ($h$ is defined on holomorphic tangent bundle) $V = \sum_{i=1}^n a_i \frac{\partial}{\partial z^i}$ and $W = \sum_{j=1}^{n} b_j \frac{\partial}{\partial z_{j}}.$ But then, since $d\overline{z_i}(\frac{\partial}{\partial z_j}) = 0,$ the $h$ would be zero.

So what is the additional piece of information I am missing here and how is the local expression for $h$ really to be understood? Thank you.

It seems that you understand the picture almost perfectly, with one missing detail. Remember that a complex inner product, as opposed to a real one, is not bilinear. Rather, it is linear in one variable and antilinear in the other. Now, if we want to describe such an inner product using tensor language, we need something bilinear. Hence, we think of it as a linear map$$h:V\otimes\overline{V}\to\mathbb{C}.$$The same holds when moving from a complex vector space to a complex vector bundle (as it always is with vector spaces and vector bundles).