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I am reading Kobayashi book entitled Differential Geometry of Complex Vector bundles.

He introduced the concept of hermitian metric on a complex vector bundle $E \to M$, with $M$ not necessarily a complex manifold, being the tensor field such that at every point it defines a complex hermitian inner product.

So, if locally there exists a local frame $s = (s_1,\ldots,s_r)$ (Assuming rank $r$) and being $h$ such tensor field, I can easily compute the matrix associated to $h$ as:

$$h_{ij} := h(s_i,s_j).$$

In particular, the same procedure applies to holomorphic vector bundles, where the local frame has the property of being a holomorphic function between the basis manifold and the bundle, where the basis manifold and the total space are complex manifolds.

Considering $M$ being a complex $n-$dimensional manifold, the tangent bundle $TM$ to $M$ can be seen as a holomorphic vector bundle. In fact, if we consider $TM_{\mathbb{C}} := TM\otimes_{\mathbb{R}}\mathbb{C}$ then it splits as $$TM_{\mathbb{C}} = TM'\oplus TM'',$$ where the spaces on the decomposition are the eigenspaces associated to the extension to $TM_{\mathbb{C}}$ of the standard complex structure on $M$.

In local coordinates $(z^ 1,\ldots,z^ n)$ the space $TM'$ can be seen as the space generated by $\{\frac{\partial}{\partial z^ k}\}_{k =1}^ n.$ In particular, $TM$ possess the structure of a $n-$vector bundle once $TM \cong T'M.$

Here is where I get stuck. The author claims:

Let $TM$ as above with a hermitian metric $h$. Then, locally we can write the components of $h$ as: $$h_{i\overline{j}} := h(\frac{\partial}{\partial z^ i}, \frac{\partial}{\partial \overline{z}^ j}).$$

It is known that a complex manifold with a Riemannian metric $g$ orthogonal with the respect to the standard complex structure $J$ is called an hermitian manifold with hermitian metric $g$. The metric $g$ can be extended to $T_\mathbb{C}M$ and its only non-null components are precisely $h_{i\overline{j}}$ defined above.

So my question is: Why if we consider $TM$ as holomorphic vector bundle the expression for these two metrics coincide? Why are the mixed terms the only non-null terms even if we consider $TM$ as an holomorphic vector bundle? It should not appear any mixed index on the metric.. What am I missing?

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Since the problem has a local nature, I'll consider only the Linear Algebra couterpart. Let $V$ be a real vector space whose dimension is $2n$ and let $$J\colon V \longrightarrow V$$ be a complex structure on $V$, $J^2 = -Id$. Then $(V,J)$ is a (n-dimensional) complex vector space for the scalar multiplication $$(a+bi)v = av+bJ(v).$$ Let $\left<,\right>$ be a (real) inner product on $V$ compatible with $J$, $$\left = \left.$$ We define an hermitian product on $(V,J)$: $$\left(v,w\right) := \left -i\left.$$ On the other hand, we have a natural hermitian extension for $\left<,\right>$ to $V\otimes\mathbb{C}$: $$\left_{\mathbb{C}} := a\bar b \left.$$ Extending $J$ to $V\otimes\mathbb{C}$ we may take the eingenspaces for $J$ and make the decomposition $$V\otimes\mathbb{C} = V'\oplus V''.$$ Fix a (real) $\mathbb{C}-$basis $\{x_1, \dots x_n\}$ for $(V,J)$. Then the $y_j := J( x_j)$ complete a real basis for $V$. Define $$z_j = \frac{1}{2}(x_j -iy_j).$$ Since $J(z_j) =iz_j$, the set $\{z_1, \dots z_n\}$ is a $\mathbb{C}-$basis for $(V',i)$, the set $\{\bar z_1, \dots \bar z_n\}$ is a $\mathbb{C}-$basis for $(V'',-i)$ and $x_j \mapsto z_j$ , $x_j \mapsto \bar z_j$ are isomorphisms. Now we head to the product.

$$ \left_{\mathbb{C}} = \frac{1}{4}\left[\left -i\left +i \left +\left \right] = \frac{1}{2}\left(x_j, x_k\right), $$ $$ \left_{\mathbb{C}} = \frac{1}{4}\left[\left -i\left -i \left -\left \right] = 0, $$ $$ \left<\bar z_j,\bar z_k\right>_{\mathbb{C}} = \overline{\left}_{\mathbb{C}}. $$ We have then the relation between the hermitian products $\left(,\right)$ on $(V,J)$ and $\left<,\right>_{\mathbb{C}}$ on $(V\otimes \mathbb{C},i)$. The last thing to remark is that on Kobayashi's notation $$ h(z_j,z_k) = \left_{\mathbb{C}}.$$