Let $X$ be a paracompact topological space. Complex line bundles on $X$ are completely determined up to isomorphism by their first Chern class. More precisely, if $\operatorname{Vect}_1^{\mathbb{C}}(X)$ denotes the collection of isomorphism classes of complex line bundles on $X$, then $c_1 : \operatorname{Vect}_1^{\mathbb{C}}(X) \to H^2(X, \mathbb{Z})$ is an isomorphism. In particular, a complex line bundle $L$ is trivial if and only if $c_1(L) = 0$. If $X$ is a compact smooth manifold, isomorphism classes of topological complex line bundles coincide with isomorphism classes of smooth complex line bundles.
Now let $X$ be a complex manifold. The collection of isomorphism classes of holomorphic line bundles on $X$ has a group structure given by tensor product (the inverse is dual). This group is called the Picard group and denoted $\operatorname{Pic}(X)$. A transition functions argument shows that $\operatorname{Pic}(X) \cong H^1(X, \mathcal{O}^*)$. Associated to any complex manifold, we have a short exact sequence of sheaves
$$0 \to \mathbb{Z} \xrightarrow{\times 2\pi i} \mathcal{O} \xrightarrow{\exp}\mathcal{O}^* \to 0$$
called the exponential sequence. The long exact sequence in cohomology gives
$$\dots \to H^1(X, \mathcal{O}) \to H^1(X, \mathcal{O}^*) \to H^2(X, \mathbb{Z}) \to H^2(X, \mathcal{O}) \to \dots$$
The map $H^1(X, \mathcal{O}^*) \to H^2(X, \mathbb{Z})$ is the composition of the isomorphism $H^1(X, \mathcal{O}^*) \to \operatorname{Pic}(X)$ with the first Chern class. As $H^k(X, \mathcal{O}) \cong H^{0,k}_{\bar{\partial}}(X)$, we see that
- if $h^{0,1} = 0$, the map $H^1(X, \mathcal{O}^*) \to H^2(X, \mathbb{Z})$ is injective, and
- if $h^{0,2} = 0$, the map $H^1(X, \mathcal{O}^*) \to H^2(X, \mathbb{Z})$ is surjective.
In particular, if $h^{0, 1} > 0$, the map $H^1(X, \mathcal{O}^*) \to H^2(X, \mathbb{Z})$ may have non-trivial kernel (in fact, it necessarily does, see the discussion below the line). That is, there could be a non-trivial holomorphic line bundle $L$ with $c_1(L) = 0$, i.e. a holomorphic line bundle which is smoothly trivial.
The simplest example of a complex manifold with $h^{0,1} > 0$ is a genus one Riemann surface which has $h^{0,1} = 1$. An explicit non-trivial holomorphic line bundle is $\mathcal{O}(x_0 - x_1)$ with $x_0, x_1$ distinct points of the surface. It has first Chern class zero so it is smoothly trivial, but it is not holomorphically trivial as it has no global holomorphic sections other than the zero section: a non-zero section would have associated divisor $x_0 - x_1$ but this is impossible as divisors associated to holomorphic sections are always effective.
By investigating the long exact sequence more carefully, we can see how many holomorphic line bundles have first Chern class zero and hence are smoothly trivial; we denote the collection of isomorphism classes of such line bundles by $\operatorname{Pic}^0(X)$.
$$\dots \to H^0(X, \mathcal{O}) \to H^0(X, \mathcal{O}^*) \to H^1(X, \mathbb{Z}) \to H^1(X, \mathcal{O}) \to H^1(X, \mathcal{O}^*) \to H^2(X, \mathbb{Z}) \to \dots$$
Note that $H^0(X, \mathcal{O}) \cong \Gamma(X, \mathcal{O}) = \mathcal{O}(X)$ is the collection of holomorphic functions on $X$, and $H^0(X, \mathcal{O}^*) \cong \Gamma(X, \mathcal{O}^*)$ is the collection of nowhere-zero holomorphic functions on $X$. If $X$ is compact, the only holomorphic functions are constant functions, so $H^0(X, \mathcal{O}) \cong \mathbb{C}$, $H^0(X, \mathcal{O}^*) \cong \mathbb{C}^*$ and the map $H^0(X, \mathcal{O}) \to H^0(X, \mathcal{O}^*)$ is nothing but the exponential map $\mathbb{C} \to \mathbb{C}^*$. As this is surjective, $H^0(X, \mathcal{O}^*) \to H^1(X, \mathbb{Z})$ is the zero map, and hence $H^1(X, \mathbb{Z}) \to H^1(X, \mathcal{O})$ is injective.
By exactness, the kernel of $H^1(X, \mathcal{O}^*) \to H^2(X, \mathbb{Z})$ is precisely the image of $H^1(X, \mathcal{O}) \to H^1(X, \mathcal{O}^*)$ which is isomorphic to the quotient of $H^1(X, \mathcal{O})$ by the kernel. Again by exactness, the kernel of this map is equal to the image of the map $H^1(X, \mathbb{Z}) \to H^1(X, \mathcal{O})$ which is isomorphic to $H^1(X, \mathbb{Z})$ as the map is injective. Therefore, if $X$ is compact,
$$\operatorname{Pic}^0(X) \cong \frac{H^1(X, \mathcal{O})}{H^1(X, \mathbb{Z})}$$
which is a complex torus of dimension $h^{0,1}$.
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 $J$, $$\left<J(v),J(w)\right> = \left<v,w\right>.$$ We define an hermitian product on $(V,J)$: $$\left(v,w\right) := \left<v,w\right> -i\left<J(v),w\right>.$$ On the other hand, we have a natural hermitian extension for $\left<,\right>$ to $V\otimes\mathbb{C}$: $$\left<v\otimes a,w \otimes b\right>_{\mathbb{C}} := a\bar b \left<v,w\right>.$$
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.
$$ \left<z_j,z_k\right>_{\mathbb{C}} = \frac{1}{4}\left[\left<x_j,x_k\right> -i\left<y_j,x_k\right> +i \left<x_j,y_k\right> +\left<y_j,y_k\right> \right] = \frac{1}{2}\left(x_j, x_k\right),
$$
$$ \left<z_j,\bar z_k\right>_{\mathbb{C}} = \frac{1}{4}\left[\left<x_j,x_k\right> -i\left<y_j,x_k\right> -i \left<x_j,y_k\right> -\left<y_j,y_k\right> \right] = 0,
$$
$$ \left<\bar z_j,\bar z_k\right>_{\mathbb{C}} = \overline{\left<z_j,z_k\right>}_{\mathbb{C}}.
$$
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 $$ h(z_j,z_k) = \left<z_j,\bar z_k\right>_{\mathbb{C}}.$$
Best Answer
It’s possible to view the Poincaré metric on the unit disk as the curvature form of a Hermitian metric on the trivial line bundle over the disk. In this trivialization the metric is identified with a smooth function on the disk. This function must be real-valued to be a Hermitian metric (because a Hermitian inner product on a one-dimensional complex vector space is just a positive real number), so it cannot be holomorphic.
More generally, if a metric $h$ on a holomorphic vector bundle is holomorphic it follows that it is flat: Pick a local holomorphic frame for the bundle and let $H$ be the matrix of $h$ in that frame. Then the entries of $H$ are holomorphic functions. However, as $H$ is Hermitian we have $H = \overline H^t$, so the conjugates of the entries of $H$ are holomorphic functions as well. It follows that the entries of $H$ are constant, so its Chern curvature tensor vanishes. Now, there are holomorphic vector bundles that do not admit flat metrics (like the tangent bundle of the projective space), so those bundles do not admit holomorphic Hermitian metrics.