If a holomorphic bundle is smoothly trivial, is it holomorphically trivial? Let $E\rightarrow X$ be a holomorphic vector bundle which is smoothly trivial (i.e. $E$ has a smooth global frame). Can we say that $E$ is holomorphically trivial? 
If yes, then what about the following generalization? If $E_{1}$ and $E_{2}$ are two holomorphic vector bundles over a complex manifold $X$ which are smoothly isomorphic, then they are holomorphically isomorphic.
If no, then does there exist an easy counterexample?
 A: 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 $h^{0,1}$.
