Let $\pi \colon E \rightarrow \mathbb{CP}^1$ be a complex vector bundle. It is a well-known fact that a Dolbeault operator on $\pi\colon E \rightarrow \mathbb{CP}^1$ gives a holomorphic structure on $E$.


My questions are derived from this fact:


1. Let $\{D\_t\}\_{t\in [0,1]}$ be a smooth family of Dolbeault operators on $\pi \colon E \rightarrow \mathbb{CP}^1$. This family induces a family of holomorphic vector bundles whose underlying complex vector bundle is the given one. Are these holomorphic vector bundles isomorphic as holomorphic vector bundles?
2. Suppose that the answer to the first question is no in general. What condition on $\{D\_t\}\_{t\in [0,1]}$ except for a constant family gives us a family of isomorphic holomorphic vector bundles?


Any comments and references are welcome. Thank you in advance.



Family of Dolbeault operators on complex vector bundles over $\mathbb{CP}^1$
The answer to the first question is in fact no: consider the family of Dolbeaut operators on the complex vectorbundle of degree 0 and rank 2 underlying $$V=\mathcal O(-1)\oplus\mathcal O(1).$$ Consider the family of operators $$\bar\partial^t=\begin{pmatrix}\bar\partial^{\mathcal O(-1)} & t\, \gamma \\0& \bar\partial^{\mathcal O(1)} \end{pmatrix}$$ parametrised by $t\in\mathbb C$, where $\gamma\in\Gamma(CP^1,\bar K K)=\Omega^2(CP^1,\mathbb C)$ is a 2-form with non-vanishing integral. By Serre-duality the bundle $\mathcal O(1)$ is a holomorphic subbundle w.r.t. the holomorphic bundle defined by the Dolbeaut operator $\bar\partial^t$ if and only if $t=0$.


Concerning 2: You should compute the dimensions $D(d)$ of $H^0(CP^1, (V,\bar\partial^t)\otimes \mathcal O(d))$ for some $d\in\mathbb Z.$ If the dimension is constant for all $d\in\mathbb Z$ the bundles are isomorphic by Grothendieck splitting. But it is enough to restrict to some $d$'s, depending on the actual situation. For example, if the holomorphic bundle is trivial at one point in the (connected) family, then all bundles are isomorphic if the dimension $D(-1)=0$ stays trivial.


