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:
- 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?
- 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.
Best Answer
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.