Ok so I am following a set of notes on complex differential geometry and there is a theorem that says the following:
If $E$ is a complex vector bundle over a complex manifold and $\overline{\partial^E}:\Omega^{p,q}E \to \Omega^{p,q+1}E$ is a linear operator that satisfies the Leibniz rule. Then $\overline{\partial^E}$ is induced from the structure of a holomorphic vector bundle on $E$ iff $(\overline{\partial^E})^2=0$
My first question is on the wording of the above theorem, is this theorem saying that $E$ is a holomorphic vector bundle iff $(\overline{\partial^E})^2=0$?
The second question is about proving the $\impliedby$ implication. My notes suggest using the Newlander-Nirenberg theorem. So I have a couple of ideas about proving it but I am not sure how to formalise it.
Idea:
The Newlander-Nirenberg (NN) gives a criterion for an almost complex manifold to be a complex manifold. In particular it says that $E$ will be a complex manifold iff $[X,Y]_p\in T_p^{(1,0)}E$ whenever $X,Y\in T_p^{(1,0)}E$
If i can show that $E$ is a complex manifold (via NN) then i can use it's charts to construct bi-holomorphic trivialisations of $E$ then the result follows. Here are the issues I am having:
I need to first equip $E$ with the structure of an almost complex manifold in order to use the NN. So what type of linear map am I meant to construct on its tangent spaces? I think it would need to be one related to the $\overline{\partial^E}$ operator in order for the $(\overline{\partial^E})^2=0$ condition to imply the criteria needed to use the NN theorem.
If someone could give me advice on how to complete the proof, construct the almost complex structure on $E$ which would allow me to use NN or even point me to a reference where the proof is established I'd be very very grateful.
Thanks for reading.
Best Answer
To answer your first question, a more precise statement should be:
For your second question, you need to relate $\bar{\partial}^2=0$ with the Newlander-Nirenberg condition.
This works in general (forget about the vevtor bundle structure), so let me start with an almost complex structure $J$ on a smooth manifold $M$, i.e., $J:TM\to TM$ is an operator such that $J^2=-\text{Id}$. One has decomposition of $TM\otimes \mathbb C=TM^{1,0}\oplus TM^{0,1}$ with respect to the eigenspaces $\pm 1$ of $J$. Let $v_1,...,v_n$ be a local basis on $T^{0,1}M$ and $v_1^*,...,v_n^*$ the dual sections in $\Omega^{0,1}M$, then the $\bar{\partial}$ operator can be written as $$\bar{\partial}=\sum_iv_i\otimes v_i^*:\Omega^{p,q}\to \Omega^{p,q+1}$$ For example, if $f$ is a smooth function, $\bar{\partial}(f)=\sum_iv_i(f)v_i^*$. Therefore $$\bar{\partial}^2=\bar{\partial}(\sum_iv_i\otimes v_i^*)=\sum_{i<j}[v_j,v_i]v_i^*\wedge v_j^*\tag{1}\label{1}$$
where $[v_j,v_i]=v_j\circ v_i-v_i\circ v_j$ is the Lie bracket of vector fields.
Based on the discussion above, we claim that:
$\textit{Proof.}$ It follows from $(\ref{1})$ that $\bar{\partial}^2=0$ implies that $v_1,...,v_n$ are $\textit{mutually commutative}$, so any $[\sum_if_iv_i,\sum_jg_jv_j]$ is a linear combination of $v_i$, so $[T^{0,1},T^{0,1}]\subset T^{0,1}$ (It's easy to check this is equivalent to the NN condition on $T^{1,0}$).
Conversely, assuming $[T^{0,1},T^{0,1}]\subset T^{0,1}$, to make sure $\bar{\partial}^2=0$, according to $(\ref{1})$, we need to find a mutually commutative basis $v_1,..,v_n$. This essentially is the Frobenius theorem. One can imitate the proof to give a complex version of that.$\tag*{$\blacksquare$}$