A complex vector bundle $E$ is a holomorphic vector bundle iff $(\overline{\partial^E})^2=0$ help with proof?

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.

If $$\bar{\partial}^{E}:\Omega^{p,q}\to \Omega^{p,q+1}$$ is an operator on comple vector bundle $$E$$ satisfying Leibniz rule and $$(\bar{\partial}^{E})^2=0$$, then $$\bar{\partial}^{E}$$ defines a holomorphic structure on $$E$$.

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

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:

Claim: $$J$$ is integrable (Newlander-Nirenberg condition holds) if and only if $$\bar{\partial}^2=0$$.

$$\textit{Proof.}$$ 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.$$\tag*{\blacksquare}$$

• I know it's not particularly important, but don't you get $\sum\limits_{i<j}[v_j, v_i]v_j^*\wedge v_i^*$? Feb 5 '21 at 22:33