If $\nabla,\nabla'$ are connections on a vector bundle $E$, then $\nabla-\nabla'\in\mathcal A^1(\operatorname{End}E)$: trouble with proof

Question

Let $X$ be a complex manifold and $E$ a complex vector bundle on $X$. Straight off my Complex Geometry professor's notes:

Translation:

Proposition 15.1. Given a connection $\nabla$ on $E$, every connection $\nabla'$ has the form $\nabla+A$, where $A\in\mathcal A^1(\operatorname{End}E)$.

Proof. Let $A:\mathcal A^0(E)\to\mathcal A^1(E)$ be defined as $A(s)=\nabla s-\nabla's$. By the Leibniz rule, $A(fs)=f(A(s))$. Given a local basis of sections $s_1,\dotsc,s_n$, one has:

$$A(x_1s_1+\dotso+x_ns_n)=x_1A(s_1)+\dotso+x_nA(s_n),$$

so $A$ matches the section of $\mathcal A^1(\operatorname{End}E)$ that sends $s_1,\dotsc,s_n$ to $A(s_1),\dotsc,A(s_n)$. $\hspace{2cm}\square$

If I understand correctly, a section of $\mathcal A^1(\operatorname{End}E)$ is locally a sum $\sum\alpha_i\otimes T_i$, where $\alpha_i$ are 1-forms on $X$ and $T_i$ are sections of $\operatorname{End}E$, that is maps $T_i:E\to E$ which are linear on the fibres, but not necessarily bundle morphisms, because the rank need not be constant.

By that proof, I can see how:

$$A(s_j)=\sum\alpha_{ij}\otimes s_i,$$

where $\alpha_{ij}$ are 1-forms and the $s_i$'s and $s_j$'s are the basis taken in the proof.

What I'm having trouble seeing is how to reduce the above form of $A$ to the general form of $\mathcal A^1(\operatorname{End}E)$, since I need to have the same forms for all $s_j$ and to have only the second component depend on $j$, whereas as of now I have the opposite situation. How do I solve that?

Answer 1

Locally, you may assume you are in a coordinate neighborhood with coordinates $dz_i,d\overline z_i$. Write:

$$\alpha_{ij}=\sum_kf_{ijk}dz^k+\sum_kg_{ijk}d\overline z^k.$$

Then you will have:

\begin{align*} A(s_j)={}&\sum\alpha_{ij}\otimes s_i=\sum_{i,k}f_{ijk}dz^k\otimes s_i+\sum_{i,k}g_{ijk}d\overline z^k\otimes s_i={} \\ {}={}&\sum_kdz^k\otimes\left(\sum_if_{ijk}s_i\right)+\sum_kd\overline z^k\otimes\left(\sum_ig_{ijk}s_i\right). \end{align*}

Now set:

\begin{align*} \beta_k={}& \begin{cases} dz^k & 1\leq k\leq\dim X \\ d\overline z^{k-\dim X} & \dim X+1\leq k\leq2\dim X \end{cases} \\ T_k(s_j)={}& \begin{cases} \sum_if_{ijk}s_i & 1\leq k\leq\dim X \\ \sum_ig_{i,j,k-\dim X}s_i & \dim X+1\leq k\leq2\dim X \end{cases}, \end{align*}

and voilà:

$$A=\sum_k\beta_k\otimes T_k.$$

And apparently, this was not the last item in the self-answer series, nor the last situation of its kind: with the present question, I was asking, and while I typed the question the above solution sprang to my mind, and I thought I'd put it up in case someone else gets stuck on the same problem. Who knows if this question here will end the self-answers… :)