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

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?

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 voilĂ :

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

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-answersâ€¦ :)