Isomorphism between $\operatorname{End}_0(E)$ and $\operatorname{End}_0(E')$ as Lie algebra bundles

Question

This may be a stupid question. I'm reading the paper "Automorphisms of moduli spaces of vector bundles over a curve" of Indranil Biswas, Tomas L. Gomez, V. Munoz (arXiv link). I have a problem in the proof of Theorem 5.3.

Let $X$ be a smooth projective curve over $\mathbb{C}$. $E$ and $E'$ are vector bundles of rank $r$ and degree $d$. $K_X$ is the canonical bundle. $\mathcal{N}_{E}$ is the nilponent cone bundle whose fiber is $$\mathcal{N}_{E,x}=\{A \in \operatorname{End}E \otimes K_X \mid x \mid A^r=0 \}\subset \operatorname{End}_0E \otimes K_X \mid x.$$

In the proof of Theorem 5.3, this paper says

There is an isomorphism $\mathcal{N}_E\rightarrow \mathcal{N}_{E'}$. By Lemma $5.2$, we get an isomorphism $\operatorname{Fl}(E) \rightarrow \operatorname{Fl}(E')$ of the corresponding flag variety bundles. Considering the global vertical fields, we have a Lie algebra bundle isomorphism $\operatorname{End}_0E\rightarrow \operatorname{End}_0{E'}$.

But I don't understand how I can deduce $\operatorname{End}_0E\rightarrow \operatorname{End}_0{E'}$ from $\operatorname{Fl}(E) \rightarrow \operatorname{Fl}(E')$ explicitly, and why it is an isomorphism as a Lie algebra bundle.
In addition, what is "global vertical fields"? I have no definition of it in this paper.

Answer 1

With this type of question, it is better to write to one of the authors directly.

$\DeclareMathOperator\Fl{Fl}\DeclareMathOperator\End{End}$First, a vertical vector field is given by a global vector field of the fiber $\Fl(E_x)$, for a point of $x$. These are determined by a nilpotent matrix in $\End_0 E$. Next, note that there is already an isomorphism of vector spaces $\End_0 E\rightarrow \End_0 E'$ (last line of page 10). So the vertical vector fields of $\Fl(E)$ and $\Fl(E')$ are isomorphic. The issue here is that the map on vector fields respects the Lie bracket, so the map is a Lie algebra isomorphism (at least on nilpotent matrices, but this implies for all matrices).