Gauge fields and restrictions of the connection one form I am working through some lecture notes on principal bundles and am stuck on the proof of a certain proposition. In the following, $\pi:P\rightarrow M$ is a principal bundle, $\omega$ is the connection one-form, $s_{\alpha}$ is a local section, $V_p = \text{ker}(\pi_*)$, and $A_{\alpha}:=s_{\alpha}^* \omega$. Furthermore, $g_{\alpha}$ is the equivariant map $\pi^{-1}U_{\alpha}\rightarrow G$ determined by the local section. So $g_{\alpha}(s_{\alpha}(m))=e$ and $g_{\alpha}(pg)=g_{\alpha}(p)g$.
Here is the proposition with proof (this pdf can be found here)

I have numerous questions about this argument, and it would probably take a while to read/answer them. I would just appreciate help on any of the following questions:
1) Why can $T_p P$ be decomposed as $\text{Im}(s_{\alpha}\circ \pi)_* \oplus V_p$? I understand that $V_p$ is the set of all vectors tangent to the fibre through $p$ (they get mapped to the zero vector under $\pi_*$) but I don't understand why we are guaranteed this decomposition of $T_p P$.
2) In the first set of equations, the author evaluates $\omega_{\alpha}$ on $v\in T_pP$. In the second set of equations, he does not input a vector. Is there any particular reason for this? Ease of notation, perhaps?
3) The equation the author is trying to prove is $\omega_{\alpha}=\text{Ad}_{g_{\alpha}^{-1}}\circ \pi^* s_{\alpha}^* \omega + g_{\alpha}^* \theta$. In the first set of equations, the author evaluates $\omega$ on $v\in T_p P$. So is it true to say they are calculating (writing $v_p=v$ to emphasise $v\in T_p P$)
$
\begin{align*}
(\omega_{\alpha})_p(v_p)&=\text{Ad}_{(g_{\alpha}(p))^{-1}}\circ( \pi^*s_{\alpha}^*\omega)(v_p)+(g_{\alpha}^*\theta)_p v_p?
\end{align*}
$
Now, in his first line of working, $\theta$ becomes $\theta_e$. Is this because of the fact that if you want to evaluate $(g_{\alpha}^* \theta)_p v_p$, then this is $(\theta_{g_{\alpha}(p)})({g_{\alpha}}_* v_p)=\theta_e ({g_{\alpha}}_* v_p)?$
4) In the first set of equations he concludes $\theta_e(({g_{\alpha}}_*)\overline{v})=\omega(\overline{v})$. I believe this is because of the definition of $\omega$ and 
(a) the fact that we can consider $v_p$ as the value of some fundamental vector field $\sigma(X)$ at $p$ and 
(b) $(g_{\alpha})_* \sigma(X)_p=(L_{g_{\alpha}(p)})_* X=(L_e)_*X=X$. 
So, using the fact that $\theta_e$ is the identity, 
$\theta_e(({g_{\alpha}}_*)\overline{v})=(g_{\alpha})_* \sigma(X)_p=(L_{g_{\alpha}(p)})_* X=(L_e)_*X=X=\omega(\overline{v})$
5) The second set of equations is where I am really confused. He doesn’t input a vector, but I am assuming he wants to show that for $v_p\in T_p P$, 
$(R_g^*(\omega_{\alpha})_{pg}))_p v_p={\omega_{\alpha}}_{pg}\left(({R_g}_*)_p v_p\right)$?
So I believe the first few lines would be
$\begin{align*}
(\omega_{\alpha})_{pg}(({R_g}_*)_p v_p)&=\text{Ad}_{(g_{\alpha}(p)g)^{-1}}\circ (\pi^*s_{\alpha}^*\omega)(({R_g}_*)_p v_p)+(g_{\alpha}^*\theta)_{pg}(({R_g}_*)_p v_p)\\
\\
&=\text{Ad}_{(g_{\alpha}(p)g)^{-1}}\circ R_g^* \pi^* s_{\alpha}^*\omega(v_p)+(R_g^*((g_{\alpha}^*\theta)_{pg})_p v_p)
\end{align*}
$
Giving reason to first line in the second set of equations. But I am stuck after that. In particular, how does the author swap the $R_g^*$ and $g_{\alpha}^*$ in the second line, and then how does he factor out the $\text{Ad}_{g^{-1}}$ to arrive at the 4th line?
6) Is there any particular reason why the author is using $\circ$ notation in some places but doesn’t in other places?
I know there are lots of questions here. If anyone can provide any help, I would really appreciate it.
 A: The proof wants to show that $\omega$ equals $\omega_{\alpha}$ above the open set $U_{\alpha}$ not that $\omega_{\alpha}=\text{ad}_{g_{\alpha}^{-1}}\circ \pi^* s_{\alpha}^* \omega + g_{\alpha}^* \theta$. That is how $\omega_{\alpha}$ is defined.


*

*Since  $(s_{\alpha})_* $  is non-singular it maps the tangent space of the base n-manifold isomorphically onto an n-dimensional subspace of TP at s(m). This subspace connot contain any vertical vectors because $(\pi\circ\ s_{\alpha})_* $ is the identity map so it has no kernel. This means that the tangent space at s(m) is a direct sum of the vertical space and the image of $(s_{\alpha})_* $. Note that the image of $(s_{\alpha})_* $ may not be horizontal. Still since it contains no vertical vectors (except zero) the direct sum decomposition is valid. 

*The first set of equations just wants to show that $\omega$ and $\omega_{\alpha}$ are equal at the tangent space to the point s(m). At that point ${ad}_{g_{\alpha}^{-1}}$ is the identity and can be dropped. The rest is a computation. You need to be convinced that $\theta_e(({g_{\alpha}}_*(\overline{v}))=\omega(\overline{v})$ on all vertical vectors. This follows since $g_{\alpha}((s(m).h.exp(tX))= h.exp(tX)$ for any h in G so $(g_{\alpha})_*(\sigma_{s(m).h}(X)) = (L_{h})_*(X) $ 

*The second set of equations shows how $\omega_{\alpha}$ transforms under right translation by the structure group,G. This shows that $\omega_{\alpha}$ and $\omega$ not only agree at the tangent space to s(m) but agree everywhere.  
Notice that
$R_{g}^*(\omega_{\alpha})_{pg}(v_{p}) = (\omega_{\alpha})_{pg}(R_{g})_*(v_{p})$ where $v_{p}$ is a vector at p. 
So $(R_{g})^*(ad_{g_\alpha}((pg)^{-1})\pi^*s^*\omega(v_{p}) = (ad_{g_\alpha}((pg)^{-1})\pi^*s^*\omega((R_{g})_*(v_{p})) = (ad_{g_\alpha}((pg)^{-1})R_{g}^*\pi^*s^*\omega(v_{p})$
This gets you the first line of the second set of equations. 
The next two lines are explained. The fourth line follow because ad is a representation of G i.e. a homomorphism of G into the group of linear transformations of its Lie algebra.
A: I went through the same path you there very long ago, now I'm out, I could not tell you all that I have written in this note, or say if any errors (can Maybe you will find them), but I had the good idea to take notes of my scrap paper:
https://www.evernote.com/shard/s318/sh/bbe0818a-4682-43bd-8046-4b7505a11535/453e89fe5cac313bc4f13d267778811e
Hope this will help you
