Question about gauge transformation I'm studying something fundamental about gauge theory and I find that many materials state(without proof) that:
for a principal bundle $P$ with correspondent connection $\omega$ and correspondent Lie group $G$. Take $\phi \in Aut(P)$, we can view $\phi$ a map $P \to G$ and write the pullback connection along $\phi$ by
$\phi^*\omega=h^{-1}\omega h+h^{-1}dh$.(1)
But I'm quite confused about this equation, if I take a vector $v \in T_p P$, what's the $\phi^*\omega(v)$ exactly means? Is it $h^{-1}\omega|_p(v) h+h^{-1}dh(v)$ or $h^{-1}\omega|_{hp}h_*(v)+h^{-1}dh(v)$ ? Also, as far as I know, we always define pullback as $\phi^*\omega(v)=\omega(\phi_*v)$, but it seems not coincide to the equation (1) above.
Meanwhile in the page 153 of taubes book differential geometry he say that
for a trivial principal bundle $M\times G$ with a connection $A$ on it and a map $h:M \to G$, we can make an automorphism of $M\times G$ as $\phi:(x,g)\to (x,h(x)g)$ then we will obtains a pullback $\phi^*A=g^{-1}dg+h^{-1}dh$, also not coincide with the (1).
And I see from a material that view $\mathbb{R}^4$ as $\mathbb{H}$ and view $SU(2)$ as $Im(\mathbb{H})$ we can pullback the connection $Im(\frac{xd \bar{x}}{1+|x|^2})$ by scaling
$\lambda$ and we can get the result as $Im(\frac{xd\bar{x}}{\lambda^2+|x|^2})$,
but I think scaling is not a map from $P$ to $G$ right? How can I apply the pullback operation? Could anyone help
me to clarify the meaning of these notation?Thanks!
 A: Let me try to explain the definition of gauge transformations more preciselly and how it acts on connection 1-forms. For this, let $\mathcal{M}$ be a smooth manifold and $\pi:P\to\mathcal{M}$ a principal $G$-bundle, where $G$ is a (finite-dimensional) Lie group. In the following, we will denote by "$\cdot$" be always the group action of $G$ on $P$.

*

*A gauge transformation is a bundle automorphism $f\in\mathrm{Aut}(P)$. As you stated, there is an isomorphism
$$\mathrm{Aut}(P)\to C^{\infty}(P,G)^{G}, f\mapsto\sigma_{f}$$
where $C^{\infty}(P,G)^{G}:=\{\sigma\in C^{\infty}(P,G)\mid \sigma(p\cdot g)=g^{-1}\sigma(p)g\}$ and where $\sigma_{f}$ is the map defined by $f(p):=p\cdot\sigma_{f}(p)$ for all $p\in P$. As an aside: If $G$ is an abelian group, there is also an isomorphism $C^{\infty}(P,G)^{G}\cong C^{\infty}(\mathcal{M},G)$. If $G$ is not abelian, then this is actually true only locally, i.e. choosing a local section $s\in\Gamma(U,P)$ (a "local gauge") where $U\subset\mathcal{M}$ open, then there is an isomorphism $C^{\infty}(P_{U},G)^{G}\cong C^{\infty}(U,G)$.

*Let $A\in\Omega^{1}(P,\mathfrak{g})$ be a connection 1-form. Now, one can show, using the properties of the connection 1-form that
$$f^{\ast}A=\mathrm{Ad}_{\sigma_{f}^{-1}}\circ A+\sigma_{f}^{\ast}\mu_{G}$$
where $\mathrm{Ad}:G\to\mathrm{Aut}(\mathfrak{g})$ denotes the adjoint representation of $G$ and where $\mu\in\Omega^{1}(G,\mathfrak{g})$ denotes the Maurer-Cartan form of $G$, i.e. the 1-form defined via $(\mu_{G})_{g}(v):=D_{g}L_{{g}^{-1}}(v)\in T_{e}G\cong\mathfrak{g}$ for all $g\in G$ and for all $v\in T_{g}G$ ($L_{g}:G\to G$ denotes left multiplication by $g$) . This is very straight-forward to show. If you have troubles, just ask in the comments for details.

*Now, lets look how this looks like locally: If you take a local section $s\in\Gamma(U,P)$ (a "local gauge"), we define the local gauge fields $A_{s}:=s^{\ast}A\in\Omega^{1}(U,\mathfrak{g})$. Now, if $s_{i}$ and $s_{j}$ are two local gauges, then we have that
$$A_{s_{i}}=\mathrm{Ad}_{g_{ji}^{-1}}\circ A_{s_{j}}+\mu_{ji}$$
where the local gauge transformation $g_{ij}:U_{i}\cap U_{j}\to G$ is defined by $s_{i}(x)=s_{j}(x)\cdot g_{ji}(x)$ and where $\mu_{ji}:=g_{ji}^{\ast}\mu_{G}$. Now, in the special case where $G$ is a matrix Lie group, i.e. $G\subset\mathrm{Gl}(n,\mathbb{R})$, it follows that
$$A_{s_{i}}=g_{ji}^{-1}\cdot A_{s_{j}}\cdot g_{ji}+g_{ji}^{-1}\cdot\mathrm{d}g_{ji}$$
which looks like the formula you stated above. So, this is more precisely where it comes from.

