how to show that ${ad}_{g_{\alpha\beta}} \circ g_{\alpha\beta}^{\star}\theta=-g_{\beta\alpha}^{\star}\theta$? I'm digging my head to prove the following relation:
${ad}_{g_{\alpha\beta}} \circ g_{\alpha\beta}^{\star}\theta=-g_{\beta\alpha}^{\star}\theta$
where $\theta$ is the Maurer-Cartan form and $g_{\alpha\beta}$ is the local transition fonction on a double overlap $U_{\alpha\beta}$
My attempt was to remark that the LHS can be written as $g_{\alpha\beta}^{\star} ({ad}_{g_{\alpha\beta}}\circ\theta)$ which in turn can be written as
$g_{\alpha\beta}^{\star} (R^{\star}_{(g_{\alpha\beta})^{-1}}\theta) = (R_{(g_{\alpha\beta})^{-1}}\circ g_{\alpha\beta})^{\star} (\theta)$
I really do not see how the $-$ (minus) sign appears!
 A: Okey, the trick was to remark the following:
Let $v_{m}\in T_{m}M$ be a vector tangent at $m$ to the base space $M$ of a $G$-principal bundle $P$ where $G$ is a Lie group, describe this vector as the derivative of a map $\gamma:[0,1]\longrightarrow M$ passing throught $m$: $v_{m} = \gamma'(0)$ such that $\gamma(0)=m$, using the transition functions $g_{\alpha\beta}:M\longrightarrow G$ one construct paths in $G$ by composing with $\gamma$, espacially one can constrcut two paths $g_{\alpha\beta}\circ\gamma$ and $g_{\beta\alpha}\circ\gamma$, but:
$\forall t\in [0,1]: (g_{\alpha\beta}\circ\gamma)(t) . (g_{\beta\alpha}\circ\gamma)(t) = e$
where $e$ is the identity element of $G$, so:
$[(g_{\alpha\beta}\circ\gamma)(t) . (g_{\beta\alpha}\circ\gamma)(t)]' = 0$
$(g_{\alpha\beta}\circ\gamma)'(t) . (g_{\beta\alpha}\circ\gamma)(t) + (g_{\alpha\beta}\circ\gamma)(t) . (g_{\beta\alpha}\circ\gamma)'(t)= 0$
$(g_{\alpha\beta}\circ\gamma)'(t) = - (g_{\alpha\beta}\circ\gamma)(t) . (g_{\beta\alpha}\circ\gamma)'(t) . (g_{\beta\alpha}\circ\gamma)(t)^{-1}$
$(g_{\alpha\beta}\circ\gamma)'(0) = - g_{\alpha\beta}(m) . (g_{\beta\alpha}\circ\gamma)'(0) . g_{\beta\alpha}(m)^{-1}$
$(g_{\alpha\beta}\circ\gamma)'(0) = - g_{\alpha\beta}(m) . (g_{\beta\alpha}\circ\gamma)'(0) . g_{\alpha\beta}(m)$
Now, the proof:
\begin{eqnarray*}
\left[ad_{g_{\alpha\beta}}\circ g_{\alpha\beta}^{\star}\theta\right]_{m}(v_{m}) & = & ad_{g_{\alpha\beta}(m)}\left[\left(g_{\alpha\beta}^{\star}\theta\right)_{m}(v_{m})\right]\\
 & = & ad_{g_{\alpha\beta}(m)}\left[\theta_{g_{\alpha\beta}(m)}\left[\left(g_{\alpha\beta}\right)_{\star\, m}(v_{m})\right]\right]\\
 & = & ad_{g_{\alpha\beta}(m)}\left[\theta_{g_{\alpha\beta}(m)}\left[\left(g_{\alpha\beta}\right)_{\star\, m}\left(\gamma'(0)_{m}\right)\right]\right]\\
 & = & ad_{g_{\alpha\beta}(m)}\left[\theta_{g_{\alpha\beta}(m)}\left((g_{\alpha\beta}\circ\gamma)'(0)_{g_{\alpha\beta}(m)}\right)\right]\\
 & = & ad_{g_{\alpha\beta}(m)}\left[\theta_{g_{\alpha\beta}(m)}\left(-g_{\alpha\beta}(m).(g_{\beta\alpha}\circ\gamma)'(0)_{g_{\beta\alpha}(m)}.g_{\alpha\beta}(m)\right)\right]\\
 & = & (Ad_{g_{\alpha\beta}(m)})_{\star\, g_{\alpha\beta}(m)}\left[\left(L_{g_{\alpha\beta}(m)^{-1}}\right)_{\star\, g_{\alpha\beta}(m)}\left(-g_{\alpha\beta}(m).(g_{\beta\alpha}\circ\gamma)'(0)_{g_{\beta\alpha}(m)}.g_{\alpha\beta}(m)\right)\right]\\
 & = & (Ad_{g_{\alpha\beta}(m)})_{\star\, g_{\alpha\beta}(m)}\left(-g_{\alpha\beta}(m)^{-1}.g_{\alpha\beta}(m).(g_{\beta\alpha}\circ\gamma)'(0)_{g_{\beta\alpha}(m)}.g_{\alpha\beta}(m)\right)\\
 & =\\
 & = & -g_{\alpha\beta}(m).g_{\alpha\beta}(m)^{-1}.g_{\alpha\beta}(m).(g_{\beta\alpha}\circ\gamma)'(0)_{g_{\beta\alpha}(m)}.g_{\alpha\beta}(m).g_{\alpha\beta}(m)^{-1}\\
 & = & -g_{\alpha\beta}(m).(g_{\beta\alpha}\circ\gamma)'(0)_{g_{\beta\alpha}(m)}\\
 & = & -g_{\beta\alpha}(m)^{-1}.(g_{\beta\alpha}\circ\gamma)'(0)_{g_{\beta\alpha}(m)}\\
 & = & \left(L_{g_{\beta\alpha}(m)^{-1}}\right)_{\star\, g_{\beta\alpha}(m)}\left[(g_{\beta\alpha}\circ\gamma)'(0)_{g_{\beta\alpha}(m)}\right]\\
 & = & \theta_{g_{\beta\alpha}(m)}\left[\left(g_{\beta\alpha}\right)_{\star\, m}(\gamma)'(0)_{m}\right]\\
 & = & (g_{\beta\alpha}^{\star}\theta)_{m}\left(v_{m}\right)
\end{eqnarray*}
EDIT
A less "sloppy with the notation" begining was developped by an author here 
