This is a question in regards to an identity in Gauge theory.
Let $\omega$ be the connection one form on a principal bundle $\pi:P\rightarrow M$ and let $A_{\alpha}:=s_{\alpha}^*\omega$ be the gauge fields. ($s_{\alpha}$ is the local section associated to the local trivialisation).
On $U_{\alpha}\cap U_{\beta}\neq \varnothing$, there holds
\begin{align*} A_{\alpha}&=\text{Ad}_{g_{\alpha\beta}}\circ(A_{\beta}-g_{\alpha\beta}^*\theta) \end{align*}
For matrix groups, this equation becomes:
\begin{align*} A_{\alpha}&=g_{\alpha\beta}A_{\beta}g_{\alpha\beta}^{-1}-dg_{\alpha\beta}g_{\alpha\beta}^{-1}. \end{align*}
My question - can someone please explain to me why the top equation reduces to the bottom equation when considering matrix groups?
Thanks!
In a matrix group, $Ad_g(A)=\displaystyle\frac{d}{dt}\bigg|_{t=0}g\exp(tA)g^{-1}=g(\displaystyle\frac{d}{dt}\bigg|_{t=0}\exp(tA))g^{-1}=gAg^{-1}$, i.e. the adjoint action, which is the derivative of conjugation in the Lie group, is really just conjugation in the larger space of all matrices. That's where the first term comes from.
From the context, I gather that $\theta$ is the Maurer-Cartan form. Let $X\in TM$. We have to compute
$Ad_{g_{\alpha\beta}}((g_{\alpha\beta})^*\theta(X))=Ad_{g_{\alpha\beta}}(\theta((g_{\alpha\beta})_*X)))=(R_{g_{\alpha\beta}^{-1}})_*(L_{g_{\alpha\beta}})_*(L_{g_{\alpha\beta}^{-1}})_*(g_{\alpha\beta})_*(X)=(R_{g_{\alpha\beta}^{-1}})_*(g_{\alpha\beta})_*(X)=(g_{\alpha\beta})_*(X)g_{\alpha\beta}^{-1}$.
Now, what is $(g_{\alpha\beta})_*(X)$? The function $g_{\alpha\beta}(x)$ is matrix-valued, so it has components $(a_{ij}(x))$ which are real-valued functions. Take some curve $\gamma(t)$ in $M$ with $\dot\gamma(0)=X$. Then
$(g_{\alpha\beta})_*(X)=\displaystyle\frac{d}{dt}\bigg|_{t=0}g_{\alpha\beta}(\gamma(t))=\displaystyle\frac{d}{dt}\bigg|_{t=0}(a_{ij}(\alpha(t)))=((a_{ij})_*(X))=(da_{ij}(X))$.
This explains the notation $dg_{\alpha\beta}$, which at first glance seems to make little sense. So by $dg_{\alpha\beta}$ is meant the matrix of one-forms $(da_{ij})$, and thus our computation in total is
$Ad_{g_{\alpha\beta}}((g_{\alpha\beta})^*\theta(X))=dg_{\alpha\beta}(X)g_{\alpha\beta}^{-1}$
which takes care of the second term.