I have a problem deriving the adjoint action $ad_X(Y)=XY-YX$ from the adjoint transformation of the group on the Lie algebra.
Background:
The adjoint action of the Lie algebra on itself is given by the differential of the adjoint action of the Lie group on the Lie algebra.
Let $\gamma(t)$ be a curve through some element $g\in G$ on the manifold $G$ with $\gamma(0)=e \in G$ and tangent vector $\gamma'(0)=X \in T_eG$. We write the adjoint action
$Ad_g (X)= \underbrace{g}_{\in G} \underbrace{X}_{\in T_eG} \underbrace{g^{-1}}_{\in G}$ as $Ad_g (Y)= Ad_{\gamma(t)}(Y) = \gamma(t) Y \gamma(t)^{-1} $
Problem
My problem is calculating the differential at the identity
$ ad_X(Y)=\frac{d}{dt} Ad_{\gamma(t)}(Y) \big |_{t=0}= \frac{d}{dt} \gamma(t) Y \gamma(t)^{-1} \big |_{t=0} = \gamma'(0) Y \gamma(0)^{-1} + \gamma(0) Y \frac{d}{dt} \gamma(t)^{-1} \big |_{t=0} $
For example in "Fulton Harris - Representation Theory" the last step is calculated as
$ ad_X(Y)=\frac{d}{dt} Ad_{\gamma(t)}(Y) \big |_{t=0}= \frac{d}{dt} \gamma(t) Y \gamma(t)^{-1} \big |_{t=0} = \gamma'(0) Y \gamma(0)^{-1} + \gamma'(0) Y (-\gamma(0)^{-1} \gamma'(0) \gamma(0)^{-1}) $ and therefore $ad_x=XY-YX$
can someone explain this last step?
What you write probably makes sense only for matrix Lie groups and algebras: the last term $\gamma'(0) Y (-\gamma(0)^{-1} \gamma'(0) \gamma(0)^{-1})$ should probably be $\gamma(0) Y (-\gamma(0)^{-1} \gamma'(0) \gamma(0)^{-1})$.
Then the rest follows from the matrix identity $$\frac{d}{dt} A^{-1}(t)=-A^{-1}(t) \big(\frac{d}{dt} A(t)\big) \,A^{-1}(t)$$ which easilly follows from $\frac{d}{dt} \big(A(t)\,A^{-1}(t)\big)=0$.