Tangent Space of $\operatorname{Aut}(T_e G)$ Let $G$ be a Lie Group, how to prove that the tangent Space of $\operatorname{Aut}(T_e G)$ at identity is $\operatorname{End}(T_e G)$? Thanks.
 A: If $V$ is a finite-dimensional vector space, considered as a manifold, then the tangent space is $V$ at any point. And $\mathrm{Aut}$ is an open subspace of the vector space $\mathrm{End}$. This doesn't have anything to do with Lie groups.
A: Assume $E$ is a Banach space. Let $B(E)$ be the Banach algebra of bounded linear operators on $E$, equipped with the induced operator norm. You can call it the algebra of endomorphisms if you prefer. Then denote $GL(E)$ the group of invertible elements in $B(E)$, that is the group of automorphisms.
Fix $T_0$ in $GL(E)$. Then for very $S\in B(E)$ such that $\|S\|<\frac{1}{\|T_0^{-1}\|}$, we have $T_0+S$ in $GL(E)$ with inverse given by
$$
(T_0+S)^{-1}=T_0^{-1}(I+ST_0^{-1})^{-1}=T_0^{-1}\sum_{n\geq 0}(-ST_0^{-1})^n.
$$
The convergence of the Neumann series is due to the fact that $\|ST_0^{-1}\|\leq\|S\|\|T_0^{-1}\|<1$.
This proves that $GL(E)$ is open in $B(E)$, therefore it is a manifold modeled on $B(E)$.
In your case $E=T_eG$, $GL(E)=Aut(T_eG)$ and $B(E)=End(T_eG)$.
Actually, the same argument proves more generally that the set of invertible elements in a Banach algebra is open, hence a manifold modeled on this Banach algebra.
