Let $G$ be a Lie group and $e \in G$ be the identity. I want to understand the following sentence.
“$\operatorname{Aut}(T_eG)$ being just an open subset of the vector space of endomorphisms of $T_eG$, its tangent space at the identity is naturally identified with $\operatorname{End}(T_eG)$.”
I don’t understand
- why $\operatorname{Aut}(T_eG)$ is an open subset.
- how to naturally identify $T_e(\operatorname{Aut}(T_eG))$ with $\operatorname{End}(T_e G)$.
I think I am lacking basic knowledge about tangent spaces. I appreciate any help.
If $G$ is a manifold of dimension $n$, then $T_eG$ is just a $n$-dimensional real vector space, so that after choosing a basis of $T_eG$, we can make an identification $T_eG\cong \mathbb{R}^n$. The endomorphisms of $\mathbb{R}^n$ (i.e., the linear maps from $\mathbb{R}^n$ to itself) can be identified with the $n\times n$ matrices, and hence so can the endomorphisms of $T_eG$. The automorphisms of $\mathbb{R}^n$ are the invertible, a.k.a. “non-singular” matrices. The space of $n\times n$ matrices has its own topology, and in this topology, the invertible matrices form an open set; that is covered in this older answer of mine.
As to your second question, for any real vector space $V$ of finite dimension $n$, and any $v\in V$, there is a natural identification of $T_vV$ with $V$ (this is Prop 3.8 in Lee’s Introduction to Smooth Manifolds, 1st ed.). That gives us an identification $$T_\text{id}(\operatorname{End}(T_eG))\cong \operatorname{End}(T_eG)$$ where $\text{id}\in\operatorname{End}(T_eG)$ is the identity map from $T_eG$ to itself (I assume this is what you intended in your problem statement). Additionally, for any manifold $M$, any open set $U\subseteq M$, and any $p\in U$, there is a natural identification of $T_pU$ with $T_pM$ (this is Prop 3.7, ibid.). Thus, we have a sequence of identifications $$T_\text{id}(\operatorname{Aut}(T_eG))\cong T_\text{id}(\operatorname{End}(T_eG))\cong \operatorname{End}(T_eG).$$