The tangent space of $\mathrm{Aut}(T_eG)$ Let $G$ be a Lie group and $e \in G$ be the identity. I want to understand the following sentence.
" $\mathrm{Aut}(T_eG)$ being just an open subset of the vector space of endomorphisms of $T_eG$, its tangent space at the identity is naturally identified with $\mathrm{End}(T_eG)$".
I don't understand 


*

*why $\mathrm{Aut}(T_eG)$ is an open subset.

*how to naturally identify $T_e(\mathrm{Aut}(T_eG))$ with $\mathrm{End}(T_e G)$.


I think I am lacking basic knowledge about tangent spaces.
I appreciate any help.
 A: 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
$$T_\mathrm{id}(\mathrm{End}(T_eG))\cong \mathrm{End}(T_eG)$$
where $\mathrm{id}\in\mathrm{End}(T_eG)$ 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
$$T_\mathrm{id}(\mathrm{Aut}(T_eG))\cong T_\mathrm{id}(\mathrm{End}(T_eG))\cong \mathrm{End}(T_eG).$$
A: Let me just expand on Zev's answer. 
Any finite dimensional vector space $V$ has a unique linear topology. This essential comes from identifying $V$ with $\mathbb{R}^{\dim V}$. Although such an identification is not unique, it turns out that the topology on $V$ that comes from such identifications is. Moreover, the determinant map $\text{det}:\text{End} V\rightarrow\mathbb{R}$ is continuous in this topology. Indeed, under an identification of $V$ with $\mathbb{R}^{\dim V}$, the determinant map becomes a polynomial. Since a linear map is invertible if and only if it has a non-zero determinant, we have $\text{Aut}V=\text{det}^{-1}(\mathbb{R}\setminus\{0\})$. Since $\mathbb{R}\setminus\{0\}$ is open, and $\text{det}$ is continuous, we conclude that $\text{Aut}V\subseteq\text{End}V$ is open. We obtain an answer to your first question by letting $V=T_eG$. 
Now, if $U\subseteq V$ is open, for every $p\in U$ we have a natural identification of $T_pU$ with V. Namely, every vector $X\in T_pU$ is defined by a curve $\gamma:\mathbb{R}\rightarrow U$ such that $\gamma(0)=p$. Indeed, we define the action of $X$ on $f\in C^\infty(U)$ by $Xf:=(f\circ\gamma)'(0)$. However, $\gamma$, being a function between two vector spaces, has a derivative. In this particular case, this dervative is of the form $$\lim_{t\rightarrow 0}\frac{\gamma(t)-p}{t}=\tilde{X}.$$ We can then associate the vector $X$ to the vector $\tilde{X}$, showing there is a natural identification between $T_pU$ and $V$. We obtain an answer to your second question by letting $V$ be $\text{End}(T_eG)$ and $U$ be $\text{Aut} T_eG$.
