Confusion about Lie groups in Fulton & Harris

Question

Near the beginning of chapter 8 (titled Lie groups and Lie algebras) authors motivate the definition of Lie algebra. I’m confused by two things in just one sentence:

($G$ is a Lie group)

The group $\operatorname{Aut} T_e G$ being just an open subset of the vector space of endomorphisms $\operatorname{End} T_e G$, its tangent space at the identity is naturally identified with $\operatorname{End} T_e G$.

First: How do we get the topology on $\operatorname{End} T_e G$? Is it just that $\operatorname{End} T_e G$ is (by a choice of basis) isomorphic to $M_n \mathbb R$ and we can identify $M_n \mathbb R$ with $R^{n^2}$? Or we can do something similar without choice of basis?

Second: How do we “naturally identify” a tangent space at the identity of $\operatorname{End} T_e G$ with $\operatorname{End} T_e G$ itself?

Answer 1

1. Yes, any finite dimensional vector space over $\mathbb R$ is isomorphic to $\mathbb R^n$ for some $n$ (by choosing a basis) so we just define that isomorphism to be a homeomorphism. Then you check that linear maps on $\mathbb R^n$ are always continuous, hence vector space isomorphisms are always homeomorphisms, and this implies that the topology you get on $\mathrm{End}(T_eG)$ did not depend on your choice of isomorphisms $\mathrm{End}(T_eG) \simeq \mathbb M_n(\mathbb R) \simeq \mathbb R^{n^2}$.

So while you do define the topology using a basis, the result is independent of your choice of basis.

2. This depends largely on what specific definition you take for "tangent vectors" and "tangent spaces". I don't have Fulton & Harris in front of me, but choosing a basis you have $\mathrm{End}(T_eG) \simeq \mathbb M_n(\mathbb R)$. Any function $f\colon \mathbb R \to \mathbb M_n(\mathbb R)$ is given by component functions $f_{ij}\colon\mathbb R \to \mathbb R$ where $f_{ij}(M)$ is the $(i, j)^\text{th}$ entry of the matrix $f(M)$. Then $\frac{\mathrm{d}f}{\mathrm{d}t}$ is the matrix whose $(i, j)^\text{th}$ entry is $\frac{\mathrm{d}f_{ij}}{\mathrm{d}t}$. In particular, $\frac{\mathrm{d}f}{\mathrm{d}t}(0)$ (which should be a tangent vector) is a matrix in $\mathbb M_n(\mathbb R) \simeq \mathrm{End}(T_eG)$.

Answer 2

In $\mathbf{R}^n$ any open set has $\mathbf{R}^n$ as tangent space at any of its points. Automorphisms of $ \mathbf{R}^n$ is an open set in the collection of all endomorphisms.