Why is a matrix lie group a closed subgroup of the general linear group?

Question

Why is a matrix lie group a closed subgroup of the general linear group over complex entries?

Answer 1

This follows from the following:

Theorem. (Von Neumann) Let $G$ be a closed subgroup of $\textrm{GL}_n(\mathbb{C})$, then $G$ is submanifold of $\mathbb{C}^{n^2}$.

Observation. It suffices to prove that $G$ is diffeomorphic to a subvector space of $\mathcal{M}_n(\mathbb{C})$ only in a neighbourhood of $I_n$. Indeed, for $X\in G$, $M\mapsto XM$ is a diffeomorphism of $\mathcal{M}_n(\mathbb{R})$ which sends $I_n$ on $X$.

Then, the proof relies on the following:

Formula. Let $A$ and $B$ be two matrices with real entries, then one has: $$\exp(A+B)=\lim_{k\to+\infty}\left[\exp\left(\frac{A}{k}\right)\exp\left(\frac{B}{k}\right)\right]^k.$$

Proof. By definition of the exponential, for an operator norm, one has the following asymptotic: $$\exp(X)=I_n+X+O(\|X\|^2).$$ Therefore, using the inverse function theorem, $\exp$ is a smooth diffeomorphism of inverse $L$ between a neighbourhood of $0$ and $I_n$ and $L$. Furthermore, $L$ has the following assymptotic: $$L(I_n+X)=X+O(\|X\|^2).$$ Therefore, one has the following equality: $$\left[\exp\left(\frac{A}{k}\right)\exp\left(\frac{B}{k}\right)\right]^k=\exp\left(kL\left(\exp\left(\frac{A}{k}\right)\exp\left(\frac{B}{k}\right)\right)\right).$$ Besides, one has the following assymptotic: $$kL\left(\exp\left(\frac{A}{k}\right)\exp\left(\frac{B}{k}\right)\right)=L\left(I_n+\frac{A+B}{k}+O(k^{-2})\right)=A+B+o(1).$$ Whence the result using the continuity of $\exp$. $\Box$

Using the above formula and the fact that $G$ is closed, it can be seen that: $$\mathfrak{g}:=\{X\in\mathcal{M}_n(\mathbb{C})\textrm{ s.t. }\forall t\in\mathbb{C},\exp(tX)\in G\}$$ is a subvector space of $\mathcal{M}_n(\mathbb{C})$. Let $\mathfrak{g}^0$ be a subvector space of $\mathcal{M}_n(\mathbb{C})$ such that one has the following: $$\mathcal{M}_n(\mathbb{C})=\mathfrak{g}\oplus\mathfrak{g}^0.$$ Can you prove that $\varphi\colon\mathfrak{g}\oplus\mathfrak{g}^0\rightarrow\mathcal{M}_n(\mathbb{C})$ defined by: $$\varphi(L+M)=\exp(L)\exp(M)$$ is a diffeomorphism between $U$ a neighborhood of $0$ and $V$ a neighborhood of $I_n$ such that: $$\varphi(U\cap\mathfrak{g})=V\cap G?$$

Hint. To do so, it will be helpful to prove that you can choose $U$ such that $\exp(U\cap\mathfrak{g}^0)=\{I_n\}$.