Prove that the commutator is a Lie bracket in $\mathrm{Lie}(G)$

Question

Let $G$ be a matrix Lie group (i.e. a closed subgroup of $\mathrm{GL}(n,\mathbb C)$), and consider the set (following the notation in the Wikipedia page): $$\mathrm{Lie}(G)\equiv\{X\in M(n,\mathbb C) : \,\,e^{tX}\in G\,\,\forall t\in\mathbb R\}.$$ I know that this turns out to be a Lie algebra with Lie bracket given by the commutator of matrices, but I'm trying to get a better understanding of why this is the case.

The case $G=SO(3)$, $\mathfrak g=\mathfrak{so}(3)$ was worked out in this other question. What about the more general scenario of $G$ an arbitrary closed subset of $\mathrm{GL}(n,\mathbb C)$? Can a similar argument be made in this case?

Answer 1

We need to check that ${\rm Lie}(G)$ is (1) a vector space, and (2) it is closed under the standard commutation operation. The commutator of matrices then always satisfies the axioms of a Lie bracket, and thus this is sufficient to show that ${\rm Lie}(G)$ is a Lie algebra.

1. Let $X,Y\in{\rm Lie}(G)$. That $\lambda X\in{\rm Lie}(G)$ for all $\lambda\in\mathbb R$ is immediate from the definition. On the other hand, closure under addition, $X+Y\in{\rm Lie}(G)$, follows from the Lie product formula (remembering that $G$ is, by definition, closed): $$e^{X+Y} = \lim_{N\to \infty} \left(e^{X/N}e^{Y/N}\right)^N.$$ In this expression, we know that $X/N,Y/N\in\mathrm{Lie}(G)$, thus $e^{X/N}e^{Y/N}\in G$, and thus we get the limit of a sequence elements of $G\subseteq M(n,\mathbb C)$, which must converge in $G$ due to closedeness.

2. To show that $[X,Y]\equiv XY-YX\in{\rm Lie}(G)$, observe that

   1. For any $g\in G$ and $Y\in{\rm Lie}(G)$, we have ${\rm Ad}(g)(Y)\equiv gYg^{-1}\in{\rm Lie}(G)$, and thus $e^{hX}Y e^{-hX}\in{\rm Lie}(G)$ for any $h\in\mathbb R$ and $X\in\mathrm{Lie}(G)$.

      This result comes from observing how $e^{t (gYg^{-1})}=g e^{tY} g^{-1}\in G$.

   2. Finally, $$[X,Y] = \partial_h|_0 e^{hX}Y e^{-hX} \equiv \lim_{h\to 0}\frac{e^{hX}Y e^{-hX}-Y}{h}.$$ This implies $[X,Y]\in{\rm Lie}(G)$ because $e^{hX}\in G$, and we already proved that ${\rm Lie}(G)$ is a vector space.