TITLE: Computing the Lie bracket on the Lie group $GL(n, \mathbb{R})$
QUESTION [12 upvotes]: Consider the Lie group $GL(n, \mathbb{R})$. Since $GL(n, \mathbb{R})$ is an open subset of the space $M_{n,n}(\mathbb{R})$ of $n \times n$ matrices, we can identify the tangent space (Lie algebra) $T_{e}GL(n, \mathbb{R})$ with $M_{n,n}(\mathbb{R})$, where $e$ is the identity matrix. I'm trying to understand the following:
For all $x, y \in M_{n,n}(\mathbb{R}) = T_{Id}GL(n, \mathbb{R})$ we have $[x, y] = xy -yx$ where $[\cdot, \cdot]$ is the Lie algebra bracket.
According to the question Is the Lie bracket of a Lie Algebra of a Matrix Lie Group always the commutator?, this seems to be the case for matrix Lie groups in general.
So far, I've worked out the following: Let $x \in M_{n,n}(\mathbb{R}) = T_{Id}GL(n, \mathbb{R})$ and let $X \in \Gamma^{inv}(TG)$ be the left invariant vector field corresponding to $x$, i.e. with $X(e) = x$. By openness, we can choose $\epsilon > 0$ such that the curve $(-\epsilon, \epsilon) \ni t \mapsto e + tx$ through the identity is contained in $GL(n, \mathbb{R})$. For $g \in GL(n, \mathbb{R})$ let $L_g$ be the left multiplication by $g$ in $GL(n, \mathbb{R})$. Using left invariance of $X$ calculate:
$$X(g) = ((L_g)_{*}X)(g) = D_eL_g(X(e)) = D_eL_g(x) = \frac{d}{dt}|_{t=0}g(e+tx) = gx$$
Now, the above implies for $f \in C^{\infty}(GL(n, \mathbb{R}))$ and $g \in GL(n, \mathbb{R})$:
$$X(g)(f) = (D_gf)(gx)$$
where we identify the tangent vector $X(g)$ with the corresponding derivation at $g$.
Now how do I calculate $[x,y] \stackrel{def.}{=} [X, Y](e) \stackrel{\text{WHY?}}{=}xy-yx$ from this?

REPLY [2 votes]: I think the sensible way to get an intuition for this is to just look at the Taylor expansion of the group commutator:
$$e^{\varepsilon x} e^{\varepsilon y} e^{-\varepsilon x} e^{-\varepsilon y}$$
Which to second order is $1+\varepsilon^2(xy-yx)$. Presumably you know how to prove that the second derivative of the above expression is equivalent to the derivative-of-the-adjoint definition.