Lie algebra of $\operatorname{GL}_n\mathbb{R}$

Question

Let $G=\operatorname{GL}_n\mathbb{R}$. I am trying to understand why $L_G=M_n\mathbb{R}$ with the usual Lie algebra structure of $M_n\mathbb{R}$.

1. I understand the canonical identification $L_G=T_1G=M_n\mathbb{R}$.
2. I understand that a tangent vector $A\in M_n\mathbb{R}=T_1G$ corresponds to a one-parameter group $t\mapsto R_{e^{tA}}$ (where $R_B$ is the linear operator of "right multiplication by $B$").
3. I understand that, therefore, the tangent vector $A\in T_1G$ corresponds to the vector field which assigns to $B\in G$ the tangent vector $BA\in T_BG$.
4. I understand that $[R_{A_1},R_{A_2}]=R_{[A_1,A_2]}$.

I don't understand how to use all this to prove the desired result. For me, the definition of the Lie bracket on $L_G$ is by thinking of $L_G$ as left invariant derivations of $C^{\infty}(G)$ with the usual Lie bracket of operators.

Answer 1

Let $M$ be a manifold, $\chi(M)$ the space of vector fields on $M$ is identified with the space of derivations of $C^{\infty}(M)$ by the formula $L_X(f)=X.f=df.X$ so you have $X.(ff')=d(ff').X=fdf'.X+f'df.X$.Under this identification, we have $[L_X,L_Y]=L_{[X,Y]}$. Where $[X,Y]=-DY.X+DX.Y$.

Now if $G$ is a Lie group and ${\cal G}$ the tangent space of the identity. For every $A\in {\cal G}, g\in G$, set $l_A(g)=d{L_g}_1(A)$,where $L_g$ is defined by $L_g(g')=gg'$. It is a vector field invariant by left multiplications, and it can be shown that $[l_A,l_B]$ is invariant by left multiplication, so there exists $[A,B]\in T_1G$ such that $[l_A,l_B]=l_{[A,B]}$.

Now suppose that $G=Gl_n(R)$. It is an open subset of the vector space $M_n(R)$. So you can identify $T_1Gl_n(R)$ with $M_n(R)$, if $A\in M_n(R), X\in Gl_n(R)$ the lelt multiplication $L_X$ is the linear map $L_X(Y)=XY$, so $dL_X(A) =l_A(X)=XA$, $[l_A,l_B](X)=-Dl_B(l_A)(X)+Dl_A(l_B)(X)=[AB-BA](X)$.