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.

  • $\begingroup$ Do you want to understand why the isomorphism $\mathfrak {gl}_n \mathbb R$ with $M_n (\mathbb R)$ induces a Lie algebra structure on $M_n (\mathbb R)$? $\endgroup$ – Aaron Maroja Aug 28 '16 at 15:52
  • $\begingroup$ @AaronMaroja: Both $M_n(\mathbb{R})$ and $L_G$ have natural Lie algebra structures. In addition, there is a natural linear isomorphism between them. I would like to understand why this isomorphism is a Lie algebra isomorphism. $\endgroup$ – Terry Aug 28 '16 at 15:55
  • $\begingroup$ Did you show that this isomorphism preserves brackets? $\endgroup$ – Aaron Maroja Aug 28 '16 at 15:57
  • $\begingroup$ @AaronMaroja: (the natural Lie algebra structure on $L_G$ is the one we get when thinking of derivations, as I wrote in my question) $\endgroup$ – Terry Aug 28 '16 at 15:57
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    $\begingroup$ @AaronMaroja: Thank you very much for this very accurate reference. $\endgroup$ – Terry Aug 28 '16 at 17:06

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 $dL_X(A) =l_A(X)=XA$, $[l_A,l_B](X)=-Dl_B(l_A)(X)+Dl_A(l_B)(X)=[AB-BA](X)$.

  • $\begingroup$ What does "D" stand for (capital d)? $\endgroup$ – Terry Aug 28 '16 at 16:15
  • $\begingroup$ It means the differential of the vector field $\endgroup$ – Tsemo Aristide Aug 28 '16 at 16:15
  • $\begingroup$ Is D missing twice on the last line? $\endgroup$ – Terry Aug 28 '16 at 16:39
  • $\begingroup$ No, the last line is the formula $[X,Y]=-DY.X+DX.Y$ applied to $X=l_A, Y=l_B$. $\endgroup$ – Tsemo Aristide Aug 28 '16 at 16:44
  • $\begingroup$ So are there too many D's in the first paragraph where $[X,Y]$ is defined? $\endgroup$ – Terry Aug 28 '16 at 16:46

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