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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.

Terry
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  • Do you want to understand why the isomorphism $\mathfrak {gl}_n \mathbb R$ with $M_n (\mathbb R)$ induces a Lie algebra structure on $M_n (\mathbb R)$? – Aaron Maroja Aug 28 '16 at 15:52
  • @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. – Terry Aug 28 '16 at 15:55
  • Did you show that this isomorphism preserves brackets? – Aaron Maroja Aug 28 '16 at 15:57
  • @AaronMaroja: (the natural Lie algebra structure on $L_G$ is the one we get when thinking of derivations, as I wrote in my question) – Terry Aug 28 '16 at 15:57
  • @AaronMaroja: No. This is what I would like to know: Why does it preserve brackets? – Terry Aug 28 '16 at 15:57
  • It is a not so difficult proposition once you've defined certain things. This result, for example, is in Lee, Jeffrey Manifolds and Differential Geometry, Chapter 5, page 207, as a Corolllary of Proposition 5.52. – Aaron Maroja Aug 28 '16 at 15:59
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    @AaronMaroja: Thank you very much for this very accurate reference. – Terry Aug 28 '16 at 17:06

1 Answers1

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

Tsemo Aristide
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