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Let $G$ be a lie group and $l_g:G\rightarrow G$ be the left multiplication map.

Let $X$ be a left invariant vector field on $G$ i.e., $X:G\rightarrow TG$ is such that $(l_g)_*X=X$ on $G$ where $(l_g)_*X$ is defined as follows : $$((l_g)_*X)_{gh}=(l_g)_{*,h}(X_h)$$ where $(l_g)_{*,h}:T_hG\rightarrow T_{gh}G$ is the differential of $l_g$ at $h$.

So, supposing $X$ is a left invariant vector field, we have $$(l_g)_{*,e}(X_e)=X_g.$$

I am trying to prove that the lie bracket $[X,Y]$ is also a left invariant vector field.

$$[X,Y]_g(f)=X_g(Y(f))-Y_g(X(f))$$ $$=(l_g)_{*,e}(X_e)(Y(f))-(l_g)_{*,e}(Y_e)(X(f)) =X_e(Y(f)\circ l_g)-Y_e(X(f)\circ l_g)$$ $$=X_e(Y(f\circ l_g))-Y_e(X(f\circ l_g))=[X,Y]_e(f\circ l_g)=(l_g)_{*,e}([X,Y]_e)(f)$$

This is true for all $f$. So, we have $$(l_g)_{*,e}([X,Y]_e)=[X,Y]_g.$$

Using same idea, we can prove that $$(l_g)_{*,h}([X,Y]_h)=[X,Y]_{gh}$$ for all $h\in G$ which is same as saying that $[X,Y]$ is a left invariant vector field.

I am sure about all equalities except one i.e., $Y(f\circ l_g)=Y(f)\circ l_g$.

I am sure this is correct but could not see. Any suggestion about the proof is welcome.

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I have not gone through all your computations, I did just check the equality you were doubting.

First, let me recall that by definition, for all $x\in G$ and all smooth map $h\colon G\rightarrow\mathbb{R}$, one has the following: $$Y(h)(x):=T_xh(Y(x))\tag{1}.$$ Therefore, using the definition $(1)$ and the chain rule formula, one gets: $$Y(f\circ l_g)(x)=T_x(f\circ l_g)(Y(x))=T_{gx}f(T_xl_g(Y(x)))\tag{2}.$$ Besides, using twice the left-invariance of $Y$ and the chain rule formula along with $l_{gx}=l_g\circ l_x$, one has: $$Y(gx)=T_el_{gx}(Y(e))=T_e(l_g\circ l_x)(Y(e))=T_xl_g(T_el_x(Y(e))=T_xl_g(Y(x))\tag{3}.$$ Hence, using the definition $(1)$ and the equality $(3)$, one gets: $$Y(f)(l_g(x))=T_{gx}f(Y(gx))=T_{gx}f(T_xl_g(Y(x)))\tag{4}.$$ Finally, the left terms of $(2)$ and $(4)$ are equal. Whence the result.


Here is another way to see that if $X$ and $Y$ are left-invariant vector fields of $G$, then $[X,Y]$ is left invariant.

Proposition. Let $M$ be a smooth manifold, $X,Y\in\Gamma(TM)$ and $f\colon M\rightarrow\mathbb{R}$ be a smooth map, then: $$[f^*X,f^*Y]=f^*[X,Y].$$

This identity is easily checked on the level of derivations of $C^\infty(\mathbb{R}^n,\mathbb{R})$, so it holds true working in charts.

To conclude, just apply this proposition to $f=l_g$ for all $g\in G$.

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  • $\begingroup$ I have checked it one time back. Thanks for writing an answer, $\endgroup$ – user312648 Dec 29 '17 at 17:02

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