Let $G$ be a Lie group and $v\in X_L(G)$. Then $v$ is complete. There is a proposition in my textbook that states,
" Let $G$ be a Lie group and   $v\in X_L(G)$. Then $v$ is complete. "
I believe that $X_L(G)$ is the left-invariant vector field, so then I interpret the theorem as if $G$ is a Lie group then all left-invariant vector fields are complete. I know that for this to be true then the flow must exist for all ??$\in\mathbb{R}$, would this be for all $v \in \mathbb{R}$?
How would I prove the proposition?
My idea:
Let $G$ be a Lie group with Lie algebra $\mathfrak g$ and $v \in \mathfrak g$. Write $X_L(G)$ as the left-invariant vector field on $\mathfrak g$ with $v \in X_L(G)$. Then
$$\phi_t(a) = a\gamma x(t) $$
is the flow of $X_L(G)$
Then for all $a\in G$ we have
$$\frac{d}{dt} |_{t=s} a\gamma x(t) = (dL_a)_{\gamma x(s)}(\frac{d}{dt}_{t=s} \gamma x(t))$$
$$= (dL_a)_{\gamma x(s)}(\frac{d}{dt}_{t=s} \gamma x(t+s))$$
$$=(dL_a)_{\gamma x(s)}(\frac{d}{dt}_{t=s} \gamma x(s) \gamma x (t))$$
$$=(dL_a)_{\gamma x(s)}(\frac{d}{dt}_{t=s}  L_{\gamma x(s)}(\gamma x (t)))$$
$$=(dL_a)_{\gamma x(s)}1(\frac{d}{dt}_{t=s} \gamma x (t))$$
$$=(dL_a)_{\gamma x(s)}(X)$$
$$=X_L(G)(a\gamma x(s))$$
which would show that the flow of $X_L(G)$ exists for all $t$?
Any help would be appreciated. I find this topic very difficult.
 A: Yes $X_L(G)$ denotes the lie algebra of the lie group , i.e , the set of left-invariant vector fields or equivalently $T_e G$. Now for a vector field to be complete we have to show that it's flow is defined for all $t\in \mathbb{R}$, by definition. So take a left-invariant vector field $X$ and for example the identity $e$ in $G$. By results from ordinary differential equations there will exists an integral curve for $X$, $c_e(t)$ for $t\in (-\epsilon,\epsilon)$ such that $c_h(0)=e$. Now the idea is to use the left-translations of a lie group which are diffeomorphism to extend this curve to the lie group that is take $t_0 \in (-\epsilon,\epsilon)$ and we have that $c_e(t_0)c_e(t)$ will define an integral curve for $X$ at $c_e(t_0)$, where we use here the fact that $X$ is left-invariant. Now this curve is extending our previous one, and so by uniquess of the Picard-Lindelof theorem we have this curve extends, i.e, we can always extend it's interval of definition. Doing this again and again for all the points you get the desired result.
Here I will actually compute the differential so that things are clear :
Let $g=c_e(t_0)$ then we have that $\frac{d}{dt}gc_e(t)=\frac{d}{dt}L_g(c_e(t))=\frac{d}{ds}|_{s=0}L_g(c_e(t+s))=d_{c_e(t)}L_g(X_{c_e(t)})=X_{gc_e(t)}$.
