The simplest way to prove that any left-invariant vector field on a Lie group is complete

Question

It's all in the question: I look for the most intuitive proof that the integral curves of any left-invaraint vector field on a Lie group can be extended for all values of "time". I realize that the argument is always based on the existence of group multiplication; what I look for is the most straightforward proof available. Thanks in advance!

Answer 1

A vector field is complete if and only if its flow $\Phi_t$ exists for all $t \in \Bbb{R}$. So we can show that a left-invariant vector field $X$ on a Lie group $G$ is complete by computing its flow.

The flow $\Phi_t$ of $X$ satisfies the differential equation $\frac{ \rm{d}}{\rm{d}t} \Phi_t(g) |_{t=0} = X_g$ for all $g \in G$. Applying the flow to the identity element, we have $\Phi_t(1) = e^{tX}$, the Lie-theoretic exponential map. Moreover since $X$ is left-invariant, $\Phi_t$ is left-invariant $\Phi_t(g) = g\Phi_t(1) = ge^{tX}.$ So we see that $\Phi_t: G \to G$ is equal to right-multiplication by $e^{tX}$.