Left Invariant smooth vector field on a Lie Group is complete

Question

How do I show the claim of the title?

I was suggested to use the fact that given a one parameter family of local diffeomorphisms that agree in the intersection of their domain of definition I can produce a unique vector field whose flows are exactly this family of local diffeomorphisms.

Answer 1

Let $G$ a Lie group and ${\cal G}$ its Lie algebra. Let $X\in {\cal G}$, the left invariant vector $\tilde X$ defiined by $X$ is the vector $\tilde X(g)=d{L_g}_e(X)$ where $d{L_g}_e$ is the differential of $L_g(x)=gx$ at the identity.

Consider $\phi_t(g)=gexp(tX)$, $\phi_{t+t'}(g)=exp(tX)exp(t'X)$ and ${d\over {dt}}_{t=0}gexp(tX)=d{L_g}_eX=\tilde X(g)$. This show that $\phi_t$ is the flow of $\tilde X$.

https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory)#Properties

Answer 2

You know that for every manifold $M$, the flow of a vector field $X \in \mathfrak{X}(M)$ is a map $\Phi_X\colon \mathcal{D}_X \subseteq \Bbb R \times M \to M$, where $\mathcal{D}_X$ is some open domain containing $\{0\}\times M$. If there is $\epsilon > 0$ such that $]-\epsilon,\epsilon[\times M \subseteq \mathcal{D}_X$, then we actually have $\mathcal{D}_X = \Bbb R \times M$, meaning that $X$ is complete. The idea of the proof is that if you take an integral curve of $X$, $\alpha\colon [0, b) \to M$, look at the point $\alpha(b-\epsilon/2)$, then the integral curve of $X$ starting at this point will give an extension of $\alpha$ to at least $[0,b+\epsilon/2)$. Repeat to conclude that $\alpha$ may be extended up to $+\infty$. Similarly, you show that the left end of the interval domain is $-\infty$. If I recall, Lee calls this "Escape Lemma", or something like this. (By the way, a trivial corollary of this is that every vector field in a compact manifold is complete)

Now, if $G$ is a Lie group and $X \in \mathfrak{X}^L(G)$ is left-invariant, we show that $X$ has the above property. This is done by showing that given $g,h \in G$, the flow satisfies $$\Phi_X(t,gh) = g\Phi_X(t,h)$$for all $t$, $g$ and $h$. To see this, show that both sides are solutions to the initial value problem $$\begin{cases} \alpha'(t) = X_{\alpha(t)} \\ \alpha(0) = gh\end{cases}$$Taking $h=e$, we see that $\Phi_X(t,g) = g\Phi_X(t,e)$. This shows that if $I_g\subseteq \Bbb R$ is the domain of the integral curve of $X$ starting at $g$, then $I_e \subseteq I_g$ for all $g \in G$. By the first paragraph, $\mathcal{D}_X = \Bbb R \times G$. Observe that knowing the definition of $\exp(tX)$ is not required for this argument.