Uniqueness of local flows of left-invariant vector fields on a Lie group

Question

The following questions suggest for any left-invariant vector field $X$ on a Lie group, we may speak of 'the' local flow of $X$.

Does local flow of left-invariant vector field commute with the left-translation operator?

Property of left invariant vector field and its local flow.

$X\in \mathfrak{g}$ means flow commutes with left-translation

I believe this is specific to left-invariant vector fields on a Lie group because of a paragraph in Mathematics For Physics: An Illustrated Handbook By Marsh Adam (I hope "flows" in this physics book are the same as "flows" in maths)

This is what I know thus far:

Definition 1. Let $M$ be a smooth manifold. Let $p \in M$. Let $(U, \varphi)$ be a chart about $p$ in $M$. Let $N$ be a neighborhood of $p$ in $U$ (and thus in $M$). Let $\varepsilon \in (0,\infty]$. A map $L: (-\varepsilon,\varepsilon) \times N \to U$ is a local flow about $p$ in $U$ if

1. $L$ is smooth

2. $L(0,n)=n$, for each $n \in N$.

3. Let $a,b \in (-\varepsilon,\varepsilon)$ and $n \in N$. If $a+b \in (-\varepsilon,\varepsilon)$ and $L(b,n)\in N$, then $L(a,L(b,n)) = L(a+b,n)$

Proposition 2. For each $n \in N$, $L$ gives rise to a map $L_t(n): (-\varepsilon,\varepsilon) \to U$, where $L_t(n) (t_0) := L(t_0,n)$, for each $t_0 \in (-\varepsilon,\varepsilon)$

Definition 3. The $L_t(n)$'s are called flow lines of the local flow $L$.

Definition 4. Additionally, $L$ is generated by $X$, for a smooth vector field on $U$, $X: U \to TU$, if for each $n \in N$, the flow line $L_t(n)$ is an integral curve of $X$ starting at $n$.

Definition 5. $X$ is complete if there exists a local flow with $\varepsilon = \infty$ and $N = U = M$

Proposition 6. If $M$ is a Lie group and $X$ is left-invariant, then $X$ is complete.

Now my questions:

First question: Let $G$ be a Lie group. For the definitions and proposition above, let $M = G$. Let $X$ be left-invariant. I believe this is more than enough to say some local flow $L$ exists. However, is this enough to say $L$ is unique?

Second question: Same as the first but $M = G = U$. (just in case the answer to the first question is negative)

Note. Just in case, I think we cannot use Prop 6 to answer either question because proving Prop 6 may first require the uniqueness of $L$ (after proving $L$ exists).

Thanks in advance!

Answer 1

Once you have formulated Definition 1, Proposition 2, Definition 3, and Definition 4, there is a further theorem that you need, which leads to a more general formulation of Definition 5.

The theorem you need is basically the existence and uniqueness theorem for solutions to first order, linear ODE's, rewritten in the setting of manifolds and vector fields. This theorem is not special to Lie groups; my best guess is that the paragraph you linked to is implicitly alluding to the more general existence and uniqueness theorem.

Theorem: For any smooth manifold $M$ and any smooth vector field $X$ on $M$, there exists an open subset $D \subset \mathbb R \times M$ and a smooth function $L : D \to M$ satisfying the following properties:

1. $D = \bigcup_{p \in M} I_p \times \{p\}$ where $I_p$ is an open inteval in $\mathbb R$ containing $p$.

2. $L$ is a flow.

3. $L$ is generated by $X$.

(2 and 3 are your Definitions 1 and 4, but appropriately rewritten using the domain $D$).

Furthermore,

4. $L$ is locally unique. To be precise, for any two choices $D,L$ and $D',L'$, and for any $p \in M$, there exist neighborhoods $N \subset U$ of $p$ in $M$, and there exists $\epsilon > 0$, such that $(-\epsilon,+\epsilon) \times N \subset D \cap D'$, and such that $L$ and $L'$ restrict to identical local flows $(-\epsilon,+\epsilon) \times N \mapsto U$: $$L \mid (-\epsilon,+\epsilon) \times N = L' \mid (-\epsilon,+\epsilon) \times N $$

From this local uniqueness it is straightforward to deduce a more global uniqueness, namely that for any two choices $D,L$ and $D',L'$, the restrictions $L \mid D \cap D'$ and $L' \mid D \cap D'$ are identical.

5. There exists a maximal choice of $D,L$, meaning that for any other choice $D',L'$ we have $D' \subset D$ and $L \mid D' = L'$.

6. A maximal choice of $D,L$ is unique.

In this very general context, one would form a generalization of Definition 5: $X$ is complete if the maximal $D$ is equal to $\mathbb R \times M$.

I believe this answers your first question affirmatively, which settles the rest of your questions as well.