# Completeness of invariant vector fields on a principal bundle

Let $$G$$ be a Lie group. A Princpal $$G$$-bundle is a surjective smooth submersion $$\pi:P \rightarrow M$$ equipped with a smooth right $$G$$-action on $$P$$ such that the following properties hold:

1. For any $$x \in M$$, we have $$P_x \cdot G \subseteq P_x$$, where $$P_x := \pi^{-1}(x)$$.
2. For any $$x \in M$$, there exists a neighbourhood $$U \subseteq M$$ of $$x$$ and a $$G$$-equivariant diffeomorphism $$\Phi: \pi^{-1}(U) \rightarrow U \times G$$ such that $$\pi = \pi_1 \circ \Phi$$, where $$\pi_1:U \times G \rightarrow U$$ is the obvious projection, and the action of $$G$$ on $$U \times G$$ is given by $$(y,g) \cdot h = (y,gh)$$.

Let $$\pi:P \rightarrow M$$ be a principal $$G$$-bundle. We say that a vector field $$X \in \mathfrak X(P)$$ is $$G$$-invariant if for any $$p \in P$$ and $$g \in G$$, we can write $$X_{pg} = d\theta_g|_p(X_p)$$, where $$\theta_g:P \rightarrow P$$ is the map given by $$p \mapsto pg$$. On the other hand, we say that a vector field $$X \in \mathfrak X(P)$$ is complete if the domains of its maximal integral curves is $$\mathbb R$$.

Question. Let $$\pi:P \rightarrow M$$ be a principal $$G$$-bundle. Suppose $$M$$ is compact. Let $$X \in \mathfrak X(P)$$ be a $$G$$-invariant vector field. Is $$X$$ complete?

Here is what I have figured out so far: since $$X$$ is $$G$$-invariant, there exists a unique vector field $$Y$$ on $$M$$ which is $$\pi$$-related to $$X$$. Since $$M$$ is compact, Corollary 9.17 of Lee's Smooth Manifolds implies that $$Y$$ is a complete vector field. I also know that if I have any $$G$$-invariant vector field on $$G$$ is complete by Theorem 9.18 of Lee.

I tried looking at the case when the bundle is trivial, i.e. $$P = M \times G$$, but I wasn't able to answer my question.

If $$M$$ is not compact, then the answer to my question is no: for example, take $$M = \mathbb R^+$$, $$G = \mathbb R$$, and $$P = \mathbb R^+ \times \mathbb R$$. Let $$X:P \rightarrow TP$$ be the vector field given by $$X_{(x,y)} := ( d/dt |_x, 0_y)$$. Then $$X$$ is not complete.

Notation Let $$G$$ be a Lie group and $$\pi\colon P\to M$$ be a principal $$G$$-bundle with principal action denoted by $$\Phi\colon G\times P\to P,\ (g,p)\mapsto \Phi(g,p)\equiv \Phi_g(p)\equiv g.p$$ Let us also consider a $$G$$-invariant vector field $$X$$ on $$P$$, i.e. a vector field $$X\in\mathfrak{X}(P)$$ such that $$(\Phi_g)_\ast X=X, \text{for all}\ g\in G.\tag{\star}$$ As you may already know (cf, e.g., Proposition 9.6 in Lee), this condition is equivalent to requiring that, for any $$g\in G$$ and $$t\in\mathbb{R}$$, the following identity holds where both sides are defined $$\text{Fl}^X_t\circ\Phi_g=\Phi_g\circ\text{Fl}^X_t.\tag{\star\star}$$ Above, as in the following, we denote the flow of $$X$$ at time $$t$$ as $$\text{Fl}_t^X\colon\mathcal{U}_t\subseteq P\to P$$, while instead, we denote the flow of $$X$$ as $$\text{Fl}\colon\cup_{t\in\mathbb{R}}(\{t\}\times\mathcal{U}_t)\subseteq\mathbb{R}\times P\to P, (t,p)\mapsto\text{Fl}^X(t,p)=\text{Fl}^X_t(p).$$

Lemma 1 For any $$x\in M$$, there exist a positive real number $$\epsilon_x\in\mathbb{R}_+$$ and an open neighborhood $$V_x$$ of $$x$$ in $$M$$ such that, each maximal integral curve of $$X$$ starting from $$\pi^{-1}(V_x)$$ is defined at least over $$[-\epsilon_x,+\epsilon_x]$$, i.e. $$[-\epsilon_x,+\epsilon_x]\times \pi^{-1}(V_x)\subset\text{dom}(\text{Fl}^X).$$

Proof Pick an arbitrary $$p\in\pi^{-1}(x)$$. Then there exist a positive real number $$\epsilon_p\in\mathbb{R}$$ and an open neighborhood $$U$$ of $$p$$ in $$P$$ such that each maximal integral curve of $$X$$ starting from $$U$$ is defined at least over the interval $$[-\epsilon_p,+\epsilon_p]$$, i.e. $$[-\epsilon_p,+\epsilon_p]\times U\subseteq\text{dom}(\text{Fl}^X).$$

Condition ($$\star\star$$) assures that, for any $$q\in U$$, if $$\gamma\colon[-\epsilon_p,+\epsilon_p]\to P$$ is an integral curve of $$X$$ with $$\gamma(0)=q$$, then $$\Phi_g\circ\gamma\colon[-\epsilon_p,+\epsilon_p]\to P,\ t\mapsto g.\gamma(t)$$, is an integral curve of $$X$$ with $$(\Phi_g\circ\gamma)(0)=g.q$$. Consequently, if one denotes by $$V$$ the open neighborhood of $$x$$ in $$M$$ given by $$V=\pi(U)$$, one gets that:

• $$\pi^{-1}(V)=G.U=\cup_{g\in G}(g.U)$$ is a $$G$$-saturated open neighborhood of $$G.p$$ in $$P$$ and
• each maximal integral curve of $$X$$ starting from $$\pi^{-1}(V)$$ is defined at least over the interval $$[-\epsilon_p,+\epsilon_p]$$, i.e. $$[-\epsilon_p,\epsilon_p]\times\pi^{-1}(V)\subset\text{dom}(\text{Fl}^X),$$

as one needed to prove.

Lemma 2 If $$M$$ is compact, there exists $$\epsilon\in\mathbb{R}_+$$ such that each maximal integral curve of $$X$$ is defined at least on the interval $$[-\epsilon,+\epsilon]$$, i.e. $$[-\epsilon,+\epsilon]\times P\subset\text{dom}(\text{Fl}^X).$$

Proof Since $$M$$ is compact and $$\{V_x\}_{x\in M}$$ is an open covering of $$M$$, there exists $$x_1,\ldots,x_N\in M$$ such that $$V_{x_1}\cup\ldots\cup V_{x_N}=M$$. Hence, setting $$\epsilon:=\min\{\epsilon_{x_1}\ldots,\epsilon_{x_N}\}\in\mathbb{R}_+$$, one gets that $$[-\epsilon,\epsilon]\times P\subseteq \text{dom}(\text{Fl}^X)$$.

Proposition If $$M$$ is compact, then $$X$$ is complete.
Proof In view of Lemma 2, for any $$p\in P$$, each $$\gamma\colon[-R,R]\to P$$, integral curve of $$X$$ with $$\gamma(0)=p$$, can be extended to an integral curve of $$X$$ defined on $$[-R-\epsilon,R+\epsilon]$$ defined as follows $$\tilde\gamma\colon[-R-\epsilon,R+\epsilon]\to P,\ t\mapsto\tilde\gamma(t):=\begin{cases}\text{Fl}_{t-R}^X(\gamma(R)),&\text{if}\ t\in[R-\epsilon,R+\epsilon];\\ \gamma(t), &\text{if}\ t\in[-R,+R];\\ \text{Fl}_{t+R}^X(\gamma(-R)),&\text{if}\ t\in[-R-\epsilon,-R+\epsilon].\end{cases}$$ Since this procedure can be iterated indefinitely, one concludes that each maximal integral curve of $$X$$ is defined on the entire $$\mathbb{R}$$, i.e. the vector field $$X$$ is complete.