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Let $\pi: P \rightarrow M$ be a principal $G$-bundle over a smooth manifold $M$. Now given any two points $p_1, p_2 \in P$ there exist a canonical function $f_{p_{1},p_2}:\bar{p_1} \rightarrow \bar{p_2}$ from the unique fibre $\bar{p_1}$ containing $p_1$ to the unique fibre $\bar{p_2}$ containing $p_2$ which is defined as the unique $G$-equivariant function which sends $p_1$ to $p_2$ i.e $p_1.g \mapsto p_2.g$ for all $g \in G$.

My question is the following:

Is $f_{p_1,p_2}$ always smooth for any $(p_1,p_2) \in P \times P$?

mathematics student
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  • Is $P\xrightarrow{\pi}M$ a smooth principal $G$-bundle? That is, do $P$ and $M$ have smooth structures, is $G$ a Lie group, and are the maps $\pi:P\to M$ and $\theta:P\times G\to P$ smooth in your definition? – Kajelad Oct 15 '20 at 19:25
  • @Kajelad Yes the answers to all your questions are affirmative. – mathematics student Oct 16 '20 at 03:22

1 Answers1

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This canonical map is smooth, and it can be derived from a simpler one: For every $p\in P$ there is a canonical equivariant diffeomorphism $\theta_p:G\to\bar{p}$. This map is simply the restriction of the action map $\theta_p(g)=p\cdot g$. Your map is then given by $f_{p_1,p_2}=\theta_{p_2}\circ\theta_{p_1}^{-1}$.

To show that $\theta_p$ is a diffeomorphism, first note that the $G$-action restricts to a smooth, free, and transitive action on each fiber (since fibers are embedded), so $\theta_p:G\to\bar{p}$ is smooth and bijective. It now suffices to show that $\theta_g$ is an immersion (see this question). This is the case, as can be shown using a condraditiction.

Suppose $\theta_p$ is not an immersion. There must be some nonzero $v\in T_gG$ with $d_g\theta_p(v)=0$. Let $V$ be the right-invariant vector field with $V(g)=v$. Since $\theta_p$ is equivariant, $d_h\theta_p(V(h))=0$ for all $h\in G$. Let $\gamma_V:(-\epsilon,\epsilon)\to G$ be an integral curve satisfying $\gamma_V(0)=e$ and $\frac{d}{dt}\gamma_V(t)=V(\gamma_V(t))$ (one can show $\gamma_V$ extends to a Lie group homomorphism $\mathbb{R}\to G$, but that isn't needed here). We then have $$ \frac{d}{dx}\theta_g(\gamma_V(g))=d_{\gamma_V(t)}\theta_g\left(\frac{d}{dt}\gamma_v(t)\right)=d_{\gamma_V(t)}\theta_g\left(V(\gamma_V(t))\right)=0 $$ thus $\theta_g(\gamma_V(t))=p$ and the action is not free; a contradiction.

Kajelad
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  • Sir, I understood all the arguments except the last line where you used the notion of 1-parameter subgroup corresponding to the right invariant extension of $v$ to get the contradiction. Can you please explain in little detail "this particular line"? Thanks in advance. – mathematics student Oct 16 '20 at 07:47
  • Are you familiar with the notion of a right-invariant vector field, and the one parameter subgroup it generates? – Kajelad Oct 16 '20 at 10:42
  • I am familiar with the definition of right invariant vector field. But I don't know much about the one parameter subgroup it generates! – mathematics student Oct 16 '20 at 12:20
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    The one parameter subgroup is just the integral curve of the vector field starting at the identity. I've added a bit more detail to that step. – Kajelad Oct 16 '20 at 13:23
  • Thank you Sir.. – mathematics student Oct 16 '20 at 14:40