# Automorphism on a trivial principal bundle

I have a simple question, related to a precedent post: Invariant vector field by group action.

$M$ is a n-manifold. $P = M \times U(1)$ is a trivial principal bundle over $M$. $X$ is a vector field over $P$. $\Phi_t$ is the flow of $X$. $u$ and $z$ are members of $U(1)$.

In the exercise, one has to prove that $\Phi_t$ is an automorphism if and only if $X$ is an $U(1)$-invariant vector field.

The problem is that in the solution, the authors writes the automorphism condition as $\Phi_t(u)\cdot z=\Phi_t(u \cdot z)$. I would have written this condition as $\Phi_t(u)\cdot \Phi_t(z)=\Phi_t(u \cdot z)$.

If I make a visual analogy, for the case $M = \mathbb{R}$, then the trivial principal bundle is an infinite cylinder. Now if I take a vector field that "spirals" around this cylinder ($\Leftrightarrow$ has non-zero components), $\Phi_t(z) \neq z$.

Can someone help clarify how $\Phi_t(u)\cdot z=\Phi_t(u \cdot z)$ characterize $\Phi(t)$ as an automorphism ?

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It can't be your version of the condition because on a general principal bundle, you can't multiply points. Also, I think $u$ should be in $P$, not in $U(1)$. – Jason DeVito Nov 24 '12 at 23:23
The nature of $u$ is not detailed in the exercise. Suppose then that $u \in P$ and $z \in U(1)$. Is $\Phi_t(u)\cdot z=\Phi_t(u \cdot z)$ an automorphism ? Because in that case two spaces ($P$ and $U(1)$) are involved in the inputs. For an automorphism one shouldn't have the same space in the inputs and output ? – vkubicki Nov 25 '12 at 0:17
For a $G$-principal bundle $P$, the group $G$ acts on the bundle $P$. So for each $g\in G$, one gets an associated automorphism of $P$ denoted by right multiplication by $g$. – Jason DeVito Nov 25 '12 at 1:17
Thank you ! Would you like to upgrade this as an answer so I can reward it ? – vkubicki Nov 25 '12 at 2:00
I could (probably tomorrow). Do you understand the rest of the solution? Or should I worry about that as well? – Jason DeVito Nov 25 '12 at 3:40

In the notation, we should have $u\in P$ and $z\in U(1)$. In general, given a $G$-principal bundle $P$, we cannot multiply points of $P$ together (as the notation $\phi_t(u)\cdot \phi_t(z)$ suggests). Instead, given each $g\in G$, there is a corresponding automorphism of $P$ (denoted by right multiplication by $g). So, at least the equation$\phi_t(u)\cdot z = \phi_t(u\cdot z)$could be true (meaning, that for any choice of$t\in \mathbb{R}$,$u\in P$, and$z\in U(1)$both the left hand and right hand sides evaluate to points in$P\$, so they at least have a chance of being equal.)

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