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 ?