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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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    I get the rest, thank you. My question originated from my confusion in the notations and the definition of an automorphism on a principal bundle.2012-11-25

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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.)