Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
    
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

1 Answer 1

up vote 1 down vote accepted

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.