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.

in a solved exercise, there is a point in the solution that I can't work out. I would be grateful if somebody could give me the detailed steps.

Consider the trivial principal bundle $P = M\times U(1)$ over a $C^\infty$-manifold $M$. Let $\Phi_t$ be the flow of a vector field $\mathfrak{X}(P)$.

Apparently, if $R_z$ designates the group action of $z \in U(1)$ on $M$, $X$ is $U(1)$-invariant ($R_z \cdot X=X$) if and only if $R_z$ commutes with $\Phi_t$ ($R_z \circ \Phi_t= \Phi_t \circ R_z$). Can somebody confirm this and help me with the proof ?

Thanks,

JD

share|improve this question

1 Answer 1

This is true.

More generally, let the Lie group $G$ act on a manifold $M$ from right. Then a smooth map $f:M \rightarrow M $ is called equivalent if $f$ preserves the action of $G$ on $M$, i.e., $f\circ R_g=R_g\circ f $ for all $g\in G$. Now, suppose $X$ is a vector field on $M$ and $\Phi_t$ denotes its one parameter group. We want to prove that $X$ is $G$-invariant if and only if every $\Phi_t$ is equivalent.

Taking derivative from $\Phi_t(p.g)=\Phi_t(p).g$ with respect to $t$ and letting $t=0$, imply that $X(p.g)=dR_g(X(p))$, which means that $X$ is $G$-invariant. The converse is also hold.

share|improve this answer
    
I believe the right term is equivariant, not equivalent. –  Chris Godsil Dec 25 '12 at 16:22

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.