# Invariant vector field by group action

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

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## 1 Answer

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 $G$-equivariant 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 $G$-equivariant.

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.

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