The Unapologetic Mathematician

Mathematics for the interested outsider

Invariance and Flows

When we discussed the Lie algebra of a Lie group we discussed “left-invariant” vector fields. More generally than this if f:M\to M is a diffeomorphism we say that a vector field X\in\mathfrak{X}M is “f-invariant” if it is f-related to itself. That is, a vector field on a Lie group G is left-invariant if it is L_g-invariant for all g\in G.

Now we want a characterization of f-invariance in terms of the flow \Phi_t of X. I say that X is f-invariant if and only if \Phi_t\circ f=f\circ\Phi_t for all t. That is, the flow should commute with f.

We’ll show this by showing that the vector field \tilde{X}=f_*\circ X\circ f^{-1} has flow \tilde{\Phi}_t=f\circ\Phi_t\circ f^{-1}. Then if X is f-related to itself we know that X=f_*\circ X\circ f^{-1}, and so by uniqueness we conclude that the flows \Phi_t and f\circ\Phi_t\circ f^{-1} are equal, as asserted.

So, what makes f\circ\Phi_t\circ f^{-1} the flow of f_*\circ X\circ f^{-1}? First of all, we have to check the initial condition that \tilde{\Phi}_0(p)=p, which is perfectly straightforward to check:


More involved is the differential condition. It will help if we rewrite \tilde{\Phi} a bit as a function of both t and p:

\displaystyle\begin{aligned}\tilde{\Phi}(t,p)&=f\left(\Phi_t\left(f^{-1}(p)\right)\right)\\&=f\left(\Phi\left(t,f^{-1}(p)\right)\right)\\&=f\left(\Phi\left(\left[1_\mathbb{R}\times f^{-1}\right](t,p)\right)\right)\\&=\left[f\circ\Phi\circ\left(1_\mathbb{R}\times f^{-1}\right)\right](t,p)\end{aligned}

Now we can start on the differential condition:

\displaystyle\begin{aligned}\tilde{\Phi}_*\left(\frac{\partial}{\partial t}(t,p)\right)&=f_*\left(\Phi_*\left((1_\mathbb{R}\times f^{-1})_*\left(\iota_{p*}\left(\frac{d}{dt}(t)\right)\right)\right)\right)\\&=f_*\left(\Phi_*\left(\left((1_\mathbb{R}\times f^{-1})\circ\iota_p\right)_*\left(\frac{d}{dt}(t)\right)\right)\right)\\&=f_*\left(\Phi_*\left(\iota_{f^{-1}(p)*}\left(\frac{d}{dt}(t)\right)\right)\right)\\&=f_*\left(X\left(\Phi\left(t,f^{-1}(p)\right)\right)\right)\\&=\left[f_*\circ X\circ\Phi\right]\left(t,f^{-1}(p)\right)\\&=\left[f_*\circ X\circ f^{-1}\circ f\circ\Phi\circ(1_\mathbb{R}\times f^{-1})\right](t,p)\\&=\left[\tilde{X}\circ\tilde{\Phi}\right](t,p)\\&=\tilde{X}\left(\tilde{\Phi}(t,p)\right)\end{aligned}

And thus \tilde{\Phi} is indeed the flow of \tilde{X}.


June 17, 2011 - Posted by | Differential Topology, Topology

1 Comment »

  1. […] we assume that the flows commute. As we just saw last time, the fact that for all means that is -invariant. That is, . But this implies that the Lie […]

    Pingback by Brackets and Flows « The Unapologetic Mathematician | June 18, 2011 | Reply

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