# The Unapologetic Mathematician

## 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:

\displaystyle\begin{aligned}\tilde{\Phi}_0(p)&=f\left(\Phi_0\left(f^{-1}(p)\right)\right)\\&=f\left(f^{-1}(p)\right)=p\end{aligned}

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 $\tilde{X}$.