The Unapologetic Mathematician

Mathematics for the interested outsider

The Lie Derivative

Let’s go back to the way a vector field on a manifold M gives us a “derivative” of smooth functions f\in\mathcal{O}M. If X\in\mathfrak{X}M is a smooth vector field it has a maximal flow \Phi which gives a one-parameter family \Phi_t of diffeomorphisms, which we can think of as “moving forward along X by t.

Now given a smooth function f we use this as if we were taking a derivative from all the way back in single-variable calculus: measure f at p\in M, flow forward by \Delta t and measure f at \Phi_{\Delta t}(p), take the difference, divide by \Delta t, and take the limit as \Delta t approaches zero:

\displaystyle\begin{aligned}\lim\limits_{\Delta t\to0}\frac{f\left(\Phi_{\Delta t}(p)\right)-f(p)}{\Delta t}&=\lim\limits_{\Delta t\to0}\frac{f\left(\Phi(\Delta t,p)\right)-f\left(\Phi(0,p)\right)}{\Delta t}\\&=\frac{\partial}{\partial t}f\left(\Phi(\Delta t,p)\right)\Big\vert_{t=0}\\&=\left[df(p)\right]\left(\Phi_*\left(\frac{\partial}{\partial t}(t,p)\Big\vert_{t=0}\right)\right)\\&=\left[df(p)\right]\left(X(\Phi(0,p))\right)\\&=\left[df(p)\right]\left(X(p)\right)\\&=Xf(p)\end{aligned}

Note that even if X is not complete we do always have some interval around t=0 on which \Phi_t is defined and this difference quotient makes sense.

So far this is just a complicated (but descriptive!) way of restating something we already knew about. But now we can take this same approach and apply it to other vector fields. So if Y\in\mathfrak{X}M is another smooth vector field, we define the “Lie derivative” of Y by X as:

\displaystyle(L_XY)_p=\lim\limits_{\Delta t\to0}\frac{\Phi_{-\Delta t*}\left(Y_{\Phi_{\Delta t}(p)}\right)-Y_p}{\Delta t}

Again we evaluate Y at both p and \Phi_{\Delta t}(p), but here’s where a trick comes in: we can’t compare these two vectors directly, since they live at different points on M, and thus in different tangent spaces. So in order to compensate we use the flow itself to move backwards from \Phi_{\Delta t}(p) back to p, and use the derivative \Phi_{-\Delta t*} to carry along the vector Y_{\Phi_{\Delta t}(p)}.

We can come up with an alternate version of this formula by using similar techniques to those above:

\displaystyle\begin{aligned}(L_XY)_p&=\lim\limits_{\Delta t\to0}\frac{\Phi_{-\Delta t*}\left(Y_{\Phi_{\Delta t}(p)}\right)-Y_p}{\Delta t}\\&=\lim\limits_{\Delta t\to0}\frac{\Phi_{-\Delta t*}\left(Y_{\Phi_{\Delta t}(p)}\right)-\Phi_{0*}\left(Y_{\Phi_0(p)}\right)}{\Delta t}\\&=\frac{\partial}{\partial t}\Phi_{-t*}\left(Y_{\Phi_t(p)}\right)\Big\vert_{t=0}\end{aligned}

That is, if we define the curve c(t)=\Phi_{-t*}\left(Y_{\Phi_t(p)}\right) in the tangent space \mathcal{T}_pM then (L_XY)_p=c'(0). This would seem to make it live in the tangent space to c(0) — that is, in \mathcal{T}_{Y_p}\mathcal{T}_pM — but remember that since \mathcal{T}_pM is a vector space we identify it with all of its tangent spaces. Thus (L_XY)_p\in\mathcal{T}_pM, just like Y_p is.

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June 15, 2011 - Posted by | Differential Topology, Topology

5 Comments »

  1. [...] all well and good to define the Lie derivative, but it’s not exactly straightforward to calculate it from the definition. For one thing, it [...]

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  2. [...] last time, the fact that for all means that is -invariant. That is, . But this implies that the Lie derivative vanishes, and we know that [...]

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  3. [...] we mentioned last time, Math Fail showed how to spot what your colleagues did over the summer,  The Unapologetic Mathematician gave a nice introduction to the Lie derivative, and on Computational Complexity there was a list of computer scientists that recently found a new [...]

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  4. [...] Armstrong: The Lie Derivative, Brackets and [...]

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  5. [...] defined the Lie derivative of one vector field by another, . This worked by using the flow of to compare nearby points, and [...]

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