# The Unapologetic Mathematician

## 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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Posted by John Armstrong | 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 [...]

Pingback by | June 16, 2011 | Reply

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 [...]

Pingback by | June 18, 2011 | Reply

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 [...]

Pingback by | June 21, 2011 | Reply

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

Pingback by | June 25, 2011 | Reply

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

Pingback by | July 13, 2011 | Reply

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