Lie Derivative, Geometrically
Steve Trettel
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The Lie derivative is a coordinate invariant, and metric-independent notion of differentiation on a smooth manifold. It differentiates with respect to a vector field, and its goal is to take some geometric object on the manifold and quantify how that object changes if you were to watch it as you traveled around the manifold, flowing along the field. The goal of this note is to give a brief formal definition. Note for vector and tensor fields I will give a definition that is slightly different than the usual one occurring in textbooks, from which I find the the connection with the Lie bracket more natural to motivate. Then of course we show equivalence with the usual form of the definition.
Lie Derivatives of Functions
We warm up by discussing the Lie derivative of a function $f\colon M\to \mathbb{R}$, which gives us opportunity to formalize the notion of watching $f$ while flowing along the vector field. Given a vector field $V$ on $M$, we can define a flow: where every point $p\in M$ is carried along an integral curve of $V$. By the existence and uniqueness of solutions to ODEs, doing this at each point generates1 a 1-parameter family of diffeomorphisms2 $\Phi_t\colon M\to M$, satisfying $$\frac{d}{dt}\Big|_{t=0}\Phi_t=V$$
If we were to start at the point $p$ and flow along $V$ for time $t$ we would end up at $\Phi_t(p)$, and so if we were watching $f$, the value we would see is $f(\Phi_t(p))$. To measure how quickly we see $f$ change as we flow care-free along the manifold, we simply differentiate this quantity with respect to $t$:
For comparison with the coming generalizations, it will be useful to write out the difference quotient defining the derivative explicitly:
$$(\mathcal{L}Vf)(p)=\lim{t\to 0}\frac{f(\Phi_t(p))-f(p)}{t}$$
Finally, while our goal here is to understand the geometric definition and not particularly to find convenient computational formulas, its worth pointing out that the Lie derivative of functions coincides3 with the directional derivative:
$$\begin{align}
(\mathcal{L}Vf)(p):&=\frac{d}{dt}\Big|{t=0}f(\Phi_t(p))\
&= (df){\Phi_0(p)}\left(\frac{d}{dt}\Big|{t=0}\Phi_t(p)\right)\
&=(df)_{\Phi_0(p)}\left(V(p)\right)\
&= df_p(V(p))\
&= V(f)
\end{align}$$
This is why you don’t often hear about the Lie derivative of functions as a separate concept.
Lie Derivatives of Vector Fields
Moving on to perhaps the most common case, what is the Lie derivative of a vector field? Imagine again that we are lazily flowing along $M$ following the integral curves of $V$, but this time instead of watching the values of a scalar function $f$ we are observing a vector field $X$ as we pass by. At time $t$, we hae flowed to the point $\Phi_t(p)$ and so we observe the vector $X_{\Phi_t(p)}$. And we may be tempted to say that the change we observe in $X$ is simply the time derivative of this quantity $\tfrac{d}{dt}|_{t=0}X_{\Phi_t(p)}$. But we must be careful! This quantity actually *does not make any sense*: if we wrote out the defining difference quotient we would find the numerator contains the term $X_{\Phi_t(p)}-X_{p}$, and these two vectors live in two different4 vector spaces!
To fix this we need to think harder about the picture we are trying to formalize, of us flowing along the vector field and comparing the vector we see alongside our boat with the original vector we saw at the start. To be able to make the comparison, we need to actually carry the original vector with us so we can place it directly alongside the vectors we flow past and make an accurate comparison.
Happily, there is a natural means5 of carrying a vector along the flow with us; which is to use the pushforward $(\Phi_t)\star$. If $X$ is any vector based at $p$ then $(\Phi_t)\ast X$ is the result carrying $X$ along the flow to the point $\Phi_t(p)$.
So now that we have brought our vector with us, we can legitimately make a comparison of two vectors in the same tangent space: $X_{\Phi_t(p)}$ and $(\Phi_t)_\ast X_p$. Thus, one may wish to attempt a definition of the form
This is both correct and matches perfectly well our mental image, but the formula itself raises some new technical difficulties. Particularly, while for each fixed $t$ the difference quotient is perfectly well defined, it is occurring in the tangent space at $\Phi_t(p)$, and so to compute our limit we cannot work in a single vector space but rather need to deal with the full machinery of the tangent bundle.
This is not a serious issue, as in differential geometry, a vector field is an object which acts on functions. So, to ‘compute6’ the Lie derivative $\mathcal{L}_VX$ means to be able to compute the number $(\mathcal{L}_VX)_p(f)$ for an arbitrary smooth function $f$ at each $p\in M$.
$$\begin{align}
(\mathcal{L}_VX)p(f)
&=\lim_{t\to 0 }\frac{X_{\Phi_t(p)}(f)-\left\ast X_p\right(\Phi_t)}{t}\
&=\lim_{t\to 0 }\frac{X_{\Phi_t(p)}(f)-X_p(f\circ \Phi_t)}{t}\
\end{align}
$$
While each term in the numerator here is being computed in a different tangent space the result of each computation is just a real number. So, this is a real-valued limit, and can be computed without thinking in tangent-bundle terms.
An Alternative Formulation
We should also note that there is a means of getting around this technical difficulty (of the vectors for each $t$ occuring in a different tangent space) by proposing an alternative definition. Instead of carrying the original vector along for the journey, why don’t we just pull the new vector back to the starting point? Now every difference quotient occurs in the fixed tangent space $T_pM$.
This results in the expression below, where we have introduced the tilde to distinguish it from our definition above
$$(\widetilde{\mathcal{L}VX})p:=\lim{t\to 0}\frac{\Phi_t^\ast X{\Phi_t(p)}-X_p}{t}$$
Here $\Phi_t^\ast$ is the pullback along the diffeomorphism $\Phi_t$. This however introduces a new complication : vector fields are objects which naturally push forward under maps rather than pull back. For a diffeomorphism like $\Phi_t$ of course, both are defined, but it is still simplest to compute the pullback as the pushforward along the inverse. And, as the inverse of the time-$t$ flow is simply to flow backwards for the same amount of time, $\Phi_t^{-1}=\Phi_{-t}$, one often sees this second definition appearing in textbooks as
$$(\widetilde{\mathcal{L}VX})p=\lim{t\to 0}\frac{(\Phi{-t})\ast X{\Phi_t(p)}-X_p}{t}$$
Also note that while the definition itself cleanly all occurs in a single tangent space, when one goes to use it, of course the computations go back to occuring in different tangent spaces, much like for our definition above. Below we apply this to a smooth function $f$:
$$\begin{align}
(\widetilde{\mathcal{L}VX})p(f)&=\lim{t\to 0}\frac{(\Phi{-t})\ast X{\Phi_t(p)}(f)-X_p(f)}{t}\
&=\lim_{t\to 0}\frac{X_{\Phi_t(p)}(f\circ \Phi_{-t})-X_p(f)}{t}
\end{align}$$
While the limit is again real valued, the first term in the numerator is computed in the tangent space at $\Phi_t(p)$ and the second is computed directly at $p$.
Of course, the only reason we are carrying on speaking about two Lie derivatives is that they are one in the same thing: either can be used depending on convenience.
Let $\Delta_t, \widetilde{\Delta_t}$ denote the difference quotients occurring in the definitions of $\mathcal{L}VX$ and $\widetilde{\mathcal{L}VX}$ respectively.
A quick computation shows $\Phi_t^\ast \Delta_t = \widetilde{\Delta_t}$:
$$\begin{align}
\Phi_t^\ast \Delta_t &=\Phi_t^\ast \left(\frac{X{\Phi_t(p)}-(\Phi_t)\ast X_p}{t}\right)\
&= \frac{\Phi_t^\ast X_{\Phi_t(p)}-\Phi_t^\ast(\Phi_t)_\ast X_p}{t}\
&= \frac{\Phi_t^\ast X_{\Phi_t(p)}-(\Phi_{-t})_\ast (\Phi_t)_\ast X_p}{t}\
&= \frac{\Phi_t^\ast X_{\Phi_t(p)}-(\Phi_{t-t})_\ast X_p}{t}\
&= \frac{\Phi_t^\ast X_{\Phi_t(p)}-X_p}{t}\
&= \widetilde{\Delta_t}
\end{align}$$
Where we have used that pulling back along $\Phi_t$ is equal to pushing forward along $\Phi_{-t}$, and that pushforwards along a 1-parameter family of diffeomorphisms compose via $(\Phi_s)_\ast(\Phi_t)_\ast = (\Phi_{s+t})_\ast$, and that $\Phi_0=\mathrm{id}$. Thus,
$$\widetilde{\mathcal{L}_VX}=\lim_{t\to 0}\widetilde{\Delta_t}=\lim_{t\to 0}(\Phi_t)_\ast \Delta_t=\lim_{t\to 0}(D\Phi_t)\Delta_t$$
Where we’ve realized the pushforward by the derivative of the diffeomorphism $\Phi_t$. As $t\to 0$, the limit of $\Delta_t$ as well as the limit of $(D\Phi_t)$ exist, so we can compute the limit termwise:
$$\begin{align}
\widetilde{\mathcal{L}VX} &=
\lim{t\to 0}(D\Phi_t)\Delta_t\
&= \left(\lim_{t\to 0}D\Phi_t\right)\left(\lim_{t\to 0}\Delta_t\right)\
&=\left(\mathrm{Id}_{T_pM}\right)\mathcal{L}_VX\
&=\mathcal{L}_VX
\end{align}$$
So the two are equal.
Lie Derivative of Tensors
The same geometric intuition carries over to general tensors $T$: starting at a point $p$ we carry the tensor $T_p$ along with us as we flow, and compare the result to the tensor $T_{\Phi_t(p)}$ we find at time $t$. The Lie derivative is the infinitesimal difference between these quantities:
As for vectors, the usual textbook definition is not as written here but rather formulated to pull all vectors back to the original tangent space, instead of push the original forward along the flow:
$$(\mathcal{L}V T)p=\lim{t\to 0 }\frac{(\Phi_t)^\ast T{\Phi_t(p)}-T_p}{t}=\lim_{t\to 0 }\frac{(\Phi_{-t})_\ast T_{\Phi_t(p)}-T_p}{t}$$
If $M$ is compact, we can define this flow globally for some interval $[0,\epsilon]$ by a compactness argument. And to calculate the Lie derivative at any point $p\in M$, its enough to have the flow on some compact neighborhood of $p$. ↩︎
To see the flow actually generates a diffeomorphism at each time step requires some analysis and topology. By the uniqueness of solutions to ODEs, for each fixed $t$ the map $\Phi_t$ is injective (if we have two points $p$, $q$ where $\Phi_t(p)=\Phi_t(q)$: call this common point $r$. Flowing backwards along $V$ is the same as flowing along the vector field $-V$, and this would imply that the point $r$ is carried under this flow to both $p$ and $q$, so by uniqueness of solutions $p=q$). Thus, on any compact subset of $M$, $\Phi_t$ is an injection from a compact space into a hausdorff space, so its a homeomorphism onto its image. And in fact, by the assumed smoothness of $V$ and $M$ its a local diffeomorphism. This implies its a covering map, but every point has a unique preimage (recall its injective), so its a 1-sheeted cover, or a global diffeomorphism. ↩︎
We have used that the flow at time zero is the identity, that the derivative of a function satisfies the identity $df(X)=X(f)$ for any vector field $X$, and the defining property of the flow $\Phi_t$, that $\frac{d}{dt}|_{t=0}\Phi_t(p)=V(p)$. ↩︎
This poses a serious problem, as there is no unique natural way to identify vectors in $T_pM$ with those in $T_qM$ for $p\neq q$. And we can’t just choose some random identification, as we are looking to define an intrinsic geometric notion, something that depends only on the vector field $X$ and the flow $\Phi$. ↩︎
If you’ve seen enough geometry you might have an idea here - we could carry the original vector along with us by parallel transport! But this won’t do here: we are looking for something metric independent. Parallel transport aims to keep everything rigid as it moves, whereas we are very much trying to go with the flow. Instead of imagining yourself and your boat as a rigid object which is pushed and pulled by the vector field $V$, we instead allow ourselves to deform freely following the flow $\Phi_t$. ↩︎
This fits well with the more practical-minded concern of computing the components of $\mathcal{L}_V X$ in a coordinate basis. Indeed, if $x,y,z$ are coordinates on $M$ near $p$ and one is looking for an expression of the form $(\mathcal{L}_V X)_p = a\partial_x + b\partial_y + c\partial_z$, one can compute the coefficients $a,b,c$ using the dual $1-$form basis $dx, dy, dz$ as $a=dx(\mathcal{L}_V X)$, $b=dy(\mathcal{L}_V X)$ and $c=dz(\mathcal{L}_V X)$. But, as $df(v)=v(f)$ for any vector $v$ and smooth function $f$ these become $a=\mathcal{L}_V X_p(x)$, $b=\mathcal{L}_V X_p(y)$ and $c=\mathcal{L}_V X_p(z)$ respectively; all quantities of the form we computed above. ↩︎