Lie derivative

Motivation and intuition

The plain, vanilla directional derivative is great! It tells us the rate of change of a function in any direction (given as a vector $v$). However, if we use the “normal” definition of directional derivative $D_{\vec{v}}f(p)={\lim }_{t\to 0}\frac{f(p+t\vec{v})-f(p)}{t}$ we run into a problem very quickly. What if $p+t\vec{v}$ isn’t on the manifold? Then the entire difference doesn’t make a lot of sense. Thus, we can improve on this definition using flows.

Formal definition

For all definitions, assume $M$ is a smooth manifold.

Functions

For a function $f:M\to \mathbb{R}$ and a vector field the Lie derivative of $f$ with respect to $X$ is ${\mathcal{L}}_{X}f(p)={\lim }_{t\to 0}\frac{f({\Phi }_X^t(p))-f(p)}{t}=\frac{d}{dt}{|}_{t=0}[(f\circ {\Phi }_X^t)(p)]=\frac{d}{dt}{|}_{t=0}({\Phi }_X^t{)}^{\ast }f$ Where ${\Phi }_X^t$ is the flow of $X$

This tells us how much the function moves in the “direction” of the vector field.

Notes

Notice, this doesn’t have the pesky issues of not being defined since we can always find a small enough neighborhood for the flow to be defined for a given point $p$.

Using the definitions, we can see ${\mathcal{L}}_{X}f=X(f)=df(X)$

Vector fields

Given vector fields $X,Y\in \mathfrak{X}(M)$, the Lie derivative of Y with respect to X is $({\mathcal{L}}_{X}Y{)}_p={\lim }_{t\to 0}\frac{(d{\Phi }_X^{-t}{)}_{{\Phi }_X^t(p)}(Y_{{\Phi }_X^t(p)})-Y_p}{t}=\frac{d}{dt}{|}_{t=0}[d({\Phi }_X^{-t}{)}_{{\Phi }_X^t(p)}(Y_{{\Phi }_X^t(p)})]$

Notes

Looking at the naive case shows why this definition is important. If we are working with a manifold ${\mathbb{R}}^n$, then $T_p{\mathbb{R}}^n\cong {\mathbb{R}}^n$. So we can look at a type of directional derivative of a vector field $X\in \mathfrak{X}({\mathbb{R}}^n)$. ${\lim }_{t\to 0}\frac{X_{p+t\vec{v}}-X_p}{t}$

But again, the tangent spaces at different points on a manifold do not always turn out to be the exact same. Using the differential of the flow gives a way to move the tangent vectors at different points into the same vector space.

Everywhere defined

For a smooth manifold $M$ and vector spaces $X,Y\in \mathfrak{X}(M)$, the Lie derivative $({\mathcal{L}}_{X}Y{)}_p$ is defined for every point $p\in M$ and ${\mathcal{L}}_{X}Y$ is a smooth vector field.

Proof

todo

Connection to Lie theory

The Lie derivative for vector fields is equivalent to the Lie bracket on the Lie algebra of vector fields on M. I.e. ${\mathcal{L}}_{V}W=[V,W]$

Tensor fields

For a smooth manifold $M$, $X\in \mathfrak{X}(M)$ and a smooth covariant tensor field, $A$ (i.e. $A_p:T_{p}M\times \cdots \times T_{p}M\to \mathbb{R}$), the Lie derivative of $A$ with respect to $X$ is $({\mathcal{L}}_{X}A{)}_p={\lim }_{t\to 0}\frac{(d{\Phi }_X^t{)}_p^{\ast }(A_{{\Phi }_X^t(p)})-A_p}{t}=\frac{d}{dt}{|}_{t=0}({{\Phi }_X^t}^{\ast }A{)}_p$

In general, this is pretty difficult to use as is. Instead, for differential forms we can use Cartan’s magic formula: ${\mathcal{L}}_X\alpha =d{\iota }_X\alpha +{\iota }_{X}d\alpha .$

Properties

• Linear

${\mathcal{L}}_{X+Y}f={\mathcal{L}}_{X}f+{\mathcal{L}}_{Y}f$

$d{\mathcal{L}}_X\omega ={\mathcal{L}}_X(d\omega )\forall \omega \in {\Omega }^{\bullet }(M)$

• ${\mathcal{L}}_X(f\omega )=({\mathcal{L}}_{X}f)\omega +f{\mathcal{L}}_X\omega$

• ${\mathcal{L}}_{fX}\omega =f{\mathcal{L}}_X\omega +df\wedge {\iota }_X\omega$