Lie Derivative

We know that the derivative compares something at two point separated infinitesimally. It works perfectly well for simple functions, for when evaluated at different points, simple functions always spits out the same thing, may it be real number, complex number etc. The problem is that sometimes things at different points are not directly comparable, even when they are infinitely close to each other. The most important example may be vector fields $X$ on a manifold $M$, we know that at different points $p\in M$ there exists different vector spaces $T_{p}M$, so at two different points $x, x+dx$ there exists two different vector spaces and vectors in one space cannot be directly compared to that in the other vector space. Hence we need to find a way to move vectors from one vector space to another, only then can we compare them. Lie derivative provides a very simple way to do that, namely the transportation of vectors. A more sophisticated version of the so-called vector transportation is achieved with the help of affine connections, which will not be discussed here.

Let $X,Y$ be vector fields on a manifold $M$ of dimension $n$, let $\phi(t) = \phi_{t}$ be the local flow generated by the field $X$. Then $\phi_{t}(x)$ is the orbit of point $x$ generated by the flow. If we want to compare $Y$ at $\phi_{t}(x)$ to that at $x$, we can either push $Y_{x}$ to $\phi_{t}(x)$ by the means of the differential map $\phi_{t\ast}$ , or push $Y_{\phi_{t}x}$ back to $x$ by the means of $\phi_{-t\ast}$, they yield the same result. Out of this spirit we define the Lie derivative of $Y$ with respect to $X$, denoted by $\mathcal{L}_{X}Y$ whose value at $x$ is

\[\mathcal{L}_{X}Y \vert_{x} := \lim_{ t\to 0} \frac{Y_{\phi_{t}x} - \phi_{t\ast}Y_{x}}{t},\]

where $(\text{something})\vert_{x}$ means “something at x”. This expression is the same as

\[\mathcal{L}_{X}Y \vert_{x} := \lim_{ t\to 0} \frac{\phi_{-t\ast}Y_{\phi_{t}x} - Y_{x}}{t}.\]

There is a useful alternative expression of Lie derivative,

\[\mathcal{L}_{X}Y = [X,Y]\]

where $[X,Y]$ is the commutative of $X$ and $Y$.

The Flow Generated by $[X,Y]$

Let $\phi(t)$ be the flow generated by $X$ and $\psi(t)$ the flow generated by Y. $[X,Y]$ is itself a vector field, so the question is, what is the flow generated by it? We can already guess that the flow must be anti-symmetric in $X$ and $Y$. We claim that the flow is in the following sense the commutator of the two flows. Let $x \in M$.

Let $\sigma$ be the curve

\[\sigma(t) := \psi_{-t}\circ \phi_{-t} \circ \psi_{t} \circ \phi_{t}x\]

Then for any smooth function $f$,

\[[X,Y]\vert_{x}(f) = \lim_{ t\to 0 } \frac{f[\sigma(\sqrt{ t })] - f[\sigma(0)]}{t}\]

A tip for proving it is using the Taylor expansion,

\[f(\phi(t)x) - f(x) = tX_{0}(f) - \frac{t^{2}}{2}X_{0}(X(f))+\mathcal{O}(t^3)\]

Or, if it helps, you can think of $f$ as a one-variable function of $t$, perform the usually Taylor expansion, then rewrite things in terms of vector field $X$.

We may write

\[\mathcal{L}_{X}Y = [X,Y] = \frac{d}{dt} \sigma(\sqrt{ t })\lvert_{t=0}.\]

The Lie Derivative of a Form

The Lie derivative $\mathcal{L}_ X$ first uses the flow generated by $X$, denoted by $\phi_{t}(\bullet)$ where $\bullet$ is the starting point, to pull back things from one point to another, then compares the pulled-back value with the original value. When applied on a scalar function, we have

\[\mathcal{L}_ {X} f = \frac{d}{dt}[\phi^\ast_{T}f]_{t=0} = \frac{d}{dt}f[\phi_{t}x]_{t=0} = X(f).\]

This simply describe how $f$ changes along the orbits of $X$. If $\alpha$ is a p-form we define, putting $\alpha_{x} = \alpha(x)$,

\[\mathcal{L}_{X} \alpha := \frac{d}{dt} [\phi_{t}^\ast \alpha]_{t_{0}} = \lim_{ t \to 0 } \frac{\phi_{t}^\ast\alpha_{\phi_{t}x} - \alpha_{x}}{t}.\]

Recall that by definition $f^\ast\alpha(X) = \alpha(f_{\ast}X)$ . Thus

\(\left[ \frac{d}{dt} \phi_{t}^\ast \alpha\right](Y_{1},\dots,Y_{p}) = \frac{d}{dt}\phi_{t}^\ast\alpha(Y_{1},\dots,Y_{N}) = \frac{d}{dt} [\alpha(\phi_{t\ast}Y_{1},\dots,\phi_{t\ast}Y_{p})]\)
In particular, if we extend the vectors $Y_{i}$ to be invariant fields along the orbit through $x$, $\phi_{t\ast} Y_x = Y_{\phi_{t}x}$, then we can write

\[\mathcal{L}_{X} \alpha (Y_{1},\dots,Y_{p}) = \frac{d}{dt} [\alpha_{\phi_{t}x}(Y_{1},\dots,Y_{p})],\]

that is,

\[\mathcal{L}_{X}\alpha(Y_{1},\dots,Y_{p})\]

measures the derivative (as one moves along the orbit of $X$) of the value of $\alpha$ evaluated on a p-tuple of vector fields $Y$ that are invariant under the flow generated by $X$.

Particularly, recall that the Lie derivative of a basis vector field with respect to another basis vector field is always zero by construction, hence if $\alpha = \text{vol}^n = \sqrt{ g }dx^1 \wedge \dots\wedge dx^n$ is the volume form for a Riemannian manifold $M^n$ and if $X$ is a vector field on the same manifold, then

\[\mathcal{L}_{X} \text{vol}^n\]

is the n-form that reads off the rate of change of volume of a parallelopiped spanned by $n$ vectors that are pushed forward by the flow $\phi_{t}$ generated by $X$.

We need convenient methods for computing Lie derivatives. You can prove the Lie derivative is indeed a “derivative”, in the sense of the following.

\[\mathcal{L}_{X} (\alpha^p \wedge \beta^q) = (\mathcal{L}_{X} \alpha^p) \wedge \beta^q + \alpha^p \wedge \mathcal{L}_{X} (\beta^q).\]

Theorem $\mathcal{L}_ {X}$ commutes with exterior differentiation $d$,

\[\mathcal{L}_{X} \circ d = d \circ \mathcal{L}_{X}.\]

To prove it, first prove it for 0-forms, then generalized this to p-forms by considering the components. As a small exercise, prove that $\mathcal{L}_{X} dx^i = d X^i$.

Formulas Involving the Lie Derivative

Let $\bigwedge^p M$ be the space of p-forms on $M^n$, this is an infinite dimensional vector space since the since the components are functions.

Definition A linear map $D:$ is said to be a derivation if $r$ is even and

\[D(\alpha \wedge\beta) = (D\alpha) \wedge\beta + \alpha \wedge(D\beta),\quad \text{e.g. } \mathcal{L}_{X},\]

where $\alpha\in\Omega^p,\beta\in\Omega^q$ are differential forms, and is said to be an ant derivation if $r$ is odd and

\[D(\alpha \wedge\beta) = (D\alpha) \wedge\beta + (-1)^p\alpha \wedge(D\beta),\quad \text{e.g. } \mathcal{L}_{X},\quad \text{e.g. } d, i_{X}.\]

If A and B are both derivation or antiderivations, then to prove $A\alpha = B \alpha, \, \alpha \in \Omega^p$ we need only prove this for $\alpha$ a function or $\alpha = d(\text{function})$.

Cartan’s formula When acting on exterior forms,

\[\boxed{ \mathcal{L}_{X} = i_{X} \circ d + d \circ i_{X}. }\]

First prove it for a function, then for $d(\text{function})$. On functions, $i_{X} f = 0$ and $i_{X} df = X(f) = \mathcal{L}_{X} f$.

When applied to forms,

\[i_{[X,Y]} = \mathcal{L}_{X} \circ i_{Y} - i_{Y} \circ \mathcal{L}_{X}\]

The following is a coordinate free expression for the exterior derivative of a 1-form.

Let $\alpha \in \Omega^1M$ and $X,Y$ be vector fields on M. Then

\[\boxed{ d\alpha(X,Y) = X(\alpha(Y)) - Y(\alpha(X)) - \alpha([X,Y]). }\]

Try to prove it yourself if you are interested. This formula can be generalized to $p$-forms, for $\alpha \in \Omega^p$,

\[d\alpha(Y_{0},Y_{1},\dots,Y_{p}) = \sum_{r} (-1)^r Y_{r} \{\alpha(Y_{0},\dots,\hat{Y}_{r},\dots,Y_{p})\} + \sum_{r<s}(-1)^{r+z}\alpha^p([Y_{r},Y_{s}],\dots,\hat{Y}_{R},\dots,\hat{Y}_{s},\dots,Y_{p})\]

where $\hat{Y}_r$ means $Y_r$ is neglected from the list. This is a rather tedious formula, hope I never have to use it.

Some Vector Analysis

Let $\text{Vol}^n$ be the n-dimensional volume form for manifold M. A terminology: the volume form is not a real form but an pseudo-form. Pseudo-forms has to do with the orientation of a manifold. Note that the orientation of a vector space is just assignment of a function $o$ on bases of the vector space whose values are the two integers $\pm 1$; $o = +1$ if and only if the basis has the given orientation. It’s nothing more than a definition. Sometime, orientation has no global meaning, for example, rotation “in the clockwise sense” has no meaning on the Mobius band.

Definition a Pseudo-p-form $\alpha$ on a vector space E assigns, for each orientation $o$ of vector space E, an exterior p-form $\alpha_{o}$ such that if the orientation is reversed the exterior form is replaced by its negative,

\[\alpha_{-o} = - \alpha_{o}.\]

A pseudo-form on an entire manifold M assigns a pseudo-form $\alpha$ to each tangent space $T_{p}M$ in a smooth fashion.

Take the volume form of $\mathbb{R}^3$ for example. Let $x,y,z$ be a Cartesian coordinate system, it may be left-handed or right-handed, up to your preference. Then the volume form is

\[\text{vol}^3 := o(\partial_{x},\partial_{y},\partial_{z}) dx \wedge dy \wedge dz\]

Thus if $o$ is the right-handed orientation of $\mathbb{R}^3$, and if the coordinate system is right-handed, then $\text{vol}_o = dx \wedge dy \wedge dz$.

Similarly we can define pseudovectors, pseudoscalars, and so on, pseudo- always referring to a change of sign with a change of orientation. For example, the magnetic field is a pseudovector field.

Let $X$ be a vector field on $M^n$, the divergence of $X$ is defined by the formula

\[\mathcal{L}_{X} \text{vol}^n = (\text{div}X) \text{vol}^n.\]

For divergent-free vector fields, the flow generated by it preserves the volume form. With the help of Cartan’s formula, we can prove that in local coordinates $\text{vol} = \rho dx^1 \wedge \dots \wedge dx^n$,

\[\text{div} X = \frac{1}{\rho} \sum_{r} \frac{ \partial }{ \partial x^r } (\rho X^r).\]

There is an easier way to remember this formula with the help of Lagrangian of scalar fields, which I first read from A. Zee’s textbook General Relativity in a Nutshell.

Differentiation of Integrals

The question is, how does one compute the rate of change of an integral when the domain of integration is also changing?

First, some terminologies. An autonomous system or autonomous differential equation is a system of ordinary differential equations which does not explicitly depend on the independent variable. When the variable is time, they are also called time-invariant systems.

Let $\alpha$ be a p-form and $V^p$ an oriented compact submanifold (whose boundary, if exists, is $\partial V$) of manifold $M^n$. We consider a “variation” of $V^p$ as follows. We suppose that there is a flow $\phi_{t}: M^n \to M^n$, that is, a 1-parameter “group” of diffeomorphisms $\phi_{t}$, defined in a neighborhood of $V^p$ for small times $t$, and we define the time-dependent submanifold $V^p(t) := \phi_{t} V^p$.

Let

\[X_x = \left. \frac{d\phi_{t}(x)}{dt} \right\rvert_{t=0}\]

be the “velocity” field of the flow. We are interested in the time variation of the integral

\[I(t) = \int_{V(t)} \alpha = \int_{V} \phi_{t}^\ast \alpha\]

The last equality comes from the definition of pullbacks of forms. It is essential here, for it make the domain of integral formally independent of time, making it possible for us to calculate the derivative. We have

\[\begin{align} \dot{I}(t) &= \lim_{ h \to 0 } \frac{I(t+h) - I(t)}{h} \\ & = \lim_{ h \to 0 } \frac{\int_{V} \phi^\ast_{t+h}\alpha - \int_{V} \phi^\ast_{t}\alpha}{h} \\ &= \lim_{ h \to 0 } \frac{\int_{V} \phi^\ast_{t}\phi^\ast_{h}\alpha - \int_{V} \phi^\ast_{t}\alpha}{h} \\ &= \lim_{ h \to 0 } \frac{\phi^\ast_{t}\left(\int_{V} \phi^\ast_{h}\alpha - \int_{V} \alpha\right)}{h} \\ &= \lim_{ h \to 0 } \frac{\left(\int_{V(t)} \phi^\ast_{h}\alpha - \int_{V(t)} \alpha\right)}{h} \\ & = \int_{V(t)} \lim_{ h \to 0 } \frac{\phi_{h}^\ast \alpha - \alpha}{h} \\ & = \int_{V(t)} \mathcal{L}_{X} \alpha \end{align}\]

I was quite impressed by the simplicity of this formula. Frankel kept on talking about the divergence theorem, including how to construct the volume form on the boundary of $V$, $\text{Vol}^{n-1}_{\partial V}$ from the volume form of $V$ itself, but it looks rather complicated and I decided to leave it out from the note. Interested readers are referred to Frankel’s textbook, Chapter 4.3a.

Now let’s turn to time-dependent fields.

Consider a nonautonomous flow of water in $\mathbb{R}^3$, that is, a flow of water where the velocity field $\mathbf{v}(t,\mathbf{x})$ depends on time explicitly. As we will see later, the word “flow” here is not really accurate, due to the time dependence of the velocity field. We define a map $\phi_{t} : \mathbb{R}^3 \to \mathbb{R}^3$ as follows. Let a molecule be at position $x$ when time $t=0$. We let $\phi_{t}(x) \equiv \phi_{t}x$ be the position of the same molecule $t$ seconds later. Similarly, $\phi_{s+t} x$ is where the current take the molecule after $s+t$ seconds. Then what is $\phi_{s}(\phi_{t}x)$? It is the position of a molecule started at position $\phi_{t}x$ when $t=0$ after $s$ seconds! I emphasis that $\phi_{s}(\phi_{t}x)$, by construction, is NOT the position of some molecule after $s+t$ seconds, but the position of some molecule after $s$ seconds! If the current is steady, meaning the velocity is time independent, then it actually doesn’t make any differences. BUT, if the velocity is time-dependent, then in general $\phi_{s+t}x \neq \phi_{s}(\phi_{t}x)$. So we claim that

A time dependent vector field on a manifold $M^n$ does not generate a flow since it doesn’t satisfy the 1-parameter group property.

Suppose we have a time-dependent vector field $v = v(t,x)$ on $M^n$. We apply a simple classical trick: any tensor field $A(x,t)$ on $M^n$ that is time-dependent should be considered as a tensor field on the product manifold $\mathbb{R} \times M^n$, where $t$ is the coordinate for $\mathbb{R}$. A time dependent vector field is now an ordinary vector field on the space-time. To parametrize the flow given by the $n+1$ dimensional vector field, we introduce a new parameter $s$ such that

\[\begin{align} \frac{d x^i}{ds} &= v(t,x),\quad x^i\rvert_{s=0} = x_{0}^i, i = 1,\dots,n, \\ \frac{dt}{ds} &= 1, \quad t|_{s=t} = t_{0}. \end{align}\]