# The Unapologetic Mathematician

## Integrable Distributions Have Integral Submanifolds

Let’s say that we have a one-dimensional distribution on a manifold $M$. Around any point $p\in M$ we can find a patch $V$ and an everywhere-nonzero vector field $X$ on $V$ so that $X_q$ spans $\Delta_q$ for every $q\in V$. Then we know we can find a chart $(U,x)$ around $p$ such that $\frac{\partial}{\partial x^1}=X$ on $U$. Then the curve with coordinates $x^1=t$ and $x^i=0$ for all other $i\neq1$ is a one-dimensional integral submanifold of $\Delta$ through $p$.

Now, this doesn’t always work for every distribution $\Delta$. It turns out that the key ingredient is that all one-dimensional distributions are integrable; we can show that any integrable $k$-dimensional distribution has an integrable submanifold through any point $p$.

To give an even more detailed statement, let $\Delta$ be a $k$-dimensional integrable distribution on $M$. for every $p\in M$ there is a chart $(U,x)$ with $x(p)=0$, $x(U)=(-1,1)^n$, and so that for any $a_{k+1},\dots,a_n\in(-1,1)$ the set $\{q\in U\vert x^i(q)=a_i\}$ is an integral submanifold of $\Delta$. Further, any connected integral submanifold of $\Delta$ comes in this way.

Since this statement is purely local we can get away with working in some set of local coordinates to start with, although obviously not the one we’re trying to find. As such, we will assume that $M=\mathbb{R}^n$, that $p=0$, and that $\Delta_0$ is spanned by the $\left(D_i\right)_0$ for $1\leq i\leq k$. We’ll also let $\pi:\mathbb{R}^n\to\mathbb{R}^k$ be the projection onto the first $k$ components, so $\pi_*\vert_{\Delta_0}:\Delta_0\to\mathcal{T}_0\mathbb{R}^k$ is an isomorphism. By parallel translation, $\pi_*\vert_{\Delta_q}:\Delta_q\to\mathcal{T}_{\pi(q)}\mathbb{R}^k$ for all $q$ in a neighborhood $V$ of $0$.

Now, like we did yesterday we can use these isomorphisms to build $k$ vector fields $X_i$ on $V$ belonging to $\Delta$ that are $\pi$-related to the $D_i$ on $\mathbb{R}^k$. Then $\pi_*[X_i,X_j]=[D_i,D_j]=0$, but since $\Delta$ is integrable we know that $[X_i,X_j]\in\Delta$. Since $\pi_*$ is an isomorphism on $\Delta$, we conclude that $[X_1,X_j]=0$. Now we can find a coordinate patch $(U,x)$ around $0$ with $\frac{\partial}{\partial x^k}=X_k$ on $U$, just as in the one-dimensional case. It’s no loss of generality to tweak it until we have $x(U)=(-1,1)^n$. This gives us an integral submanifold through the origin.

But our assertion goes further! Let $f=\pi_2\circ x:U\to(-1,1)^{n-k}$, where $\pi_2$ is the complementary projection to $\pi$. This map has maximal rank everywhere, so we know that for each $a\in(-1,1)^{n-k}$ the preimage $f^{-1}(a)$ is an $n-(n-k)=k$-dimensional submanifold $N\subseteq M$. The tangent space to $N$ consists of exactly those vectors in the kernel of $f_*$, but since $x$ is a diffeomorphism these are exactly those vectors $v$ such that $x_*(v)$ is in the kernel of $\pi_{2*}$ — those $v\in\Delta$.

Conversely, if $N$ is a connected integral manifold of $\Delta$ contained in $U$. If $v\in\mathcal{T}_qN$, then $\iota_*v$ is in $\Delta_q$, which is spanned by the $\frac{\partial}{\partial x^i}$ for $1\leq i\leq k$. Thus $\iota_*v(x^{k+j})=0$. And so $\left((x^{k+j}\circ\iota)_*\right)_q=0$ for all $q\in N$. Since $N$ is connected, $x^{i+j}\circ\iota$ is constant, and thus $N$ comes from picking a value $a_{k+j}$ for each $1\leq j\leq n-k$.

June 30, 2011

## Integral Submanifolds

Given a $k$-dimensional distribution $\Delta$ on an $n$-dimensional manifold $M$, we say that a $k$-dimensional submanifold $\iota:N\hookrightarrow M$ is an “integral submanifold” of $\Delta$ if $\iota_*\mathcal{T}_pN=\Delta_{\iota(p)}$ for every $p\in N$. That is, if the subspace of $\mathcal{T}_{\iota(p)}M$ spanned by the images of vectors from $\mathcal{T}_pN$ is exactly $\Delta_p$.

This is a lot like an integral curve, with one slight distinction: in the case on an integral curve we also demand that the length of $c'(t)$ match that of $X_{c(t)}$, not just the direction (up to sign).

Now, if for every $p\in M$ there exists an integral submanifold $N(p)$ of $\Delta$ with $p\in N(p)$, then $\Delta$ is integrable. Indeed, let $X$ and $Y$ belong to $\Delta$. Since $\iota_{*q}:N(p)_q\to\Delta_{\iota(q)}$ is an isomorphism of vector spaces at every point, we can find $\tilde{X}$ and $\tilde{Y}$ that are $\iota$-related to $X$ and $Y$, respectively. That is, $X_{\iota(q)}=\iota_*\tilde{X}_q$ for all $q\in N$, and similarly for $Y$ and $\tilde{Y}$. But then we know that $[X,Y]_{\iota(q)}=\iota_*[\tilde{X},\tilde{Y}]_q$, and so $[X,Y]_{\iota(q)}\in\iota_*\mathcal{T}_qN=\Delta_{\iota(q)}$.

June 30, 2011

## Distributions

A vector field $v$ defines a one-dimensional subspace of $\mathcal{T}_pM$ at any point $p$ with $v_p\neq0$: the subspace spanned by $v_p\in\mathcal{T}_pM$. If $v$ is everywhere nonzero, then it defines a one-dimensional subspace of each tangent space. A distribution generalizes this sort of thing to higher dimensions.

To this end, we define a $k$-dimensional distribution $\Delta$ on an $n$-dimensional manifold $M$ to be a map $p\mapsto\Delta_p\subseteq\mathcal{T}_pM$, where $\Delta$ is a $k$-dimensional subspace of $\mathcal{T}_pM$. Further, we require that this map be “smooth”, in the sense that for any $q\in M$ there exists some neighborhood $U$ of $q$ and $k$ vector fields $X_1,\dots,X_k$ such that the vectors $X_i(r)\in\mathcal{T}_rM$ span $\Delta_r$ for each $r\in U$.

Notice here that the $X_k$ don’t have to work for the whole manifold $M$. Indeed, we will see that in many cases there are no everywhere-nonzero vector fields on a manifold $M$. But over a small patch $U$ we might more easily find $k$ vector fields that are linearly independent at each point, and thus define a smooth $k$-dimensional distribution over $U$. Then more general smooth distributions come from patching these sorts of smooth distributions together.

A vector field $X$ on $M$ “belongs to” a distribution $\Delta$ — which we write $X\in\Delta$ — if $X_p\in\Delta_p$ for all $p\in M$. We say that $\Delta$ is “integrable” if $[X,Y]\in\Delta$ for all $X$ and $Y$ belonging to $\Delta$.

Every one-dimensional manifold is integrable. To see this, we note that if $X$ and $Y$ belong to $\Delta$ then $X_p=f(p)Y_p$ for some constant $f(p)$, at least at those points $p\in M$ where $Y_p\neq0$. Thus we see that

$\displaystyle[X,Y]=[fY,Y]=f[Y,Y]-(Yf)Y=-(Yf)Y$

and so $[X,Y]$ is proportional to $Y$, and thus belongs to $\Delta$. To handle points where $Y_p=0$, we can put the scalar multiplier on the other side.

June 28, 2011

## What Does the Bracket Measure? (part 2)

Today we’ll prove the assertions we made last time: if $X$ and $Y$ are vector fields with flows $\Phi$ and $\Psi$, respectively in some neighborhood $U$ of some point $p\in M$, we define the curve

$\displaystyle c(t)=\left[\Phi_{-t}\circ\Phi_{-t}\circ\Psi_t\circ\Phi_t\right](p)$

We assert that for any smooth $f\in\mathcal{O}U$ we have

\displaystyle\begin{aligned}(f\circ c)'(0)&=0\\\frac{1}{2}(f\circ c)''(0)&=[X,Y]_p(f)\end{aligned}

To show the first, we define the three “rectangles”

\displaystyle\begin{aligned}V_1(s,t)&=\left[\Psi_s\circ\Phi_t\right](p)\\V_2(s,t)&=\left[\Phi_{-s}\circ\Psi_t\circ\Phi_t\right](p)\\V_3(s,t)&=\left[\Psi_{-s}\circ\Phi_{-t}\circ\Psi_t\circ\Phi_t\right](p)\end{aligned}

Notice that $c(t)=V_3(t,t)$, $V_3(0,t)=V_2(t,t)$, and $V_2(0,t)=V_1(t,t)$. The chain rule lets us then calculate:

\displaystyle\begin{aligned}(f\circ c)'(0)=&\frac{d}{dt}f(c(t))\Big\vert_{t=0}\\=&\frac{d}{dt}f(V_3(t,t))\Big\vert_{t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_3(s,t))\Big\vert_{s=t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{d}{dt}f(V_3(0,t))\Big\vert_{t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{d}{dt}f(V_2(t,t))\Big\vert_{t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial s}f(V_2(t,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_2(t,t))\Big\vert_{s=t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial s}f(V_2(t,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_2(t,t))\Big\vert_{s=t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial s}f(V_2(t,t))\Big\vert_{s=t=0}\\&+\frac{\partial}{\partial s}f(V_1(t,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_1(t,t))\Big\vert_{s=t=0}\\=&-Y_pf-X_pf+Y_pf+X_pf=0\end{aligned}

as asserted. As for the other assertion, we start by observing

$\displaystyle(f\circ c)''(0)=\frac{\partial^2(f\circ V_3)}{\partial s^2}(0,0)+2\frac{\partial^2(f\circ V_3)}{\partial s\partial t}(0,0)+\frac{\partial^2(f\circ V_3)}{\partial t^2}(0,0)$

Using the fact that $\frac{\partial(f\circ V_3)}{\partial s}=-(Yf)\circ V_3$ we can turn the first term on the right into

$\displaystyle\frac{\partial^2(f\circ V_3)}{\partial s^2}(0,0)=\frac{\partial(-(Yf)\circ V_3)}{\partial s}(0,0)=Y_pYf$

Now, similar tedious calculations that make the big one above look like idle doodling give us two more identities:

\displaystyle\begin{aligned}\frac{\partial^2(f\circ V_3)}{\partial s\partial t}(0,0)&=-Y_pYf\\\frac{\partial^2(f\circ V_3)}{\partial t^2}(0,0)&=Y_pYf+2[X,Y]_pf\end{aligned}

and we conclude that

$\displaystyle(f\circ c)''(0)=2[X,Y]_pf$

as asserted.

June 27, 2011

## What Does the Bracket Measure? (part 1)

We again pick up the question we posed earlier about what the bracket measures. Clearly it should have something to do with flows and their failure to commute. So we consider vector fields $X$ and $Y$ with flows $\Phi$ and $\Psi$, respectively.

Now, starting with a point $p$ and given some small time displacement $t$ we can flow along $X$ and then $Y$$\left[\Psi_t\circ\Phi_t\right](p)$ — or we can flow along $Y$ and then $X$$\left[\Phi_t\circ\Psi_t\right](p)$ — and get two (potentially different) points in $M$. Unfortunately, we can’t just take their difference as we could when measuring the failure of algebra elements to commute.

However, we do have something rather closely related: the commutator in the sense of a group. That is, we can flow forwards along $X$, then along $Y$, then backwards along $X$ and then backwards along $Y$ to get the point $c(t)$:

$\displaystyle c(t)=\left[\Psi_{-t}\circ\Phi_{-t}\circ\Psi_t\circ\Phi_t\right](p)$

If $\Phi_t$ and $\Psi_t$ were to commute this would just be the constant curve $c(t)=p$, but in general it’s a smooth curve passing through $p$ at $t=0$. The bracket $[X,Y]_p$ will indicate the direction in which $c(t)$ passes through $p$, and something about its speed.

Now, it turns out that the derivative $c'(t)$ always vanishes at $t=0$. In fact, $[X,Y]_p$ is more closely related to the “second derivative” of $c$, though it’s not immediately clear what this means. Indeed, we can define $c'(t)$ for any $t$ in an interval around $0$, but $c'(t)\in\mathcal{T}_{c(t)}M$, and there is in general no good way to compare these vectors for different values of $t$.

Instead, we will prove the following two facts for any smooth $f\in\mathcal{O}U$ — here “smoothness” entails twice differentiability — where $U$ is any neighborhood of $p$:

\displaystyle\begin{aligned}(f\circ c)'(0)&=0\\\frac{1}{2}(f\circ c)''(0)&=[X,Y]_p(f)\end{aligned}

The first indicates that $c'(0)=0\in\mathcal{T}_pM$. Indeed, if $c'(0)\neq0$ then we could surely find some $f$ which changes in the direction it points, which would give a nonzero value to $(f\circ c)'(0)$. The second gives the result we’re really after, as Taylor’s theorem applied to $f\circ c$ shows us that

$\displaystyle[X,Y]_p(f)=\frac{1}{2}(f\circ c)''(0)=\lim\limits_{t\to0}\frac{\left[f\circ c\right](t)-\left[f\circ c\right](0)}{t^2}$

And thus the bracket indeed is the second (and lowest) order term in describing the direction of $c$ as it passes through $p$.

June 23, 2011

## Building Charts from Vector Fields

Sorry for the delay; I’ve been swamped at my actual job the last couple days.

Anyway, I left off last time by pointing out that coordinate vector fields commute:

$\displaystyle\left[\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right]=0$

Today I want to show a certain converse: if we have $k$ vector fields $X_1,\dots,X_k$ on some open region $V\subseteq M^n$ that are linearly independent at some $p\in V$ and which commute — $[X_i,X_j]=0$ for all $i$ and $j$ — then we can find some coordinate chart $(U,x)$ around $p$ so that the $X_i$ are the first $k$ coordinate vector fields. That is,

$\displaystyle\frac{\partial}{\partial x^i}=X_i\vert_U$

In particular, if $X$ is a vector field with $X_p\neq0$ then there is a coordinate chart $(U,x)$ around $p$ with $\frac{\partial}{\partial x^1}=X\vert_U$.

If $(U,x)$ is any chart, then we can describe the $i$th coordinate vector field by saying it’s the unique vector field on $U$ that is $x$-related to the $i$th partial derivative in $\mathbb{R}^n$. That is, we’re trying to prove that: $x_*\circ X_i\circ x^{-1}=D_i$ on $x(U)$.

In fact, we can further simplify our claim by assuming that $M=\mathbb{R}^n$, $p=0$, and $X_{i0}=D_{i0}$ — that the vector fields agree at the point $0\in\mathbb{R}^n$. Indeed, if $z$ is any coordinate map taking $p$ to $0$ then we can define the vector fields $Y_i=z_*\circ X_i\circ z^{-1}$ on $\mathbb{R}^n$. These must have vanishing brackets because we can calculate:

\displaystyle\begin{aligned}{}[Y_i,Y_j]&=[z_*\circ X_i\circ z^{-1},z_*\circ X_j\circ z^{-1}]\\&=z_*\circ[X_i,X_j]\circ z^{-1}\\&=z_*\circ0\circ z^{-1}=0\end{aligned}

What’s more, if $y$ is a local diffeomorphism of $\mathbb{R}^n$ with $y_*\circ Y_i\circ y^{-1}=D_i$, then $x=y\circ z$ is a coordinate map satisfying our assertion.

Now, let $\Phi^i_t$ be the flow of $X_i$, and let $W$ be a small enough neighborhood of $0$ that we can define $f:W\to\mathbb{R}^n$ by

$\displaystyle f(a_1,\dots,a_n)=\left[\Phi^1_{a_1}\circ\dots\circ\Phi^k_{a_k}\right](0,\dots,0,a_{k+1},\dots,a_n)$

The order that the flows come in here doesn’t matter, since we’re assuming that the $X_i$ — and thus their flows — commute. Anyway, given any smooth test function $\phi$ on $\mathbb{R}^n$ we can check

\displaystyle\begin{aligned}\left(f_*D_1\right)_a\phi&=\left(D_1\right)_a(\phi\circ f)\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\left[\phi\circ f\right](a_1+h,a_2,\dots,a_n)-\left[\phi\circ f\right](a_1,a_2,\dots,a_n)\right]\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\left[\phi\circ\Phi^1_{a_1+h}\circ\dots\circ\Phi^k_{a_k}\right](0,\dots,0,a_{k+1},\dots,a_n)-\left[\phi\circ f\right](a)\right]\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\left[\phi\circ\Phi_h\circ\Phi^1_{a_1}\circ\dots\circ\Phi^k_{a_k}\right](0,\dots,0,a_{k+1},\dots,a_n)-\left[\phi\circ f\right](a)\right]\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\left[\phi\circ\Phi_h\right](a)-\left[\phi\circ f\right](a)\right]\\&=\left(X_1\right)_{f(a)}(\phi)\end{aligned}

That is, $f_*D_1=X_1\circ f$. To see this for any other $X_i$, simply swap around the flows to bring $\Phi^i_t$ to the front.

We can also check that

\displaystyle\begin{aligned}\left(f_*D_{k+i}\right)_0\phi&=\left(D_{k+i}\right)_0(\phi\circ f)\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\left[\phi\circ f\right](0,\dots,0,h,0,\dots,0)-\left[\phi\circ f\right](0,\dots,0)\right]\\&=\lim\limits_{h\to0}\frac{1}{h}\left[\phi(0,\dots,0,h,0,\dots,0)-\phi(0,\dots,0)\right]\\&=\left(D_{k+i}\right)_0\phi\end{aligned}

Thus $f_{*0}$ is the identity transformation on $\mathcal{T}_0\mathbb{R}^n$. The inverse function theorem now tells us that there is a chart $(U,x)$ around $0$ with $x=f^{-1}$, which will then satisfy our assertions.

June 22, 2011

## Brackets and Flows

Now, what does the Lie bracket of two vector fields really measure? We’ve gone through all this time defining and manipulating a bunch of algebraic expressions, but this is supposed to be geometry! What does the bracket actually mean? It turns out that the bracket of two vector fields measures the extent to which their flows fail to commute.

We won’t work this all out today, but we’ll start with an important first step: the bracket of two vector fields vanishes if and only if their flows commute. That is, if $X$ and $Y$ are vector fields with flows $\Phi_s$ and $\Psi_t$, respectively, then $[X,Y]=0$ if and only if $\Phi_s\circ\Psi_t=\Psi_t\circ\Phi_s$ for all $s$ and $t$.

First we assume that the flows commute. As we just saw last time, the fact that $\Phi_s\circ\Psi_t=\Psi_t\circ\Phi_s$ for all $t$ means that $Y$ is $\Phi_s$-invariant. That is, $\Phi_{-t*}Y\circ\Phi_t=Y$. But this implies that the Lie derivative $L_XY$ vanishes, and we know that $L_XY=[X,Y]$.

Conversely, let’s assume that $[X,Y]=0$. For any $p\in M$ we can define the curve $c_p$ in the tangent space $\mathcal{T}_pM$ by $c_p(s)=\Phi_{-s*}Y\circ\Phi_s(p)$. Since the Lie derivative vanishes, we know that $c_p'(0)=0$, and I say that $c_p'(s)=0$ for all $s$, or (equivalently) that $c_p(s)=Y_p$.

Fixing any $s$ we can set $q=\Phi_s(p)$. Then we calculate

\displaystyle\begin{aligned}c_p'(s)&=\lim\limits_{\Delta s\to0}\frac{c_p(s+\Delta s)-c_p(s)}{\Delta s}\\&=\lim\limits_{\Delta s\to0}\frac{1}{\Delta s}\left[\Phi_{-(s+\Delta s)*}\circ Y\circ\Phi_{s+\Delta s}(p)-\Phi_{-s*}\circ Y\circ\Phi_s(p)\right]\\&=\lim\limits_{\Delta s\to0}\frac{1}{\Delta s}\Phi_{-s*}\left[\left[\Phi_{-\Delta s*}\circ Y\circ\Phi_{\Delta s}\right]\left(\Phi_s(p)\right)-Y\left(\Phi_s(p)\right)\right]\\&=\Phi_{-s*}\lim\limits_{\Delta s\to0}\frac{1}{\Delta s}\left[\left[\Phi_{-\Delta s*}\circ Y\circ\Phi_{\Delta s}\right](q)-Y(q)\right]\\&=\Phi_{-s*}c_q'(0)=\Phi_{-s*}0=0\end{aligned}

Now this means that $Y$ is $\Phi_s$-invariant for all $t$, meaning that $\Phi_s$ and $\Psi_t$ commute for all $s$ and $t$, as asserted.

As a special case, if $(U,x)$ is a coordinate patch then we have the coordinate vector fields $\frac{\partial}{\partial x^i}$. The fact that partial derivatives commute means that the brackets disappear:

$\displaystyle\left[\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right]=0$

This corresponds to the fact that adding $s$ to the $i$th coordinate and $t$ to the $j$th coordinate can be done in either order. That is, their flows commute.

June 18, 2011

## Invariance and Flows

When we discussed the Lie algebra of a Lie group we discussed “left-invariant” vector fields. More generally than this if $f:M\to M$ is a diffeomorphism we say that a vector field $X\in\mathfrak{X}M$ is “$f$-invariant” if it is $f$-related to itself. That is, a vector field on a Lie group $G$ is left-invariant if it is $L_g$-invariant for all $g\in G$.

Now we want a characterization of $f$-invariance in terms of the flow $\Phi_t$ of $X$. I say that $X$ is $f$-invariant if and only if $\Phi_t\circ f=f\circ\Phi_t$ for all $t$. That is, the flow should commute with $f$.

We’ll show this by showing that the vector field $\tilde{X}=f_*\circ X\circ f^{-1}$ has flow $\tilde{\Phi}_t=f\circ\Phi_t\circ f^{-1}$. Then if $X$ is $f$-related to itself we know that $X=f_*\circ X\circ f^{-1}$, and so by uniqueness we conclude that the flows $\Phi_t$ and $f\circ\Phi_t\circ f^{-1}$ are equal, as asserted.

So, what makes $f\circ\Phi_t\circ f^{-1}$ the flow of $f_*\circ X\circ f^{-1}$? First of all, we have to check the initial condition that $\tilde{\Phi}_0(p)=p$, which is perfectly straightforward to check:

\displaystyle\begin{aligned}\tilde{\Phi}_0(p)&=f\left(\Phi_0\left(f^{-1}(p)\right)\right)\\&=f\left(f^{-1}(p)\right)=p\end{aligned}

More involved is the differential condition. It will help if we rewrite $\tilde{\Phi}$ a bit as a function of both $t$ and $p$:

\displaystyle\begin{aligned}\tilde{\Phi}(t,p)&=f\left(\Phi_t\left(f^{-1}(p)\right)\right)\\&=f\left(\Phi\left(t,f^{-1}(p)\right)\right)\\&=f\left(\Phi\left(\left[1_\mathbb{R}\times f^{-1}\right](t,p)\right)\right)\\&=\left[f\circ\Phi\circ\left(1_\mathbb{R}\times f^{-1}\right)\right](t,p)\end{aligned}

Now we can start on the differential condition:

\displaystyle\begin{aligned}\tilde{\Phi}_*\left(\frac{\partial}{\partial t}(t,p)\right)&=f_*\left(\Phi_*\left((1_\mathbb{R}\times f^{-1})_*\left(\iota_{p*}\left(\frac{d}{dt}(t)\right)\right)\right)\right)\\&=f_*\left(\Phi_*\left(\left((1_\mathbb{R}\times f^{-1})\circ\iota_p\right)_*\left(\frac{d}{dt}(t)\right)\right)\right)\\&=f_*\left(\Phi_*\left(\iota_{f^{-1}(p)*}\left(\frac{d}{dt}(t)\right)\right)\right)\\&=f_*\left(X\left(\Phi\left(t,f^{-1}(p)\right)\right)\right)\\&=\left[f_*\circ X\circ\Phi\right]\left(t,f^{-1}(p)\right)\\&=\left[f_*\circ X\circ f^{-1}\circ f\circ\Phi\circ(1_\mathbb{R}\times f^{-1})\right](t,p)\\&=\left[\tilde{X}\circ\tilde{\Phi}\right](t,p)\\&=\tilde{X}\left(\tilde{\Phi}(t,p)\right)\end{aligned}

And thus $\tilde{\Phi}$ is indeed the flow of $\tilde{X}$.

June 17, 2011

## Calculating the Lie Derivative

It’s 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 requires knowing the flow $\Phi$ of the vector field $X$, which requires solving a differential equation that might be difficult in practice. Luckily, there’s an easier way.

But first, a lemma: if $I$ is an interval containing $0\in\mathbb{R}$, $U\subseteq M$ is an open set, and $f:I\times U\to\mathbb{R}$ is a differentiable function with $f(0,p)=0$ for all $p\in U$, then there is another differentiable function $g:I\times U\to\mathbb{R}$ such that

\displaystyle\begin{aligned}f(t,p)&=tg(t,p)\\\frac{\partial f}{\partial t}(t,p)&=g(0,p)\end{aligned}

This is basically just like a lemma we proved for functions on star-shaped neighborhoods. Indeed, it suffices to set

$\displaystyle g(t_0,p)=\int\limits_0^1\frac{\partial f}{\partial t}(st_0,p)\,ds$

Now let $p\in M$, $f:M\to\mathbb{R}$, and $\Phi:I\times U\to M$ is the local flow of a vector field $X$ on a region $U$ containing $p$. Consider the function $(t,q)\mapsto f\left(\Phi(t,q)\right)-f(q)$, which satisfies the condition of the above lemma. We can thus write

$\displaystyle\left[f\circ\Phi_t\right](q)-f(q)=tg(t,q)$

for some $g:I\times U\to\mathbb{R}$. Or if we write $g_t(q)=g(t,q)$ we can write

$\displaystyle f\circ\Phi_t=f+tg_t$

We also can see that $g_0=Xf$. And so we can write

\displaystyle\begin{aligned}\Phi_{-t*}Y_{\Phi_t(p)}(f)&=Y_{\Phi_t(p)}\left(f\circ\Phi_{-t}\right)\\&=Y_{\Phi_t(p)}\left(f-tg_{-t}\right)\\&=\left[(Yf)\circ\Phi_t\right](p)-t\left[(Yg_{-t})\circ\Phi_t\right](p)\end{aligned}

So now we calculate

\displaystyle\begin{aligned}(L_XY)_pf&=\lim\limits_{t\to0}\frac{\Phi_{-t*}Y_{\Phi_{t}(p)}(f)-Y_pf}{t}\\&=\lim\limits_{t\to0}\frac{\left[(Yf)\circ\Phi_t\right](p)-t\left[(Yg_{-t})\circ\Phi_t\right](p)-Yf(p)}{t}\\&=\lim\limits_{t\to0}\frac{\left[(Yf)\circ\Phi_t\right](p)-Yf(p)}{t}-\lim\limits_{t\to0}\left[(Yg_{-t})\circ\Phi_t\right](p)\end{aligned}

Last time we wrote the action of $X$ on a smooth function $\phi$ as

$\displaystyle X_p(\phi)=\lim\limits_{t\to0}\frac{\left[\phi\circ\Phi_t\right](p)-\phi(p)}{t}$

which pattern we can recognize in our formula. We thus continue

\displaystyle\begin{aligned}(L_XY)_pf&=\lim\limits_{t\to0}\frac{\left[(Yf)\circ\Phi_t\right](p)-Yf(p)}{t}-\lim\limits_{t\to0}\left[(Yg_{-t})\circ\Phi_t\right](p)\\&=X_pYf-Yg_0(p)\\&=X_pYf-Y_pXf\\&=[X,Y]_pf\end{aligned}

That is, for any two vector fields the Lie derivative $L_XY$ is actually the same as the bracket $[X,Y]$.

June 16, 2011

## 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.

June 15, 2011