Homework 2 Problem 5

#### Problem 5

Compute the Lie derivative $\mathcal{L}_X \gamma$.

#### Solution

From Problem 4, $X = \frac{\partial}{\partial \theta} + \cos^2 \theta \frac{\partial}{\partial \phi}$ and $\gamma = -\sin \theta \cos^2 \theta d \, \theta + \sin \theta d\,\phi$.

By Cartan's Magic Formula, $\mathcal{L}_X \gamma = d(X \overset \lrcorner \, \, \gamma) + X \overset \lrcorner \, \, d \gamma$, so we'll compute each term individually and then plug them in.

First we'll compute $X \overset \lrcorner \, \, \gamma$. Since $\gamma$ is a one form and $X$ is a vector field, hooking them is the same as pairing them. That is,

(1)
\begin{align} X \overset \lrcorner \, \, \gamma = \langle \gamma, X \rangle = \langle -\sin \theta \cos^2 \theta d \, \theta + \sin \theta d\,\phi , \frac{\partial}{\partial \theta} + \cos^2 \theta \frac{\partial}{\partial \phi} \rangle = -\sin \theta \cos^2 \theta + \sin \theta \cos^2 \theta = 0 \end{align}
(2)
\begin{align} X \overset \lrcorner \, \, \gamma = 0 \end{align}
(3)
\begin{align} d(X \overset \lrcorner \, \, \gamma) = 0 \end{align}

Next we compute $d \gamma$

(4)
\begin{align} d \gamma = d(-\sin \theta \cos^2 \theta d \, \theta + \sin \theta d\,\phi) = (3 \sin^2 \theta - 1) \cos \theta d \theta \wedge d \theta + \cos \theta d\theta \wedge d\phi = \cos \theta d\theta \wedge d\phi \end{align}
(5)
\begin{align} d \gamma = \cos \theta d\theta \wedge d\phi \end{align}

(Note that $d \theta \wedge d \theta = 0$.)

Now that we have $d \gamma$, we can compute $X \overset \lrcorner \, \, d \gamma$ by using an arbitrary vector field $m\frac{\partial}{\partial\theta} + n\frac{\partial}{\partial\phi}$ as follows:

(6)
\begin{align} X \overset \lrcorner \, \, d \gamma = d \gamma (X , m\frac{\partial}{\partial\theta} + n\frac{\partial}{\partial\phi}) = \cos \theta d\theta \wedge d\phi (\frac{\partial}{\partial \theta} + \cos^2 \theta \frac{\partial}{\partial \phi} , m\frac{\partial}{\partial\theta} + n\frac{\partial}{\partial\phi}) \end{align}

Note that for one forms $\alpha_o$ and $\beta_o$, and vector fields $u$ and $v$,
$(\alpha_o \wedge \beta_o)(u,v) = \alpha_o (u) \beta_o (v) - \alpha_o (v) \beta_o (u)$
(I use the $_o$ subscript on $\alpha_o$ and $\beta_o$ to differentiate them from the $\alpha$ and $\beta$ used elsewhere on Homework 2.)

Following this property of the wedge product, we continue Equation (6):

(7)
\begin{align} \cos \theta d\theta \wedge d\phi (\frac{\partial}{\partial \theta} + \cos^2 \theta \frac{\partial}{\partial \phi} , m\frac{\partial}{\partial\theta} + n\frac{\partial}{\partial\phi}) = \cos \theta n - \cos \theta m \cos^2 \theta = \cos \theta n - \cos^3 \theta m \end{align}

Similar to Problem 4, we can then reverse engineer $X \overset \lrcorner \, \, d \gamma$

(8)
\begin{align} X \overset \lrcorner \, \, d \gamma = - \cos^3 \theta d \theta + \cos \theta d \phi \end{align}

Plugging all those elements into Cartan's Magic Formula yields

(9)
\begin{align} \mathcal{L}_X \gamma = d(X \overset \lrcorner \, \, \gamma) + X \overset \lrcorner \, \, d \gamma = 0 + (- \cos^3 \theta d \theta + \cos \theta d \phi) \end{align}
(10)
\begin{align} \mathcal{L}_X \gamma = - \cos^3 \theta d \theta + \cos \theta d \phi \end{align}

#### Discussion

Can one of you smart CS people check my work? I'm a differential geometry rookie.

There's no need! Scott agrees!

To get the hook symbol $\overset \lrcorner \, \,$, use " \overset \lrcorner \, \, ".
Has anyone else noticed that $\cos^2 \theta$ has very different spacing than $\cos \theta$? "\negmedspace" seems to help inline text (e.g. $\cos^2 \negmedspace \theta$), but I can't seem to fix it in the equations.