Homework 2 Problem 4

Problem 4

Consider the vector field $X = \frac{\partial}{\partial \theta} + \cos^2 \theta \frac{\partial}{\partial \phi}$ and the two form $\beta = \sin \theta \, d \theta \wedge d \phi$ on $M$. Compute $\gamma = i_X\beta$.

Solution

To determine $\gamma = i_X\beta$ consider pairing $\beta$ with $X$ and an arbitrary vector field $m\frac{\partial}{\partial\theta} + n\frac{\partial}{\partial\phi}$. Using the definitions of $\beta$ and $X$ we find,

(1)
\begin{align} \sin \theta \, d \, \theta \wedge d \phi \left(\frac{\partial}{\partial \theta} + \cos^2 \theta \frac{\partial}{\partial \phi}, m\frac{\partial}{\partial\theta} + n\frac{\partial}{\partial\phi}\right) \end{align}

This can be simplified using the definiton of wedge product and then applying linearity of forms

(2)
\begin{align} \sin \theta \, d \theta\left(\frac{\partial}{\partial \theta} + \cos^2 \theta \frac{\partial}{\partial \phi}\right) d \phi\left(m\frac{\partial}{\partial\theta} + n\frac{\partial}{\partial\phi}\right) - \sin \theta \, d \theta\left(m\frac{\partial}{\partial\theta} + n\frac{\partial}{\partial\phi}\right) d\,\phi \left(\frac{\partial}{\partial \theta} + \cos^2 \theta \frac{\partial}{\partial \phi}\right) \end{align}
(3)
\begin{align} (\sin \theta) \, n - (\sin \theta \cos^2 \theta) \, m \end{align}

Using this, we can reverse engineer $\gamma$

(4)
\begin{align} \gamma = i_X\beta = -\sin \theta \cos^2 \theta d \, \theta + \sin \theta d\,\phi \end{align}

Scott agrees!

page revision: 5, last edited: 03 Apr 2007 05:16