1

I've been thinking about this problem for some time, yet I am unable to solve it or find a resource online. I know that $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta}$ are coordinate vector fields and their bracket should be zero, but that reasoning doesn't seem rigorous enough. Here is the problem:

Let $\frac{\partial}{\partial r}$ be the normalized position field on $\mathbb R^2 \setminus \{ 0 \} $. That is, when viewed as a map $\mathbb R^2 \setminus \{ 0 \} \longrightarrow \mathbb R^2 \setminus \{ 0 \}$, $\frac{\partial}{\partial r}$ is

$$ \frac{\partial}{\partial r} (x,y) = \frac{(x,y)}{|(x,y)|}. $$

Let $\frac{\partial}{\partial \theta}$ be the velocity field of the $S^1$-action on $\mathbb R^2$ given by the counterclockwise rotation

$$ S^1\times\mathbb R^2 \longrightarrow \mathbb R^2 $$

where

$$ \left((\cos\theta,\sin\theta),(x,y)\right) \longmapsto \begin{pmatrix} \cos\theta & -\sin\theta \\\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x \\\ y \end{pmatrix}.$$

That is,

$$ \frac{\partial}{\partial \theta}(x,y) = \frac{\partial}{\partial \theta} \begin{pmatrix} \cos\theta & -\sin\theta \\\ \sin\theta & \cos\theta \end{pmatrix} \left.\begin{pmatrix} x \\\ y \end{pmatrix} \right|_{\theta=0}. $$

Show that $ [\frac{\partial}{\partial \theta},\frac{\partial}{\partial r}] = 0 $.

1 Answers1

1

$$[{\partial _\theta},{\partial_r}](x,y)={\partial_ \theta}\left({\partial_r}(x,y)\right)-\partial_r\left(\partial_\theta(x,y)\right)$$

$$=\partial_\theta{(x,y)\over|(x,y)|}-\partial_r\left(-x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right)$$

$$={1\over |(x,y)|}\left(-x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right)-{1 \over \scriptsize|\left(-x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right)|}\left(-x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right) $$

$$={1\over |(x,y)|}\left(-x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right)-{1 \over |(x,y)|}\left(-x\sin(\theta)-y\cos(\theta), x\cos(\theta)-y\sin(\theta)\right) $$

$$=0$$

Volk
  • 1,805