Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to understand how to derive forms that invariant under action of some group.

For example 2-form on $S^2$ and on $\mathbb{R}^3$ is very interesting for me (because I have troubles with it).

Suppose I have 2-form on $S^2$. And $SO(3)$ acts rotate sphere by $\alpha$ about some axis $e_z$. i.e. $$\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}=\begin{pmatrix} \cos(\alpha) & -\sin(\alpha) & 0\\ \sin(\alpha) & \cos(\alpha) & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}$$

As I understand n-form is invariant under action of group if $f^*w=w$.

If I try $w=dx \wedge dy$: $$f^*w=\left(\sum^n_{i_1=1}\frac{\partial f_1}{\partial x_{i_1}}dx_{i_1}\right)\wedge\left(\sum^n_{i_1=1}\frac{\partial f_2}{\partial x_{i_2}}dx_{i_2}\right)=\left(\frac{\partial f_1}{\partial x}\frac{\partial f_2}{\partial y}-\frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial x}\right)dx \wedge dy=dx \wedge dy$$ , where $f_i$ is $x'_i$. $i=1,2,3$

So this 2-form on $S^2$ is invariant under $SO(3)$.

I would be glad if someone would answer some of my questions:

  1. Am I doing it in right way?
  2. Is there any others 2-form on $S^2$ with this property?
  3. What is differences between here between 2-forms on $S^2$ and $\mathbb{R}^3$?
share|improve this question
    
You only verified a sub-group of SO(3). SO(3) is 3 dimensional. And your rotation with respect to (0,0,1) is 1 dimensional( i.e., with only one parameter $\alpha$. –  Xipan Xiao Nov 8 '13 at 13:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.