Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a left invariant 3-form, $\sigma$ on an simply connected Lie group, $G$ whose value at the identity is $\sigma=\langle[x,y],z\rangle$, where $\langle\cdot,\cdot\rangle$ denotes an invariant bilinear form on $\mathfrak{g}$. Let $i_\alpha$ denote the map $i_\alpha: SU_2 \to G$ induced by the root $\alpha$. When I pull back $\sigma$ by $i_\alpha$ I am supposed to get $\frac{1}{2}\langle h_\alpha,h_\alpha\rangle\sigma_0$ where $\sigma_0$ denotes the invariant 3 form on $SU_2$.

When I do this at the identity I must be missing something because I do not get the $\frac{1}{2}$. At the identity we should have: $$i_\alpha^*(\sigma)(e,f,h)=\sigma(e_\alpha,f_\alpha,h_\alpha)=\langle [ e_\alpha, f_\alpha ], h_\alpha \rangle=\langle i h_\alpha, h_\alpha\rangle$$ Could someone please point out where it comes from?

share|cite|improve this question
Note that the commands \langle and \rangle give angle brackets, while < and > are intended for less than and greater than symbols. – Zev Chonoles Jun 3 '12 at 2:39
Thank you, I had forgotten about those. – Sven Jun 3 '12 at 4:05
Invariant forms are only unique up to scale-- is $\sigma_0$ the pullback of $\sigma$? – Eric O. Korman Jun 3 '12 at 15:55
$\sigma_0$ is the invariant 3 form on $SU_2$ $\sigma$ is supposed to pull back to $\sigma_0$ times a scalar. I can show that this is the case but I missed something with the scalar. – Sven Jun 4 '12 at 1:36

Your Answer


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

Browse other questions tagged or ask your own question.