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

Assuming that for each root α there is only one linearly independent root vector, show that if $\alpha$, $\beta$, and $\alpha+\beta$ are roots, then [$e_\alpha$ , $e_\beta$ ] not equal to 0. Here $\alpha$, $\beta$ are roots of a simple lie algebra, and $e_\alpha$ , $e_\beta$ are the corresponding root vectors.

share|cite|improve this question
I switched the roots $\alpha,\beta$ into subscript position, because otherwise it looked too much like multiplication. Feel free to roll back, if you disagree. – Jyrki Lahtonen Mar 6 '12 at 16:31
up vote 2 down vote accepted

This fact follows from $sl_2$-theory. The sum of root spaces $L_{\beta+i\alpha}$, $i$ any integer, forms an $sl_2(\alpha)$-submodule. Then $sl_2$-theory tells us that a higher weight space (if non-zero) can be gotten from lower ones by acting on them with $e_\alpha$.

share|cite|improve this answer
If you are unfamiliar with $sl_2$-theory, then as an alternative to the above you may consider the following: the element $e_\alpha$ acts (via adjoint action) as a ladder operator in the sum of those root spaces. The $sl_2$-argument then basically says that any weight vector that is not at the bottom weight space lies in the image of the ladder operator. – Jyrki Lahtonen Mar 6 '12 at 20:03

Your Answer


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