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 had problems understanding the following proof. Maybe someone could help me with this?

Let {$\ x_i, y_i $} be the standard generators of the Lie Algebra L.

Let $\ V'(\lambda)=U(L)/J'(\lambda)$.

Let $\ V'(\lambda)$ be an irreducible standard cyclic module.

Let U(L) be the universal enveloping algebra.

Let $\ J'(\lambda)$ be the left ideal generated by $\ I(\lambda)$ along with all $\ y_i^{m_i+1}.$ $\ I(\lambda)$ is the left ideal of U(L) generated by all $\ x_\alpha, (\alpha \succ 0)$, and by all $\ h_\alpha - \lambda(h_\alpha)1, (\alpha \in \Phi).$

Show that $\ V'(\lambda)$ is finite dimensional.

For this it would suffice to show that it is a sum of finite dimensional $\ S_i$-submodules.

To show this we have to show that each $\ y_i $ is locally nilpotent on $\ V(\lambda) $.

This is obvious for the $\ x_i $, since we cannot have $\ \mu+k\alpha_i \prec \lambda$ for all $\ k \geq 0 $.

The coset of 1 in $\ V'(\lambda)$ is killed by a suitable power of $\ y_i$ (namely, $\ m_i+1$).

$\ V'(\lambda)$ is spanned by the cosets of all $\ y_{i_1}...y_{i_t}. (1\leq i_j \leq l) $.

If the coset of this monomial is killed by $\ y_i^k$, the coset of the longer monomial $\ y_{i_0}y_{i_1}...y_{i_t}$ is killed by $\ y_i^{k+3}$.

Induction on length of monomials, starting at 1, then proves the local nilpotence of $\ y_i$.

We have already proven before that $\ J(\lambda)$ is generated by $\ I(\lambda)$ along with all $\ y_i^{m+1}. m_i=\langle \lambda,\alpha_i \rangle, 1\leq i \leq l$, \lambda is a dominant integral linear function. For this we have assumed that $\ V'(\lambda)$ is finite dimensional, which remains to show for the proof to be completed.

The whole proof is based upon the following theorem, which we have already proven before: If $\ \lambda \in H*$ is dominant integral, then the irreducible L-module $\ V=V(\lambda)$ is finite dimensional, and its set of weights $\ \Pi(\lambda)$ is permuted by the Weyl group, with $\ dimV_\mu=dimV_{\sigma\mu}$ for $\ \sigma \in$ Weyl group.

share|cite|improve this question
up vote 0 down vote accepted

Ok, now I think I can follow your notation. Presumably the hard part is the inductive step. So if you know that $y_i^k$ kills the coset of $y_{i_1}\cdots y_{i_t}$, then you need to show that $y_i^{k+3}$ kills the coset of $y_{i_0}y_{i_1}\cdots y_{i_t}$. To do that, you need to write $y_i^{k+3}y_{i_0}$ as a sum of terms of the form $z y_i^{\ell}$, where $\ell\ge k$ for each term. Achieving this really boils down to the relation $(ad\; y)^4(z)=0$. This is because in the universal enveloping algebra we have the relation $$ y_i z = z y_i+ (ad\; y)(z), $$ for all $z\in U(L)$. We shall be needing the derivation rule $$ (ad\;x)(x_1\cdots x_t)=\sum_i)x_1\cdots x_{i-1}(ad\;x)(x_i)x_{i+1}\cdots x_t $$ a lot. So we first get $$ y_i y_{i_0}=y_{i_0} y_i + (ad\; y_i) (y_{i_0}), $$ where $z=y_{i_0}$. Applying this rule for another time we get first $$ y_i^2 y_{i_0}=(y_i y_{i_0})y_i+(ad\;y_i)(y_i y_{i_0}). $$ Here the first term we already sort of handled with our earlier calculation. By the derivation rule the second term is (using the rule $(ad\;y_i)(y_i)=0$) $$ (ad\;y_i)(y_i y_{i_0})=y_i (ad\; y_i)(y_{i_0})=(ad\; y_i)(y_{i_0})y_i+(ad\;y_i)^2(y_{i_0}). $$ Altogether we get $$ y_i^2y_{i_0}=y_{i_10}y_i^2+2[(ad\; y_i)(y_{i_0})]y_i+(ad\; y_i)^2(y_{i_0}). $$ Here I invite you to go on and compute $y_i^3y_{i_0}$, $y_i^4y_{i_0}$ et cetera. Always keep pushing the new $y_i$-factor `to the right'. You see the general structure that the result for $y_i^my_{i_0}$ consists of terms of something$_\ell$ times $y_i^{m-\ell}$, where something$_\ell$ has a factor of the form $(ad\;y_i)^\ell (y_{i_0})$. Because the latter is zero, whenever $\ell\ge4$, the induction step follows from this.

I really hope that a calculation making this clearer has been done in your textbook.

Hopefully TeX comes out right. The damn auto-preview makes this painfully slow by now, so I have to stop.

share|cite|improve this answer
Thank you very much for your detailed explanation! This induction step was just what I was looking for. The only remaining question I have is why the generator $\ x_i=x_{\alpha_i}$ is locally nilpotent. I can't see the connection between this and the given fact that the sum of a weight $\ \mu$ and $\ k\alpha_i$ cannot be smaller than the highest weight $\ \lambda$ for all $\ k \geq 0$. – Jim Helbert Jul 11 '11 at 16:37
@Jim: $\mu$ is a weight of the standard cyclic module of highest weight $\lambda$. Therefore $\mu=\lambda-\sum_{i=1}^nn_i\alpha_i$ for some non-negative integers $n_i$. If $k>n_i$, then it is impossible for the relation $\mu+k\alpha_i\prec\lambda$ to hold. Therefore $x_i^{n_i+1}$ kills anything of weight $\mu$ in this module. – Jyrki Lahtonen Jul 11 '11 at 18:55
Oh, that's right! Thanks a lot for all your help! – Jim Helbert Jul 11 '11 at 19:38

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.