Show that irreducible standard cyclic module is finite dimensional 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.
 A: 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.
