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

Recall that given a representation $\pi$ of $\mathfrak{sl}_n$, a weight $\mu$ is said to be of highest weight if its corresponding weight vector is annihilated by all the positive root spaces (1). Now suppose I give another definition for what it means for a weight to be of highest weight:

(2) Given weights $\mu_1 = a_1L_1 +\ldots a_nL_n$ and $\mu_2 = b_1L_1 + \ldots b_nL_n$ of some representation of $\mathfrak{sl}_n$, we say that $\mu_1$ is higher than $\mu_2$ (denoted $\mu_1 > \mu_2$) if the first $i$ for which $a_i - b_i$ is non-zero (if any) is positive. A weight $\mu$ is then said to be of highest weight if for any other weight $\nu$, $\nu \leq \mu$.

Now I can see that not (1) implies not (2), but how can I see that (1) implies (2) to show that the definitions are equivalent? I can show this in the case that $\pi$ is an irreducible representation and hence a highest weight representation, for then the weights of such a representation have a very nice description. They are all of the form

$$\text{(highest weight) minus (some positive roots)} $$

and I can see why (1) will imply (2) in this case. However in the more general case that $\pi$ is not irreducible, how will (1) imply (2)?


share|cite|improve this question
Have you try this on Verma modules instead. Then you can use universality of Verma modules as highest weight modules. – Aaron Nov 5 '12 at 12:38
@Aaron My representation theory is not so advanced, I don't know what Verma modules are! – user38268 Nov 5 '12 at 12:39
They are of the form $M(\lambda):=U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\mathbb{C}_\lambda$, where the $\lambda$ is the highest weight. Read Section 1.3, 1.4 of Humphrey's Representations of Semisimple Lie algebras in the BGG category O. (The result you want, after some corrections, is in Section 1.2 of the same book) – Aaron Nov 5 '12 at 23:04
@Aaron Would you like to post an answer to this question so that it does not go unanswered? Thanks. – user38268 Nov 8 '12 at 10:25
up vote 1 down vote accepted

Note: This is taken from Humphrey's book

We work over the universal enveloping algebra $U(\mathfrak{g})$ instead of $\mathfrak{g}$. Here $\mathfrak{g}$ is f.d. complex s.s. Lie algebra. Note (irred/h.w.) $U(\mathfrak{g})$-modules corresponds to (irred/h.w.) $\mathfrak{g}$-modules naturally.

Let the basis of $\mathfrak{n}^-$ be $\{y_1,\ldots,y_m\}$, basis of $\mathfrak{h}$ be $\{h_1,\ldots, h_l\}$ with $h_i=h_{\alpha_i}$ for simple roots $\alpha_i$, basis of $\mathfrak{n}^+$ be $\{x_1,\ldots, x_m\}$. Then the PBW basis of $U(\mathfrak{g})$ is given by $$ y_1^{r_1} \cdots y_m^{r_m} h_1^{s_1}\cdots h_l^{s_l} x_1^{t_1}\cdots x_m^{t_m}$$

Let $M$ be a highest weight module of weight $\lambda$, with maximal vector $v\in M_\lambda$, then $M=U(\mathfrak{g})\cdot v=U(\mathfrak{n}^-)\cdot v$ by PBW Theorem. (This is the key fact you missed) Now we can write basis of $M$ as $\{ y_1^{i_1}\cdots y_m^{i_m}\cdot v | i_j\in \mathbb{Z}_+\}$, which imply the non-zero weights in $M$ are of form $\lambda - \sum_j i_j \alpha_j$.

share|cite|improve this answer
Thanks for your answer. I have not yet studied Universal Enveloping Algebras and the Poincaré - Birkhoff - Witt Theorem, but when I do hopefully I can understand your answer. I have accepted it anyway so that this question does not go unanswered. Thanks for your answer! – user38268 Nov 9 '12 at 22:38

Your Answer


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