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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)?

Thanks.

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    @Aaron Would you like to post an answer to this question so that it does not go unanswered? Thanks.2012-11-08

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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 $\lambda - \sum_j i_j \alpha_j$.

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    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!2012-11-09