Fulton-Harris: Action by negative roots generates and irreducible representation

49 Views Asked by At

Fulton and Harris claim, on p. $203$ of my edition (Readings in Mathematics) the following:

For any semi-simple complex Lie Algebra $\mathfrak{g}$...

(ii) the subspace $W$ of $V$ generated by the images of a highest weight vector $v$ under successive applications of root spaces $\mathfrak{g}_\beta$ for $\beta \in R^-$ is an irreducible sub representation.

And for proof they offer the following:

Part (ii) may be proved by the same argument as in the two cases we have already discussed: we let $W_n$ be the subspace spanned by all $w_n \cdot v$ where $w_n$ is a word of length at most $n$ in elements of $\mathfrak{g}$ for negative $\beta$. We then claim that for any $X$ in any positive root space, $X \cdot W_n \subseteq W_n$ To see this, write a generator of $W_n$ in the form $Y \cdot w$, $w \in W_{n-1}$ and use the commutation relation $X \cdot Y \cdot w = Y \cdot X \cdot w + [X.Y] \cdot w$; the claim follows by induction, since $[X,Y]$ is always in $\mathfrak{h}$

where $\mathfrak{h}$ is the Cartan Subalgebra. It is this last assertion that I do not think is true. $[X,Y]$ should not be in $H$ unless $Y$ is in the weight space for the weight negative of that of $X$. For instance, in $\mathfrak{sl}_3(\mathbb{C})$ we have $[E_{1,2}, E_{3,1}] = E_{3,2} \neq 0$, where $E_{i,j}$ is the $3 \times 3$ matrix with a single $1$ in row $i$ column $j$, and zeros elsewhere.

Can someone salvage the proof in Fulton and Harris, or at least tell me where I am going wrong in my understanding of it?