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I am trying to understand some integrality properties of root spaces from Humphrey's Lie algebra, section 8.4. One or two steps in the argument are not clicking to me. These are shown below in bold-faced statements; can anyone explain these steps?

(1) $L$ is semi-simple Lie algebra, finite dimensional, over $F$, Char$F=0$ and $F=\bar{F}$.

(2) $H\subset L$ is maximal abelian subalgebra; for $\alpha\in H^*$, $L_{\alpha}:=\{x\in L : [h,x]=\alpha(h)x \forall h\in H\}$.

(3) $\Phi:=\{\alpha\in H^* : L_{\alpha}\neq 0\}$=roots of $L$ relative to $H$.

(4) $S_{\alpha}:=\langle x_{\alpha}, y_{\alpha},h_{\alpha}\rangle \leq L$, isomorphic to $\mathfrak{sl}(2,F)$, where $x_{\alpha}\in L_{\alpha}$, $y_{\alpha}\in L_{-\alpha}$ and $h_{\alpha}=[x_{\alpha},y_{\alpha}]$.

So $S_{\alpha}$ acts on $L$ via adjoint representation.

(5) Fix $\alpha\in \Phi$. Let $M=\langle H,L_{c\alpha}: L_{c\alpha}\subset L \mbox{ for some } c\in F^*\rangle$. This is $S_{\alpha}$- submodule.

(6) Weights of $h_{\alpha}$ on $M$ are $0,c\alpha(h_{\alpha})=2c$. Since weights of $h_{\alpha}$ are integers, $2c\in\mathbb{Z}$.

(7) (Easy to show) $S_{\alpha}$ acts trivially on Ker $\alpha$, and $S_{\alpha}\leq L$ is irreducible $S_{\alpha}$ submodule. [OK up to this]

(8) Taken together, Ker $\alpha$ and $S_{\alpha}$ exhaust the occurances of weight $0$ for $h_{\alpha}$. [quite not clear to me]

(9) So the only even weights occurring in $M$ are $0,\pm 2$. So $2\alpha$ is not a root. [not clear]

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I think the statement "we want to decompose $L$ into irreducible $S_{\alpha}$-submodules" is a bit misleading. Indeed, the main point is to understand the representation $M$ of $S_{\alpha}$, and you analyze this using the fact that it decomposes into a direct sum of irreducible components.

Now you know right away that the $0$-weight space in $M$ coincides with $H$. Moreover, the zero-weight space of $M$ is the direct sum of the zero weight space of all irreducible components. Now $\ker(\alpha)$ is a trivial representation of dimension $\dim(H)-1$, while $S_{\alpha}\subset L$ intersects $H$ in the one-dimensional subspac spanned by $h_{\alpha}$ (which is transversal to $\ker(\alpha)$ since $\alpha(h_{\alpha})\neq 0$. But this exactly shows that these irreducible components "use up" all the weight space $H$ of weight $0$. Since an irreducible representation of even highest weight has a non-trivial (indeed, one-dimensional) weight space of weight $0$, there cannot be any further component of even highest weight. Since irreducible representations of odd highest weight do not contain weight vectors of even weights, the weight space $L_{2\alpha}$ of weight $4$ has to be trivial.