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I have problems in understanding the proof that the root spaces of a semisimple Lie algebra are all 1-dimensional and that the only multiples of a root $\alpha \in \Phi$ which occur in $\Phi$ are $\pm \alpha$.

The proof which I am referring to is that of page 101 of the book Introduction to Lie Algebras

The problems begin with the discussion of the two cases $s$ even/odd. I agree with the authors that in the even case there must be an $h_\alpha$-eigenvector $v \in V\simeq V_s$ of zero eigenvalue (I think this follows from the fact that $h \in \mathfrak{sl}_2(\mathbb{C})$ acts diagonalizably on $V_s$ with eigenvalues $-s,-s+2,\dots, s-2,s$ so $0$ is an eigenvalue because $s$ is even by hypothesis). But how do they deduce from this that $\alpha(v)=0$? It seems to me—but it is perhaps a wrong conjecture—that they use the fact that the $0$-eigenspace of the action of $h_\alpha$ on $L$ is all contained in the direct sum $K \oplus \mathfrak{sl}(\alpha)$, but I don't understand neither why this fact is true nor how the result can be derived from it.

Xander Henderson
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Diogenes
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1 Answers1

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If $x\in M$ is such that $h_{\alpha}\cdot x = [h_{\alpha}, x] = 0$, then $x\in K\bigoplus\mathfrak{sl}(\alpha)$, otherwise $x\in\mathfrak{g}_{c\alpha}$ for some $c\alpha\in\Phi$ and $c\neq 0$, but then $[h_{\alpha},x]= 2cx\neq 0$ giving us the contradiction.

To show that $v\in K$, note that $\mathfrak{h} = K\bigoplus span(\{h_{\alpha}\})$ (exercise), it is also easy to show that $v\in\mathfrak{h}$. Finally, since $M=K\bigoplus \mathfrak{sl}(\alpha)\bigoplus W$ and $v\in V\subseteq W$, $v$ must lie in $K$.

Sid Caroline
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