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Let $\mathscr{W}$ be the Weyl group of a root system $\Phi$ with basis $\Delta$. If $\sigma\in \mathscr{W}$, $\sigma = \sigma_{\alpha_1} .. \sigma_{\alpha_t}$ where $\alpha_1, ...,\alpha_t \in \Delta$, and $t$ is as small as possible, then $\sigma(\alpha_t)$ is a negative root.

snsunx
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1 Answers1

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This is the Corollary to Lemma $C$ in Humphreys text "Introduction on Lie algebras and Representation Theory" in section $10.2$. The proof involves some lemmas on simple roots, which are elementary, but a bit lengthy. In particular we need that for a simple root $\alpha$ the Weyl group element $\sigma_{\alpha}$ permutes the positive roots other than $\alpha$, and, writing $\sigma_i$ for $\sigma_{\alpha_i}$, if $\sigma_1\cdots \sigma_{t-1}(\alpha_t)$ is negative, then $ \sigma_1\cdots \sigma_t=\sigma_1\cdots \sigma_{s-1}\sigma_{s+1}\cdots \sigma_{t-1} $ for some index $1\le s<t$.

Dietrich Burde
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  • Actually I understand that lemma but don't know how the corollary follows from that lemma – snsunx Mar 28 '16 at 19:20
  • Now I kind of understand it. If $\alpha_t<0$, then $\sigma_1 ... \sigma_{t-1} (\alpha_t)<0$, then by Lemma B we have $t$ is not minimal. So $\alpha_t>0$, then $\sigma(\sigma_t(\alpha_t)) = \sigma_1...\sigma_{t-1}(\alpha_t)>0 \implies \sigma(\alpha_t)<0$. Could you check it for me? – snsunx Mar 28 '16 at 19:57
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    Actually if $\sigma(\alpha_t)>0$ then $\sigma_1...\sigma_{t-1}(\alpha_t)<0$ then by Lemma C $t$ is not minimal. – snsunx Mar 28 '16 at 20:29