I am trying to solve an exercise from Humphreys Lie algebra; please check if solution is correct.

Let $\Phi$ be a root system in Euclidean space $E$ wit inner product $(\cdot,\cdot)$. Denote$\langle \alpha,\beta\rangle:=\frac{2(\alpha,\beta)}{(\beta,\beta)}$.

Let $\Delta=\{\alpha_1,\cdots,\alpha_l\}$ be a base (fundamental root system) of $\Phi$ and $\Lambda$ the weight space $$\Lambda:=\{ \lambda\in E : \langle\lambda,\alpha\rangle\in\mathbb{Z}, \forall \alpha\in\Delta\}.$$ Then dominant weight set is defined by $$\Lambda^+:=\{ \lambda\in \Lambda : \langle\lambda,\alpha\rangle\geq 0, \forall \alpha\in\Delta\}.$$ Let $\{ \lambda_1,\cdots,\lambda_l\}$ be basis dual to basis of co-roots .e. $\{ \check{\alpha_1},\cdots,\check{\alpha_l} \}$, where $\check{\alpha_i}=\frac{2\alpha_i}{(\alpha_i,\alpha_i)}$; i.e. $$(\lambda_i,\check{\alpha_j})=\delta_{ij}.$$ Finally $\delta:=\sum_i \lambda_i$.

By Lemma 13.3 A, p. 70 (Humphreys Lie algebra), $\delta$ is strongly dominant weight.

Exercise: Let $\lambda\in \Lambda^+$ and $\sigma\in W$, the Weyl group of $\Phi$. Then $\sigma(\lambda+\delta)-\delta$ is dominant only when $\sigma=1$.

My answer

(1) Since $\delta$ is strongly dominant, so is $\lambda+\delta$ because $\lambda$ is dominant.

(2) If $\sigma(\lambda+\delta)-\delta$ is dominant, its sum with $\delta$ is strongly dominant, i.e. $\sigma(\lambda+\delta)$ is strongly dominant.

(3)Thus, $\lambda+\delta$ and $\sigma(\lambda+\delta)$ are strongly dominant.

(4) But strongly dominant elements belong to the fundamental (open) Weyl chamber, say $C_0$.

(5) It is a well known property of Weyl group that given a Weyl chamber $C$ (w.r.t. $\Delta$) and element $\sigma$ of Weyl group such that $\sigma(C)\cap C\neq \phi$, then $\sigma$ must be necessarily identity.

So we conclude by (4), (5) that $\sigma=1$.


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