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Let $G$ be a connected, reductive linear algebraic group with semisimple rank one. Let $H = (G,G)$, $T_1$ a maximal torus of $H$, and $T$ a maximal torus of $G$ containing $T_1$. Let $\lambda: k^{\ast} \rightarrow T_1$ be an isomorphism, and let $\alpha$ be a character of $T$. Let $U$ be the unipotent part of a Borel subgroup of $H$, and fix an isomorphism $u: k \rightarrow H$. Then there is a character $\alpha$ of $T$ such that $$tu(a)t^{-1} = u(\alpha(t)a)$$ for all $t \in T$. Fix an isomorphism $\lambda: k^{\ast} \rightarrow T_1$, and interpret $\lambda$ as a cocharacter of $T$. Then $\langle \alpha, \lambda \rangle$ is an integer $m$, which can be shown to be equal to $\pm 1$ or $\pm 2$.

Let $n \in N_H(T_1) \subseteq N_G(T)$, but $\not\in T$, so $n$ induces an automorphism $s$ of the vector space $$V = \mathbb{R} \otimes_{\mathbb{Z}} X(T)$$ One can show that $s$ is a Euclidean reflection about $\alpha$ with respect to any Weyl group-invariant bilinear form on $V$. The statement I'm trying to understand is that $$s^{\wedge}(\lambda) = -\lambda$$ where $s^{\wedge}$ is the dual map of $s$, where the dual of $V$ is identified with $\mathbb{R} \otimes_{\mathbb{Z}} Y(T)$.

This is a claim made in Springer, Linear Algebraic Groups (see picture below) enter image description here

The justification for this seems to have a typo. $s(t) = t^{-1}$ doesn't make any sense notationally, as $s$ is an automorphism of $V$, not a map defined on $T_1$. However, it is the case that $s(\alpha)$ is a character of $T$, with $s(\alpha)(t) = \alpha(t^{-1})$ for $t \in T$.

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Typing up questions on stackexchange is really helping me out. I spent 3 hours trying to prove this yesterday. Five minutes after posting this question, I figured it out. Let $\chi$ be any character of $T$. Then $s(\chi)$ is the character $t \mapsto \chi(ntn^{-1})$. For $t \in T_1$, $ntn^{-1} = t^{-1}$, so $s(\chi) + \chi$ is trivial on $T_1$.

It follows that $[s(\chi) + \chi] \circ \lambda : k^{\ast} \rightarrow k^{\ast}$ is the trivial map, or in other words, $$\langle s(\chi) + \chi, \lambda \rangle = 0$$ Hence $$ \langle \chi, -\lambda \rangle = \langle -\chi, \lambda \rangle = \langle s(\chi), \lambda \rangle = \langle \chi, s^{\wedge}(\lambda) \rangle$$ Since $-\lambda$ and $s^{\wedge}(\lambda)$ take the same value on all characters, they must be equal.

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