# Why is $s_{\alpha}^{\wedge}(\lambda) = -\lambda$?

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)

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 $$t \in T$$.

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.