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Let $V$ be a finite dimensional real vector space, and $\langle-,-\rangle$ a symmetric, nondegenerate positive definite bilinear form on $V$. If $v \in V$, the Euclidean reflection about $v$ is defined to be the unique linear transformation $\phi: V \rightarrow V$ which sends $v$ to $-v$ and fixes pointwise all the elements in the orthogonal complement of $v$.

Let $G$ be a linear algebraic group, $T$ a maximal torus with character group $X$ and Weyl group $W$. View $X$ as an additive subgroup of $V = \mathbb{R} \otimes_{\mathbb{Z}} X$. Then $W$ extends to a group of automorphisms of $V$. Let $\langle - ,-\rangle$ be a bilinear form on $V$ as above which is fixed by the elements of the Weyl group.

Let $\alpha$ be a weight of $T$ such that $G_{\alpha} = Z_G((\textrm{Ker } \alpha)^0)$ is not solvable, so $W_{\alpha} = W(G_{\alpha},T)$ is a subgroup of $W$ with order $2$. Let $x \in N_{G_{\alpha}}(T)/Z_{G_{\alpha}}(T)$, and let $s_{\alpha}$ be the image of $x$ in $W$. Then $s_{\alpha}$ extends to an automorphism of $V$.

I'm trying to understand why $s_{\alpha}$ is the Euclidean reflection about $\alpha$. This is what Springer claims in Linear Algebraic Groups, (7.1.6). I just don't know how to use the condition that $(\alpha,\chi) = 0$, since I don't know much about the bilinear form. It is not difficult to see that $s_{\alpha}$ has to send $\alpha$ to $-\alpha$ (edit: not really, I would love for you to explain this as well), but I also need to show that if $\chi$ is a character of $T$ with $\langle \alpha, \chi \rangle = 0$, then $s_{\alpha}(\chi) = \chi$, or in other words, $\chi(y) = \chi(xyx^{-1})$ for all $y \in T$.

D_S
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1 Answers1

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There exists $\lambda \in Y(T)$ (a morphism $G_m \to T$) such that $s_{\alpha} \lambda = - \lambda$. (since $Y=X^{\vee}$)

$s_{\alpha}$ action on $Y(T)$ is defined by $s_{\alpha} \lambda(t) = n_{\alpha} \lambda(t) n_{\alpha}^{-1}$.

we can assume $<\alpha, \lambda> \neq 0$ if we know $s_{\alpha} \alpha = - \alpha$.

For $\chi \in X$ satisfy $<\chi, \lambda>=0$,

$s_{\alpha} \chi (t) = \chi(t)$, if $t \in (\text{Ker } \alpha)^{o}$, ($t$ commutative with $n_\alpha$)

$s_{\alpha} \chi (t) =\chi(n_{\alpha}^{-1}\lambda(s)n_{\alpha}) = \chi(\lambda(s^{-1})) = 1 = \chi(t)$, if $t=\lambda(s)$.

$s_{\alpha} \chi =\chi$ on Im $\lambda$ $\cdot$ $(\text{Ker } \alpha)^{o}$ which is a closed connected subgroup of $T$, must equal to $T$ if Im $\lambda$ $\not\subset$ Ker $\alpha$ (i.e. $<\alpha, \lambda> \neq 0$).

$\\$

Proof for $s_{\alpha} \alpha = - \alpha$.

$s_{\alpha} \alpha = \alpha \circ \text{Int } n_{\alpha}^{-1}$, $\text{Int } n_{\alpha}^{-1}$ induce $T/(\text{Ker} \alpha)^o \to T/(\text{Ker} \alpha)^o$, an automorphism of $G_m$, and $\text{Aut } G_m = \mathbb{Z}/2\mathbb{Z}$.

$\text{Int } n_{\alpha}^{-1}(t) \equiv t^{-1} \pmod{(\text{Ker} \alpha)^o}$, since $n_{\alpha}$ is not in $Z_G(T)$, can't induce the identity morphism.

Hence $s_{\alpha} \alpha (t) = (-\alpha)(t)$.

yisishoujo
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  • What is meant by $s_{\alpha} \lambda$, if $\lambda$ is a cocharacter? Also, I am actually not sure why $s_{\alpha} \alpha = -\alpha$, can you explain? – D_S May 18 '16 at 18:30
  • Are you sure that $s_{\alpha}(\lambda(t)) = n \lambda(t)n^{-1}$ is the right map? Don't we have for all characters $\chi$ $$\chi \circ s_{\alpha}(\lambda)(t) = \chi(n\lambda(t)n^{-1}) = \chi(n) \chi(\lambda(t)) \chi(n^{-1}) = \chi \circ \lambda(t)$$ which implies $\chi \circ s_{\alpha}(\lambda) = \chi \circ \lambda $ which implies $(\chi,\lambda) = (\chi, s_{\alpha}(\lambda))$ for all $\chi$ (where $(-,-)$ is the perfect pairing between $Y$ and $X$), which implies $\lambda = s_{\alpha}(\lambda)$? – D_S May 23 '16 at 00:09
  • Thanks for taking the time to answer. I'm also confused on something else: is there a connection you are using between the Weyl-group invariant bilinear form on $\mathbb{R} \otimes_{\mathbb{Z}} X$ and the perfect pairing between $X$ and $Y$? If so, what is the relationship? – D_S May 23 '16 at 00:17
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    $\chi$ is defined on $T$, but $n$ is not in $T$, $\chi(n)$ is not defined. – yisishoujo May 23 '16 at 01:57
  • Weyl group preserve pairing since $(s_{\alpha} \chi, s_{\alpha} \lambda) = s_{\alpha} \chi \circ s_{\alpha} \lambda = \chi \circ \text{Int } n_{\alpha}^{-1} \circ \text{Int } n_{\alpha} \circ \lambda = (\chi, \lambda)$. – yisishoujo May 23 '16 at 02:01
  • This pairing identifies $\mathbb{Q} \otimes_{\mathbb{Z}} Y$ and dual of $\mathbb{Q} \otimes_{\mathbb{Z}} X$, and an inner product identifies $\mathbb{Q} \otimes_{\mathbb{Z}} X$ with its dual. – yisishoujo May 23 '16 at 02:11
  • So, GIVEN a weyl group-invariant form $(-,-)$ on $V = \mathbb{R} \otimes_{\mathbb{Z}} X$, we get an isomorphism $V \rightarrow V^{\ast}$ by $v \mapsto (v,-)$. Now $V^{\ast} \cong V^{\wedge} := \mathbb{R} \otimes_{\mathbb{Z}} Y$, so we get an isomorphism $V \rightarrow V^{\wedge}$. – D_S May 23 '16 at 04:16
  • I still don't get how you are relating the inner product to the pairing. I need to show that if $\chi$ is a character, and $(\alpha, \chi) = 0$, then $s_{\alpha}(\chi) = \chi$. – D_S May 23 '16 at 04:17
  • What I want to say is that the pairing and the inner product can be exactly the same thing. The W-invariant pairing only tells that $s_{\alpha}$ preserves a hyperplane, it's in fact identity on the hyperplane (but not identity on the whole space) follows from the property of $\text{Int } n_{\alpha}$ (i.e. the specific $s_{\alpha}$ action), which reduces the problem to "rank 1" case. – yisishoujo May 23 '16 at 12:57
  • Let us continue this discussion in chat. – D_S May 23 '16 at 14:50
  • I am awarding you the bounty, although I am still confused about a few things. Thanks so much for your help. – D_S May 25 '16 at 15:07