Weyl group, bilinear form, and character/cocharacter pairing. Many questions!

Question

Let $G$ be a connected linear algebraic group, $T$ a maximal torus of $G$, and $\alpha$ a weight of $T$ such that $G_{\alpha} = Z_G(S)$ is not solvable, where $S = (\textrm{Ker } \alpha)^0$. I have been really confused about a few things concerning the Weyl group $W(G,T) = N_G(T)/Z_G(T)$, which injects into the group of automorphisms of the group of characters of $T$, as $(nZ_G(T) \cdot \chi)(t) = \chi(ntn^{-1})$.

I had asked a question earlier (Why is $s_{\alpha}$ a Euclidean reflection?) and user yisishoujo had given me a detailed answer. Unfortunately I was not able to understand all of his answer, although most of it was very helpful. I feel like there is something very subtle about this topic which everyone else realizes but that I am not aware of. So I want to break this question into several questions in the hope someone can point out to me what it is I'm missing.

Notation: $X, Y$ are the groups of characters/cocharacters of $T$.

$V = \mathbb{R} \otimes_{\mathbb{Z}} X$

$V ^\wedge{} = \mathbb{R} \otimes_{\mathbb{Z}} Y$.

We can identify $X$ as an additive subgroup of $V$ as $\chi \mapsto 1 \otimes \chi$. Similarly for $Y$ and $V^{\wedge}$.

Relevant facts:

1 . There is a canonical pairing $\langle -, -\rangle:X \times Y \rightarrow \mathbb{Z}$, where for $\chi \in X, \gamma \in Y$, $\langle \chi, \gamma \rangle$ is the unique integer for which $\chi \circ \gamma(x) = x^{\langle \chi, \gamma \rangle}$ for all $x \in k^{\ast}$.

2 . If $\gamma \in Y$, $\gamma$ induces an abelian group homomorphism of $X$ into $\mathbb{Z}$ by $\chi \mapsto \langle \chi, \gamma \rangle$. The assignment $\gamma \mapsto \langle -, \gamma \rangle$ is an isomorphism of abelian groups $Y \rightarrow \textrm{Hom}_{\mathbb{Z}}(X,\mathbb{Z})$. We identify $Y$ with $\textrm{Hom}_{\mathbb{Z}}(X,\mathbb{Z})$.

3 . Identifying $Y$ as above, there is a canonical isomorphism of real vector spaces from $V^{\wedge}$ onto the dual of $V$, where to any generator $\lambda \otimes \gamma$, we associate the map $V \rightarrow \mathbb{R}$ defined on generators by $r \otimes \chi \mapsto r \lambda \langle \chi, \gamma \rangle$.

4 . Any abelian group automorphism of $X$ extends to a linear isomorphism of $V$, the same for $Y$ and $V^{\wedge}$. The subgroup of the Weyl group, $W_{\alpha} := Z_G(S) \cap N_G(T)/Z_G(T)$, can be shown to have order exactly two. Choosing an $n \in Z_G(S) \cap N_G(T)$ which is not in $Z_G(T)$, we obtain an automorphism $s$ of $V$ which has order exactly $2$.

5 . The Weyl group also injects into the group of automorphisms of $Y$, as $(nZ_G(T) \cdot \gamma)(x) = n^{-1}\gamma(x)n$. Under this action, we have $\langle \chi, \gamma \rangle = \langle w \cdot \chi, w \cdot \gamma \rangle$ for any $w$ in the Weyl group, $\chi \in X, \gamma \in Y$. The actions of the Weyl group on $X, Y$ extend respectively to actions on $V, V^{\wedge}$. The pairing $\langle -, - \rangle$ extends to a Weyl group-invariant pairing $V \times V^{\wedge} \rightarrow \mathbb{R}$

Questions:

1 . Why does $s(\alpha) \neq \alpha$? I know it is not the case that $s(\chi) = \chi$ for all characters $\chi$, but it is not clear what happens to $\alpha$. If I can show that $s(\alpha) \neq \alpha$, yisishoujo explained why we must have $s(\alpha) =- \alpha$. Indeed, conjugation by $n$ (the $n$ as in Fact 4) induces an automorphism of algebraic groups $T/S$. But $T/S$ is isomorphic to $k^{\ast}$, so there are only two such automorphisms, the identity map and the inverse map. If $\textrm{Int } n$ induced the identity, then we would have $ntn^{-1}S = tS$ for all $t \in T$, i.e. $\alpha(ntn^{-1}) = \alpha(t)$ for all $t$, which implies $s \alpha = \alpha$, contradiction. It follows that $ntn^{-1}S = t^{-1}S$ for all $t$, whence $s(\alpha) = - \alpha$.

2 . Take $s, n$ as in Fact 4. We can think of $s$ as an automorphism on $Y$, even of $V^{\wedge}$, as in Fact 5. Why does there exist a cocharacter $\gamma$ such that $s \cdot \gamma = - \gamma$? Here is my attempt at a proof. We know $s$ extends to an automorphism of $V^{\wedge}$, and still $s^2 = 1_{V^{\wedge}}$. Choosing a basis for $Y$, we can think of $s$ as a matrix with entries in $\mathbb{Z}$. It follows that the only eigenvalues of $s$ as $\pm 1$. If I could show $-1$ was an eigenvalue, then the nullspace of $s + 1_{V^{\wedge}}$ would be nontrivial. Since solving a system of linear equations doesn't depend on the ambient field, we would find a solution with entries in $\mathbb{Z}$, i.e. a solution $\gamma \in Y$ with $s \cdot \gamma = - \gamma$. But why must $-1$ be an eigenvalue?

3 . Take $s, n$ as in Question 2. Supposing that Question 2 has been answered, is it possible to choose $\gamma \in Y$ such that $s \cdot \gamma = -\gamma$ as well as such that $\langle \alpha, \gamma \rangle \neq 0$? yisishoujo explains, given such a cocharacter, how to show the following:

($\ast$) If $\chi$ is a character of $T$, and $\langle \chi, \lambda \rangle = 0$, then $s \cdot \chi = \chi$.

4 . The Weyl group acts as a group of automorphisms of $V$ (Fact 5). By the standard averaging process, one can show there exists a symmetric, positive define bilinear form on $V$ which is invariant under the action of the Weyl group $(s \cdot v, s \cdot w) = (v,w)$ for $s$ in the Weyl group, $v, w \in V$. Is such a bilinear form unique up to a scalar?

5 . Similar to question 4, we identified a canonical Weyl group-invariant pairing $V \times V^{\wedge} \rightarrow \mathbb{R}$ as in Facts 1, 5. Is such a pairing unique up to scalar?

6 . What is the relationship between a Weyl group-invariant form on $V$ and the canonical pairing $V \times V^{\wedge}$? Can they in any sense be 'identified?' If so, does any possible nonuniqueness of the form on $V$ (as in Question 4) come into play? I know that GIVEN a Weyl group invariant form $(-,-)$ on $V$, there is a isomorphism of $V$ into its dual by sending $v$ to the functional $(v,-)$. Composing that with the isomorphism identified in Fact 3, we get an isomorphism $V \rightarrow V^{\wedge}$. This isomorphism induces an action of the Weyl group on $V^{\wedge}$ (which I do not believe is necessarily the canonical action as in Fact 5, however). Letting $\Phi$ be the inverse map $V^{\wedge} \rightarrow V$, we get a Weyl group invariant pairing $\langle -,-\rangle_1 V \times V^{\wedge} \rightarrow \mathbb{R}$ by setting $\langle v, \eta \rangle_1 = (v, \Phi(\eta))$.

And finally, the main question which all of this is intended to be put together to answer: given any Weyl group invariant bilinear form $(-,-)$ on $V$, and given $n,s$ as in Fact 4, why is it the case that $s$ is a Euclidean reflection of $V$ on $\alpha$? In other words, why is it the case that $s(\alpha) =- \alpha$? And if $v$ is any element of $V$ with $(\alpha, v) = 0$, why is it the case that $s \cdot v = v$?

This main question which has been so difficult for me is an offhand comment in Springer, Linear Algebraic Groups, 7.17. User yisishoujo indicated that the result $s \cdot v = v$ for all $v$ satisfying $(\alpha, v) = 0$ is a result of the statement $(\ast$) given in Fact 3, but I do not understand why. Maybe I need to understand the relationship between the bilinear form and the pairing (Fact 6) better. Thank you for your time.

Answer 1

1. Since $s$ has finite order (and acts on a characteristic-0 vector space), its action on $V$ is semisimple. Since the element $s$ of $G$ preserves $S$, the linear transformation $s$ of $X$ preserves $X^*(T/S)$; so there is an $s$-stable complement $V'$ to $\mathbb R\alpha = X^*(T/S) \otimes_{\mathbb Z} \mathbb R$ in $V$. The composition $V \to X^*(S) \otimes_{\mathbb Z} \mathbb R$ is surjective and trivial on $\alpha$, hence induces a surjection $V' \to X^*(S) \otimes_{\mathbb Z} \mathbb R$, which, by dimension-counting, is an isomorphism. It is clearly $s$-equivariant, and $s$ acts trivially on the codomain, so it also acts trivially on the domain $V'$. Since $s^2 = 1$ and $s$ preserves $\mathbb R\alpha$, we have that $s\alpha = \pm\alpha$. If $s\alpha = \alpha$, then $s$ acts trivially on $V' \oplus \mathbb R\alpha = V$, which you have agreed that it does not.

2. As in (1), $s$ is semisimple; so, if 1 were its only eigenvalue, then we would have that $s = 1$ (on $\hat V$, but equivalently on $V$), which is a contradiction.

3. If the ($-1$)-eigenspace of $s$ acting on $\hat V$ were contained in $\ker \alpha$, then $s$ would act trivially on $\hat V/\ker \alpha$, hence on its dual $\mathbb R\alpha$ in $V$.

4. Yes in case $G$ is semisimple (EDIT: I meant ‘(absolutely) simple’, so that the root system is irreducible), in which case $V$ is spanned by the roots of $T$ in $G$ and we may use the fact that the reflection representation of an irreducible root system is irreducible, but not in general. For example, if $G$ is Abelian, then $W = 1$ and any inner product on $V$ is Weyl-invariant.

5. See (4) and (6).

6. Since $V$ and $\hat V$ are isomorphic as representations of $W$, the space of $W$-invariant pairings $V \otimes \hat V \to \mathbb R$ may be identified with the space of $W$-invariant pairings $V \otimes V \to \mathbb R$. The $W$-invariant isomorphism of $V$ with $\hat V$, hence the identification of these two spaces that it determines, is unique up to scalars if $G$ is semisimple, but not in general.

Answer 2

Thanks very much to L Spice for explaining this to me. Here is what I wanted to originally find out in my own words:

Let $n \in Z_G(S) \cap N_G(T) \setminus Z_G(T)$, and let $s$ be the corresponding element in the Weyl group $W(G,T)$. Then with respect to any Weyl-group invariant, symmetric, positive definite bilinear form $(-,-)$, $s$ is a Euclidean reflection.

The character group $X(T/S)$ injects into $X = X(T)$; a character $\chi \in X$ lies in the image of $X(T/S)$ if and only if it is trivial on $S$. In particular, $\alpha$ lies in the image. Identifying $X(T/S)$ as a subgroup of $X(T)$, we can identify $\mathbb{R} \otimes_{\mathbb{Z}} X(T/S)$ as a subspace of $V$. This subspace is one dimensional, because the torus $T/S$ is isomorphic to $k^{\ast}$.

Now $\alpha$ is not the trivial character of $T$, hence it is nontrivial as a character of $T/S$. Hence $\alpha$ is generates the subspace $\mathbb{R} \otimes_{\mathbb{Z}} X(T/S)$, i.e. $\mathbb{R} \alpha = \mathbb{R} \otimes_{\mathbb{Z}} X(T/S)$.

Since $n \in Z_G(S)$, we see that $\mathbb{R} \otimes_{\mathbb{Z}} X(T/S)$ is stable under the action of $s$.

Now we know $s^2 = 1_V$ and, $V$ being a vector space over a field with characteristic $0$, we conclude that $s$ is semisimple, in the sense that any $s$-stable subspace of $V$ has an $s$-stable complement (write $V$ as a representation of the group $\mathbb{Z}/2\mathbb{Z}$). Let $V'$ be any $s$-stable complement of $\mathbb{R}\alpha$.

Every character of the torus $S$ extends to a character of $T$, i.e. the restriction homomorphism $X \rightarrow X(S)$ is surjective. Hence the induced linear transformation $\varphi: V \rightarrow \mathbb{R} \otimes_{\mathbb{Z}} X(S)$ is surjective. Now $V = V' \oplus \mathbb{R} \alpha$. The restriction of $\alpha$ to $S$ is the trivial character, so $\varphi(\alpha) = 0$. Thus, the surjectivity of $\varphi$ must be accomplished by $V'$ alone, i.e. the restriction of $\varphi$ to $V'$ is still surjective. But $V'$ and $\mathbb{R} \otimes_{\mathbb{Z}} X(S)$ have the same dimension, so $\varphi|V'$ is an isomorphism.

Also, it is clear that $\varphi(s \cdot v) = s \cdot \varphi(v)$ for all $v \in V$. But the action of $s$ on $\mathbb{R} \otimes_{\mathbb Z} X(S)$ is trivial, as $n \in Z_G(S)$. Thus the action of $s$ on $V'$ is also trivial. So $s$ is fixes $V'$ pointwise, stablizes $\mathbb{R} \alpha$, satisfies $s^2 = 1_V$, and is not the identity map. We must conclude that $s \cdot \alpha = -\alpha$.

Now let $(-,-)$ be any Weyl group invariant bilinear form on $V$. Let $V'$ be the orthogonal complement of $\alpha$ with respect to $(-,-)$. Then $V'$ is stable under $s$, because for any $v \in V'$, $$(s \cdot v, \alpha) = (s^2 \cdot v, s \cdot \alpha) = (v,-\alpha) = -(v,\alpha) = 0$$ By the argument above, $V'$ must be fixed pointwise by $s$. Since $s$ sends $\alpha$ to $-\alpha$ and fixes pointwise the orthogonal complement to $\alpha$, it must be a Euclidean reflection. $\blacksquare$