Let $G$ be a connected, reductive group over a $p$-adic field $k$, with maximal split torus $A_0$ contained in a maximal torus $T$. Let $X = X(T)$, and $V = X \otimes \mathbb R$.

Let $W = W(G,T)$ be the Weyl group of $T$ in $G$ (realized as $N_G(T)/T$), and let $(-,-)$ be a nondegenerate, symmetric $W$-invariant bilinear form on $V$. I have two questions:

1 . Is this bilinear form always Galois invariant?

2 . Can it be used to construct a nondegenerate, symmetric $W(G,A_0)$-invariant form on $X(A_0) \otimes \mathbb R$?

The first thing we can do is consider our three actions (of $W$ on $V$, of $\Gamma$ on $V$, and of $\Gamma$ on $W$) and see how they interact. If $\rho: W \rightarrow \operatorname{GL}(V)$ is the usual Weyl group action, given on characters by $\rho(nT)(\chi) = \chi \circ \operatorname{Int}n^{-1}$, and $\pi:\Gamma \rightarrow \operatorname{GL}(V)$ is given its usual action on characters by $\sigma.\chi = \sigma \circ \chi \circ \sigma^{-1}$, and $\Gamma$ is given its action on $W$ by $\sigma.(nT) = \sigma(n)T$, then

$$\pi(\sigma) \circ \rho(w) \circ \pi(\sigma^{-1}) = \rho(\sigma.w)$$

for all $w \in W$ and $\sigma \in \Gamma$.


1 . I still don't know the answer to this, but you can always choose the form on $V$ to be both $W$ and $\Gamma$-invariant: since the images of $W$ and $\Gamma$ in $\operatorname{GL}(V)$ are finite groups, and the image of $\Gamma$ normalizes the image of $W$, the group $H$ generated by these two images is also finite, so one can use the above procedure with $W$ replaced by $H$.

2 . Yes. Assume the form on $V$ is chosen to be $W(G,T)$ and $\Gamma$ invariant. Let $K$ be a Galois extension of $k$ over which $T$ splits, so we can replace $\Gamma$ by $\Gamma_K = \operatorname{Gal}(K/k)$. Let $d = [K : k]$.

Let $$X_0 = \{ \chi \in X : \sum\limits_{\sigma \in \Gamma_K} \sigma.\chi = 0\}$$ Then $X_0$ is orthogonal to $X^{\Gamma}$ with respect to our form, since for $\lambda \in X^{\Gamma}$ and $\chi \in X_0$, we have

$$d(\chi, \lambda) = (\chi, d \lambda) =(\chi, \sum\limits_{\sigma \in \Gamma_K} \sigma^{-1}. \lambda)$$

$$ = \sum\limits_{\sigma \in \Gamma_K} (\chi,\sigma^{-1}.\lambda) = \sum\limits_{\sigma \in \Gamma_K} (\sigma.\chi,\lambda) = (\sum\limits_{\sigma \in \Gamma_K} \sigma.\chi,\lambda) = 0$$

One further sees that $X_0 \cap X^{\Gamma} = 0$ and $X_0 + X^{\Gamma}$ is of finite index in $X$. Thus $V$ is the orthogonal direct sum of $X^{\Gamma} \otimes \mathbb R$ and $X_0 \otimes \mathbb R$, and that $X^{\Gamma} \rightarrow X(A_0)$ is an injection onto a subgroup of finite index. Hence $X^{\Gamma} \otimes \mathbb R \rightarrow X(A_0) \otimes \mathbb R$ is an isomorphism.

Now the action of $W(G,A_0)$ on $X(A_0) \otimes \mathbb R$ coincides with the action of a corresponding element of $W(G,T)$ on $X^{\Gamma} \otimes \mathbb R$. Thus the restriction of $(-,-)$ to $X^{\Gamma} \otimes \mathbb R$, identified with $X(A_0) \otimes \mathbb R$, is $W(G,A_0)$-invariant. This restriction is also nondegenerate: if $v \in X^{\Gamma} \otimes \mathbb R$, and $(v,w) = 0$ for all $w \in X^{\Gamma} \otimes \mathbb R$, then $(v,w) = 0$ for all $w \in V$, since $w = w_1 +w_2$ in the orthogonal direct sum, and $(v,w_2) = 0$.


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