Let $G$ be semisimple algebraic group over a field $k$. Let $T$ be a maximal torus of $G$ which is defined over $k$, and let $K$ be a finite Galois extension of $k$ over which $T$ is split. Let $\Gamma = \textrm{Gal}(K/k)$. Then $\Gamma$ acts on $G(K)$, as well as on the group $X$ of characters of $T$: specifically, if $\sigma \in \Gamma, \chi \in X$, then $\chi^{\sigma}$ is the unique character of $X$ satisfying $\chi^{\sigma}(t) = \sigma \chi(\sigma^{-1} t)$ for all $t \in T(K)$.

Let $V = \mathbb{Q} \otimes X$, and $\Phi \subseteq X$ the set of roots of $T$ in $G$. Then $(V,\Phi)$ is root system. Let $W$ be its Weyl group. Let

$$X_0 = \{ \chi \in X : \sum\limits_{\sigma \in \Gamma} \chi^{\sigma} = 0\}$$ $$V_0 = \mathbb{Q} \otimes X_0$$ $$A = \bigcap\limits_{\chi \in X_0} \textrm{Ker } \chi$$

Then $A$ is connected and defined over $k$, and is the largest subtorus of $T$ which is $k$-split. Let $\Phi_0 = \Phi \cap X_0$, and let $W_0$ be the subgroup of $W$ generated by all reflections $w_{\alpha}$ with $\alpha \in \Phi_0$. Then $W_0$ stabilizes $X_0$ and $\Phi_0$, and hence restricts to a group of automorphisms of $V_0$.

Proposition: $(\Phi_0,V_0)$ is a root system with Weyl group $W_0$.

This is what I am having trouble seeing. The only nonobvious thing is why $\Phi_0$ spans $V_0$. All we have to go on is that $\Phi$ spans $V$ (since $G$ is semisimple). I've been trying to prove this for awhile with no success. If $\chi \in X_0$, we can write

$$\chi = c_1\alpha_1 + \cdots + c_n\alpha_n$$ for some $\alpha_i \in \Phi$ and $c_i \in \mathbb{Q}$. Let's say that only $\alpha_1, ... , \alpha_t$ lie in $\Phi_0$. Applying $\sigma$ and summing, we get that

$$ 0 =\sum\limits_{i=t+1}^n c_i \sum\limits_{\sigma} \alpha_i^{\sigma}$$


The trouble comes from the fact that the proposition is not true: In general, $\Phi_0 := \Phi \cap X_0$ does not span $X_0$ (or, a fortiori, $V_0$).

For example, in quasi-split forms, $\Phi_0$ is empty, whereas $X_0$ is not just $\lbrace 0 \rbrace$. E.g. take a quasi-split form of $\mathfrak{sl}_3$ (root system of type $A_2$), where the Galois action switches the two basis roots $\alpha_1$ and $\alpha_2$. Then $V_0$ is the one-dimensional subspace $\mathbb{Q} \cdot (\alpha_1-\alpha_2)$, which however contains no roots.

As another example, over $\mathbb{R}$ as well as over $p$-adic fields, there exists a form of type $A_3$ with quadratic splitting field such that the generator $\sigma$ of the Galois group operates via

$\alpha_1 \mapsto \alpha_2+\alpha_3$

$\alpha_2 \mapsto -\alpha_2$

$\alpha_3 \mapsto \alpha_1+\alpha_2$.

on the three basis roots $\alpha_1, \alpha_2, \alpha_3$. (Over the reals, it's the form with Lie algebra called $\mathfrak{su}_{1,3}$ here; likewise, the group would be the (projective) special unitary group w.r.t. to a hermitian form of index 1 on $\mathbb{C}^4$.)

Then $V_0$ is the two-dimensional space spanned by $\alpha_2$ and $\alpha_1-\alpha_2-\alpha_3$, but $\Phi_0$ just consists of $\pm \alpha_2$.

What is true is that $\Phi_0$ is a root system in the subspace it spans (if $\Phi_0$ is not empty): It is the root system of the "anisotropic kernel" of the form.


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