I asked this question on stackexchange (<https://math.stackexchange.com/questions/1964415/maximal-k-split-torus-and-the-weyl-group-why-is-this-a-root-system>), but did not get an answer.


$G$ is a semisimple algebraic group defined over a field $F$, maximally split maximal torus $T$ defined over $F$ and maximal $F$-split subtorus $A\_0 \subseteq T$. $R$ is the set of roots of $T$ in $G$, $X$ is the group of characters of $T$, and $X\_0 = A\_0^{\perp}$. The notes I'm reading claim that $R\_0 := R \cap X\_0$ is such that $(\mathbb{Q} \otimes\_{\mathbb{Z}} X\_0,R\_0)$ is a root system. Moreover, it is claimed that the $\mathbb{Q}$-span of $R\_0$, intersected with $X$, is equal to $X\_0$.


I haven't been able to verify these claims. In fact, I believe that $R\_0$ may even be the empty set even when $X\_0 \neq 0$. Here is the example I used:


**Example**: $F = \mathbb{R}$, $G$ is the derived group of $U(2,1)$, where $U(2,1)(\mathbb{C}) = \textrm{GL}\_3(\mathbb{C})$, and $U(2,1)(F) = \{ x \in \textrm{GL}\_3 : w \space ^t\overline{x}^{-1}w = x \}$. Here $w = \space ^t\textrm{Diag}(1,-1,1)$, and bar denotes element wise complex conjugation. Then


$$ T = \{ \begin{pmatrix} a & & \\ & b & \\ & & \frac{1}{ab} \end{pmatrix} \}$$ is a maximal torus of $G$, defined over $F$, and I believe


$$A\_0 = \{ \begin{pmatrix}x & & \\ & 1 & \\ & & \frac{1}{x} \end{pmatrix} \}$$ is a maximally split $F$-split subtorus of $G$, contained in $T$. We can take $\chi\_1, \chi\_2$ as a basis of $X$ by restricting the standard characters on the usual maximal torus of $\textrm{GL}\_3$. Then $$X\_0 = \{ n \chi\_2 : n \in \mathbb{Z} \}$$ 


$$R = \{ \pm (\chi\_1 - \chi\_2), \pm(2 \chi\_1 + \chi\_2), \pm (\chi\_1 + 2\chi\_2) \}$$ and we see that $R \cap X\_0$ is empty. $\blacksquare$


Is my example incorrect? This fact about $(X\_0, R\_0)$ being a root system is used in the notes I'm reading to prove facts about a "relative root system" inside $X/X\_0$. But it seems to be like $(X\_0,R\_0)$ need not be a root system. 



$(X_0,R_0)$ is a root system
I think the writer of the notes is mostly right. The sets $X\_0$, $R\_0$ consists of all weights, roots which restrict to $0$ on $A\_0$. So $(X\_0,R\_0)$ is indeed a root system, namely that of the anisotropic kernel, say $L$, of $G$ but in the more general sense that $R\_0$ is not required to rationally span $X\_0$. That would be the case if and only if $L$ is semisimple, as well. In your example $L$ is a $1$-dimensional anisotropic torus so $R\_0$ is empty while $X\_0\cong\mathbb Z$. In general, the center of $L$ comes from conjugate non-compact simple roots in the Satake-Tits diagram of $G$. So $L$ is semisimple if e.g. $G$ is of inner type. Maybe that's what the author had in mind.


