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.