I have had a lot of questions lately about root systems and rationality questions. Several people have referred me to the article *Groupes Reductifs* by Borel and Tits, which I am slowly reading. In the meantime, I had a question about relative roots for which I hope there is a simple explanation.


$G$ is a connected reductive group defined over a field $F$, $T$ is a maximal torus of $G$, $A\_0$ the maximal $F$-split subtorus of $T$, $X\_0$ the kernel of the restriction map $X = X(T) \rightarrow X(A\_0)$, and $\Phi \subseteq X$ the set of roots of $T$ in $G$. Let $\mathfrak g$ be the Lie algebra of $G$.


Let $\Phi\_0 = \Phi \cap X\_0$, and let $\overline{\Phi}$ be the image of $\Phi - \Phi\_0$ in $X/X\_0 \cong X(A\_0)$. Then $\overline{\Phi}$ is called the set of restricted roots of $T$ in $G$.


On the other hand, I have seen the restricted roots defined to be $\Phi(A\_0,G)$. Here $\alpha \in X(A\_0)$ lies in $\Phi(A\_0,G)$ if and only if there exists a nonzero $X \in \mathfrak g$ (or maybe it is required to be in $\mathfrak g(F)$, I'm not sure) such that $\textrm{Ad } t X = \alpha(t)X$ for all $t \in A\_0$.


It is clear that $$\overline{\Phi} \subseteq \Phi(A\_0,G)$$ On the other hand, the converse inclusion seems to be asserting the following: for each $\alpha \in \Phi(A\_0,G)$, and any $\chi \in X$ whose restriction to $A\_0$ is $\alpha$, there exists an element $\chi\_0 \in X\_0$ such that $\chi + \chi\_0 \in \Phi$. Should this be the case?



Two definitions of restricted roots
As far as I can tell, there isn't anything fancy you need to prove the equivalence of these definitions beyond basic facts about rational representations of tori. Let $T$ be a torus, $S$ a closed subgroup, and $\pi: T \rightarrow \textrm{GL}(V)$ a rational representation. Then $V$ breaks up into a direct sum


$$V = \bigoplus\limits\_{\chi \in X(T)} V\_{\chi}$$


where $V\_{\chi} = \{ v \in V: \pi(t)v = \chi(t)v \textrm{ for all } t \in T\}$. The restriction of $\pi$ to $S$ also breaks up $V$ into a direct sum


$$V = \bigoplus\limits\_{\alpha \in X(S)} V\_{\alpha}$$


where $V\_{\alpha} = \{ v \in V: \pi(s)v = \alpha(s)v \textrm{ for all } s \in S\}$.


**Lemma**: for $\alpha \in X(S)$, we have 


$$V\_{\alpha} = \bigoplus\limits\_{\chi \in X(T), \chi|S = \alpha} V\_{\chi}$$


Proof: It is clear that the right hand side is contained in the left. Taking the direct sum over all $\alpha \in X(S)$, both sides are equal to $V$, hence each inclusion must be an equality. $\blacksquare$


The characters $\chi$ of $T$ for which $V\_{\chi} \neq 0$ are called the weights of $T$ for $\pi$. The lemma implies that a character $\alpha$ of $S$ is a weight if and only if it is the restriction of a weight of $T$.


Now let $G$ be a connected, reductive group over $F$. Let $S$ be an $F$-split subtorus of a torus $T$ which is defined over $F$. 


**Proposition**: Let $\alpha \in X(S)$. Then the following are equivalent:


1 . There exists a nonzero element $X \in \mathfrak g$, and a character $\chi \in X(T)$, such that $\chi|S = \alpha$, and $\textrm{Ad }t X = \chi(t)X$ for all $t \in T$.


2 . There exists a nonzero $X \in \mathfrak g$ such that $\textrm{Ad } s X = \alpha(s) X$ for all $s \in S$.


3 . There exists a nonzero $X \in \mathfrak g(F)$ such that $\textrm{Ad } s X = \alpha(s)X$ for all $s \in S$.


(1) $\Rightarrow$ (2) $\Rightarrow$ (3) is clear. (3) $\Rightarrow$ (2) comes from the rationality statement in 3.2.12(ii) (Springer, Linear Algebraic Groups), and the fact that the morphism $\textrm{Ad}:G \rightarrow \textrm{GL}(\mathfrak g)$ is defined over $F$. (2) $\Rightarrow$ (1) follows from the lemma. $\blacksquare$.


In particular, when $T$ is maximally split, and $S = A\_0$, the maximal $F$-split subtorus of $T$, this gives three equivalent definitions of restricted roots.


