Proposition 28

Let $B$ be a basis of $R$, $B'$ a subset of $B$, $V'$ the vector subspace of $V$ generated by $B'$, $R' = R \cap V'$ (which is a root system in $V'$*),* $R'\tilde{}$ the inverse root system (which is identified with the canonical image of $R'$ in $R'\tilde{}$), $V_1$ the orthogonal complement of $R'\tilde{}$ in $V$, and $p$ the projection of $V$ onto $V'$ parallel to $V_1$. Then, $Q(R') = Q(R) \cap V'$, $P(R') = p(P(R))$. The set of dominant weights of $R'$ is the image under $p$ of the set of dominant weights of $R$.

Indeed, $Q(R)$ is the subgroup of $V$ with basis $B$, $Q(R')$ is the subgroup of $V'$ with basis $B'$ (no. 7, Cor. 4 of Prop. 20) from which $Q(R') = Q(R) \cap V'$ is immediate. If $\overline{\omega} \in P(R)$ and $\alpha \in R'$, $\langle p(\overline{\omega}), \alpha^\vee \rangle = \langle \overline{\omega}, \alpha^\vee \rangle \in \mathbf{Z}$, so $p(\overline{\omega}) \in P(R')$, and hence $p(P(R)) \subseteq P(R')$. If $\overline{\omega}' \in P(R')$, $\overline{\omega}'$ extends to a linear form $\overline{\omega}$ on $V^*$ vanishing on $(B - B')\tilde{}$; we have $\langle \overline{\omega}, \alpha^\vee \rangle \in \mathbf{Z}$ for all $\alpha \in B$, so $\overline{\omega} \in P(R)$, and $\overline{\omega}' = p(\overline{\omega})$; hence $P(R') \subseteq p(P(R))$. Thus $P(R') = p(P(R))$, and the assertion about dominant weights is proved in the same way.