The relative weights form a basis of $\mathfrak a_P^{G \ast}$

Question

Let $G$ be a connected, reductive group over $\mathbb Q$ with minimal parabolic $P_0 = M_0 N_0$. For $P = MN$ a standard parabolic subgroup of $G$, let $A_P$ be the split component of $M$ and let $\mathfrak a_P^{\ast} = X(A_P) \otimes_{\mathbb Z} \mathbb R$, and $\mathfrak a_P = \operatorname{Hom}_{\mathbb R}(\mathfrak a_P^{\ast},\mathbb R)$.

For $P_1 \subset P_2$, we let $(\mathfrak a_{P_1}^{P_2})^{\ast}$ be the kernel of the natural surjection $\mathfrak a_{P_1}^{\ast} \rightarrow \mathfrak a_{P_2}^{\ast}$. This exact sequence splits canonically, so we can write $\mathfrak a_{P_1}^{\ast} = \mathfrak a_{P_2}^{\ast} \oplus (\mathfrak a_{P_1}^{P_2})^{\ast}$ and dually $\mathfrak a_{P_1} = \mathfrak a_{P_2} \oplus \mathfrak a_{P_1}^{P_2}$.

For the minimal parabolic, we have the roots $\Delta_0$ and weights $\hat{\Delta}_0$ which are both bases of $\mathfrak a_0^{G \ast}$, and their respective dual bases $\hat{\Delta}_0^{\vee}$ and $\Delta_0^{\vee}$ of $\mathfrak a_0^G$.

Let $\theta_P$ be a subset of $\Delta_0$ corresponding to a standard parabolic subgroup $P = MN$. Specifically, $A_P$ is the connected component of the intersection of the kernels of the $\alpha \in \theta_P$, and $M$ is the centralizer of $A_P$. If we set $\Delta_P$ to be the projection of $\Delta_0 - \theta_P$ onto $\mathfrak a_P^{G \ast}$, then $\Delta_P$ is a basis of $\mathfrak a_P^{G \ast}$.

Arthur's notes on the trace formula identify another basis of $\mathfrak a_P^{G \ast}$, the so called "relative weights" $\hat{\Delta}_P$. By definition,

$$\hat{\Delta}_P = \{ \varpi_{\alpha} : \alpha \in \Delta_0 - \theta_P \}.$$

A priori, the elements of $\hat{\Delta}_P$ lie in $\mathfrak a_0^{G \ast}$. We may write $\mathfrak a_0^{G \ast} = \mathfrak a_0^{P \ast} \oplus \mathfrak a_P^{G \ast}$. The definition of the relative weights implicitly assumes that $\hat{\Delta}_P$ is contained in $\mathfrak a_P^{G \ast}$. Why is this the case?

Answer 1

In Arthur's notes, $\Delta_0^P=\theta_P$ is a basis for $(\mathfrak{a}_0^P)^*$, $\Delta_0=\Delta_0^G$ is the set of simple roots, and it is a basis for $(\mathfrak{a}_0^G)^*$. So $\Delta_0-\Delta_0^P$ is in one-to-one correspondence with a basis for $(\mathfrak{a}_P^G)^*$, given by the fact the sequence $$0\longrightarrow (\mathfrak{a}_P^G)^*\longrightarrow (\mathfrak{a}_0^G)^*\longrightarrow (\mathfrak{a}_0^P)^*\longrightarrow0$$ is split exact(as $\mathbb{R}$-vector spaces).

On the other hand, we also have $$0\longrightarrow \mathfrak{a}_P^G\longrightarrow \mathfrak{a}_0^G\longrightarrow \mathfrak{a}_0^P\longrightarrow 0$$ is exact, and $\widehat{\Delta}_0^P=\{\beta^\vee:\beta\in \Delta_0^P\}$ is a basis for $\mathfrak{a}_0^P$. Hence $\langle \varpi_\alpha,\beta^\vee\rangle=0$ if $\alpha\in \Delta_0-\Delta_0^P$ and $\beta^\vee\in \widehat{\Delta}_0^P$, since simple weights and simple coroots are in dual pair. Therefore all elements in $\widehat{\Delta}_P=\{\varpi_\alpha: \alpha\in \Delta_0-\Delta_0^P\}$, a priori as elements in $(\mathfrak{a}_0^G)^*$, are in fact in $(\mathfrak{a}_P^G)^*$.