I was reading James Arthur's book *An Introduction to the Trace Formula* and had a couple of questions. Here $A_0$ is a maximal split torus of a reductive group $G$, $P_0 \supset A_0$ is a minimal parabolic, $\Delta_0$ the set of simple roots, $P = MN$ a standard parabolic, $A_P = A_M$ the split component of $M$, and $\mathfrak a_M = \operatorname{Hom}(X(A_M),\mathbb R)$.

The weights $\hat{\Delta}_0 = \{ \varpi_{\alpha} \in (\mathfrak a_{0})^{G \ast} \subset \mathfrak a_0^{\ast} : \alpha \in \Delta\}$ are by definition the dual basis of the basis $\alpha^{\vee} : \alpha \in \Delta$ of $\mathfrak a_0^G \subset \mathfrak a_0$.

First, I believe that it should say "order preserving bijection $P \leftrightarrow \Delta_0^P$," not "order reversing bijection."

Second, I am confused when Arthur says "We obtain a second basis of $(\mathfrak a_M^G)^{\ast}$ by taking the subset

$$\hat{\Delta}_P = \{ \varpi_{\alpha} : \alpha \in \Delta_0 - \Delta_0^P\}$$

of $\hat{\Delta}_0$." Is he saying that the weights $\varpi_{\alpha} : \alpha \in \Delta_0 - \Delta_0^P$, which are elements of $\mathfrak a_0^{\ast}$, actually lie in $(\mathfrak a_P^G)^{\ast}$? Or is he saying that we project them to $(\mathfrak a_P^G)^{\ast}$, via the direct sum decomposition $\mathfrak a_0^{\ast} = \mathfrak a_G^{\ast} \oplus (\mathfrak a_P^G)^{\ast} \oplus (\mathfrak a_0^P)^{\ast}$?