# Some confusion about weights and roots in parabolic root systems

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 $$\mathfrak a_0^{\ast} = \mathfrak a_G^{\ast} \oplus (\mathfrak a_P^G)^{\ast} \oplus (\mathfrak a_0^P)^{\ast}$$?

## 2 Answers

I only find one paper (a book chapter, not a book itself) with the indicated title, Arthur - An introduction to the trace formula, and I can't find in it the sentences you quote, so it's hard to speak exactly to your questions. Could you give the exact reference?

There are two natural choices of how to parameterise standard parabolics: by the simple roots that appear in their Levis, and by the simple roots that appear in their unipotent radicals. The first choice, which Arthur denotes by $$P \leftrightarrow \Delta_0^P$$, gives an order-preserving bijection, sending $$P_0$$ to $$\emptyset$$ and $$G$$ to $$\Delta_0$$. The second, which Arthur denotes by $$P \leftrightarrow \Delta_P$$ (EDIT: this is not quite true; I meant $$P \leftrightarrow \Delta_0 \setminus \Delta_0^P$$, of which $$\Delta_P$$ consists of the restrictions to the central torus of $$M_P$$), gives an order-reversing bijection. (One can tell which bijection is which by the fact that $$\mathfrak a_P$$, the real-ised cocharacter lattice of the standard Levi component $$M_P$$ of $$P$$, is annihilated by $$\Delta_0^P$$; in particular, $$\mathfrak a_{P_0} = \mathfrak a_0$$ is the full real-ised cocharacter lattice of the maximal split torus $$A_0$$, whose annihilator in $$\Delta_0$$ is empty.)

Elements of $$\mathfrak a_0^*$$ can be restricted to $$\mathfrak a_P^G$$; this furnishes a natural projection $$\mathfrak a_0^* \to (\mathfrak a_P^G)^*$$. The same trick is applied both to elements of $$\Delta_P \subseteq \mathfrak a_0^*$$ and to the fundamental weights $$\varpi_\alpha \in \mathfrak a_0^*$$; since the first application of this trick doesn't seem to bother you, probably the second shouldn't either.

• The part I cited is on page 26 of the paper you linked. – D_S Jan 17 '19 at 20:56
• Granting that the "coroots" $\Delta_P^{\vee}$ are defined to be the basis of $\mathfrak a_P^G$ which is dual to the basis of $\mathfrak a_P^{G \ast}$ obtained by taking the image of the weights $\varpi_{\alpha} : \alpha \in \Delta_0 - \Delta_0^P$ into $\mathfrak a_P^{G \ast}$, it isn't clear to me why $\Delta_P^{\vee}$ is the same as the restrictions to $\mathfrak a_P^G$ of the ordinary coroots $\beta^{\vee} : \beta \in \Delta_0 - \Delta_0^P$ as Arthur says. In general, taking dual bases and projecting to subspaces does not result in dual bases. – D_S Jan 20 '19 at 0:23
• @D_S, my explanation on dual bases was confusing, if not downright wrong, so I deleted it. I will try to post a better one tomorrow. – LSpice Jan 25 '19 at 2:46
• Thank you. I don't know why this part of Arthur's notes is so hard for me! – D_S Jan 25 '19 at 4:51

I think that in the complex case, the set $$\Delta_0^P$$ would consists of roots whose root spaces generate the center of the Levi part $$M$$. (These are sometimes denoted in Dynkin diagram notation by crosses.) If $$P_1 \leq P_2$$ then $$M_1 \leq M_2$$ but the central part is actually bigger for $$M_1$$ and contains the center of $$M_2$$. So this relations is indeed order reversing. (Also note that $$N_2 \leq N_1$$.) I am not that familiar with the real case but I think it works in pretty much the same way (modulo some coverings and discrete extensions).

It's a bit hard to understand your second question without definition of $$\mathfrak{a}^G_P.$$