Let $P = MN$ be a parabolic subgroup of a $p$-adic reductive group $G$ with split component $A_M$. There is bijection from the set of parabolic subgroups of $G$ with Levi $M$ and the chambers of $\mathfrak a_M$ with respect to the hyperplanes $H_{\alpha} = \{ h \in \mathfrak a_M : \alpha \in \Phi(A_M,G) \}$, where $P$ corresponds to the chamber

$$\{ h \in \mathfrak a_M : \langle h,\alpha \rangle > 0 \textrm{ for all } \alpha \in \Phi(A_M,N)\}$$

A parabolic subgroup $P' = MN'$ is said to be adjacent to $P$ if the chambers of $P$ and $P'$ are separated by a single hyperplane $H_{\alpha}$. Then there is a unique simple root $\alpha$ of $A_M$ in $N$ for which $-\alpha$ is a simple root of $A_M$ in $N'$.

Let $K$ be a maximal compact open subgroup of $G$ in good position relative to $P$ and $P'$, so that the Harish-Chandra map $H_P$ extends to $G$ via the equation $G = PK$. Let $\alpha^{\vee} \in \mathfrak a_M$ be the coroot corresponding to $\alpha$, and suppose $n' \in \overline{N} \cap N'$. Is it true that $H_P(n')$ is proportional to $\alpha^{\vee}$? This is claimed in Waldspurger's writeup on Harish-Chandra's notes on the proof of the convergence of intertwining operators:

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To compute $H_P(n')$, we need to write $n' = nmk$ for $n \in N, m \in M, k \in K$, so that $H_P(n') = H_P(m)$. As a start, I could try to show that $\langle \beta, H_P(m)\rangle$ for all $\beta \in \Delta(A_M,N)$ except for $\alpha$. But I don't see how to use the fact that $n' \in \overline{N} \cap N'$ to tell anything about $m$.


1 Answer 1


Probably you have already know the answer. For those who are interested in it, a proof is given by

Lemme 3.3.1 of La formule des traces tordue d’après le Friday Morning Seminar by Labesse and Waldspurger.

In split case, an enlightening proof can be found in

Lemme 8.16.8 of Le lemme fondamental Pondéré I. Constructions Géométriques by Chaudouard and Laumon.


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