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:

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 $m$.