Let $G$ be a connected, reductive group over a $p$-adic field. Let $A_0$ be a maximal split torus of $G$ and $P = MU$ a parabolic subgroup with Levi $M$ containing $A_0$, and opposite parabolic $\overline{P} = M \overline{U}$. I'm trying to understand where the integration formula

$$\int\limits_G f(g) dg = \gamma(P)^{-1} \int\limits_P \int\limits_{\overline{U}} f(p \bar{u}) d\bar{u} d_lp$$

comes from, where $\gamma(P) = \int\limits_{\overline{U}} \delta_P(m_p(\bar{u})) \, d\bar{u}$. This is claimed in writeup (page 6) by Waldspurger on Harish-Chandra's unpublished notes on the Plancherel formula for $p$-adic groups, in the second equality of (2).

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I know that the product map $U \times M \times \overline{U} \rightarrow G$ gives an open immersion into $G$. So the right hand side of this equality seems like it is integration over some open set in $G$. When $P$ is minimal, I believe this open set is dense. Can the second equality in (2) be derived from the first equality?


1 Answer 1


The integral $$\int\limits_{U \times M \times \overline{U}} f(um\bar u) \delta_P(m)^{-1} d\bar u dm du = \int\limits_{P \times \overline{U}} f(p \bar{u}) d \bar u d_lp $$ is integration with respect to the product measure $d_lp d \bar{u}$ on the open dense set $P \times \overline{U} \cong P \overline{U}$ of $G$. One has to check that the Haar measure on $G$ restricts to a Haar measure on the group $P \times \overline{U}$, even though $(p, \bar u) \mapsto p \bar u$ is not a group homomorphism.

Therefore $\int\limits_{P \times \overline{U}} f(p \bar u) d \bar ud_lp$ is a scalar multiple of $\int\limits_G f(g) dg$, and to find the scalar, it suffices to compute the integral over $f = \operatorname{Char} K$. The measure $dg$ is chosen so that $\int\limits_G f(g)dg = 1$.

We extend $\delta_P$ to a continuous function on $G$ by making it trivial on $K$, so that $$\gamma(P) = \int\limits_{\overline{U}} \delta_P( \bar{u}) d \bar u$$ For each $\bar{u} \in \overline{U}$, write $\bar{u} = p_u k_u$ for $p_u \in P$ and $k_u \in K$ (nonuniquely), so that $\delta_P(\bar{u}) = \delta_P(p_u)$. Now for $f = \operatorname{Char} K$,

$$\int\limits_{\overline{U}} \int\limits_P f(p \bar{u})d\bar{u} d_lp = \int\limits_{\overline{U}} \int\limits_P f(p p_u k_u) d_lp d \bar{u}$$

$$ = \int\limits_{\overline{U}} \int\limits_P f(pk_u) \delta_P(p_u) d_lp d \bar{u} = \int\limits_{\overline{U}} \int\limits_{P \cap K} \delta_P(\bar{u}) d_lp d \bar{u} = \gamma(P)$$

since $d_lp(P \cap K) = 1$. This gives the second equality in (2) and also guarantees the convergence of the integral defining $\gamma(P)$, since the Radon measure of the open compact set $K \cap P \overline{U}$ is finite and positive.


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