# Integration over a reductive group $G$ using the constant $\gamma(P)$

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 This is claimed in writeup (page 6) by Waldspurger on Harish-Chandra's unpublished notes on the Plancherel formula for $$p$$-adic groups, in the second equality of (2). 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?

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.