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For simplicity, let $G_n=GL(n,\mathbb{R})$, $N_n$ be the upper trianguler unipotent subgroup, $P_{n-i}$ be the standard parabolic subgroup associated to partition $n=(n-i,i)$, and finally let $K=O(n)$ be the maximal compact subgroup.

Consider the left coset space $N_n\backslash G_n$ with a right invariant Haar measure, denoted as $dg$ on it. Decompose $N_n\backslash G_n=N_n\backslash P_{n-1} \bullet P_{n-1}\backslash G_n$, correspondingly there is a decomposition of $dg$ in terms of right Haar measures $dp$ on $ N_n\backslash P_{n-1}$ and $dh$ on $P_{n-1}\backslash G_n$, if we write $g=ph$ with $p\in P_{n-1}$ and $h\in P_{n-1}\backslash G_n$

$$\int_{N_n\backslash G_n}f(g)dg=\int_{P_{n-1}\backslash G_n}\int_{N_n\backslash P_{n-1}}f(ph)|det p|^{-1}dpdh$$

The first question I have is that why do we have the factor $|det p|^{-1}$ on the right side? Don't we just have a formula like $$\int_{N_n\backslash G_n}f(g)dg=\int_{P_{n-1}\backslash G_n}\int_{N_n\backslash P_{n-1}}f(ph)dpdh \ \ \ \ ?$$

And what's the formular if we use decomposition $G_n=P_{n-1}K$?

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