# Haar Measure and Automorphic Forms

Let $G = SL_{2}(\mathbb{R})$ and $\Gamma = \Gamma_{0}(N)$. Every element $g =\begin{pmatrix}a & b\\ c& d\end{pmatrix}\in G$ can be written as $$\begin{pmatrix} y^{1/2} & xy^{-1/2} \\ 0 & y^{-1/2}\end{pmatrix}\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$ for some $x, y, \theta$. Therefore we can associate each $g \in G$ with $(x, y, \theta)$ with $x \in \mathbb{R}$, $y > 0$, and $\theta \in [0, 2\pi]$. With $g$ as defined above, $z = x + iy = g(i)$ and $\theta = \arg(ci + d)$. For each $f \in S_{k}(\Gamma)$, define $\phi_{f}(g)$ on $G$ by $\phi_{f}(g) = f(g(i))j(g, i)^{-k}$ where $j(g, i) = (ci + d)(\det g)^{-1/2}$. We consider the Haar measure on $\Gamma$ and $\Gamma\backslash G$.

My question is: Why can we normalize the Haar measure on $G$ through the formula $$\int_{G}\phi_{f}(g)\, dg = \frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^{\infty}\int_{-\infty}^{\infty} \phi_{f}(x, y, \theta)\, \frac{dxdy}{y^{2}}\, d\theta,$$ what is the reasoning behind this formula? Also why does this imply that $$\int_{\Gamma\backslash G} |\phi_{f}(g)|^{2}\, dg = \iint_{F} |f(z)|^{2} y^{k}\frac{dxdy}{y^{2}}$$ where $F$ is the fundamental domain for $\Gamma$ in the upper half plane.

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Does it help if I tell you that the inner integrals $\int_{0}^{\infty}\int_{-\infty}^{\infty}$ correspond to the (essentially unique) translation invariant measure on the upper half plane $\mathbb{H} = G/K$ and the integral over $[0,2\pi]$ corresponds to integrating over the compact stabilizer $K = SO_2$ of $i$? See the linked thread for an outline of the verification of the invariance of the inner two integrals. Notice also that $F = \Gamma\backslash G/K$ from a measure-perspective. – t.b. Oct 9 '11 at 8:50

The book has also a section about the Selberg trace formula (Chapter 9), where they proof a bunch of integral formulas for $SL(2, \mathbb{R})$, but I do not remember, if the above is contained in there.
Lang $SL(2, R)$ is another place, where you might want to look (pg.37).