The bi-invariant Haar measure on the quotient $\mathrm{SL}_2(\mathbb{Z}) \backslash \mathrm{SL}_2(\mathbb{R}) / \mathrm{SO}_2(\mathbb{R})$ represents the moduli space of rank two real lattices modulo rotation and is easy to write down. An element of the quotient can be written (using the Iwasawa decomposition) as: \[\begin{pmatrix} y^{1/2} & xy^{-1/2} \\ 0 & y^{-1/2} \end{pmatrix}\] with $z=x+iy$ in a fundamental domain of the action of $\mathrm{SL}_2(\mathbb{Z})$ over the upper-half plane, such as $D = \{(x,y) : x^2+y^2 \geq 1,|x| \leq 1/2,y > 0\}$. In these coordinates, the invariant measure is a scale of the hyperbolic measure, namely \[\frac{3}{\pi} \frac{dx \, dy}{y^2}.\] Is there a similar nice parametrisation for the rank 3 case (namely explicating the bi-invariant Haar measure of $\mathrm{SL}_3(\mathbb{Z}) \backslash \mathrm{SL}_3(\mathbb{R}) / \mathrm{SO}_3(\mathbb{R})$?
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2$\begingroup$ It's: "Iwasawa" $\endgroup$– YCorCommented Feb 19, 2018 at 13:35
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3$\begingroup$ In general for double coset spaces $\Gamma\backslash G/K$ ($G$ unimodular, $K$ compact, $\Gamma$ discrete) the main step is to describe a $G$-invariant Radon measure on $G/K$, and then modding out by $\Gamma$ yields the given space ($G/K$ already known the local form of the measure). The main difficulty is not to describe the measure, but rather describe a fundamental domain. There exists a large literature on this, starting with Siegel. See Peter's answer. $\endgroup$– YCorCommented Feb 19, 2018 at 13:42
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$\begingroup$ It's edited, thanks for noticing and explaining. $\endgroup$– user120980Commented Feb 20, 2018 at 6:58
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$\begingroup$ How do you single out a (family of) measure on this space, which does not carry any obvious group action, as 'Haar measure'? Do you just mean the measure for which $$\int_{\operatorname{SL}_3(\mathbb R)} f(x)\mathrm dx = \int_{\operatorname{SL}_3(\mathbb Z)\backslash{\operatorname{SL}_3(\mathbb R)}/{\operatorname{SO}_3(\mathbb R)}} \Bigl(\int_{\operatorname{SL}_3(\mathbb Z)x{\operatorname{SO}_3(\mathbb R)}} f(y)\mathrm dy\Bigr)\mathrm dx?$$ $\endgroup$– LSpiceCommented May 17, 2019 at 17:08
1 Answer
Sure, and you can even do this for $\mathrm{SL}_n$; I believe this goes back to Siegel. A good hands-on reference is Chapter 1 of Automorphic Forms and $L$-Functions for the group $\mathrm{GL}(n,\mathbb{R})$ by Dorian Goldfeld.
More precisely, we have the Iwasawa decomposition $z = xy$ for $z \in \mathrm{SL}_n(\mathbb{R}) / \mathrm{SO}_n(\mathbb{R})$, where \[x = \begin{pmatrix} 1 & x_{1,2} & x_{1,3} & \cdots & x_{1,n} \\ & 1 & x_{2,3} & \cdots & x_{2,n} \\ & & \ddots & & \vdots \\ & & & 1 & x_{n-1,n} \\ & & & & 1 \end{pmatrix}, \qquad y = \begin{pmatrix} y_1 \cdots y_{n - 1} t & & & \\ & \ddots & & \\ & & y_1 t & \\ & & & t \end{pmatrix} \] where $x_{j,k} \in \mathbb{R}$, $y_{\ell} \in (0,\infty)$, and $t = \prod_{\ell = 1}^{n - 1} y_{\ell}^{\frac{\ell}{n} - 1}$ (so that $\det z = \det x \det y = 1$). Then the Haar measure is \[dz = \prod_{1 \leq j < k \leq n} dx_{j,k} \prod_{\ell = 1}^{n - 1} y_{\ell}^{-\ell(n - \ell) - 1} \, dy_{\ell},\] which gives $\mathrm{SL}_n(\mathbb{Z}) \backslash \mathrm{SL}_n(\mathbb{R}) / \mathrm{SO}_n(\mathbb{R})$ volume \[n 2^{n - 1} \prod_{m = 2}^{n} \frac{\zeta(m)}{\mathrm{vol}(S^{m - 1})},\] where $\zeta(s)$ denotes the Riemann zeta function and $S^{n - 1}$ denotes the $n - 1$-sphere, which has volume $2\pi^{n/2} / \Gamma(n/2)$.
In particular, the volume of $\mathrm{SL}_2(\mathbb{Z}) \backslash \mathrm{SL}_2(\mathbb{R}) / \mathrm{SO}_2(\mathbb{R})$ with respect to the Haar measure $dz = y^{-2} \, dx \, dy$ is \[4 \frac{\zeta(2)}{\mathrm{vol}(S^1)} = \frac{\pi}{3},\] as $\zeta(2) = \pi^2/6$, while the volume of $\mathrm{SL}_3(\mathbb{Z}) \backslash \mathrm{SL}_3(\mathbb{R}) / \mathrm{SO}_3(\mathbb{R})$ is \[\frac{\zeta(3)}{4},\] since $\Gamma(3/2) = \sqrt{\pi}/2$; note that there is no nice closed-form expression for $\zeta(3)$.
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1$\begingroup$ That very clear! Thanks for the reference. $\endgroup$ Commented Feb 20, 2018 at 6:58