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How does one parametrize those spaces in order to do integration over them?

What's a good reference for doing integral a with Haar measures over matrix groups?

BigM
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2 Answers2

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An element of $SL_2(\mathbb{R})$ can be written as ($NAK$) $$ \left(\begin{array}{cc} 1&x\\ 0&1\\ \end{array} \right) \left(\begin{array}{cc} y^{1/2}&0\\ 0&y^{-1/2}\\ \end{array} \right) \left(\begin{array}{cc} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\ \end{array} \right) $$ with $z=x+iy$ in the upper half-plane. In these coordinates, the invariant measure is $$ d\theta \frac{dx dy}{y^2}, $$ the product of the invariant measures on the two subgroups, hyperbolic area (upper half-plane model) and arclength on the circle. See Lang's book $SL_2(\mathbb{R})$ for more details.

yoyo
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You can parameterize $PSL(2,\mathbb{R})$ by its action on the unit tangent bundle of the upper half-space model $\mathbb{H}$ of the hyperbolic plane. Under this parameterization, Haar measure is the Liouville measure.

You can then get the Haar measure on $SL(2,\mathbb{R})$ by pulling back under the double covering map $SL(2,\mathbb{R}) \mapsto PSL(2,\mathbb{R})$.

Added: To get the Haar measure on $SL(2,\mathbb{R}) / SL(2,\mathbb{Z})$, you use the orbit map $$f : SL(2,\mathbb{R}) \mapsto SL(2,\mathbb{R}) \, / \, SL(2,\mathbb{Z}) = Q $$ This map $f$ is a regular covering map with deck transformation group $SL(2,\mathbb{Z})$ acting by multiplication from one side (the notation should perhaps be $Q = SL(2,\mathbb{Z}) \, \backslash \, SL(2,\mathbb{R})$ because usually one thinks of the orbits as defined by left multiplication of elements of $SL(2,\mathbb{Z})$, and in that case one should be using the left invariant Haar measure). Therefore, $Q$ has a basis consisting of all open subsets $U \subset Q$ that are evenly covered by $f$.

For each such $U$, one uses the definition of an even covering to write $f^{-1}(U)$ as a disjoint union $$f^{-1}(U) = \cup V_i $$ such that for each $i$ the restriction $f \bigm| V_i$ is a homeomorphism from the open subset $V_i \subset SL(2,\mathbb{R})$ onto $U$. Since the covering is a regular covering, and since Haar measure on $SL(2,\mathbb{R})$ is invariant under the deck transformation subgroup $SL(2,\mathbb{Z})$, all of the $V_i$'s have the same Haar measure, and one defines the Haar measure of $U$ to equal the Haar measure of any of the $V_i$'s. Then there's a formal process in measure theory that one goes through, to extend this to a Haar measure defined on all Borel subsets of $Q = SL(2,\mathbb{R}) \, / \, SL(2,\mathbb{Z})$.

Lee Mosher
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    What about the quotient with a lattice $\Gamma$ like $SL(2, Z)$? –  Aug 09 '15 at 23:31
  • @user98246: I added an explanation. – Lee Mosher Aug 11 '15 at 18:06
  • Actually, for measures/integrals on quotients, there is a general existence-and-uniqueness for (e.g.) discrete subgroups $\Gamma$ of (e.g., unimodular) topological groups $G$, that there is a unique $G$-invariant measure/integral on $G/\Gamma$ such that for $f\in C^o_c(G)$, $\int_{G/\Gamma}\sum_{\gamma\in\Gamma} f(g\gamma),dg=\int_G f(g),dg$. No need to partition or fool with cosets, really. – paul garrett Jun 18 '20 at 21:48