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Why is it that the upper half plane $\mathbb{H}$ can be thought of as equivalent to $SL_{2}(\mathbb{R})/SO(2)$?

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You can find this as Proposition 2.1d of Milne's notes. –  Dylan Moreland Oct 3 '11 at 21:16

2 Answers 2

up vote 3 down vote accepted

If $S$ is a set on which a group $G$ acts transitively, then as a $G$-set, $S$ can be identified with $G/\text{Stab}(s)$ for any $s \in S$. This follows from the fact that there is an obvious morphism of $G$-sets

$$G \ni g \mapsto gs \in S$$

which factors through a morphism $G/\text{Stab}(s) \to S$, and by the definition of the stabilizer and transitivity this is an isomorphism.

Now $\text{SL}_2(\mathbb{R})$ naturally acts transitively on $\mathbb{H}$ by hyperbolic isometries. I claim that $\text{Stab}(i) = \text{SO}(2)$ (regarded as a subgroup of $\text{SL}_2(\mathbb{R})$ in the usual way). To see this, write

$$\frac{ai + b}{ci + d} = i \Leftrightarrow ai + b = di - c$$

where $z \mapsto \frac{az + b}{cz + d} \in \text{PSL}_2(\mathbb{R})$. Since $ad - bc = 1, a = d, b = -c$, it follows that $a^2 + b^2 = 1$, so we can write

$$a = \cos \theta, b = \sin \theta$$

for some $\theta$, and then $d = \cos \theta, c = - \sin \theta$; precisely a rotation matrix. Conversely, any such element of $\text{SL}_2(\mathbb{R})$ fixes $i$. So $\mathbb{H}$ can be identified with $\text{SL}_2(\mathbb{R})/\text{SO}(2)$ as a $\text{SL}_2(\mathbb{R})$-space as desired.

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That certainly looks fancier than my suggestion -- which presumably indicates that you can do something useful with this construction, but what? Is $SL_2(\mathbb R)/SO(2)$ to be taken as just a flat collection of cosets, or does any structure carry over from $SL_2(\mathbb R)$ to it? –  Henning Makholm Oct 3 '11 at 22:16
    
@Henning: see mathoverflow.net/questions/19740/… . The description of $\mathbb{H}$ as a coset space is used in the theory of automorphic forms but I'm not familiar with the details. –  Qiaochu Yuan Oct 3 '11 at 22:31

Each member of $SL(2,\mathbb R)/SO(2,\mathbb R)$ is an equivalence class of linear mappings modulo rotations. Choose for each equivalence class the unique member $f$ such that $f(1,0)$ lies on the positive $x$ axis and $f(0,1)$ lies in the upper half plane. Because the determinant must be $1$, knowing the coordinates of $f(0,1)$ allows you to determine the length of $f(1,0)$.

By the way, $SO(2)$ is not normal in $SL(2)$, as far as I can figure out, so I'm not sure how meaningful this construction is. It's not as it it will give you a group structure on the upper half-plane, for example. If you're taking left cosets instead of right ones, you probably need to work with the inverse maps to establish the equivalence.

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