Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is maybe a dumb question. $SL_2(\mathbb{R})$ has a natural action on the upper half plane $\mathbb{H}$ which is transitive with stabilizer isomorphic to $SO_2(\mathbb{R})$. For this reason, people sometimes write $\mathbb{H}$ as the coset space $SL_2(\mathbb{R})/SO_2(\mathbb{R})$.

Now, it's clear how this description recovers the topology of $\mathbb{H}$: it's just the quotient topology. But can you recover either the Riemann surface structure or the hyperbolic metric on $\mathbb{H}$ from this description? How much of the structure of $SL_2(\mathbb{R})$ and $SO_2(\mathbb{R})$ do you need to do this, if it's possible?

share|improve this question
3  
@Qiaochu: that's a highly non-dumb question! It comes up in the theory of moduli spaces. It's not hard to check that the points of SL_2(Z) \ H parametrise elliptic curves over the complexes, but why should such a set be a complex manifold? I think there's a very fancy answer to do with variation of Hodge structures but I don't think I ever understood the details of that point of view well enough to be able to explain them :-( . I somehow feel that Deligne's axioms for a Shimura variety should somehow help, but on some level I've never understood these either :-( –  Kevin Buzzard Mar 29 '10 at 17:49
    
"t's not hard to check that the points of SL_2(Z) \ H parametrise elliptic curves over the complexes, but why should such a set be a complex manifold?" I am not sure if I understood you correctly. What's wrong with the usual proof (say, in Silverman's Advanced Topics in Elliptic Curves") that shows that after giving the appropriate charts and adding an extra point at infinity (one needs to be careful about the elliptic points) this is biholomorphic to the Riemann sphere? –  Idoneal Mar 29 '10 at 18:21
1  
Sorry Idoneal, I didn't explain myself well. I claim that there is a canonical bijection between the set of isomorphism classes of elliptic curves over the complexes, and SL_2(Z) \ H. If you're prepared to believe that H is a complex manifold, then this set becomes a complex manifold. However if you didn't know H existed, and just had a set of isomorphism classes of elliptic curves, however would you put a complex structure on it? It can be done! You need to consider holomorphic families of ell curves (i.e. maps M-->S of cx mfds whose fibres are ell curves)... –  Kevin Buzzard Mar 29 '10 at 18:27
1  
...and the point is that you want M-->S to be a holomorphic family of elliptic curves iff the induced map from the complex manifold S to the set of iso classes of ell curves (sending s to the fibre above s) is holomorphic. This gives you the structure of a complex manifold on SL_2(Z) \ H. But now one has to prove that it's the same one as the one coming from the upper half plane, and this genuinely needs proof. –  Kevin Buzzard Mar 29 '10 at 18:28
1  
@Mariano: it's not obvious that SL_2(R)/SO_2(R) has a complex structure, and it's not obvious that SL_2(Z)\SL_2(R)/SO_2(R) has a complex structure, but I'm sure you can see that one has a complex structure iff the other one does. He asked about the first but I mentioned the second because I am pretty sure he knows SL_2(Z)\H is a moduli space and I'm less sure if he knows that H is (it's also another moduli space but it's of elliptic curves plus more structure). –  Kevin Buzzard Mar 29 '10 at 20:27
show 2 more comments

3 Answers 3

up vote 3 down vote accepted

Edit: I should have put a short version of the answer in the beginning, so here is how the various structures are recovered. To get a smooth manifold structure on the quotient, you use the fact that $SL_2(\mathbb{R})$ is a real Lie group and $SO_2(\mathbb{R})$ is a closed subgroup. To get a hyperbolic structure, you use the fact that $SL_2(\mathbb{R})$ is isomorphic to an orthogonal group of signature (n,1) for some n (giving a transitive action on hyperbolic n-space). To get a complex structure, you use the fact that $SL_2(\mathbb{R})$ is isomorphic to an orthogonal group of signature (2,m) for some m (giving an action on a hermitian symmetric space).

As others have noted, you can get a bijection on points using the Iwasawa decomposition, and you can get a hyperbolic structure using the exceptional isomorphism $PSL_2(\mathbb{R}) \cong SO_{2,1}^+(\mathbb{R})$. First, I'd like to clean up the Iwasawa treatment a bit. Any element of $SL_2(\mathbb{R})$ can be uniquely decomposed as BK, where K is a rotation and B is upper triangular with positive diagonal. Any rotation K fixes i, so we should consider what elements B do. A bit of fiddling shows that $\begin{pmatrix} \sqrt{y} & x/\sqrt{y} \\ 0 & 1/\sqrt{y} \end{pmatrix} \cdot i = x+iy$.

We can view the exceptional isomorphism in another way that makes the complex structure more apparent, by viewing the hyperbolic plane as the Grassmannian $O_{2,1}(\mathbb{R})/(O_2(\mathbb{R}) \times O_1(\mathbb{R}))$. From the standpoint of special relativity, this is the space of timelike lines through the origin in $\mathbb{R}^{2,1}$. Taking a quotient of the total space of these lines (minus origin) by positive rescaling, we find that this space is isomorphic to the space of pairs of antipodal points of norm -1. In particular, we have an isomorphism of the Grassmannian with the quotient of the hyperboloid with two sheets (i.e., solutions of the equation $x^2 + y^2 - z^2 = -1$) by the antipodal automorphism.

One way to explain the origin of the complex structure is by the fact that all Grassmannians of the form $O(2,n)/(O(2) \times O(n))$ are hermitian symmetric spaces, and the hyperbolic plane is just the case $n=1$. The 2 in $O(2)$ is essential, because the orthogonal group action is what yields the ninety degree rotation in the tangent space of any point, and this is what endows the quotient with an almost complex structure. If you want to see more about hermitian symmetric spaces than the Wikipedia blurb, I recommend looking in chapter 1 of Milne's introduction to Shimura varieties.

Finally, I'd like to point out Deligne's description of the upper half plane as a moduli space of structured elliptic curves. Points on H parametrize elliptic curves with an oriented basis of first homology (as mentioned a few times in our class). If you want to say it is a fine moduli space, you need a functor that it represents, and it is unfortunately a bit complicated. The functor takes as input the category of complex analytic spaces, and for any such space S, it gives the set of isomorphism classes of elliptic curves over S (i.e., diagrams $E \underset{\pi}{\leftrightarrows} S$ of complex analytic spaces, where $\pi$ is smooth and proper with one-dimensional genus one fibers and the leftward map is a section) equipped with an isomorphism $R^1\pi_*\underline{\mathbb{Z}} \cong \underline{\mathbb{Z} \times \mathbb{Z}}$ that induces the canonical identity $R^2\pi_*\underline{\mathbb{Z}} \cong \underline{\mathbb{Z}}$ on exterior squares. Here, the underscore indicates a constant sheaf. The functor also takes morphisms to "the evident diagrams". To be honest, I have never seen a complete proof that this functor is represented by the complex upper half plane, although it seems to be more a question of doing lots of writing than an honest theoretical problem. You can probably do it using the fact that H is a classifying space of polarized Hodge structures, as Kevin Buzzard mentioned in the comments.

share|improve this answer
add comment

Its easy to check that every matrix in $SL_2(\mathbb{R})$ can be wrtten uniquely as $\begin{pmatrix} \lambda & \alpha \\ 0 & \lambda^{-1}\end{pmatrix}\begin{pmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{pmatrix}$ where $\lambda>0$. This is exactly written then as a coset representative of the quotient you wrote down above. You can arrive at the hyperbolic metric by following the definition of the pushforward metric from a left invariant metric on $SL_2(\mathbb{R})$. Notice $(\alpha,\lambda)$ is a point in the upperhalf plane.

The latex misbehaved, those should be $2\times 2$ matrices.

share|improve this answer
    
Thanks for your answer! What's an example of a left-invariant metric on SL_2(R)? Does it matter which one I choose if there are more than one? –  Qiaochu Yuan Mar 29 '10 at 17:14
1  
Quaochu, pick any metric on the Lie algebra---that is, the tangent space at the identity element---and translate it. –  Mariano Suárez-Alvarez Mar 29 '10 at 17:26
    
Cool. So there remains another question which is maybe better asked as a separate question: what is the connection between the metric structure on H and the Riemann surface structure? –  Qiaochu Yuan Mar 29 '10 at 18:15
2  
A Riemann surface structure is the same thing as a conformal class of metric structures. –  Mariano Suárez-Alvarez Mar 29 '10 at 18:49
    
In general left-invariant metrics on a Lie group, even a simply connected one, can be non-isometric -- Tam Nguyen Phan showed me a nice example of different left-invariant metrics on S^3. But the pushforward metric on SL(2,R)/SO(2) is homogeneous (acted on transitively by isometries), and thus of constant curvature, and so is either S^2, R^2, or H^2. You can then check (in any number of ways) that it's not flat, so must be H^2. –  Tom Church Mar 29 '10 at 20:21
add comment

There is a very physical picture to this, if you are willing to work with the disk model of hyperbolic space, instead of the upper half plane, to which it is related by an isometry.

The Lie group $\mathrm{SL}(2,\mathbb{R})$ is a double cover of the identity component $\mathrm{SO}_0(2,1)$ of $\mathrm{O}(2,1)$, which is the Lorentz group in 3 dimensions. In other words, $\mathrm{O}(2,1)$ is the subgroup of $\mathrm{GL}(3,\mathbb{R})$ which preserves a symmetric inner product $\eta$ of signature $(2,1)$: $$\eta = \begin{pmatrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & -1 \end{pmatrix}.$$

Now consider the two-sheeted hyperboloid in $\mathbb{R}^3$ defined by $x^2 + y^2 - z^2 = -1$. The upper sheet -- let's call it $\mathbb{D}$ -- with $z>0$ is topologically a disk. It inherits a riemannian metric from the Minkowski metric on $\mathbb{R}^3$ defined by $\eta$. The identity component of $\mathrm{O}(2,1)$ acts on $\mathbb{D}$ as isometries.

The isotropy at the point $(0,0,1)$ consists of rotations in the $x,y$-plane, whence it is isomorphic to $\mathrm{SO}(2)$. Hence $\mathbb{D} = \mathrm{SO}_0(2,1)/\mathrm{SO}(2)$.

Notice that it is is $\mathrm{SO}_0(2,1)$ (a.k.a. $\mathrm{PSL}(2,\mathbb{R})$) which acts effectively on $\mathbb{D}$ and not $\mathrm{SL}(2,\mathbb{R})$.


Added I forgot to relate the disk to the upper half plane. If you think of $\mathbb{D}$ as the unit disk in the complex plane, then the map $\mathbb{D} \to \mathbb{H}$ is given by the following Möbius transformation: $$ z \mapsto \frac{z-i}{z+i}$$

share|improve this answer
    
I like this explanation very much - I remember Scott Carnahan explaining this to us once - but one has to produce the action of O(2, 1) on a "unit ball" in order to do this. Given only the abstract structure of O(2, 1) (in whatever category is necessary), can one recover its action on Minkowski space? –  Qiaochu Yuan Mar 29 '10 at 18:33
    
What is O(2,1) except the linear isometries of Minkowski space? –  Tom Church Mar 29 '10 at 20:00
    
I guess I'm not explaining myself well. Suppose I'm given a nice topological group G and a nice subgroup H. Then G/H is a topological space. What are general situations in which G/H comes equipped with extra geometric structure, and how much about G and H do I have to know to find that structure? (In other words, is it enough to have a black box which tells you group-theoretic things about G and H such as their other subgroups?) –  Qiaochu Yuan Mar 29 '10 at 20:49
2  
Yes, if you include the structure of Lie algebras as part of your group theoretic things. When I was a student I really enjoyed Helgason's Differential Geometry, Lie Groups and Symmetric Spaces, and Joe Wolfe's Spaces of Constant Curvature. –  Charlie Frohman Mar 29 '10 at 21:05
1  
The adjoint representation of O(2,1) on its Lie algebra preserves the Killing form, which has signature (2,1). This is as canonical as it gets. This is very particular to this signature, though. –  José Figueroa-O'Farrill Mar 29 '10 at 21:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.