Affine Functions, Möbius and Hyperbolic

Question

I haven't been able to figure out the relationship between objects in an Exercise of Do Carmo's book on Riemannian Geometry and this may be a bit vague, but I am posting out of despair.

I want to understand how the hyperbolic metric arises on the upper half plane $\mathbb{H}$, but struggle to understand why Do Carmo starts with $g(t) = yt +x$.

The following steps aim to make my question more precise, but not well defined:

1. The functions $f(t) = yt +x$ ($x,y \in \mathbb{R}, y>0$) define a group under composition, which can be understood as the upper half plane.

2. Under composition, these functions define a group. We use this to define a left invariant metric, that at $e = (0,1)$ coincides with the usual scalar product on $\mathbb{R}^2$ ($\langle u,v \rangle_{(f_2t+f_1)} = f_2^{-2}\langle u, v \rangle_e$).

3. So far, we have $(G,\circ)$ and $\mathbb{R} \times \mathbb{R}_{>0}$ and our defined metric.

4. Now, in part (b) of the Exercise, we add the complex upper half plane (which I write $\mathbb{H}$). It inherits complex multiplication and addition. We are asked to prove that $f_1 + if_2 = z \mapsto \frac{az + b}{cz + d}$ is an isometry of $G$.

Q: Why introduce the metric $g_{ij} = \delta_{ij}/y^2$ through these affine functions ($G,\circ)$?

It seems Do Carmo motivates the appearance of this metric as "the one left-invariant under composition of real affine Transformations" and then drops it and uses it on the complex upper half plane, which has a completely different structure to these affine transformations.

Through my approaches so far:

• I've had the Idea of looking at the equivalence relation $(f,g) \sim (\lambda f, \lambda g), \lambda \in \mathbb{C}^\times$. In this case: $(f_2t + f_1, g_2t + g_1) \sim (\frac{f_2t + f_1}{g_2t + g_1},1)$ which looks like a Möbius transformation, only $t \in \mathbb{R}$ in the Exercise. This didn't satisfy me but looks promising as $\mathrm{SL}_2(\mathbb{Z})$ is very close (acting on each component).
• If we add $z \mapsto -1/z$ we can then express the Möbius transformation as a composition of transformations of $G$. But then, why start with $G$ first? (random guess, Does adding this inversion correspond to adding a point at infinity?)
• In euclidian spaces, such affine transformations are our "change of coordinates", which leave the metric invariant. Here, we show that the Möbius transormations in a more general setting when we add that inversion.
• These actions seem to give us a way to "straighten" two points to a "line".

Pointing in some direction for me to investigate would make me happy.

The context is a talk that I will give on closed geodesics in $\mathbb{H}/\mathrm{SL}_2(\mathbb{Z})$ and their relationship to class Numbers.

P.S. Having solved the Exercise, the discussion is not meant to be about a specific solution. Just the motivation behind the structure of the exercise.

Answer 1

To my mind this is a wonderful and weird coincidence that I cannot really explain, but which, as time passes, becomes more and more interesting to me (it has creeped into my most recent work-in-progress on the Dehn function of the group $\text{Aut}(F_n)$.) All I can suggest mathematically is that you work through the exercise.

But I suppose there are two ways that I look at this coincidence from an intuitive perspective, depending on whether you come at this first from the Riemannian geometry point of view, or first from the Lie group point of view.

And, by the way, I'll take the point of view that your Lie group is the semidirect product $\mathbb R \rtimes \mathbb R_>$, with respect to the homomorphism $\mathbb R_> \mapsto \text{Aut}(\mathbb R)$ that takes $t \in \mathbb R_>$ to the operation of multiplication by $t$ on $\mathbb R$.

One way goes like this (and this is how I first ran across this coincidence):

Okay, here's the hyperbolic plane $\mathbb H$. It's pretty remarkable. It has an interesting upper half plane model. It has constant negative curvature. Its group of orientation preserving isometries is the group $\text{PSL}(2,\mathbb R)$ acting by fractional linear transformations. That action is transitive and isotropic, although not free. Hmm... the isometry group of $\mathbb H$ has an interesting subgroup, namely the subgroup that fixes the point at infinity of the upper half plane. Hmm... that subgroup is a semidirect product $\mathbb R \rtimes \mathbb R_>$. OMG... that subgroup acts transitively and freely on $\mathbb H$... Wait, doesn't this sort of mean that $\mathbb H$ IS this Lie group, in some weird way?

The other way, which is my own souped up interpretation of what Do Carmo does, goes like this:

Okay, here's this Lie group $\mathbb R \rtimes \mathbb R_>$. Like every Lie group, it has a left invariant metric that acts freely and transitively. So the Lie group itself is a subgroup of its own isometry group... hmm... I wonder what it's full group is isometries is ... OMG the full group of isometries not only acts transitively, it acts isotropically!! OMG the full group of isometries is the same as $\text{PSL}(2,\mathbb R)$ acting on $\mathbb H$! Wait... doesn't that sort of mean that this Lie group IS $\mathbb H$, in some weird sense?

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One last comment: I actually stumbled across this coincidence in an even more backhanded way. There's a very cool theory of homogeneous Riemannian manifolds of negative curvature, relating them to certain Lie groups. I first learned about this in 3-dimensions: there's $\mathbb H^3$ but then there's a whole bunch of other ones that are not isotropic and that turn out to be solvable Lie groups of the form $\mathbb R^2 \rtimes \mathbb R_>$, using different actions of $\mathbb R_>$ on $\mathbb R^2$.

Only later did I realize that there's just one example in dimension 2, which is the topic of this whole post.