Isometries of the hyperbolic plane in the Beltrami–Klein disk model I am interested in the isometries of the hyperbolic plane in the Beltrami–Klein disk model. (https://en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model) The Wikipedia article does not say anything about the structure of the isometries in this model.
Since the isometries in the upper half-plane model are well-known, I did a change of variables to go from upper half-plane to the Beltrami–Klein disk. After some calculations, I determined that every isometry in the Beltrami-Klein disk model must be a projective transformation $\mathbb R  \mathbb P^2 \to \mathbb R  \mathbb P^2$ which maps the "unit disk" $\{ [x:y:1] \in \mathbb R \mathbb P^2 ~|~ x^2 + y^2 < 1 \}$ to itself. (The unit disk is, after all, the disk in the Beltrami-Klein disk model of hyperbolic space.)
Some questions I have:


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*Is every projective transformation of this form an isometry of the hyperbolic plane? (Is there an easy way to see this?)

*The set of such projective transformations is a subgroup of $SL(3,\mathbb R)$. Is there a nice characterization of this group?


Update: I just realized that every projective transformation which maps the unit disk to itself must be an element of $SO(2,1)$. Here is the reason: such a projective transformation must map the unit circle to itself, so the corresponding linear transformation $\mathbb R^3 \to \mathbb R^3$ must preserve the quadratic form $x^2+y^2-z^2$. Therefore, if the answer to Question 1 is "yes," then the answer to Question 2 should be $SO(2,1)$.
 A: As you have discovered, the answers to your questions are "yes". You can read about this in Thurston's notes which explains in detail the relations between various models of the hyperbolic plane, including a step-by-step way to get between any two of those models such as the upper half plane model and the Beltrami-Klein disc model.
There is also a more direct but more abstract way to go between those two models. You need a theorem of differential geometry which says that any two complete, simply connected, Riemannian manifolds of dimension 2 and of curvature -1 are isometric. 
One of the two models is the upper half plane $\mathbb{H}$ with its metric $\frac{dx^2+dy^2}{y^2}$. The group $PSL(2,\mathbb{R})$ acts on $\mathbb{H}$ by orientation preserving isometries, and this action is transitive on positively oriented orthonormal frames, hence it is the entire isometry group.
The other one is the hyperboloid model. The underlying space of the hyperboloid model is one sheet of the two-sheeted hyperbola $x^2+y^2-z^2=-1$. The metric on that space is the restriction of the Lorentz matrix $dx^2 + dy^2 - dz^2$: although that is not a Riemannian metric on $x,y,z$ space, when you restrict it to the tangent planes of the one-sheeted hyperboloid it does become a Riemannian metric on that hyperboloid, and it is complete and of sectional curvature $-1$. One can show pretty easy that the action of $SO(2,1)$ restricts to an isometry action on the one-sheeted hyperboloid. That action is transitive on oriented orthonormal frames, hence it is the entire isometry group.
The differential geometry theorem I quoted above then produces an isometry between the upper half plane model and the one-sheeted hyperboloid model (as said, Thurston's notes gives a more explicit construction of an isometry).
Finally, to get from the hyperboloid model to the Beltrami-Klein disc model, one places that disc in the $z=1$ plane, and one uses rays through the origin to project from the one-sheeted hyperboloid to the disc.
A: In addition to Lee Mosher's very nice answer, here are some texts that some of my friends have recommended:


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*John Ratcliffe: Foundations of Hyperbolic Manifolds 

*Cannon, Floyd,
Kenyon, Parry: Hyperbolic Geometry
