# When can I conclude two models of the hyperbolic plane are isomorphic?

I have a question regarding the hyperbolic plane. As is well known, there are various models in use for describing this space.

Right now I'm considering the 'hyperboloid model' and the 'Poincaré upper-half model' for the hyperbolic plane.

For the first one, we consider a Lorentzian metric $ds^2 = (dx^0)^2 - (dx^1)^2 - (dx^2)^2$, and define the hyperbolic plane as $$\mathcal{H}^2 := \left\{ (x^0, x^1, x^2) \in \mathbb{R}^{3} \mid (x^0)^2 - (x^1)^2 - (x^2)^2 =1, x^0 > 0 \right\}.$$ In the literature, this is called like the 'forward sheet' of the hyperboloid, because the equation $(x^0)^2 - (x^1)^2 - (x^2)^2 =1$ describes a two-sheeted hyperboloid, and we only want the one with $x^0 > 0$. I have found out that the group of isometries of this space is $SO^{+} (1,2)$, i.e. the proper ortochronous Lorentz matrices.

Then, we also have the other model (of Poincaré), which is defined as $$\mathbb{H}^2 := \left\{(x,y) \in \mathbb{R}^2 \mid y > 0 \right\}.$$ The group of isometries here is described by linear fractional transformations (Möbius transformations), and it is given by $PSL(2, \mathbb{R})$.

Now, my question is: suppose I have proven the groups of isometries are the same, i.e. that $PSL(2, \mathbb{R}) \cong SO^{+} (1,2)$, can I conclude from this that the spaces (described by these models) are also isomorphic?

My tutor said you cannot conclude this, and he gave the following example: the sphere $S^n(r)$ for different radii $r$ has the same isometries, but the spaces are not isomorphic (you need to rescale it).

So what extra conditions (local or global) would I have to assume in order to obtain an isomorphism?

Thank you kindly for any help.

• You mean Isometric? Observe that isometry preserve (Gaussian) curvature. Commented Dec 1, 2017 at 12:28
• The usual isometry goes through the unit disk model. Commented Dec 1, 2017 at 14:53

## 1 Answer

You can use the following theorem of differential geometry:

If $X,Y$ are $n$-dimensional, complete, simply connected, Riemannian manifolds with constant sectional curvatures $\kappa_X$, $\kappa_Y$ respectively, then $X$ is isometric to $Y$ if and only if $\kappa_X=\kappa_Y$.