Spherical, hyperbolic and parabolic geometry

Let's consider the space $\mathbb{M}^4$, that is, the manifold $\mathbb{R}^4$ with coordinates $(x,y,z,t)$ but with the pseudo-Riemannian metric given in each tangent space by
$$ds^2=dx^2+dy^2+dz^2-dt^2.$$
This metric is left invariant by the group $O(3,1)$ when acting on $\mathbb{M}^4$. It also leaves invariant the cone
$$t^2-x^2-y^2-z^2=0$$
so is logical that we pay attention to it.

Now, following TRTR from Penrose page 423, we can consider the family of hyperplanes inside $\mathbb{M}^4$ given by
$$z+t+\lambda(t-z)=2$$
intersecting our cone in different 2-dimensional submanifolds $P_{\lambda}$ of $\mathbb{M}^4$.

Here we have a picture from that book, where we consider only $(x,z,t)$ for purposes of drawability.

If we analyse the case $\lambda=0$, we observe that we obtain the submanifold $P_0$, $E$ in the picture with a shape of a parabola, given by the embedding

$$\begin{align*}   \phi \colon \mathbb{R}^2 &\longrightarrow \mathbb{M}^4\\   (u_1,u_2) &\mapsto (u_1, u_2, 1-\frac{u_1^2+u_2^2}{4},1+\frac{u_1^2+u_2^2}{4}) \end{align*}$$

We can look for the (possibly pseudo) metric inherited by $P_0$. Since
$$d\phi=\begin{pmatrix} 1 & 0 \\ 0 & 1\\ -\frac{u_1}{2} &-\frac{u_2}{2}\\ \frac{u_1}{2} &\frac{u_2}{2} \end{pmatrix}$$
we conclude that $P_0$ consists of $\mathbb{R}^2$ with the euclidean metric:
$$g=\begin{pmatrix} 1 & 0 \\ 0 & 1\\ \end{pmatrix}$$

That is, although we see it like a parabola, intrinsically is only the Euclidean plane, and that is the reason the latter is sometimes called parabolic geometry.

But, what happens for others $\lambda$? Let's see. We have a far more complicated parametrization of our embedded manifold:
$$\phi(u_1,u_2)=(u_1,u_2, \frac{-1+{ \lambda }+\sqrt{-(1+{ \lambda })^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)}}{2{ \lambda }},$$
$$\frac{2+{ \lambda }+\sqrt{-(1+{\lambda })^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)+\frac{1-\sqrt{-(1+{\lambda})^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)}}{{ \lambda }}}}{2(1+{ \lambda })})$$
(computations made with Mathematica).

If we compute the (possibly pseudo) metric in this chart we obtain the inherited metric in $\mathbb{R}^2$

$$g_{\lambda}=\begin{pmatrix} \frac{1-{ \lambda } u_2^{2}}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} & \frac{\lambda u_1 u_2}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} \\ \frac{\lambda u_1 u_2}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} & \frac{1-{ \lambda } u_1^{2}}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)}\\ \end{pmatrix}$$

Next step is to show that for $\lambda>0$ we have an isometry into the usual sphere, and that for $\lambda<0$ we have an isometry into the hyperbolic space. This would justify the terms elliptic and hyperbolic for those geometries, as you can see at the picture above ($S$ and $H$ respectively). In order to keep it simple we can work with $\lambda=1$ and $\lambda=-1$.

For $\lambda=1$, we have
$$g=\begin{pmatrix} \frac{1- u_2^{2}}{1-\left(u_1^{2}+u_2^{2}\right)} & \frac{ u_1 u_2}{1-\left(u_1^{2}+u_2^{2}\right)} \\ \frac{u_1 u_2}{1-\left(u_1^{2}+u_2^{2}\right)} & \frac{1- u_1^{2}}{1-\left(u_1^{2}+u_2^{2}\right)}\\ \end{pmatrix}$$

But if we look at a typical chart of the usual sphere in $\mathbb{E}^3$
$$\psi: (x,y)\longmapsto (x,y,\sqrt{1-x^2-y^2})$$
and compute the metric we obtain just
$$g=\begin{pmatrix} \frac{1- y^{2}}{1-\left(x^{2}+y^{2}\right)} & \frac{ xy}{1-\left(x^{2}+y^{2}\right)} \\ \frac{xy}{1-\left(x^{2}+y^{2}\right)} & \frac{1- x^{2}}{1-\left(x^{2}+y^{2}\right)}\\ \end{pmatrix}$$
so they are the same.

And for $\lambda=-1$ we have
$$g=\begin{pmatrix} \frac{1+ u_2^{2}}{1+u_1^{2}+u_2^{2}} & \frac{ -u_1 u_2}{1+u_1^{2}+u_2^{2}} \\ \frac{-u_1 u_2}{1+u_1^{2}+u_2^{2}} & \frac{1+ u_1^{2}}{1+u_1^{2}+u_2^{2}}\\ \end{pmatrix}$$

But if we look at the typical model for hyperbolic plane that consists of the pseudosphere embedded in $\mathbb{M}^3$:
$$x^2+y^2-z^2=-1$$
with a chart given by
$$\psi: (x,y)\longmapsto (x,y,\sqrt{1+x^2+y^2})$$
we obtain the inherited metric in $\mathbb{R}^2$ (keep an eye: inherited from the Minkowski metric, not the Euclidean one):
$$g=\begin{pmatrix} \frac{1+ y^{2}}{1+x^{2}+y^{2}} & \frac{ -xy}{1+x^{2}+y^{2}} \\ \frac{-xy}{1+x^{2}+y^{2}} & \frac{1+ x^{2}}{1+x^{2}+y^{2}}\\ \end{pmatrix}$$
so we are done.