### Euclidean, Hyperbolic and Elliptic Geometry

#### Posted by John Baez

There are two famous kinds of non-Euclidean geometry: hyperbolic geometry and elliptic geometry (which almost deserves to be called ‘spherical’ geometry, but not quite because we identify antipodal points on the sphere).

In fact, these two kinds of geometry, together with Euclidean geometry, fit into a unified framework with a parameter $s \in \mathbb{R}$ that tells you the curvature of space:

when $s \gt 0$ you’re doing elliptic geometry

when $s = 0$ you’re doing Euclidean geometry

when $s \lt 0$ you’re doing hyperbolic geometry.

This is all well-known, but I’m trying to explain it in a course I’m teaching, and there’s something that’s bugging me.

It concerns the precise way in which elliptic and hyperbolic geometry reduce to Euclidean geometry as $s \to 0$. I know this is a problem of deformation theory involving a group contraction, indeed I know all sorts of fancy junk, but my problem is fairly basic and this junk isn’t helping.

Here’s the nice part:

Give $\mathbb{R}^3$ a bilinear form that depends on the parameter $s \in \mathbb{R}$:

$v \cdot_s w = v_1 w_1 + v_2 w_2 + s v_3 w_3$

Let $SO_s(3)$ be the group of linear transformations $\mathbb{R}^3$ having determinant 1 that preserve $\cdot_s$. Then:

when $s \gt 0$, $SO_s(3)$ is isomorphic to the symmetry group of elliptic geometry,

when $s = 0$, $SO_s(3)$ is isomorphic to the symmetry group of Euclidean geometry,

when $s \lt 0$, $SO_s(3)$ is isomorphic to the symmetry group of hyperbolic geometry.

This is sort of obvious except for $s = 0$. The cool part is that it’s still true in the case $s = 0$! The linear transformations having determinant 1 that preserve the bilinear form

$v \cdot_0 w = v_1 w_1 + v_2 w_2$

look like this:

$\left( \begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ a & b & 1 \end{array} \right)$

And these form a group isomorphic to the Euclidean group — the group of transformations of the plane generated by rotations and translations!

So far, everything sounds pleasantly systematic. But then things get a bit quirky:

**Elliptic case.**When $s \gt 0$, the space $X = \{v \cdot_s v = 1\}$ is an ellipsoid. The 1d linear subspaces of $\mathbb{R}^3$ having nonempty intersection with $X$ are the**points**of elliptic geometry. The 2d linear subspaces of $\mathbb{R}^3$ having nonempty intersection with $X$ are the**lines**. The group $SO_s(3)$ acts on the space of points and the space of lines, preserving the obvious incidence relation.

Why not just use $X$ as our space of points? This would give a sphere, and we could use great circles as our lines—but then distinct lines would always intersect in *two* points, and two points would not determine a unique line. So we want to identify antipodal points on the sphere, and one way is to do what I’ve done.

**Hyperbolic case.**When $s \lt 0$, the space $X = \{v \cdot_s v = -1\}$ is a hyperboloid with two sheets. The 1d linear subspaces of $\mathbb{R}^3$ having nonempty intersection with $X$ are the**points**of hyperbolic geometry. The 2d linear subspaces of $\mathbb{R}^3$ having nonempty intersection with $X_s$ are the**lines**. The group $SO_s(3)$ acts on the space of points and the space of lines, preserving the obvious incidence relation.

This time $X$ is hyperboloid with two sheets, but my procedure identifies antipodal points, leaving us with a single sheet. That’s nice.

But the obnoxious thing is that in the hyperbolic case I took $X$ to be the set of points with $v \cdot_s v = -1$, instead of $v \cdot_s v = 1$. If I hadn’t switched the sign like that, $X$ would be the hyperboloid with one sheet. Maybe there’s a version of hyperbolic geometry based on the one-sheeted hyperboloid (with antipodal points identified), but nobody seems to talk about it! Have you heard about it? If not, why not?

Next:

**Euclidean case.**When $s = 0$, the space $X = \{v \cdot_s v = 1\}$ is a cylinder. The 1d linear subspaces of $\mathbb{R}^3$ having nonempty intersection with $X$ are the**lines**of Euclidean geometry. The 2d linear subspaces of $\mathbb{R}^3$ having nonempty intersection with $X$ are the**points**. The group $SO_s(3)$ acts on the space of points and the space of lines, preserving their incidence relation.

Yes, any point $(a,b,c)$ on the cylinder

$X_0 = \{(a,b,c) : \; a^2 + b^2 = 1 \}$

determines a line in the Euclidean plane, namely the line

$a x + b y + c = 0$

and antipodal points on the cylinder determine the same line. I’ll let you figure out the rest, or tell you if you’re curious.

The problem with the Euclidean case is that points and lines are getting switched! Points are corresponding to certain 2d subspaces of $\mathbb{R}^3$, and lines are corresponding to certain 1d subspaces.

You may just tell me I just got the analogy backwards. Indeed, in elliptic geometry every point has a line orthogonal to it, and vice versa. So we can switch what counts as points and what counts as lines in that case, without causing trouble. Unfortunately, it seem for hyperbolic geometry this is not true.

There’s got to be some way to smooth things down and make them nice. I could explain my favorite option, and why it doesn’t quite work, but I shouldn’t pollute your brain with my failed ideas. At least not until you try the exact same ideas.

I’m sure someone has figured this out already, somewhere.

## Re: Euclidean, Hyperbolic and Elliptic Geometry

Are you sure that it’s the space of

all2d linear subspaces of $\mathbb{R}^3$ having nonempty intersection with $X$ (which is all of them) that gives you the points of Euclidean geometry?A geometric way to say “a point $(a,b,c)$ on the cylinder determines a line $a x + b y + c = 0$” is to take the line from the origin through that point, take the plane through the origin perpendicular to it, and then intersect that plane with the plane $z=1$ (which we regard as our “Euclidean plane”). The unique line through the origin not intersecting the cylinder is $x=y=0$, which is perpendicular to the plane $z=0$, which doesn’t intersect the plane $z=1$.

The corresponding thing for a 2d linear subspace (a plane through the origin) would be to take the line through the origin perpendicular to it and intersect that line with $z=1$; then a line through the origin lies on a plane through the origin exactly when the corresponding line in $z=1$ contains the corresponding point in $z=1$. But there are plenty of planes through the origin whose orthogonal complement doesn’t intersect $z=1$ at all, namely those containing $z=0$ — yet these planes

dointersect the cylinder, each in a pair of parallel lines (whereas all the other planes intersect the cylinder in an ellipse).Of course, this is just the usual way to identify $z=1$ with the finite points of $\mathbb{R}P^2$.