Here is some fun about the curvature of curves in the hyperbolic plane. It’s nothing new and well-known to experts, but perhaps isn’t really emphasised when students first learn about hyperbolic geometry. It was a fun thing to see, or be reminded of, I can’t really tell which, when reading a research article recently.
For the purposes of this article, I’m going to assume you know a bit about the hyperbolic plane, and a bit about what curvature is. If so, great. If not, well you may get something out of just reading on anyway.
By curvature here I don’t mean anything very complicated, I just mean the curvature of a curve on a surface. If you’ve taken a first course in differential geometry, like the one I teach at Monash, about curves and surfaces in \( \mathbb{R}^3 \), I’m basically talking about the notion of geodesic curvature. Although, what I’m talking about here is a bit more than that: it’s the the curvature of a curve in the hyperbolic plane, which can’t be globally isometrically embedded in \( \mathbb{R}^3 \). But, it’s basically the same thing. If this means nothing to you, for the purposes of this article that’s probably not a big deal. This article being mostly for the purposes of light education and diversion and entertainment, we won’t use need any rigorous notions or prove anything anyway.
Basically, curvature is just how much a curve curves. A curve with curvature zero “goes straight”, it’s what we call a geodesic. A curvature with high curvature “turns fast”, and will have a “tight turning circle”.
In the Euclidean plane, the one we all know and (maybe?) love from high school geometry (is there any high school geometry any more?), a straight line has curvature \( 0 \), and a circle of radius \( r \) has curvature \( 1/r \). So, the higher your curvature, the lower your radius, and the tighter your turning circle. Indeed, you can imagine the curvature of a curve as how much you’ve turned the steering wheel, as you drive along your curve. If you start from zero curvature, you’re going straight. If you start from a very small but positive curvature, you’re going in a big circle. And as curvature increases, the circle gets smaller.
In the hyperbolic plane, things are a bit more interesting. Let’s pick a point in the hyperbolic plane, and a direction there, to start from. We’ll use the conformal disc model of the hyperbolic plane.
(This and all pictures drawn in Geogebra.)
Well, what happens if we proceed from here with curvature \( 0 \) ? We’ll go along a straight line, or geodesic, of course! We go all the way out to infinity. And if we extend our curve in both directions, we’ll obtain a line starting at one point at infinity, all the way through the plane, through our point, and off to another point at infinity.
So far, so normal. But now, what happens if instead we proceed from that same point, in that same direction, but with a little curvature? Just a tiny amount. Remember that in the Euclidean plane, we’d get a circle: a very large one, since the radius is the reciprocal of the curvature. In the hyperbolic plane, that’s not the case. It’s not a circle!
Instead this is a curve known as a hypercycle. It looks like an arc of a circle in the model. But in fact this is an infinitely long curve in the hyperbolic plane, proceeding from one endpoint at infinity, through our point, and all the way off to another endpoint at infinity. It’s not a circle, it has no radius! But as it’s got nonzero curvature, it’s not a geodesic either. (As it turns out, it is a curve at constant distance from the geodesic connecting its endpoints at infinity.)
We can consider the same idea, starting from our given point and direction, but with bigger values for the curvature. The endpoints of our curve move around the boundary circle of the hyperbolic plane.
And as the curvature increases more, the endpoints of the hypercycle move further around the boundary…
Perhaps you can see, something is going to give when those two endpoints approach each other. Indeed, as it turns out, when the curvature reaches a critical value, our curve is a circle…. sort of. It’s a “circle” tangent to the boundary of the hyperbolic plane. The point of tangency isn’t in the plane, it’s on the boundary, infinitely far away. So what we really have is another infinitely long curve in the hyperbolic plane, beginning and ending at the same point at infinity.
Such a curve is called a horocycle. I like horocycles. I’ve written papers about them (and also about their 3-dimensional counterparts, horospheres.) As it turns out, the curvature of any horocycle is precisely \( 1 \). So all the hypercycles occurred when our curvature was between \( 0 \) and \( 1 \).
What happens if we now increase the curvature to be more than \( 1 \) ? As you might suspect, the “circle” now detaches from the boundary of the hyperbolic plane, and becomes a bona fide circle. It looks like a (Euclidean) circle, and it’s an actual hyperbolic circle too.
So we observe the fun fact that, in the hyperbolic plane, all circles have curvature greater than \( 1 \). As the curvature increases, the circle gets smaller, its radius decreases. And just as in Euclidean geometry, as the curvature goes to infinity, the radius goes to zero.
The precise relationship between the curvature \( \kappa \) of the circle, and its radius \( r \) (by which I mean the actual hyperbolic distance from its centre to its boundary, which will be very different from what the radius appears to be in the disc model) is given by the equation \[ \kappa = \frac{1}{\tanh r}.\]
(Here’s a math.stackexchange discussion about it. It must be in some books somewhere, but I can’t remember a reference off the cuff. If you know one let me know!) That is, the curvature is given by the reciprocal of the hyperbolic tangent function \( \tanh r \), sometimes known as the hyperbolic cotangent function \( \coth r \). Written in terms of exponentials, this is \[ \kappa = \frac{e^r + e^{-r}}{e^r – e^{-r}}. \] From which you can observe algebraically, the numerator is always greater than the denominator, so the curvature is always more than \( 1 \)…