Curvature of curves on the hyperbolic plane

Solution 1:

The natural notion to use is Geodesic curvature which makes sense for curves on any Riemannian manifolds. The name comes from the fact that geodesics have zero curvature.

For example, on the hyperbolic plane with Gaussian curvature $-1$, horocycles have geodesic curvature $1$. Indeed, let's consider Poincaré half-plane model with metric $(dx^2+dy^2)/y^2$. The line $y=1$ is a horocycle that is naturally parametrized by arclength, $\alpha(t) = (t, 1)$. The tangent vector is the unit vector that points to the right. This makes it appear as if the horocycle doesn't curve but we should use parallel transport to judge whether two vectors are parallel.

Take two points $A=(t, 1)$ and $B=(t+h, 1)$ on the horocycle and draw a geodesic between them: it's an arc of a circle with Euclidean radius $\sqrt{1+(h/2)^2}$. The tangent vector to $\alpha$ makes the angle $\sin^{-1}(h/2)$ with the geodesic at both $A$ and $B$, but it's in opposite directions. So, transporting the vector $\alpha'(t)$ from $A$ to $B$ along the geodesic, we see that at the point $B$ it makes the angle $2\sin^{-1}(h/2)$ with $\alpha'(t+h)$. Since the unit tangent rotates by $2\sin^{-1}(h/2)$ over the distance $h$, the geodesic curvature is $$ \lim_{h\to 0} \frac{2\sin^{-1}(h/2)}{h} = 1 $$

Disk model

On second thought, it's easier to use the disk model; the metric will be $4ds^2/(1-x^2-y^2)^2$ so the curvature is still $-1$. The diameter $(-1,1)\times \{0\}$ is a geodesic, and near the center $(0,0)$ its arclength parameterization moves approximately as $t\mapsto (t/2,0)$ when $t\approx 0$. So the parallel transport along this geodesic for small distances near center will be Euclidean, which implies that the geodesic curvature of any curve tangent to this geodesic at $(0,0)$ will be just $1/2$ of its Euclidean curvature at that point. (Here $1/2$ comes from the aforementioned speed of parameterization).

Summary: to compute geodesic curvature in the hyperbolic disk model, move the point of interest to the center by a Möbius transformation, and take $1/2$ of Euclidean curvature there. Examples:

• Horocycles have geodesic curvature $1$, as shown by the horocycle $x^2+(y-1/2)^2=1/4$ that passes through $(0,0)$ and has Euclidean curvature $2$.
• A hyperbolic circle of hyperbolic radius $R$ has geodesic curvature $1/\tanh R \in (1,\infty)$. Indeed, when such a circle passes through $(0,0)$, its point furthest from $(0,0)$ is at hyperbolic distance $2R$ from $(0,0)$. Solving $2\tanh^{-1} d = 2R$ yields $d=\tanh R$ for the Euclidean diameter of this circle, so its Euclidean curvature is $2/\tanh R$ and geodesic curvature is $1/\tanh R$

Solution 2:

Using Hyperbolic geometry to calculate the geodesic curvature of a circle

If the circle has centre M and a radius $r$

Imagine 2 points on the circle $A$ and $B$

let $C$ be the midpoint of the segment (chord) $AB$

Then the triangle $\triangle ACM$ is a right angled triangle (right angle is $\angle ACM$) and $\angle M = \angle AMC = 1/2 \angle AMB $

For this triangle we have the following formula's

• $ \tan(\angle CAM) = \cot(\angle M) / \cosh(r) $
• $ \sinh(AC) = \sin(\angle M) \sinh(r) $
• $ \tanh(MC) = \cos(\angle M) \tanh(r) $ (not needed)

The angle between the tangent at A and $AB$ is $$ \frac{\pi}{2} -\angle CAM = \frac{\pi}{2} -\arctan \left( \frac{\cot(\angle M)}{\cosh(r)} \right) = \arctan \left( \tan ( \angle M)\cosh(r) \right) $$

And the curvature is $ \lim_{\angle M \to 0} \frac{\arctan \left( \tan(\angle M)\cosh(r)\right)} {\sin(\angle M) \sinh(r)} $

Because $ \lim_{\angle M \to 0} \frac{ \arctan \left( (\tan(\angle M) C \right) }{\sin(\angle M)} = C $

The limit is $\frac{\cosh(r)}{\sinh(r)} =\frac{1}{\tanh(r)}$