In Euclidean geometry the curvature of a line or curve in the reciprocal of the radius of the kissing circle
In hyperbolic geometry we cannot use this as measure of curvature (what is the curvature of an hypercycle?)
What measure can we use instead as measure of curvature?
Best Answer
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: