Geodesic curvature of a curve in the hyperbolic plane

Question

Consider the curve $\gamma$ given by $y=b$ in the upper half-plane equipped with the hyperbolic metric $$\dfrac{dx^2+dy^2}{y^2}$$ Calculate the geodesic curvature of $\gamma$.

The problem I'm having is that every way I know of calculating the geodesic curvature of a curve in a surface involves knowledge of a normal vector and I'm not sure how one would go about defining the tangential derivative for such abstract smooth surfaces. Any clarification would be appreciated.

Answer 1

HINT: You should have a formula for geodesic curvature just in terms of the first fundamental form (and the curve, of course) if you're working in an orthogonal parametrization. (You don't say what material you know and what tools you have at your disposal. It's also a very straightforward computation using differential forms and moving frames.)

Answer 2

An easy way to calculate the geodesic curvature of a curve $\gamma$ in hyperbolic plane is to use a hyperbolic geodesic $\Gamma$ that is also a Euclidean line and is tangent to $\gamma$. The parallel transport along $\Gamma$ is Euclidean on small scales, and so the geodesic curvature of $\gamma$ at the point of tangency is proportional to its Euclidean curvature, the coefficient being the Euclidean speed of the hyperbolic-arclength parameterization of $\gamma$.

The disk model is more convenient for the above approach, because all geodesics through $(0,0)$ are Euclidean lines, regardless of direction. As a result, the geodesic curvature of any curve $\gamma$ at the point $(0,0)$ of hyperbolic disk is exactly $1/2$ of its Euclidean curvature. This is assuming Gaussian curvature $-1$, which corresponds to disk model with $ds = \frac{4\,ds^2}{(1-x^2-y^2)^2}$; the factor $1/2$ comes from the Euclidean metric being $1/2$ of hyperbolic metric at $(0,0)$.

I used this approach here to calculate the geodesic curvature of horocycles such as $y=b$ (it's $1$) and of circles of hyperbolic radius $R$ (it's $\coth R$). For a general curve $\gamma$ in the upper half-plane, fix a point $w\in \gamma$ and consider the Möbius-transformed curve $\zeta(t) = (\gamma(t)-w)/(\gamma(t)-\overline{w})$ in the hyperbolic disk. The Euclidean curvature of $\zeta$ at $0$ can be computed by the standard formula $$ \kappa = \frac{\operatorname{Im} (\overline{\zeta'}\zeta'')}{|\zeta'|^3} $$ Hence, the geodesic curvature of $\zeta$ at $0$ is $\kappa/2$, and this is also the geodesic curvature of $\gamma$ at $w$, since Möbius transformations are isometries.

One can also use this approach directly in the half-plane model, which has vertical geodesics $x=a$. But then the curve $\gamma$ would have to be rotated so it's tangent to the vertical geodesic.