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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.

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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.)

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  • $\begingroup$ The formula doesn't appear to be provided in my course, could you possibly send me a link to somewhere I could read about it? (If it's not too much trouble) $\endgroup$ – Rob Rockwood Jun 10 '16 at 19:39
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    $\begingroup$ Download my differential geometry text. The link is on my profile page. $\endgroup$ – Ted Shifrin Jun 10 '16 at 19:41
  • $\begingroup$ Can you please give the disk model formula of geodesic curvature in terms of first fundamental form coefficients? I like to see a sketch of increasing $k_g$ in hyperbolic plane. Thanks. $\endgroup$ – Narasimham Jun 19 at 8:57
  • $\begingroup$ Particularly referring to Prob 17 c pp 100 of your DG textbook is it possible to sketch all possible real values of $k_g$? Is it possible this way to draw an in-circle to three sides of a hyperbolic triangle? $\endgroup$ – Narasimham Jun 19 at 9:07
  • $\begingroup$ @Narasimham: I don't see any reason to work in the disk model. It's very inconvenient for this. Work in the half-space model and then transfer over. You only get curves with constant $\kappa_g>-1$. You can generalize the computation in Exercise 7 in that same section of my notes to get arcs of circles. The constant value of $\kappa_g$ determines where the center of the circle is. $\endgroup$ – Ted Shifrin Jun 19 at 17:41
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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.

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