# Computing Gaussian curvature of hyperbolic space using Thurston

I'm currently reading through Thurston's book Three-Dimensional Geometry and Topology, and I am working on an exercise that has you compute the Gaussian curvature of the Poincare ball model of hyperbolic space. The way he has you compute this is by computing the hyperbolic length of a circle of radius $r$ centered at the origin, and then using the following fact that he gave earlier: he claims that $-1/3$ times the Gaussian curvature is equal to the second derivative of the ratio $c/2\pi r$ at $r=0$, where $c$ is the hyperbolic circumference of a circle of radius $r$, (this is found on page 47 of Thurston).

Here is what I have so far. To find the circumference of the circle, I computed $$c=2\int_0^{2\pi} \frac{r}{1-r^2}dt = \frac{4\pi r}{1-r^2}.$$

Then if we let $f(r) = c/2\pi r$, then $$f''(r) = 2\left(\frac{8r^2}{(1-r^2)^3} + \frac{2}{(1-r^2)^2}\right),$$ so $f''(0) = 4$. Then according to Thurston, if $K$ is the Gaussian curvature, we should get that $-1/3 K = 4$, meaning $K = -12$. This is clearly not correct since I know the curvature is supposed to be $-1$.

If anyone sees a mistake I am making or something that I am misunderstanding, your insight would be much appreciated!

The problem is that the coordinate $r$ does not tell you the radius of the circle: the length of a geodesic from the origin of the coordinate system to a point with parameter $r$ is $$R(r) = \int_0^r ds(\rho) = \int_0^r \sqrt{\frac{4}{(1-r'^2)^2}dr'^2} = \int_0^r \frac{2}{1-r'^2} \, dr' = 2\arg\tanh{r}.$$ (You do actually know the length can't be just $r$, since the lengths of radial lines are unbounded, rather than bounded above by $1$) So $r = \tanh{(R/2)}$. Therefore, the circumference of a disc of radius $R$ in terms of $R$ is $$2\pi\frac{2r}{1-r^2} = \dotsb = 2\pi\sinh{R},$$ and $\sinh{R}/R = 1 + R^2/6 + O(R^4)$, so the second derivative is $1/3$, whence the Gaussian curvature is $-1$.