I'm an undergraduate going through Schlag's A concise course in complex analysis and Riemann surfaces, and I'm stuck on part (c) of this question, which reads

Determine all geodesics of hyperbolic space as well as its scalar curvature (we are using the terminology of Riemannian geometry).

I've found the geodesics of hyperbolic space, but I am stuck on the part about the scalar curvature. The solution suggests this—among other things it says that we can compute the Gaussian curvature and Cristoffel symbols, both concepts I am unfamiliar with, and am unsure how to compute.

My experience with hyperbolic geometry is with the hyperboloid model, so the half plane model is new to me. My background on complex analysis more or less amounts to Chapter 1 from the book, but doesn't seem hugely relevant here (I am happy to summarize if that happens to not be the case).


For surfaces, ${\rm s} = 2K$, where $K$ is the Gaussian curvature. The Gaussian curvature, in turn, is the sectional curvature $K(\partial_x,\partial_y)$. If you already found the geodesics, I assume you already found the Christoffel symbols to be $$\Gamma_{xx}^x = 0, \quad \Gamma_{xy}^x = \Gamma_{yx}^x = -\frac{1}{y}, \quad \Gamma_{yy}^x = 0, \quad \Gamma_{xx}^y = \frac{1}{y}, \quad \Gamma_{xy}^y = \Gamma_{yx}^y = 0 \quad\mbox{and}\quad \Gamma_{yy}^y = -\frac{1}{y}.$$This means that $$\begin{align} R(\partial_x,\partial_y)\partial_y &= \nabla_{\partial_x}\nabla_{\partial_y}\partial_y - \nabla_{\partial_y}\nabla_{\partial_x}\partial_y - \nabla_{[\partial_x,\partial_y]}\partial_y \\ &= \nabla_{\partial_x}\left(-\frac{1}{y}\partial_y\right) - \nabla_{\partial_y}\left(-\frac{1}{y}\partial_x\right) - \nabla_0\partial_y \\ &= -\frac{1}{y}\nabla_{\partial_x}\partial_y - \frac{1}{y^2}\partial_x + \frac{1}{y}\nabla_{\partial_y}\partial_x - 0 \\ &= -\frac{1}{y}\left(-\frac{1}{y}\partial_x\right) - \frac{1}{y^2}\partial_x +\frac{1}{y}\left(-\frac{1}{y}\partial_x\right) \\ &= -\frac{1}{y^2}\partial_x.\end{align}$$So the curvature is: $$K(\partial_x,\partial_y) = \frac{\langle R(\partial_x,\partial_y)\partial_y,\partial_x\rangle}{\langle \partial_x,\partial_x\rangle\langle\partial_y,\partial_y\rangle - \langle \partial_x,\partial_y\rangle^2} = \frac{(-1/y^2)(1/y^2)}{(1/y^2)(1/y^2) - 0^2} = -1.$$Thus ${\rm s} = -2$.

  • $\begingroup$ I didn't use the Christoffel symbols to find the geodesics; I argued that vertical lines were geodesics and all others are obtained by applying the automorphisms, which are conformal, hence the other geodesics are circles that meet the real line at a right angle. The solutions in the book say 'since the isometries act transitively, the Gaussian curvature agrees with the value at zero which can be computed', which I don't follow. $\endgroup$
    – bookworm
    Jul 14 '20 at 6:12
  • $\begingroup$ I'd need to know what definition of Gaussian curvature is the book using then (I searched for "Gaussian curvature" on the pdf and only got three unhelpful hits). But the isometry argument is standard: isometries preserve curvature, so if there is an isometry taking one point to other, the value of the curvature is the same on both points. Now transitivity kicks in to say that any two points are related by an isometry. So the value of the curvature is the same at all points. In particular it is constant, and you can find the value by considering just one point. $\endgroup$
    – Ivo Terek
    Jul 15 '20 at 0:32
  • $\begingroup$ There are lots and lots of formulas for curvatures of metrics conformal to the Euclidean one, for instance, the curvature of $({\rm d}x^2+{\rm d}y^2)/h(x,y)^2$ is $K = h\,\triangle h - \|\nabla h\|^2$, so for $h(x,y) = y$ we have $\triangle h = 0$ and $\nabla h = (0,1)$, which gives $K=-1$ again. $\endgroup$
    – Ivo Terek
    Jul 15 '20 at 0:34
  • $\begingroup$ I'm not entirely sure what definition is used in the book either. Which did you use to get $K = h\Delta h - ||\nabla h||^2$? $\endgroup$
    – bookworm
    Jul 15 '20 at 4:34
  • $\begingroup$ You need some definition of Gaussian curvature if you want to prove these formulas. The only options I see are 1) ask your instructor; 2) accept these formulas on faith for now; 3) take a differential geometry class. One possible definition you can take for the curvature of $\rho(z)|{\rm d}z|$ is $$\kappa(z) = -\frac{\triangle \log \rho(z)}{\rho(z)^2}.$$The formula in terms of $h$ follows from this definition by relating $h$ and $\rho$. $\endgroup$
    – Ivo Terek
    Jul 16 '20 at 2:44

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