Compute the Gaussian curvature and Cristoffel symbols of hyperbolic space

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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).

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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 ${\rm s} = -2$.