I am reading the book: "Analysis and Algebra on Differentiable Manifolds: A Workbook for Students and Teachers" (P.M.Gadea, J.Munoz Masque) Problem 6.5.1. Page 252-253:
Find the Riemann curvature tensor of the Riemannian manifold $(U,g)$, where $U$ denotes the unit open disk of the plane $\mathbf{R}^2$ and $$g=\frac{1}{(1 - x^2 - y^2)}(dx^2 + dy^2)$$
The Solution is given as follows: $$g^{-1}= \begin{pmatrix} 1-x^2 - y^2 & 0 \\\\ 0 & 1-x^2-y^2 \end{pmatrix}$$ $$x=x^1, y=x^2$$, the Christoffel symbols are:
$$\Gamma_{11}^{1} = \Gamma_{12}^{2}= \Gamma_{21}^{2}= - \Gamma_{22}^{1}= \frac{x}{1-x^2-y^2}$$
$$- \Gamma_{11}^{2}= \Gamma_{12}^{1}= \Gamma_{21}^{1}= \Gamma_{22}^{2}= \frac{y}{1-x^2-y^2}$$
Therefore, $$R \left( \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right) = g \left( R(\frac{\partial}{\partial x},\frac{\partial}{\partial y})\frac{\partial}{\partial y},\frac{\partial}{\partial x} \right)$$ $$= - \frac{2}{(1-x^2-y^2)^2}$$ The above is the full solution given in the book. Below is my computation: But, I could not obtain the same result, instead I get the following: $$R \left( \frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right) = R_{1212} = g_{1 \xi} R_{212}^{\xi}=g_{11}R_{212}^{1}$$
$$\frac{1}{(1-x^2-y^2)} \left[ \partial_{x} \left( \frac{-x}{1-x^2-y^2} \right) - \partial_{y} \left( \frac{y}{1-x^2-y^2} \right) + \Gamma_{11}^{1} \Gamma_{22}^{1} + \Gamma_{12}^{1} \Gamma_{22}^{2} - \Gamma_{21}^{1}\Gamma_{12}^{1} - \Gamma_{22}^{1} \Gamma_{12}^{2}\right]$$ which gives after some computations the following with two extra terms: $$\frac{-2}{(1-x^2-y^2)^2} + \frac{-2x^2}{(1-x^2-y^2)^3} + \frac{-2y^2}{(1-x^2-y^2)^3}$$ I simply can't detect where I have gone wrong. May someone help me ? Thank you so much.