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I was given this exercise but I don't even know where to start: to compute the Riemann tensor of the 2-dimensional sphere. The tensor acts on vector fields X,Y,Z like this: $R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$ where $\nabla$ is the affine connection defined considering extensions $X_1$,$Y_1$ on $\mathbb{R}^3$ of the fields $X$,$Y$ and the Gauss map $N$ so that $\nabla_XY=\nabla_{X_1}Y_1-<\nabla_{X_1}Y_1,N>N$ (here $\nabla$ is the standard flat connection). Thanks for any help

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    To explain why I am asking for your background (say, what are you reading or what book is your course based on), I'd like to add that in fact the result is almost obvious for those who know that the Riemannian curvature of 2-dimensional surfaces has only **one** independent component, and in coordinates is [expressed](http://en.wikipedia.org/wiki/Riemann_curvature_tensor) as $R_{abcd} = K (g_{ac}g_{db} - g_{ad}g_{cb})$, where $K$ is _the Gaussian curvature_.2012-05-07

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Hint. Try to understand why the Weingarten map $L: T_{p}{\mathbb{S}^2} \rightarrow T_{p}{\mathbb{S}^2}$ on the sphere is given by $L=-\frac{1}{r}\operatorname{id}$ (I assume that $r$ is the radius of your sphere), and then use the Gauss equation $ R(X,Y,Z,W) = - $ where $II(X,Y) = $ is the second fundamental form.

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    @balestrav The Gauss map of the sphere is (almost) the identity map. What can you say about the differential of the identity map?2012-05-07
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If you do not want to use Vytakin's nice suggestion, you can directly compute the Christoffel symbols for the sphere $\Gamma^\mu_{\alpha \beta}=\frac{g^{\mu\gamma}}{2}(g_{\gamma \alpha, \beta} + g_{\gamma \beta, \alpha} - g_{\alpha \beta, \gamma})$ . Few of them survive, and then you can plug them into d$x^\rho(R(\partial_{\mu},\partial_{\nu})\partial_{\sigma})=:R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma} - \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$.