Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
Since you have not given to us your background, nor showed any your effort to solve your problem, it is extremely unclear what kind of help would be the most beneficial to you. –  Yuri Vyatkin May 7 '12 at 12:12
Also, because you have used (homework) tag, I assumed that a hint would be sufficient to start the discussion. From your comment to my answer I realize that you need to understand quite a few things before you will be able to approach the calculation. –  Yuri Vyatkin May 7 '12 at 12:16
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 as $R_{abcd} = K (g_{ac}g_{db} - g_{ad}g_{cb})$, where $K$ is the Gaussian curvature. –  Yuri Vyatkin May 7 '12 at 13:33

2 Answers 2

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 $$ R(X,Y,Z,W) = <II(Y,Z),II(X,W)> - <II(X,Z),II(Y,W)> $$ where $II(X,Y) = <L(X),Y>$ is the second fundamental form.

share|improve this answer
Sorry, but i don't know how your R is related to mine.. –  balestrav May 7 '12 at 8:10
@balestrav Your (1,3)-tensor $R(X,Y)Z$ is in fact the curvature endomorphism of the Levi-Civita connection $\nabla$ on sphere $\mathbb{S}^2$, while mine $R(X,Y,Z,W) = <R(X,Y)Z,W>$ is the corresponding Riemannian curvature (0,4)-tensor. –  Yuri Vyatkin May 7 '12 at 12:09
@balestrav The Gauss map of the sphere is (almost) the identity map. What can you say about the differential of the identity map? –  Yuri Vyatkin May 7 '12 at 13:35

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$x^\rho(R(\partial_{\mu},\partial_{\nu})\partial_{\sigma})=:R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma} - \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.