Components of Riemann Tensor on sphere

Question

Since the surface of a sphere is a 2D manifold, the Riemann Tensor should have only one component. But if I calculate according to $$R^i_{~rkj}= \frac{\partial\Gamma^i_{~jr}}{\partial u^k} -\frac{\partial\Gamma^i_{~kr}}{\partial u^j} +\Gamma^i_{~ks}\Gamma^s_{~jr} -\Gamma^i_{~js}\Gamma^s_{~kr}$$ I get $$R^\theta_{~\phi\theta\phi}=\sin^2\theta$$ and $$R^\phi_{~\theta\phi\theta}=1$$ which are two seemingly independent components (plus of course their negative counterparts because of symmetry reasons).

So, I assume, they are not independent and I should be able to transform one into the other, but I have no idea how.

Can somebody please give me a hint (for dummies, please!)?

Thanks in advance

Fuzzy

Answer 1

Yep, I did the calculations:

Lowering the first index with the metric of the sphere $( g_{\vartheta\vartheta}=R^2, g_{\vartheta\varphi}= g_{\varphi\vartheta}= 0, g_{\varphi\varphi}= R^2\sin^2\vartheta )$ (sorry, "array" does not work here)

gets the correct result: $$ \begin{align*} R_{~\varphi\vartheta\varphi\vartheta}&=g_{\alpha\varphi}R^\alpha_{~\vartheta\varphi\vartheta} = g_{\vartheta\varphi}R^\vartheta_{~\vartheta\varphi\vartheta}+g_{\varphi\varphi}R^\varphi_{\vartheta\varphi\vartheta} =R^2\sin^2\vartheta \\ R_{~\vartheta\varphi\vartheta\varphi}&=g_{\vartheta\alpha}R^\alpha_{~\varphi\vartheta\varphi} =g_{\vartheta\vartheta}R^\vartheta_{~\varphi\vartheta\varphi}+g_{\vartheta\varphi}R^\varphi_{~\varphi\vartheta\varphi} =R^2\sin^2\vartheta \end{align*} $$ Symmetry takes care of the rest: $$ \begin{align*} R^\vartheta_{~\varphi\varphi\vartheta}&=-R^\vartheta_{~\varphi\vartheta\varphi}=- \sin^2\vartheta \rightarrow R_{~\vartheta\varphi\varphi\vartheta}=-R^2\sin^2\vartheta \\ R^\varphi_{~\vartheta\vartheta\varphi}&=-R^\varphi_{~\vartheta\varphi\vartheta}=-1 \rightarrow R_{~\varphi\vartheta\vartheta\varphi}=-R^2\sin^2\vartheta \end{align*} $$ Thanks to all who helped!