Riemann curvature tensor on a unit radius 2D sphere

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Summary:
How is it possible that non-zero components of the curvature tensor depend on theta
I have worked out (and then verified against some sources) that ##R^\theta_{\phi\theta\phi} = sin^2(\theta)##. The rest of the components are either zero or the same as ##R^\theta_{\phi\theta\phi} ## some with the sign flipped.
I was surprised at this, because it implies that the curvature tensor tends to zero as we approach either pole. Being a sphere I thought that the curvature tensor would have the same value everywhere. How can this be?
 

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Ibix
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According to my calculations, ##R^\phi{}_{\theta\phi\theta}=-1##. I'd check your algebra.
 
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Dale
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Summary:: How is it possible that non-zero components of the curvature tensor depend on theta

Being a sphere I thought that the curvature tensor would have the same value everywhere. How can this be?
Because of the symmetry all of the invariants of the tensor must have the same value everywhere. But the coordinates are not symmetrical, so the components of the tensor in those coordinates may be non-constant.
 
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Ibix
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To add to my last, the coordinate system you are using isn't isotropic so it shouldn't be particularly surprising that the curvature tensor "looks" different at different places on the sphere. A small circular path around the pole has a very different description in these coordinates (constant ##\theta##) than the same path around an arbitrary point. (Edit: beaten to it by Dale, I see.)

You would expect any scalar you can construct from the Riemann tensor (for example ##R^{abcd}R_{abcd}##, or ##R^{ab}R_{ab}##) to be independent of coordinates. Also, if you can find a transform to move the pole to an arbitrary point, then you should be able to use this to transform the Riemann at one point/basis on the sphere to the Riemann at any other point.
 
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I see. If I had used unit basis instead of the usual coordinate basis I would not get this dependence on theta. Thanks!
 

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