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When we calculate Riemann Tensor for different curvature we have lots of components. However, there are many components that are zero. How can we argue, based on the symmetry of connection , that those elements are zero?

For example if I am calculating the Riemann Tensor of $S^2$ sphere, I get only one non zero component i.e. $R_{\phi,r,\phi}^{\theta}$ = $sin^2 {\theta}$ and other components are zero. So, How can I argue, without calculating that all other components are zero.

Edit: (Dimension, No. of independent Riemann Components) = (2,1; 3,6 ; 4,20)

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For a two dimensional manifold, the Riemann curvature can be completely recovered from the Scalar curvature. In three dimension, the Riemann curvature can be completely recovered from the Ricci curvature. So those cases are special. –  Willie Wong Feb 16 '12 at 14:50
A tetrad formalism can sometimes help with this when you have a lot of symmetry. –  Jerry Schirmer Feb 16 '12 at 15:33
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2 Answers

The number of independent components for the Riemann curvature tensor $R_{ijk\ell}$ for the Levi-Civita connection is greatly reduced because of symmetries. The last two indices $k\neq \ell$ have to be different, because of antisymmetry $$R_{ijk\ell}~=~-R_{ij\ell k}.$$ Interchange symmetry $$R_{ijk\ell}~=~R_{k\ell ij}$$ then fixes the first two indices $i\neq j$ to be different as well. In two dimensions, if the metric $g_{ij}$ is diagonal, then there is essentially only one non-zero possibility for $R_{ijk\ell}$ and $R^i{}_{jk\ell}$ up to symmetries.

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This is not really what the OP is after--- the question is about Killing vectors. In 2D, there is only 1 component of R anyway, so this is much too special. –  Ron Maimon Feb 16 '12 at 6:28
Well, I interpret the question(v1) as OP is talking about the generic symmetries of the Levi-Civita connection. –  Qmechanic Feb 16 '12 at 6:53
@Qmechanic : Sorry if my wordings were vague. I knew the stuffs that you posted. I wanted to know a easier to figure out which goes component goes to zero if you are given any curve (i.e. connection or line element). –  Qubit Feb 16 '12 at 7:21
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Two tricks that I can think of:

  1. If your manifold admits a point in which it is locally isotropic (more precisely, suppose you have a manifold with $n$ dimensions; suppose you have a point $p$ and a neighborhood $N$ of $p$ such that the group $SO(n)$ acts transitively on $N$) then at that point your Ricci curvature is proportional to the metric there. Something similar can be done by thinking about the components of the Weyl curvature in a orthonormal frame there, but it is a bit more messy. (This would reflect that all sectional curvature are the same at that point.)

    a. If your manifold admits many points at which it is locally isotropic (say, a connected open set worth), then on the domain that this is true, you actually have that the manifold is Einstein, that is the proportion factor between the Ricci curvature and the metric is constant on that domain.

  2. A lot of times your symmetry has the nice property that its orbits are hypersurface orthogonal. In this case you can write your manifold as a warped product (a special case being the Cartesian product). Then there are known formulae relating the curvature of the full metric to that of the warp factors (see, for example, chapter 7 of O'Neill, Semi-Riemannian Geometry).

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