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Riemannian curvature of maximally symmetric spaces

  1. Sep 25, 2014 #1
    A maximally symmetric is a Riemannian n-dimensional manifold for which there is n/2 (n+1) linearly independent (as solutions) killing vectors. It is well known that in such a space
    $$R_{abcd} \propto (g_{ab}g_{cd} - g_{ac}g_{bd}) .$$

    How is this formula derived for a general maximally symmetric space?
  2. jcsd
  3. Sep 25, 2014 #2

    George Jones

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    Last edited by a moderator: May 7, 2017
  4. Sep 27, 2014 #3
    Steven Weinberg's Gravitation and Cosmology has a nice derivation. Here's a summary.

    The Killing vector ξ is defined as satisfying these two equivalent equations:
    (covariant derivative of ξ) + (reversed indices) = 0
    Lie derivative by ξ of the metric = 0

    Take the covariant derivative of this equation, rearrange indices, add and subtract, and use the Riemann curvature tensor R's definition to get
    Second covariant derivative of ξ = R.ξ (very schematic)

    The Riemann tensor is defined from
    Commutator of covariant deriative of X = R.X (very schematic)
    Commutator = (second derivative) - (reversed indices)

    From the second derivative of ξ, one can find it at any point using only its value and its first derivatives' value at that point. That must be the antisymmetric part of its first derivative, to satisfy its defining equation. For n dimensions, that's n values of ξ and n(n-1)/2 values of its antisymmetrized first derivative, giving n(n+1)/2 possible Killing vectors. In practice, there may be fewer than that, so n(n+1)/2 is the theoretical maximum.

    With the second covariant derivative of ξ, take another covariant derivative, and subtract out the last two indices reversed. Go from covariant-derivative commutators to the Riemann tensor as appropriate. One ends up with an equation that is
    Lie derivative by ξ of R = 0

    At some point, take ξ = 0 but with nonzero first derivatives. One gets R.(first derivatives of ξ) = 0 (very schematic)

    Going through all the possible first derivatives of ξ, one finds the OP's result: ## R_{ijkl} = K (g_{ik}g_{jl} - g_{il}g_{jk} ) ##.

    Contracting to get the Ricci tensor, one finds ## R_{ik} = (n-1)K g_{ik} ##, and contracting further to get the Ricci scalar, one finds ## R = n(n-1)K ##.

    Since the Lie derivative of the Ricci scalar is (gradient of Ricci scalar).ξ, and since ξ can be arbitrary at a point, one finds that the Ricci scalar is constant, and thus that K is constant. The Riemann and Ricci tensors, however, are not constant in the ordinary sense, but covariantly constant, because the covariant derivative of them is zero.
  5. Sep 27, 2014 #4
    One can get a metric of a maximally-symmetric space in an interesting way. Using its value of the Riemann tensor, one can show that a maximally-symmetric metric is conformally flat. That is, its metric = (conformal function) * (flat-space metric).

    One can use a manifestly flat metric like a constant metric for the flat-space metric, and then find the Riemann tensor from the complete metric. One can then solve for that metric's conformal function.

    Once one has that metric, one can then solve for all its Killing vectors.
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