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A Proving constant curvature

  1. Jun 12, 2016 #1
    I'm currently on section 5.1 in Wald's book. He is trying to prove that the cosmological principle implies that space has constant curvature.

    Given a spacelike hypersurface ##\Sigma_t## for some fixed time ##t##, we say that it is homogeneous if given ##p,q \in \Sigma_t##, there is an isometry, ##\phi##, of the metric ##g## such that ##\phi(p) = q##.

    Now at a given point ##p \in \Sigma_t##, ##g## induces a Riemannian metric ##h## on ##\Sigma_t## simply by restricting ##g## to spacelike tangent vectors. The Riemann curvature tensor ##R_{ab}{}^{cd}## (using ##h## to raise the third index) can be viewed as a linear map from ##\mathcal{A}^2(T_p \Sigma_t)## into ##\mathcal{A}^2(T_p \Sigma_t)## (the vector space of antisymmetric ##2##-tensors defined on the tangent space to ##\Sigma_t## at ##p##). Let ##L## denote this linear map. Viewed as a linear map, ##L## is symmetric, or equivalently, self-adjoint. Thus ##T_p \Sigma_t## has an orthonormal basis of eigenvectors of ##L##. If the eigenvalues were distinct then we would be able to construct a preferred tangent vector, violating isotropy. Hence all eigenvalues are the same and ##L = KI## for some constant ##K## and where ##I## is the identity operator. Another way to write this is ##R_{ab}{}^{cd} = K\delta^c{}_{[a} \delta^d{}_{b]}## where the square brackets denote antisymmetrization (recall that the Riemann tensor is antisymmetric in its first two indices, this is why we have the antisymmetric brackets there). Lowering the last two indices gives ##R_{abcd} = K h_{c[a}{}h_{b]d}##.

    Now here is the part that is bothering me. Wald says that homogeneity implies that ##K## must be constant, i.e. cannot vary from point to point of ##\Sigma_t##. I get that homogeneity is supposed to mean that everything is the same at each point, but we are trying to prove this mathematically, and we have a mathematical definition of homogeneity. I don't see how our mathematical definition of homogeneity shows that ##K## is constant from point to point.
     
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  3. Jun 12, 2016 #2

    PeterDonis

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    An isometry is a transformation that leaves the metric invariant; that implies that it also leaves ##K## invariant, since ##K## is part of the metric. So homogeneity, which requires that there is an isometry taking any spatial point into any other, also requires that ##K## is the same at every spatial point, since if ##K## were different for any pair of points there would not be an isometry taking one to the other.
     
  4. Jun 12, 2016 #3
    Sorry but I don't see how ##K## is part of the metric. The only relationship with ##K## and the metric that we have is ##R_{abcd} = K h_{c[a}h_{b]d} ##.
     
  5. Jun 12, 2016 #4

    PeterDonis

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    The metric encodes all information about the spacetime geometry. ##K## is part of the spacetime geometry. So it's part of the metric. It might not appear explicitly in the expression for the metric tensor; that depends on how you choose coordinates. But it will be uniquely determined by the metric.

    Quite honestly, the statement I have just made should be blindingly obvious to you. If it isn't, then you are not prepared for an "A" level thread on this subject matter.

    Note that ##h## is the spatial metric, not the spacetime metric. The spacetime metric is ##g##.

    Also, the equation you wrote down here is not the basis for Wald's assertion that homoegeneity implies that ##K## is constant. The basis for that is the mathematical definition of what an isometry is.
     
    Last edited: Jun 12, 2016
  6. Jun 12, 2016 #5
    Couldn't we use this argument for constant curvature: Let ## {}^3 R_{abc}{}^d## denote the Riemann tensor on a hypersurface ##\Sigma_t##. Since ##R_{abc}{}^d## can be expressed entirely in terms of the spacetime metric ##g##, then ##{}^3 R_{abc}{}^d## can be expressed entirely in terms of the spatial metric ##h##. Since ##h## is position invariant, then ##{}^3R_{abc}{}^d## must be the same everywhere and hence the curvature is constant.

    I was thinking about this argument for constant curvature before I started this thread, but it seems that if it were this simple then Wald would have used it. Is it incorrect?
     
  7. Jun 12, 2016 #6

    PeterDonis

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    How do you know that ##h## is position invariant?
     
  8. Jun 12, 2016 #7
    What I mean by position invariant is that given ##p,q \in \Sigma_t##, there is a diffeomorphism ##\phi## such that ##\phi(p) = q## and for two tangent vectors ##v,w \in T_p \big(\Sigma_t\big)##, ##\phi(v,w) = g\big(d\phi_p v, d\phi_p w\big)##, where ##d\phi_p## is the differential of ##\phi## at ##p##. That is the definition of homogeneous. Since ##h## is obtained by restricting ##g## to ##\Sigma_t##, then this property holds for ##h## as well.
     
  9. Jun 12, 2016 #8

    PeterDonis

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    In other words, your definition of "homogeneous" is (a) the same as Wald's (you have basically just restated Wald's definition, that there is an isometry between any two spatial points, in different words, although your restatement seems a little sloppy), and (b) sufficient to prove that homogeneity requires constant ##h## and therefore constant curvature ##K## over a spacelike hypersurface. In other words, you have now answered your own question by showing how homogeneity mathematically implies constant ##K##.
     
  10. Jun 12, 2016 #9
    Thanks for the help.
     
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