- #1

- 228

- 25

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.