TITLE: Proving constant curvature
QUESTION [3 upvotes]: 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.

REPLY [3 votes]: I think you are supposed to simply argue that since the metric is "constant" in the sense of Eq. (C.2.3) on $\Sigma_t$, the curvature should be a constant as well.
However, here is a more sophisticated point of view. Our definition of homogeneity is that the isometry group $\mathrm{Iso}(\Sigma_t,h)$ is transitive, i.e. given $p,q\in\Sigma_t$ there exists $\phi\in\mathrm{Iso}(\Sigma_t,h)$ such that $\phi(p)=q$. A very important theorem of Riemannian geometry which never shows up in GR books (or even Riemannian geometry books) is this$^1$: 

Let $\phi:(M,g)\to (\tilde M,\tilde g)$ be an isometry. Then $\phi^*\mathrm{Riem}[\tilde g]=\mathrm{Riem}[g]$, where $\mathrm{Riem}[-]$ is the $(0,4)$ Riemann tensor of the respective metrics. 

Now, if $\tilde M=M$, then really $g$ and $\tilde g$ are the same metric. Thus the pullback of the Riemann tensor at $q$ is the Riemann tensor at $p$. Contract Wald's equation ${}^{(3)}R_{ab}{}^{cd}=K\delta^c{}_{[a}\delta^d{}_{b]}$ to get ${}^{(3)}R=3K$ where ${}^{(3)}R$ is the induced scalar curvature.$^2$ Also, contract the above theorem to get $\phi^*R=R$, which means that ${}^{(3)}R(\phi(p))={}^{(3)}R(q)={}^{(3)}R(p)$ for any $p,q\in\Sigma_t$. Thus ${}^{(3)}R$ is constant and therefore $K$ is as well.
It should be noted that Wald's definition of isotropy is sufficiently powerful that homogeneity is unneeded. In fact, Wald proves this in the middle of page 94. However, the reader familiar with Riemannian geometry will recognize this as a special case of Schur's Lemma. I believe Straumann was the first to make this observation in Straumann, N. Helva. Phys. Acta. 47, 379 (1972).


It's an exercise in J.M. Lee, Introduction to Riemannian Manifolds (1997) on page 119 and an exercise in R.K. Sachs and H. Wu, General Relativity for Mathematicians (1977) on page 19, but see this blog post of mine for the solution.
Because it confused me for a second, here is the contraction of $\delta^c{}_{[a}\delta^d{}_{b]}$:
$$\delta^a{}_{[a}\delta^b{}_{b]}=\frac{1}{2}(\delta^a{}_a\delta^b{}_b-\delta^a{}_b\delta^b{}_a)=\frac{1}{2}(3\cdot 3-\delta^a{}_a)=\frac{1}{2}(3^2-3)=3.$$
{"set_name": "stack_exchange", "score": 3, "question_id": 262244}