An Einstein manifold in dimension 3 has a constant sectional curvature.

Question

Let $M$ be an Einstein manifold, I know that if dim $M\ge 3$, $M$ has a constant scalar curvature $\rho$, and $Ric=\frac{\rho}{m}g$, but I can't see how can $M$ possess a constant sectional curvature in dimension 3. can someone provide a proof?
thanks in advance!

Answer 1

A sketch of the main ideas:

1. On a three dimensional pseudo-Riemannian manifold, the Riemann curvature tensor is entirely determined by the Ricci tensor.

   First, the algebraic symmetries of the Riemann tensor implies that at a point it can be identified with a symmetric mapping from the space of antisymmetric two forms to itself via $$ \omega_{ab} \mapsto \mathrm{Riem}_{ab}{}^{cd} \omega_{cd}.$$ This means that the space of possible Riemann tensors is 6 dimensional.

   The Ricci tensor is a symmetric two tensor; pointwise this means that it also has 6 dimensions of freedom.

   Now, let $g_{ab}$ denote the metric tensor, and let $S_{ab}$ be a symmetric two tensor, and denote by $S = g_{ab} S^{ab}$ its trace. Consider the linear mapping $$ \Psi: S_{ab} \mapsto S_{ac} g_{bd} + S_{bd} g_{ac} - S_{ad} g_{bc} - S_{bc} g_{ad} - \frac12 S (g_{ac}g_{bd} - g_{ad} g_{bc}) $$ One checks that the output is an "algebraic Riemann tensor", meaning that it satisfies all the algebraic symmetries to quality it as a Riemann tensor. Giving an algebraic Riemann tensor, the mapping $T: R_{abcd} \mapsto g^{ac} R_{abcd}$ outputs a symmetric two tensor. It is a simple computation to check that $$ \Psi \circ T = \mathrm{Id}, T \circ \Psi = \mathrm{Id} $$ Since the space of algebraic Riemann tensor and that of Ricci tensors have the same number of dimension (with the manifold dimension = 3), this means that every Riemann tensor is uniquely determined (rank-nullity theorem) by its Ricci tensor via the mapping $\Psi$.

2. Using the mapping $\Psi$, it is simple to check that since we already know that the for our Einstein manifold $\mathrm{Ric}_{ab} = \Lambda g_{ab}$, and $R = 3 \Lambda$, we conclude $$ \mathrm{Riem}_{abcd} =\frac12 \Lambda (g_{ac}g_{bd} - g_{ad}g_{bc}) $$ and therefore the manifold has constant sectional curvature.

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Remark: the construction used in writing down the mapping $\Psi$ is the Kulkarni-Nomizu product.