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Constant-curvature three-manifolds: equivalence of constant sectional curvature, the tensor identity and the Ricci condition

Statement

This item assumes ${A C}_{ω}$ as inherited through Constant sectional curvature and space form; the pointwise tensor arguments below use no further choice principle (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $Σ$ be a smooth three-manifold and let $γ$ be a Riemannian metric on $Σ$ (Riemannian metric and riemannian manifold), with curvature operator $R$ , Riemann four-tensor $Rm$ (Riemann curvature four-tensor), Ricci curvature $Ric$ (Ricci curvature) and scalar curvature $S$ (Scalar curvature). Let $k \in R$ . The following three statements are equivalent:

(a) The metric $γ$ has constant sectional curvature $k$ : every tangent two-plane has sectional curvature $k$ (Constant sectional curvature and space form, Sectional curvature).

(b) $Ric (γ) = 2 k γ$ at every point of $Σ$ ; that is the curvature condition for the real parameter $k$ . When $Σ$ is connected and boundaryless and $k \in {- 1, 0, 1}$ , it is precisely the normalized spatial-model condition of The constant-curvature spatial model of a homogeneous cosmology.

(c) $R (X, Y) Z = k (γ (Y, Z) X - γ (X, Z) Y)$ for all smooth vector fields $X, Y, Z$ on $Σ$ (Curvature tensor of constant sectional curvature).

Whenever these hold, the scalar curvature is $S = 6 k$ , and the trace-free Ricci tensor vanishes, ${Ric}_{0} = 0$ (Kulkarni–Nomizu product, trace-free Ricci tensor, and Weyl curvature).

Facts & Assumptions

Given: A smooth three-manifold $Σ$ with Riemannian metric $γ$ and a real number $k$ .

[L1]

$γ$ has constant sectional curvature $k$ if and only if $R (X, Y) Z = k (γ (Y, Z) X - γ (X, Z) Y)$ ; equivalently $Rm (X, Y, Z, W) = k (γ (Y, Z) γ (X, W) - γ (X, Z) γ (Y, W))$ (Curvature tensor of constant sectional curvature, Constant sectional curvature and space form).

[L2]

Ricci curvature is $Ric (X, Y) = tr (Z \mapsto R (Z, X) Y)$ , and in an orthonormal basis $(e_{1}, e_{2}, e_{3})$ of $T_{p} Σ$ it is $Ric (X, Y) = \sum_{i = 1}^{3} Rm (e_{i}, X, Y, e_{i})$ ; it is a smooth symmetric covariant two-tensor (Ricci curvature, Ricci curvature is symmetric and basis independent). Scalar curvature is $S = {tr}_{γ} Ric$ (Scalar curvature). Every finite-dimensional inner-product space has an orthonormal basis (Every finite-dimensional real or complex inner product space has an orthonormal basis).

[L3]

The Riemann four-tensor is $Rm (X, Y, Z, W) = γ (R (X, Y) Z, W)$ (Riemann curvature four-tensor).

[L4]

In dimension three, ${Ric}_{0} = Ric - \frac{1}{3} S γ$ by definition, the Weyl tensor vanishes, and the Ricci decomposition reads $Rm = {Ric}_{0} ⊙ γ + \frac{S}{12} (γ ⊙ γ)$ ; the Kulkarni-Nomizu product is the one of Kulkarni–Nomizu product, trace-free Ricci tensor, and Weyl curvature, under which $(γ ⊙ γ) (X, Y, Z, T) = 2 (γ (X, T) γ (Y, Z) - γ (X, Z) γ (Y, T))$ and the constant-sectional-curvature- $K$ tensor is $\frac{K}{2} (γ ⊙ γ)$ (Ricci decomposition of the Riemann tensor in dimension at least three, Kulkarni–Nomizu product, trace-free Ricci tensor, and Weyl curvature).

Proof

technique · direct; the three clauses are linked by a cycle of implications, with the dimension-three decomposition carrying the converse

1.1L1

(Clause (a) implies clause (c).) If every tangent two-plane has sectional curvature $k$ , then [L1] gives $R (X, Y) Z = k (γ (Y, Z) X - γ (X, Z) Y)$ at every point of $Σ$ , which is exactly clause (c); conversely clause (c) is the tensor identity of [L1] from which the sectional curvature $k$ of every two-plane is read off.

2.1L1L2L4step 1.1algebra

(Clause (c) implies clause (b), and $S = 6 k$ whenever clause (b) holds.) Assume clause (c) and fix $p \in Σ$ . Choose a $γ_{p}$ -orthonormal basis $(e_{1}, e_{2}, e_{3})$ of $T_{p} Σ$ , which exists by [L2]. Substituting [L1] into the trace formula of [L2] gives $Ric (X, Y) = \sum_{i = 1}^{3} k (γ (X, Y) γ (e_{i}, e_{i}) - γ (e_{i}, Y) γ (X, e_{i})) = k (3 γ (X, Y) - γ (X, Y)) = 2 k γ (X, Y)$ for all $X, Y \in T_{p} Σ$ , using $γ (e_{i}, e_{i}) = 1$ and $\sum_{i} γ (e_{i}, Y) γ (X, e_{i}) = γ (X, Y)$ ; this is clause (b). Conversely, under clause (b) the metric trace gives $S = \sum_{i, j} γ^{i j} {Ric}_{i j} = 2 k \sum_{i, j} γ^{i j} γ_{i j} = 6 k$ and ${Ric}_{0} = Ric - \frac{1}{3} S γ = 2 k γ - 2 k γ = 0$ by [L4].

3.1L1L3L4step 2.1algebra

(Clause (b) implies clause (a).) Assume clause (b). By step 2.1, $S = 6 k$ and ${Ric}_{0} = 0$ ; since the dimension is three, [L4] gives $Rm = \frac{6 k}{12} (γ ⊙ γ) = \frac{k}{2} (γ ⊙ γ)$ , and [L4] identifies $\frac{k}{2} (γ ⊙ γ)$ as the constant-sectional-curvature- $k$ tensor, that is, $Rm (X, Y, Z, W) = k (γ (Y, Z) γ (X, W) - γ (X, Z) γ (Y, W))$ ; by the definition [L3] of the Riemann four-tensor this reads $R (X, Y) Z = k (γ (Y, Z) X - γ (X, Z) Y)$ , which is clause (c). By the equivalence recalled in [L1], clause (c) is clause (a).

4.1step 1.1step 2.1step 3.1∎

(Conclusion.) Step 1.1 gives (a) $\Rightarrow$ (c), step 2.1 gives (c) $\Rightarrow$ (b), and step 3.1 gives (b) $\Rightarrow$ (a); hence the three clauses are equivalent. The scalar identity $S = 6 k$ and the vanishing of ${Ric}_{0}$ were proved in step 2.1 under the equivalent hypotheses. No step assumes a value of $k$ beyond $k \in R$ , and the empty and low-dimensional cases do not arise because $Σ$ is a three-manifold.

Remarks

Why the converse is available here. In dimension three the Weyl part of the curvature vanishes identically, so the full curvature tensor is determined by the Ricci tensor and the scalar curvature; the converse (b) $\Rightarrow$ (a) of the lemma is a genuine dimension-three statement and fails in higher dimensions, where trace-free Ricci and Weyl data are independent. The lemma is used on this page only for three-dimensional spatial metrics.
Normalization. The condition $Ric (γ) = 2 k γ$ with $k \in {- 1, 0, 1}$ is the normalization adopted in The constant-curvature spatial model of a homogeneous cosmology; by the lemma it is the same as saying that the spatial model has constant sectional curvature $k$ with the three standard signs, which is the form in which the constant-curvature models are usually presented.

Constant-curvature three-manifolds: equivalence of constant sectional curvature, the tensor identity and the Ricci condition

Statement

Facts & Assumptions

Proof

Remarks

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