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Hyperbolic space has negative constant sectional curvature

Statement

This item assumes ${A C}_{ω}$ , namely countable choice. In the propagated dependency chain, that assumption is required through Sectional curvature; after those interfaces are fixed, the remaining local or finite argument makes no additional countable-family choice.

Let $r > 0$ and $n \geq 1$ , and put

$U^{n} = {(x^{1}, \dots, x^{n}) \in R^{n} : x^{n} > 0}, g = \frac{r^{2}}{(x^{n})^{2}} \sum_{a = 1}^{n} d x^{a} \otimes d x^{a} .$

For $n \geq 2$ , this metric has constant sectional curvature $- 1 / r^{2}$ . For $n = 1$ , its sectional-curvature domain is empty. Apart from the stated inherited ${A C}_{ω}$ , the calculation makes no additional countable-family choice.

Facts & Assumptions

Given: ${A C}_{ω}$ , a real number $r > 0$ , an integer $n \geq 1$ , and the global coordinates on $U^{n}$ ; when $n \geq 2$ , also a point and a tangent two-plane there.

[A1]

${A C}_{ω}$ is countable choice and is required here through Sectional curvature; after those supplied interfaces are fixed, the remaining local or finite calculation makes no additional countable-family choice.

[F1]

Smooth symmetric positive-definite coordinate matrices define Riemannian metrics. Coordinate criterion for a riemannian metric.

[F2]

The Levi–Civita Christoffel symbols are obtained from the metric and its first derivatives by the standard coordinate formula. Christoffel formula for the levi civita connection.

[F3]

With the page's index order, $R^{ℓ}_{k i j} = \partial_{i} Γ^{ℓ}_{j k} - \partial_{j} Γ^{ℓ}_{i k} + Γ^{m}_{j k} Γ^{ℓ}_{i m} - Γ^{m}_{i k} Γ^{ℓ}_{j m}$ . Coordinate formula for the curvature tensor.

[F4]

Curvature is a smooth type $(1, 3)$ tensor, so an identity on coordinate basis vectors extends multilinearly to all tangent vectors. Curvature is a type (1,3) tensor.

[F5]

The four-tensor is $Rm (X, Y, Z, W) = g (R (X, Y) Z, W)$ , and sectional curvature divides $Rm (u, v, v, u)$ by the positive Gram determinant. Riemann curvature four-tensor, Sectional curvature.

Verification

technique · direct coordinate calculation

1.1

Write $h = x^{n} > 0$ . The metric and inverse matrices are $g_{i j} = r^{2} h^{- 2} δ_{i j}$ and $g^{i j} = r^{- 2} h^{2} δ^{i j}$ . The first is smooth and symmetric, and $g_{i j} v^{i} v^{j} = r^{2} h^{- 2} \sum_{i} (v^{i})^{2} > 0$ for $v \neq 0$ , so [F1] makes $g$ Riemannian.

F1algebra

2.1

Put $ϵ_{i} = δ_{i n}$ and $ϵ^{k} = δ_{n}^{k}$ . Since $\partial_{p} g_{i j} = - 2 r^{2} h^{- 3} ϵ_{p} δ_{i j}$ , substitution in [F2] gives $Γ^{k}_{i j} = - h^{- 1} (ϵ_{i} δ_{j}^{k} + ϵ_{j} δ_{i}^{k} - δ_{i j} ϵ^{k})$ .

F2step 1.1algebra

3.1

Define $A^{ℓ}_{j k} = ϵ_{j} δ_{k}^{ℓ} + ϵ_{k} δ_{j}^{ℓ} - δ_{j k} ϵ^{ℓ}$ , so step 2.1 says $Γ^{ℓ}_{j k} = - h^{- 1} A^{ℓ}_{j k}$ and $\partial_{i} Γ^{ℓ}_{j k} = h^{- 2} ϵ_{i} A^{ℓ}_{j k}$ . The derivative difference in [F3] expands to $h^{- 2} (ϵ_{i} ϵ_{k} δ_{j}^{ℓ} - ϵ_{i} δ_{j k} ϵ^{ℓ} - ϵ_{j} ϵ_{k} δ_{i}^{ℓ} + ϵ_{j} δ_{i k} ϵ^{ℓ})$ , while the two contracted quadratic terms expand to $h^{- 2} (δ_{i}^{ℓ} ϵ_{j} ϵ_{k} - δ_{j}^{ℓ} ϵ_{i} ϵ_{k} - δ_{i k} ϵ_{j} ϵ^{ℓ} + δ_{j k} ϵ_{i} ϵ^{ℓ} - δ_{j k} δ_{i}^{ℓ} + δ_{i k} δ_{j}^{ℓ})$ . The first four terms cancel pairwise, leaving $R^{ℓ}_{k i j} = h^{- 2} (δ_{i k} δ_{j}^{ℓ} - δ_{j k} δ_{i}^{ℓ})$ .

F3step 2.1algebra

4.1

Since $g_{i k} = r^{2} h^{- 2} δ_{i k}$ , step 3.1 is $R (\partial_{i}, \partial_{j}) \partial_{k} = - (1 / r^{2}) (g_{j k} \partial_{i} - g_{i k} \partial_{j})$ . Tensoriality [F4] yields $R (X, Y) Z = - (1 / r^{2}) (g (Y, Z) X - g (X, Z) Y)$ for arbitrary tangent vectors. Pairing with $X = u$ after setting $Y = Z = v$ , [F5] gives $Rm (u, v, v, u) = - (1 / r^{2}) (g (u, u) g (v, v) - g (u, v)^{2})$ . For a basis $(u, v)$ of any tangent two-plane, the Gram determinant is positive, so [F5] yields $K = - 1 / r^{2}$ at every point and on every plane.

A1F4F5step 3.1algebra

5.1

The half-space is nonempty, for example at $(0, \dots, 0, 1)$ , and is open and boundaryless. Dimension zero is inapplicable because the defining coordinate $x^{n}$ requires $n \geq 1$ ; when $n = 1$ , step 3.1 gives zero curvature as it must, but there is no tangent two-plane, so the constant-sectional-curvature predicate is vacuous. For $n \geq 2$ , [F5] excludes degenerate Gram denominators. The conditions $h > 0$ and $r > 0$ exclude the singular height endpoint and the degenerate scale $r = 0$ ; no limiting assertion is made. Every coordinate, tensor, point, and plane basis is explicit or supplied, so no further family choice is made beyond the stated inherited assumption. The result assigns one value to every plane and states no biconditional.

F1F5step 1.1step 2.1step 3.1step 4.1∎

Depends on

Used by

Zero scalar curvature does not imply flatness Counterexample

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ved Datar, Lectures on Riemannian Geometry (standard reference, not scraped)
John M. Lee, Riemannian Manifolds: An Introduction to Curvature (standard reference, not scraped)