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Gaussian curvature structure equation

Statement

Let $(M, g)$ be an oriented Riemannian surface, $U \subseteq M$ open, and $(E_{1}, E_{2})$ a smooth positively oriented $g$ -orthonormal frame on $U$ with connection form $ω (X) = g (\nabla_{X} E_{1}, E_{2})$ . Write $R (X, Y) Z = \nabla_{X} \nabla_{Y} Z - \nabla_{Y} \nabla_{X} Z - \nabla_{[X, Y]} Z$ for the curvature of the Levi–Civita connection and let $Rm$ be the Riemann curvature four-tensor, so that $K = g (R (E_{1}, E_{2}) E_{2}, E_{1})$ is the sectional curvature of the tangent plane. Then, with $d A$ the Riemannian volume form of the oriented surface $U$ ,

$d ω = - K d A .$

The only choice principle involved is the countable choice ${A C}_{ω}$ already present in the published sectional-curvature interface; it is not used in the frame computation on $U$ .

Facts & Assumptions

Given: An oriented Riemannian surface, a smooth positive orthonormal frame on an open set $U$ , its connection one-form $ω$ , and the Levi–Civita connection of $g$ .

[F1]

The frame equations $\nabla_{X} E_{1} = ω (X) E_{2}$ and $\nabla_{X} E_{2} = - ω (X) E_{1}$ hold for the connection form $ω (X) = g (\nabla_{X} E_{1}, E_{2})$ (Connection one-form of an oriented orthonormal frame).

[F2]

The curvature of a connection is $R^{\nabla} (X, Y) Z = \nabla_{X} \nabla_{Y} Z - \nabla_{Y} \nabla_{X} Z - \nabla_{[X, Y]} Z$ (Curvature of an affine connection).

[F3]

The invariant formula for the exterior derivative of a one-form is $d η (X_{0}, X_{1}) = X_{0} η (X_{1}) - X_{1} η (X_{0}) - η ([X_{0}, X_{1}])$ (The exterior derivative by the invariant vector-field formula).

[F4]

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

[F5]

The sectional curvature of a two-plane with ordered basis $(X, Y)$ is $K (σ) = Rm (X, Y, Y, X) / (g (X, X) g (Y, Y) - g (X, Y)^{2})$ , and its interface assumes ${A C}_{ω}$ , declared through The Axiom of Countable Choice ( ${A C}_{ω}$ ) (Sectional curvature).

[F6]

On an oriented Riemannian $n$ -manifold, the Riemannian volume form is ${vol}_{g} = \sqrt{\det G_{x}} d x^{1} \land \dots \land d x^{n}$ in positively oriented charts (Riemannian volume form on an oriented manifold).

Proof

technique · Compute the curvature tensor on the frame, identify the result with the exterior derivative of the connection form, and compare with the area form on that frame

1.1F1givenalgebra

From [F1], $\nabla_{E_{2}} E_{2} = - ω (E_{2}) E_{1}$ and $\nabla_{E_{1}} E_{2} = - ω (E_{1}) E_{1}$ ; differentiating these two expressions along $E_{1}$ and $E_{2}$ with the product rule gives $\nabla_{E_{1}} \nabla_{E_{2}} E_{2} = - E_{1} (ω (E_{2})) E_{1} - ω (E_{2}) ω (E_{1}) E_{2}$ and $\nabla_{E_{2}} \nabla_{E_{1}} E_{2} = - E_{2} (ω (E_{1})) E_{1} - ω (E_{1}) ω (E_{2}) E_{2}$ , while $\nabla_{[E_{1}, E_{2}]} E_{2} = - ω ([E_{1}, E_{2}]) E_{1}$ .

1.2F6givenalgebra

Let $(θ^{1}, θ^{2})$ be the dual coframe of $(E_{1}, E_{2})$ , so that $θ^{i} (E_{j}) = δ_{j}^{i}$ ; since the frame is $g$ -orthonormal, the Gram matrix in this frame is the identity, and [F6] gives $d A = θ^{1} \land θ^{2}$ on $U$ ; hence $d A (E_{1}, E_{2}) = θ^{1} (E_{1}) θ^{2} (E_{2}) - θ^{1} (E_{2}) θ^{2} (E_{1}) = 1$ .

2.1F2F3step 1.1algebra

Subtracting the three identities of step 1.1 in the order of [F2] cancels the $E_{2}$ -components and gives $R (E_{1}, E_{2}) E_{2} = - (E_{1} ω (E_{2}) - E_{2} ω (E_{1}) - ω ([E_{1}, E_{2}])) E_{1} = - d ω (E_{1}, E_{2}) E_{1}$ , the last equality being the invariant formula of [F3] with $X_{0} = E_{1}$ , $X_{1} = E_{2}$ .

3.1F4F5step 2.1

The pair $(E_{1}, E_{2})$ is an ordered orthonormal basis of each tangent plane, so [F4] and [F5] give $K = g (R (E_{1}, E_{2}) E_{2}, E_{1}) = Rm (E_{1}, E_{2}, E_{2}, E_{1})$ at every point of $U$ ; step 2.1 then yields $K = - d ω (E_{1}, E_{2})$ . The countable-choice assumption carried by [F5] enters only through that published interface and is declared by The Axiom of Countable Choice ( ${A C}_{ω}$ ); the frame computation on $U$ uses none of it.

4.1step 1.2step 3.1algebra∎

At each point of $U$ , the two-forms $d ω$ and $- K d A$ agree on the basis $(E_{1}, E_{2})$ of the tangent plane by steps 1.2 and 3.1; since a two-form is determined by its value on any basis, $d ω = - K d A$ on $U$ .

Source locator

Lee, Riemannian Manifolds: An Introduction to Curvature, Chapter 9, §“The Gauss–Bonnet Formula,” printed p. 165, equation (9.4) and the following structure equation, writes the frame equations with $ω_{s t d} = - ω$ ; Datar, Lectures on Riemannian Geometry, Lecture 2, §2.1, printed pp. 11–12, does the same in Lemma 2.1.1. The two-frame computation above is carried out in the sign convention of this page, so that $d ω = - K d A$ ; the sign is checked again by the direct spherical and hyperbolic metric computations in the examples of this pair.

Depends on

Used by

Corner terms are required even in the plane Counterexample
Wrong boundary orientation reverses the disk term Counterexample
Area defect of a hyperbolic geodesic triangle Example
Area excess of a spherical geodesic triangle Example
Euclidean annulus boundary signs Example
Euclidean disk boundary curvature Example
Flat torus and zero Euler characteristic Example
Gauss-Bonnet for a spherical cap Example
Projective-plane curvature via a hemisphere Example
Total curvature of a round sphere Example
A boundary term is necessary False statement
A surface connection form depends on its frame False statement
The Gauss-Bonnet expression is independent of the metric Lemma
Local Gauss-Bonnet for a frameable disk region Theorem

Dependency tree · two levels

24 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

John M. Lee, Riemannian Manifolds: An Introduction to Curvature (1997) (standard reference, not scraped)
Ved Datar, Lectures on Riemannian Geometry (standard reference, not scraped)