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Gauss-Bonnet for closed nonorientable surfaces

Statement

Assume the axiom of choice through the triangulation and density-integration suppliers. Let $M$ be a closed compact Riemannian surface that is nonorientable, with Riemannian metric $g$ , Gaussian curvature $K$ , and orientation-free Riemannian area density $μ_{g}$ . Then $\int_{M} K μ_{g} = 2 π χ (M),$ where $χ (M)$ is the Euler characteristic of the smooth surface. No orientation of $M$ , no orientation double cover and no surface classification is used.

Facts & Assumptions

Given: A closed compact nonorientable Riemannian surface $(M, g)$ , its Gaussian curvature $K$ , and the orientation-free area density $μ_{g}$ .

[A1]

Full AC is assumed through the finite curvilinear triangulation supplier and its arbitrary-Jordan-curve inputs; density integration needs only its countable-choice consequence (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Every compact smooth Riemannian surface admits a finite face-to-face curvilinear triangulation whose closed faces lie in frameable coordinate disks (Finite curvilinear triangulation of a compact Riemannian surface).

[F2]

For a closed, possibly nonorientable compact Riemannian surface with arbitrary orientations on the frameable faces of a finite curvilinear triangulation, one has $\int_{M} K μ_{g} = 2 π (V - E + F)$ (Summing local Gauss-Bonnet over a supplied triangulation).

[F3]

Any two finite face-to-face piecewise $C^{2}$ curvilinear triangulations of a compact smooth surface have the same $V - E + F$ , and this common metric-independent value is written $χ (M)$ (Topological well-definedness of the surface Euler characteristic).

[F4]

The Riemannian density $μ_{g}$ and the orientation-free integral of a continuous function against it are defined without a choice of orientation (Riemannian volume density, Orientation-free density integration and its properties).

Proof

technique · choose a finite curvilinear triangulation, orient its faces arbitrarily, apply the summation lemma and substitute the well-defined Euler characteristic

1.1A1F1given

The surface $M$ is a compact smooth surface with empty boundary, so [F1] produces a finite face-to-face curvilinear triangulation $T = (V, E, F, ϕ)$ ; every closed face lies in a frameable coordinate disk and is a compact regular disk region with ordinary corners.

2.1A1F2F4step 1.1

Since $M$ is closed, orient each face arbitrarily and apply [F2] to obtain $\int_{M} K μ_{g} = 2 π (V - E + F)$ . The density integral of [F4] needs no global orientation, and reversing one face orientation reverses both its positive boundary tangent and its quarter-turn, leaving the inward conormal and the cancellation intact.

2.2F3step 1.1

The triangulation $T$ is a finite face-to-face curvilinear triangulation of the compact smooth surface $M$ , so [F3] identifies its count with the surface invariant: $V - E + F = χ (M)$ , independently of the triangulation and of any metric.

3.1F2F3step 2.1step 2.2algebra∎

Substituting step 2.2 into step 2.1 gives $\int_{M} K μ_{g} = 2 π χ (M)$ , which is the asserted identity; no orientation, orientation cover, or classification statement was used.

Source locator

Lee, Riemannian Manifolds: An Introduction to Curvature, Chapter 9, printed pp. 167-172, proves the global theorem by finite-face summation; Datar, Lectures on Riemannian Geometry, Lecture 2, Section 2.2, printed pp. 13-15, gives the same summation with facewise choices. Here the supplier is the curvilinear triangulation Finite curvilinear triangulation of a compact Riemannian surface, and the orientation-independent cancellation is proved in Summing local Gauss-Bonnet over a supplied triangulation.

Depends on

Used by

Projective-plane curvature via a hemisphere Example
Gauss-Bonnet alone does not classify surfaces False statement

Dependency tree · two levels

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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)