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A boundary term is necessary

Statement

Assume the axiom of choice. False: for a compact oriented Riemannian surface with nonempty smooth boundary the Gauss-Bonnet formula has no boundary term, that is $\int_{M} K d A = 2 π χ (M)$ . The Euclidean disk of radius $R > 0$ has $K \equiv 0$ , $χ (D) = 1$ and $\int_{\partial D} k_{g} d s = 2 π \neq 0$ , so omitting the boundary integral would assert $0 = 2 π$ .

Facts & Assumptions

Given: The claim that the closed-surface form $\int_{M} K d A = 2 π χ (M)$ holds verbatim on surfaces with boundary, to be refuted by an explicit Euclidean disk.

[A1]

full AC is assumed; it is inherited through the smooth-boundary Gauss-Bonnet corollary quoted below and is used nowhere else in this Euclidean computation (The Axiom of Choice).

[F1]

For a compact oriented Riemannian surface with smooth boundary, $\int_{M} K d A + \int_{\partial M} k_{g} d s = 2 π χ (M)$ with the outward-normal-first boundary orientation (Gauss-Bonnet with smooth boundary).

[F2]

On each smooth boundary arc the tangent is oriented by the outward-normal-first rule: for an outward transverse vector $ν$ , the selected unit tangent $T$ is the one for which $(ν, T)$ is positive; $J T$ is then the inward unit conormal (Regular oriented surface regions with corners, Oriented Riemannian surface and positive quarter-turn).

[F3]

For a regular $C^{2}$ unit-speed curve with tangent $T$ , the signed geodesic curvature is $k_{g} = g (\nabla_{T} T, J T)$ , and $\nabla_{T} T$ is the covariant acceleration (Signed geodesic curvature).

[F4]

The Euclidean metric on $R^{2}$ has Levi-Civita derivative $\nabla_{X} Y = \sum_{j} X (Y^{j}) \partial_{j}$ in Cartesian coordinates, with vanishing Christoffel symbols: ordinary directional differentiation is torsion free and metric compatible for the constant Euclidean metric, so uniqueness identifies it as the Levi-Civita connection; the covariant derivative along a curve is the ordinary derivative of the vector field, and the coordinate frame $(\partial_{1}, \partial_{2})$ is orthonormal with $\nabla \partial_{i} = 0$ (Fundamental theorem of riemannian geometry).

[F5]

For a smooth positive orthonormal frame with connection form $ω$ , $d ω = - K d A$ (Gaussian curvature structure equation).

[F6]

A finite face-to-face piecewise $C^{2}$ curvilinear triangulation of a compact smooth surface with smooth boundary has a well-defined count $χ (M) = V - E + F$ (Curvilinear face-to-face triangulation, Topological well-definedness of the surface Euler characteristic).

Refutation

technique · compute the boundary term of a Euclidean disk and show that the boundary-free identity would be false

1.1F4F5givenalgebra

Let $D = {x \in R^{2} : ∣ x ∣ \leq R}$ with $R > 0$ , the standard orientation and the Euclidean metric. By [F4] the orthonormal frame $(\partial_{1}, \partial_{2})$ has $\nabla \partial_{i} = 0$ , so its connection form $ω = 0$ and $d ω = 0$ ; [F5] gives $K d A = 0$ , and since $d A \neq 0$ at every point, $K \equiv 0$ on $D$ . Thus $\int_{D} K d A = 0$ .

1.2F6given

The boundary circle is divided by the three vertices $R (1, 0)$ , $R (- 1 / 2, \sqrt{3} / 2)$ , $R (- 1 / 2, - \sqrt{3} / 2)$ into three regular $C^{2}$ arcs, and the single closed face they bound has those three vertices and three edges, with link an interval at each boundary vertex and non-antipodal one-sided velocities; this is a finite curvilinear triangulation of $D$ with $V = 3$ , $E = 3$ , $F = 1$ , so by [F6] $χ (D) = 3 - 3 + 1 = 1$ .

1.3F2F3F4algebra

Parametrize the boundary circle by $γ (t) = R (\cos t, \sin t)$ , $0 \leq t \leq 2 π$ . Its unit tangent is $T = (- \sin t, \cos t)$ , the outward unit normal is $ν = (\cos t, \sin t)$ and $(ν, T)$ is positively oriented, so this is the outward-normal-first boundary orientation of [F2]; the inward unit conormal is $J T = - (\cos t, \sin t)$ . By [F4] the covariant acceleration is the ordinary acceleration of the unit-speed parametrization, $\nabla_{T} T = - (1 / R) (\cos t, \sin t)$ , and [F3] gives $k_{g} = g (- (1 / R) (\cos t, \sin t), - (\cos t, \sin t)) = 1 / R$ .

2.1step 1.1step 1.3algebra

By step 1.3 the boundary integral is $\int_{\partial D} k_{g} d s = (1 / R) \cdot 2 π R = 2 π$ , while $\int_{D} K d A = 0$ by step 1.1; the boundary contribution is therefore nonzero.

3.1F1step 1.1step 1.2step 2.1algebra

Omitting the boundary integral would make the Gauss-Bonnet identity read $\int_{D} K d A = 2 π χ (D)$ , that is $0 = 2 π χ (D) = 2 π \cdot 1$ , which is false. With the boundary term retained, [F1] reads $0 + 2 π = 2 π χ (D)$ , consistent with $χ (D) = 1$ . Hence the asserted boundary-free identity is false, and the boundary term is indispensable.

4.1A1step 3.1∎

No new choice is made: the Euclidean disk, its three-arc triangulation and the circle parametrization are explicit, and full AC entered only through the inherited smooth-boundary corollary of [F1].

Source locator

Lee, Riemannian Manifolds: An Introduction to Curvature, Chapter 9, Theorem 9.3, printed pp. 165-167, gives the Gauss-Bonnet formula with the boundary geodesic-curvature integral, and Datar, Lectures on Riemannian Geometry, Lecture 2, Theorem 2.0.1, printed pp. 10-13, states the same formula. The disk computation follows Lee's worked boundary case: the circle of radius $R$ has signed geodesic curvature $1 / R$ in the outward-normal-first orientation, contributing $2 π$ , which no closed-surface statement can produce. The Euclidean frame and connection are the library item Fundamental theorem of riemannian geometry, and $χ (D) = 1$ is counted here from an explicit three-arc curvilinear triangulation rather than imported from classification.

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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)