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Euclidean disk boundary curvature

Example

Assume the axiom of choice. Let $D = {x \in R^{2} : ∣ x ∣ \leq R}$ with $R > 0$ , the standard orientation and the Euclidean metric. Then $K \equiv 0$ , the positively oriented boundary is the counterclockwise circle, $k_{g} = 1 / R$ along it, and $\int_{D} K d A = 0, \int_{\partial D} k_{g} d s = 2 π, χ (D) = 1,$ so the Gauss-Bonnet identity reads $0 + 2 π = 2 π χ (D) = 2 π$ . The boundary term is exactly the missing $2 π$ : a flat disk has vanishing curvature but a positive boundary contribution.

Facts & Assumptions

Given: The radius $R > 0$ , the region $D = {x \in R^{2} : ∣ x ∣ \leq R}$ with the standard orientation of $R^{2}$ and the Euclidean metric.

[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, $\int_{D} K d A + \int_{\partial D} k_{g} d s = 2 π χ (D)$ with the outward-normal-first boundary orientation (Gauss-Bonnet with smooth boundary).

[F2]

On each boundary arc the outward-normal-first rule selects the unit tangent $T$ with $(ν, T)$ positive for an outward transverse vector $ν$ , and then $J T$ is the inward unit conormal; the signed geodesic curvature of a regular $C^{2}$ unit-speed curve is $k_{g} = g (\nabla_{T} T, J T)$ (Regular oriented surface regions with corners, Oriented Riemannian surface and positive quarter-turn, Signed geodesic curvature).

[F3]

The Euclidean Levi-Civita derivative on $R^{2}$ is $\nabla_{X} Y = \sum_{j} X (Y^{j}) \partial_{j}$ 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; in particular $\nabla_{T} T$ along a curve is the ordinary derivative of the unit tangent (Fundamental theorem of riemannian geometry).

[F4]

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

[F5]

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

Verification

technique · compute the curvature, the Euler characteristic and the boundary term of the Euclidean disk

1.1F3F4algebra

The frame $(\partial_{1}, \partial_{2})$ is orthonormal with $\nabla \partial_{i} = 0$ by [F3], so its connection form vanishes and $d ω = 0$ ; [F4] gives $K d A = 0$ , hence $K \equiv 0$ on $D$ and $\int_{D} K d A = 0$ .

1.2F5givenconstruct

The vertices $R (1, 0)$ , $R (- 1 / 2, \sqrt{3} / 2)$ , $R (- 1 / 2, - \sqrt{3} / 2)$ divide $\partial D$ into three regular $C^{2}$ arcs bounding a single closed face with three edges and three vertices, and the links are intervals at the boundary vertices; this is a finite curvilinear triangulation of $D$ with $V = 3$ , $E = 3$ , $F = 1$ , so by [F5] $χ (D) = 3 - 3 + 1 = 1$ .

1.3F2F3algebra

Parametrize $γ (t) = R (\cos t, \sin t)$ , $0 \leq t \leq 2 π$ . The outward unit normal is $ν = (\cos t, \sin t)$ , the unit tangent selected by the outward-normal-first rule is $T = (- \sin t, \cos t)$ with $(ν, T)$ positive, and $J T = - (\cos t, \sin t)$ . By [F3], $\nabla_{T} T = - (1 / R) (\cos t, \sin t)$ , so $k_{g} = g (- (1 / R) (\cos t, \sin t), - (\cos t, \sin t)) = 1 / R$ and $\int_{\partial D} k_{g} d s = (1 / R) \cdot 2 π R = 2 π$ .

2.1F1step 1.1step 1.2step 1.3algebra

By steps 1.1, 1.2 and 1.3, the Gauss-Bonnet identity [F1] holds in the form $0 + 2 π = 2 π \cdot 1$ ; the boundary integral supplies the entire right-hand side because the flat disk has $K \equiv 0$ .

3.1A1step 2.1∎

No new choice is made: the disk, its 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 and the boundary convention preceding it, printed pp. 163-167, gives the disk as the model computation: the counterclockwise circle of radius $R$ has signed geodesic curvature $1 / R$ and contributes $2 π$ . Datar, Lectures on Riemannian Geometry, Lecture 2, Theorem 2.0.1, printed pp. 10-13, states the same formula. The Euclidean connection is the published library item Fundamental theorem of riemannian geometry, and $χ (D) = 1$ is counted here from the explicit three-arc triangulation using Topological well-definedness of the surface Euler characteristic, not 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)