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Total curvature of a round sphere

Example

Assume the axiom of choice. Let SR2 be the round sphere of radius R>0 with its induced metric and standard orientation. Then K≡1/R2, the area is 4πR2, and ∫SR2K dA=1R2⋅4πR2=4π=2πχ(SR2) with χ(SR2)=2 counted from the octahedral curvilinear triangulation V=6, E=12, F=8. The polar chart used for the curvature computation misses only the two poles and the seam, where K is obtained by continuity.

Facts & Assumptions

Given: The round sphere SR2 of radius R>0 with the metric induced from R3, its standard orientation, and the spherical polar chart X(θ,φ)=R(sin⁡θcos⁡φ,sin⁡θsin⁡φ,cos⁡θ).

[A1]

full AC is assumed; it is inherited through the global Gauss-Bonnet theorem quoted below and also covers the countable-choice hypothesis inherited by the curvature structure equation; the explicit computations add no choice (The Axiom of Choice).

[F1]

For every closed oriented Riemannian surface, ∫SR2K dA=2πχ(SR2) (Global Gauss-Bonnet for closed oriented surfaces).

[F2]

For a regular embedded surface patch the induced metric coefficients are the Euclidean Gram products of its parameter tangent vectors, and its area density is the square root of the Gram determinant (The first fundamental form, Gram matrix, and area density of a surface patch).

[F3]

The area of a regular patch is the integral of its Gram area density over its parameter domain; changing bounded integrands on a content-zero parameter boundary does not change that integral (Surface area and scalar surface integrals on a regular patch, Content-zero parameter-boundary exceptions do not affect surface integrals).

[F4]

In coordinates the Levi-Civita symbols are Γkij=12∑ℓgkℓ(∂igjℓ+∂jgiℓ−∂ℓgij) (Christoffel formula for the levi civita connection).

[F5]

For a smooth positive orthonormal frame with connection form ω(X)=g(∇Xe1,e2) one has ∇Xe1=ω(X)e2 and dω=−K dA; the area form satisfies dA=e1∧e2 for the dual coframe and is the unique positive unit top form of the orientation (Connection one-form of an oriented orthonormal frame, Gaussian curvature structure equation, The riemannian volume form is the unique positive unit top form, Riemannian volume form on an oriented manifold).

[F6]

A finite face-to-face piecewise C2 curvilinear triangulation of a compact smooth surface has the well-defined count χ(SR2)=V−E+F (Curvilinear face-to-face triangulation, Topological well-definedness of the surface Euler characteristic).

Verification

technique · compute the round metric, connection form and curvature in polar coordinates, integrate the constant curvature against the sphere's area, and count an explicit triangulation
1.1F2F4algebra

Direct differentiation of X(θ,φ)=R(sin⁡θcos⁡φ,sin⁡θsin⁡φ,cos⁡θ) gives ⟨Xθ,Xθ⟩=R2, ⟨Xφ,Xφ⟩=R2sin⁡2θ, and ⟨Xθ,Xφ⟩=0. By [F2] these are the induced metric coefficients, so e1=(1/R)∂θ and e2=(1/(Rsin⁡θ))∂φ are a smooth positive orthonormal frame with dual coframe e1=R dθ, e2=Rsin⁡θ dφ. By [F4], Γθφφ=−sin⁡θcos⁡θ and Γφθφ=Γφφθ=cot⁡θ are the only nonzero symbols, so ∇e1e1=0 and ∇e2e1=(cot⁡θ/R)e2.

1.2F6givenconstruct

Let u1,u2,u3 be the standard unit coordinate vectors in R3. The six points ±Ru1,±Ru2,±Ru3 are the vertices of the regular octahedron inscribed in SR2; radial projection of its boundary gives a face-to-face curvilinear triangulation of SR2 with V=6, E=12 (the octahedron's edges, each a great-circle arc) and F=8 (the spherical triangles cut out by the coordinate octants), so by [F6] χ(SR2)=6−12+8=2.

2.1F5step 1.1algebra

By step 1.1, ω(e1)=0 and ω(e2)=cot⁡θ/R, hence ω=cos⁡θ dφ; then dω=−sin⁡θ dθ∧dφ, while dA=e1∧e2=R2sin⁡θ dθ∧dφ by [F5]. Comparing with dω=−K dA gives K=1/R2 on the chart. The chart domain is dense in SR2 (its complement is the two poles together with the seam φ=0), and both K and the constant 1/R2 are continuous on SR2, so K≡1/R2 on all of SR2.

2.2F2F3F5step 1.1algebra

The Gram determinant of step 1.1 is R4sin⁡2θ, so [F2] gives area density R2sin⁡θ on 0<θ<π, 0<φ<2π. The excluded poles and longitude seam are the parameter-boundary exceptions of [F3] and contribute zero area. Hence Area⁡(SR2)=∫02π∫0πR2sin⁡θ dθ dφ=4πR2. This is also the Riemannian area form of [F5].

3.1F1step 2.1step 2.2step 1.2algebra

By steps 2.1 and 2.2, ∫SR2K dA=(1/R2)⋅4πR2=4π, and by step 1.2, 2πχ(SR2)=2π⋅2=4π; the global identity [F1] is therefore verified on the round sphere, ∫SR2K dA=2πχ(SR2).

4.1A1step 3.1∎

No new choice is made: the polar chart, the octahedral vertices and the triangulation are explicit, and AC licenses the inherited global Gauss-Bonnet theorem of [F1] and the countable-choice assumption of the structure equation in [F5].

Source locator

Lee, Riemannian Manifolds: An Introduction to Curvature, Chapter 9, Theorem 9.7, printed pp. 167-172, gives the global identity, and Datar, Lectures on Riemannian Geometry, Lecture 2, Theorem 2.2.4, printed pp. 14-15, states it. The round metric and total area are computed directly above from the Gram coefficients and patch area density The first fundamental form, Gram matrix, and area density of a surface patch, while χ(SR2)=2 is counted from the octahedral triangulation.

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