Differential Forms 07: Gauss-Bonnet and Curvature Forms

Differential forms / 07

Gauss-Bonnet and curvature forms

Gaussian curvature is a scalar, but total curvature is the integral of the 2-form $$K\,dA$$.

Calculus root: total turning

Curvature begins in plane-curve calculus. If a unit-speed plane curve has tangent angle \(\theta(s)\), then its signed curvature is

\[\kappa(s)=\theta'(s).\]

Therefore

\[\int \kappa\,ds=\Delta\theta,\]

the total turning of the tangent. Gauss-Bonnet replaces the curve by a surface and replaces curvature times arclength by the 2-form \(K\,dA\).

Example: circle

A circle of radius \(R\) has curvature \(1/R\) and arclength \(2\pi R\). Hence

\[\int \kappa\,ds=2\pi.\]

This is the one-dimensional ancestor of total curvature formulas.

Example: polygon angle defect

For a convex polygon, curvature is concentrated at corners. The total turning is the sum of exterior angles, equal to \(2\pi\). Gauss-Bonnet spreads this same accounting over a smooth surface, with curvature density \(K\,dA\) and topological total \(2\pi\chi(M)\).

Section 4.5 of the PDF treats the area element and Gaussian curvature as invariant data on a surface. In an oriented chart, a parametrization \(\sigma(u,v)\) gives

\[dA=\sqrt{EG-F^2}\,du\wedge dv,\]

and the curvature contribution is the 2-form

\[K\,dA.\]

The key theorem for a compact oriented surface without boundary is

\[\int_M K\,dA=2\pi\chi(M).\]

The unit sphere

For the unit sphere, use the outward-oriented coordinate order \((\phi,\theta)\):

\[\sigma(\phi,\theta)=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi).\]

The first fundamental form coefficients in this order are

\[E=1,\qquad F=0,\qquad G=\sin^2\phi,\]

so the outward area form is

\[dA=\sin\phi\,d\phi\wedge d\theta.\]

Since \(K\equiv1\),

\[\int_{S^2}K\,dA =\int_0^\pi\int_0^{2\pi} \sin\phi\,d\theta\,d\phi =4\pi.\]

This agrees with \(2\pi\chi(S^2)=2\pi\cdot2\).

Example 1: sphere of radius \(R\)

For radius \(R\), the outward area element is \(R^2\sin\phi\,d\phi\wedge d\theta\) and the Gaussian curvature is \(K=1/R^2\). Hence

\[K\,dA=\sin\phi\,d\phi\wedge d\theta,\]

and the total curvature is still \(4\pi\). Scaling changes local area and local curvature inversely.

The torus of revolution

Let \(R>r>0\) and

\[\sigma(\theta,\phi)=((R+r\cos\phi)\cos\theta,(R+r\cos\phi)\sin\theta,r\sin\phi).\]

Then

\[E=(R+r\cos\phi)^2,\qquad F=0,\qquad G=r^2,\]

so

\[dA=r(R+r\cos\phi)\,d\theta\wedge d\phi.\]

The standard curvature formula is

\[K(\phi)={\cos\phi\over r(R+r\cos\phi)}.\]

Therefore

\[K\,dA=\cos\phi\,d\theta\wedge d\phi,\]

and

\[\int_{T^2}K\,dA =\int_0^{2\pi}\int_0^{2\pi}\cos\phi\,d\phi\,d\theta=0.\]

This agrees with \(\chi(T^2)=0\).

Example 2: where the torus curvature changes sign

Because \(r(R+r\cos\phi)>0\), the sign of \(K\) is the sign of \(\cos\phi\). The outer band has \(K>0\), the inner band has \(K<0\), and the top and bottom circles have \(K=0\). Gauss-Bonnet says the positive and negative contributions cancel.

What the theorem says and does not say

Gauss-Bonnet does not say curvature is determined pointwise by topology. It says the integral of the curvature 2-form is. A sphere can have nonconstant curvature under a non-round metric, but the integral remains \(4\pi\). A genus \(g\) surface satisfies

\[\int_M K\,dA=2\pi(2-2g).\]

Proof mechanism

In a moving-frame proof one chooses local orthonormal coframes, writes a connection 1-form \(\omega_{12}\), and has a structure equation

\[d\omega_{12}=-K\,dA\]

up to convention. On overlapping coordinate patches, Stokes’ theorem converts local boundary terms into transition-angle contributions. Those transition terms sum to the Euler characteristic. The proof is therefore a global bookkeeping refinement of Stokes.