9.8: The Intrinsic Form of the Gauss-Bonnet Theorem

Applying Green's Theorem

In the previous section, we showed, using orthogonal coordinates, that

∂Dκgds + ∑ext.angles = 2π + ∂DE1'(t) · E2(t))dt

In order to derive the Gauss-Bonnet Theorem from this equation, we need to show that ∂DE1'(t) · E2(t))dt = -DΚ(u,v)dA. When we decomposed the vectors T(t) and U(t) in the tangent space, we did so using an arbitrary pair of orthonormal coordinate vectors E1(t) and E2(t). Here we consider the case where the partial derivative vectors are perpendicular, so that g12 = 0 for the surface. For such a surface, we can use E1(t) = Xu(t)/√g11 and E2(t) = Xv(t)/√g22 as our orthonormal frame. So,

E1'(t) = [Xuu(t)u'(t) + Xuv(t)v'(t)]/√g11(t)   +   Xu(t)(1/√g11(t))'

Then, taking the dot product with E2(t) gives us:

E1'(t) · E2(t) = [(Xuu(t) · Xv(t))u'(t) + (Xuv(t) · Xv(t))v'(t)] / √g

Recall that,

Xuu · Xv = g22Γ112 = -(1/2)(∂g11/∂v)
Xuv · Xv = g22Γ122 = (1/2)(∂g22/∂u)

So,

E1'(t) · E2(t) = (-1/2√g)(∂g11/∂v)u'(t) + (1/2√g)(∂g22/∂u)v'(t)

At this point, we make use of a very important result from multivariable calculus called Green's Theorem. Let D be a region on a surface and let ∂D be its boundary. Green's Theorem relates area integrals over the region D to line integrals over the boundary ∂D.

∂D(P(u,v)du + Q(u,v)dv) = D[(∂Q(u,v)/∂u) - (∂P(u,v)/∂v)]dudv

Let (-1/2√g)(∂g11/∂v) = P(u,v) and let (1/2√g)(∂g22/∂u) = Q(u,v). Then,

∂DE1'(t) · E2(t))dt = ∂D(P(u,v)du + Q(u,v)dv)                                      
       = D[(∂Q(u,v)/∂u) - (∂P(u,v)/∂v)]dudv
                                          = D(1/2)[∂/∂u(1/√g · (∂g22/∂u)) + ∂/∂v(1/√g · (∂g11/∂v))]dudv

It turns out that this last result is exactly what we need in order to derive the Gauss-Bonnet Formula for the case of surfaces with orthogonal partial derivative vectors. In lab 8, we wrote down the intrinsic formula for Guassian curvature:

Κ(u,v) = - (1/2√g)[∂/∂u(1/√g · (∂g22/∂u)) + ∂/∂v(1/√g · (∂g11/∂v))]

Using this equation to calculate the total Gaussian curvature over a region D gives us

DΚ(u,v)dA = DΚ(u,v)√gdudv                                               
                                        = -D(1/2)[∂/∂u(1/√g · (∂g22/∂u)) + ∂/∂v(1/√g · (∂g11/∂v))]dudv
= -∂DE1'(t) · E2(t))dt                         

Combining this last calculation with the result from the previous section, we get

DΚdA + ∂Dκgds + ∑ αi = 2π ,

and thereby complete our derivation of the Gauss-Bonnet Theorem using Green's Theorem and the intrinsic formula for Gaussian curvature.

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