Gauss' Theorem Egregium   

The mathematician Karl Gauss was the first to study Gaussian curvature, and in doing so, he proved that K is an intrinsic property of the surface. That is, the Gaussian curvature of a surface can be measured by ''inhabitants'' of the surface without regard to the larger space the surface is embedded in. For example, a person could measure the curvature of the earth's surface even if he was blind and could not see that the earth was situated in a larger cosmos. He would simply need the formula we are about to derive for the curvature in terms of the fundamental form of the surface.

Before doing so, however, let us define the notation

g11,u =  ¶g11 ¶u     and        g22,v =  ¶g22 ¶v

This notation often leads us to call differentiation by either u or v a comma derivative of a coefficient.       

Theorem Egregium: Let r( u, v) be an orthogonal parameterization of a surface and let g = g₁₁g₂₂. Then the curvature of the surface is given in terms of the metric coefficients by

K =  -1 2Ög é ë  ¶ ¶v æ è  g11,v Ög ö ø +  ¶ ¶u æ è  g22,u Ög ö ø ù û

(3)

       

At the end of this section, we have a very accessible proof of the Theorem Egregium. However, before proving the theorem, let's examine its value as a means of exploring the geometry of a surface given only its fundamental form.       

EXAMPLE 6    What is the curvature of the hyperbolic plane, which is the upper half plane with the Poincare fundamental form

ds2 =  du2+dv2 v2

Solution: Since ds² = v^-2du²+v^-2dv², the metric coefficients are g₁₁ = g₂₂ = v^-2. Thus,

g11,v = -2v-3,                g22,u = 0

and g = g₁₁g₂₂ = v^-4. The theorem Egregium thus yields

K =  -1 2 v-4 é ê ë  ¶ ¶v æ ç è  -2v-3 v-4 ö ÷ ø +  ¶ ¶u æ ç è  0 v-4 ö ÷ ø ù ú û =  -1 2v-2 é ë  ¶ ¶v (-2v-1) +  ¶ ¶u ( 0) ù û =  -1 2v-2 ( 2v-2) = -1

Thus, the curvature of the hyperbolic plane is K = -1. That is, the hyperbolic plane is a surface of constant negative curvature, and as a result, it cannot be studied as a surface in ordinary 3 dimensional space. Instead, all information about the hyperbolic plane must come from the intrinsic properties derived from its fundamental form.

       

Proof of the Theorem Egregium: Let's show that (3) holds at a given point P on the surface. Moving the surface around in space doesn't change the curvature at P, so let's orient the surface so that P is at the origin of the xyz coordinate system and the unit normal is n = á 0,0,1 ñ (i.e., the unit vector k on the z-axis). Also, let's orient the surface so that r_u is in the x-direction and r_v is in the y-direction. The key is to show that the curvature is independent of the normal vector n. However, if r( u,v) = á x(u,v) ,y( u,v) ,z( u,v) ñ , then

ruu = á xuu,yuu,zuu ñ ,    ruv = á xuv,yuv,zuv ñ ,   and    rvv = á xvv,yvv,zvv ñ

so that our assumptions lead us to

K =  ( ruu· n) ( rvv· n) -( ruv· n)2 || ru|| 2|| rv|| 2 =  zuu zvv - zuv2 g11g22

But by definition of the inner product, we also have

ruu· rvv-ruv· ruv = xuuxvv-xuv2+yuuyvv-yuv2+zuuzvv-zuv2 zuuzvv-zuv2 = ruu· rvv-ruv· ruv-xuuxvv+xuv2-yuuyvv+yuv2

Combined with g = g₁₁g₂₂, we thus obtain

K =  1 g ( ruu· rvv - ruv· ruv - xuuxvv + xuv2 - yuuyvv + yuv2)

(4)

At this point, we have eliminated the dependence on the extrinsic quantity n. In fact, (4) is exactly (3), but with vector notation instead of metric coefficients. We simply need to translate.

To do so, let's first expand the derivatives in (3):

K =  -1 2Ög é ë  g11,vv Ög -  1 2  g11,v g3/2 gv+  g22,uu Ög -  1 2  g22,u g3/2 gu ù û =  -g11,vv 2g +  g11,v 2g  gv 2g -  g22,uu 2g +  g22,u 2g  gu 2g

The product rule implies that g_v = g_11,v g₂₂ + g₁₁ g_22,v and g_u = g_11,u g₂₂+g₁₁ g_22,u.  Thus, we have

gv 2g =  g11,vg22+g11g22,v 2g11g22 =  g11,v 2g11 +  g22,v 2g22  gu 2g =  g11,ug22+g11g22,u 2g11g22 =  g11,u 2g11 +  g22,u 2g22

Substituting into the expression for K yields

K =  1 g æ è  -g11,vv 2 +  ( g11,v) 2 4g11 +  g11,vg22,v 4g22 -  g22,uu 2 +  g22,ug11,u 4g11 +  ( g22,u) 2 4g22 ö ø

        Now let's look at each term individually. First, g₁₁ = r_u· r_u implies that

g11,v = ruv· ru+ru· ruv = 2ruv· ru

Likewise, g₂₂ = r_v· r_v implies that g_22,u = 2r_uv· r_v. Thus, we now have

K =  1 g æ è  -g11,vv 2 +  ( 2ruv·ru) 2 4ru· ru +  g11,vg22,v 4g22 -  g22,uu 2 +  g22,ug11,u 4g11 +  ( 2ruv·rv) 2 4rv· rv ö ø

However, r_u at the origin points in the i direction, so that r_u· r_u = x_u² and r_uv· r_u = x_uvx_u. Similarly, r_v· r_v = y_v² and r_uv· r_v = y_uvy_u. Thus,

K =  1 g æ è  -g11,vv 2 +  4xuv2xu2 4xu2 +  g11,vg22,v 4g22 -  g22,uu 2 +  g22,ug11,u 4g11 +  4yuvyv2 4yv2 ö ø =  1 g æ è  -g11,vv 2 +xuv2+  g11,vg22,v 4g22 -  g22,uu 2 +  g22,ug11,u 4g11 +yuv2 ö ø

We have not obtained 2 of the terms in (4). Similar manipulations translate the remaining terms. For example, 

g11,u = ruu· ru + ru· ruu = 2ruu· ru = 2xuuxu

since r_u points in the x-direction at the origin. Likewise, g_22,v = 2y_vvy_v, so that

K =  1 g æ è  -g11,vv 2 +xuv2+  g11,v2yvvyv 4yv2 -  g22,uu 2 +  g22,u2xuuxu 4xu2 +yuv2 ö ø

Now let's notice that r_u· r_v = 0 implies that

¶ ¶u ( ru·rv) = 0        and             ¶ ¶v ( ru·rv) = 0 ruu·rv+ru·ruv = 0        and        ruv·rv+ru·rvv = 0

from which we obtain r_u· r_uv = -r_uu· r_v and r_uv· r_v = -r_u· r_vv. As a result,

g11,v = 2ruv· ru = -2ruu· rv = -2yuuyv g22,u = 2rv· ruv = -2ru· rvv = -2xuxvv

so that

K =  1 g æ è  -g11,vv 2 +xuv2-  2yuuyv2yvvyv 4yv2 -  g22,uu 2 -  2xuxvv2xuuxu 4xu2 +yuv2 ö ø =  1 g æ è  -g11,vv 2 +xuv2-yuuyvv-  g22,uu 2 -xuuxvv+yuv2 ö ø

Finally, let's notice that g_11,v = 2r_uv· r_u and g_22,u = -2r_u· r_vv implies that

g11,vv = 2ruvv·ru+2ruv·ruv g22,uu = -2ruu·rvv-2ru·rvvu

Substitution then leads to

K =  1 g ( -ruvv· ru-ruv· ruv+xuv2-yuuyvv+2ruu· rvv+2ru· rvvu-xuuxvv+yuv2) =  1 g ( ruu· rvv-ruv· ruv+xuv2-xuuxvv+yuv2-yuuyvv+ru· rvvu-ruvv· ru)

Since r_uvv = r_vvu, we have shown that (3) is the same as is the same as (4).