Gaussian Curvature   

The product of the principal curvatures is known as the Gaussian curvature of the surface, which is denoted by K. For example, the Gaussian curvature of the cylinder in example 2 is K = -1·0 = 0 and the Gaussian curvature of the Enneper minimal surface is

K =  -2 ( u2+v2+1) 2 ·  2 (u2+v2+1) 2 =  -4 ( u2+v2+1) 4

As another example, consider that since the geodesics of a sphere with radius R are great circles of the sphere, they each have a curvature of k = 1/R. Thus, the principal curvatures must be k₁ = 1/R and k₂ = 1/R, so that the curvature of a sphere of radius R is

K =  1 R2

We can use (2) to obtain a formula for the Gaussian curvature. Indeed, the difference of 2 squares identity leads to

k1k2 =  ( L+N) 2 4 - é ë æ è  L-N 2 ö ø 2   +M2 ù û =  L2+2LN+N2 4 -  L2-2LN+N2 4 -M2 =  4LN 4 -M2 = LN-M2

Substituting for L, M, and N then leads to

K =  ruu·n || ru|| 2 ·  rvv·n || rv||2 - æ è  ruv·n || ru||   || rv|| ö ø 2

Thus, if r( u,v) is an orthogonal parameterization, then

K =  ( ruu·n) ( rvv·n) -( ruv·n)2 || ru|| 2|| rv|| 2

EXAMPLE 4    What is the Gaussian curvature of the surface

r( u,v) = á cosh( v) cos(u) ,cosh( v) sin( u) ,v ñ ,   v ¹ 0

Solution: To begin with, r_u = á -cosh(v) sin( u) ,cosh( v) cos( u),0 ñ and r_v = á sinh( v)cos( u) ,sinh( v) sin( u),1 ñ . Thus,

ruu = á -cosh( v) cos( u),-cosh( v) sin( u) ,0 ñ ,    rvv = á cosh( v) cos( u),cosh( v) sin( u) ,0 ñ ruv = á -sinh( v) sin( u),sinh( v) cos( u) ,0 ñ

since r_u×r_v = á cosh(v) cos( u) ,cosh( v) sin( u),-sinh( v) cosh( v) ñ has a length of cosh²( v) , the unit normal is

n =   á cos( u) ,sin( u),-sinh( v) ñ cosh( v) =    1 cosh( v) á cos(u) ,sin( u) ,-sinh( v) ñ

Since r_vv = cosh( v) á cos(u) ,sin( u) ,0 ñ , we have

ruu·n =  -1 cosh( v) cosh( v) ( cos2( u) +sin2( u)+0) = -1

Note that r_uu = -r_vv and that r_uv·n = 0. Also, notice that || r_u|| = cosh( v) and || r_v|| = ( sinh²( v) +1) ^1/2 = cosh( v) . Thus, the Gaussian curvature of the surface is

K =  -1·1 cosh( v) cosh( v) = -sech2( v)

When a surface has a Gaussian curvature of 0 at every point, then we say that the surface is Gaussian flat. It can be shown that surfaces that are flat with respect to Gaussian curvature can be constructed by deforming a section of a plane without tearing, folding, or stretching the plane. For example, the cylinder is Gaussian flat because it can be formed by rolling up a sheet of paper. 

However, a sphere is not Gaussian flat because it cannot be constructed from a plane without some type of stretching and distorting of distances.

Finally, it can be shown that if r( u,v) is not orthogonal, then the curvature is

K =  ( ruu· n) ( rvv· n) - ( ruv· n)2 || ru|| 2|| rv||2-( ru· rv) 2

Let's use this to show that Gaussian curvature of a surface z = f(x,y) is closely related to the discriminant of f( x,y) .       

EXAMPLE 5    What is the Gaussian curvature of z = f(x,y) ?       

Solution: Since z = f( x,y) is parameterized by r( u,v) = á u,v,f( u,v) ñ , we have r_u = á 1,0,f_u ñ and r_v = á 0,1,f_v ñ . As a result,

|| ru|| 2|| rv||2-( ru·rv) 2 = (1+fu2) ( 1+fv2)-fu2fv2 = 1+fu2+fv2

Also, we have

ruu = á 0,0,fuu ñ ,    ruv = á 0,0,fuv ñ ,    and    rvv = á 0,0,fvv ñ

Moreover, U( x,y,z) = z-f( x,y) implies that ÑU = á -f_u,-f_v,1 ñ , so that

n =   á -fu,-fv,1 ñ 1 + fu2 + fv2  = P-1 á -fu,-fv,1 ñ

where P = ( 1 + f_u² + f_v²) ^-1/2. It follows that r_uu·n = Pf_uu, that r_vv·n = Pf_vv, and that r_uv = Pf_uv. Thus, the Gaussian curvature is

K =  PfuuPfvv - P2 fuv2 1+fu2+fv2 =  P2( fuufvv - fuv2) 1+fu2+fv2

Finally, since x = u and y = v, this reduces to

K =  fxxfyy-fxy2 ( 1+fx2+fy2) 2