The Fundamental Form of a Surface

Properties of a curve or surface which depend on the coordinate space that curve or surface is embedded in are called extrinsic properties of the curve. For example, the slope of a tangent line is an extrinsic property since it depends on the coordinate system in which rises and runs are measured. In contrast, intrinsic properties of surfaces are properties that can be measured within the surface itself without any reference to a larger space.  

For example, the length of a curve is an intrinsic property of the curve, and thus, the length of a curve r( t) = r( u( t) ,v( t) ) , t in [ a,b] , on a surface r( u,v) is an intrinsic property of both the curve itself and the surface that contains it.  As we saw in the last section, the square of the speed of r(t) is

æ è  ds dt ö ø 2   = g11 æ è  du dt ö ø 2   +2g12 æ è  du dt ö ø æ è  dv dt ö ø +g22 æ è  dv dt ö ø 2

in terms of the metric coefficients

g11 = ru· ru,    g12 = rv· ru,    and    g22 = rv· rv

Thus, very short distances ds on the surface can be approximated by

( ds)2  = g11( du)2  +  2g12 dudv  +  g22 (dv)2

(1)

That is, if du and dv are sufficiently small, then ds is the length of an infinitesimally short curve on the surface itself. Equation (1) is the fundamental form of the surface, which intrinsic to a surface because it is related to distances on the surface itself.        

EXAMPLE 1    Find the fundamental form of the right circular cylinder of radius R, which can be parameterized by

r( u,v) = á Rcos( u), Rsin( u), v ñ

Solution: Since r_u = á -Rsin( u),Rcos( u), 0 ñ and r_v = á0, 0, 1ñ , the metric coefficients are

g11 = ru·ru = R2sin2( u)+R2cos2( u) +02 = R2 g12 = ru·rv = 0+0+0 = 0 g22 = rv·rv = 02+02+12 = 1

Thus, ds² = R²du²+dv².       

If the parameterization is orthogonal, then g₁₂ = r_u·r_v = 0, so that

ds2 = g11du2  +  g22dv2

For example, the xy-plane is parameterized by r( u,v) = á u,v,0 ñ ,which implies that r_u = i and r_v = j and that g₁₁ = g₂₂ = 1, g₁₂ = 0.  The fundamental form for the plane is

ds2 = du2+dv2

which is, in fact, the Pythagorean theorem.  Moreover, distances are not altered when a sheet of paper is rolled up into a cylinder, which means that a cylinder should have the same fundamental form as the plane.   Indeed, if R = 1 in example 1, then ds² = du² + dv².              

EXAMPLE 2    Find the fundamental form of the sphere of radius R centered at the origin in the spherical coordinate parameterization

r( f,q) = á Rsin( f) cos( q), Rsin( f) sin(q), Rcos( f) ñ

Solution: To do so, we first compute the derivatives r_f and r_q:

rf = á Rcos( f) cos(q) ,Rcos( f) sin( q) ,-Rsin( f) ñ rq = á -Rsin( f) sin(q) ,Rsin( f) cos( q),0 ñ

It then follows that

rf·rf = R2cos2( f) cos2( q) +R2cos2( f)sin2( q) +R2sin2( f) = R2cos2( f) +R2sin2( f) = R2

and thus, g₁₁ = R². Moreover, spherical coordinates is an orthogonal parameterization, which means that r_f·r_q = 0. Thus, g₁₂ = 0. Finally, g₂₂ is given by

g22 = rq·rq = R2sin2(f) sin2( q) +R2sin2( f) cos2( q) = R2sin2(f)

As a result, the fundamental form of the sphere of radius R is given by

ds2 = R2df2  +  R2sin2(f) dq2

(2)

That is, the ''hypotenuse'' of a spherical ''right triangle'' corresponds to a ''horizontal'' arc of length Rsin(f) dq and a ''vertical'' arc of length Rdf.