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## Surfaces

As in the case of curves we will begin with the study of a surface S defined by a single equation f (x, y, z) = 0 and eventually treat more general surfaces. If p = (x0, y0, z0) is a point on S, and a line l through p is given in parametric form (x0 + tx1, y0 + ty1, z0 + tz1), then we can define the order of contact r(l, S, p) as the highest power of t which divides the the function F(t) = f (x0 + tx1, y0 + ty1, z0 + tz1). A tangent line is then one which has the largest order of contact among all lines. The locus of all tangent lines describes a conical surface with vertex p. In fact,

Exercise 10   Either r(l, S, p) 2 for all lines through p or there is a plane P so that lines l with r(l, S, p) 2 are precisely lines lying in the plane and passing through p. The latter case occurs if and only if the vector (fx(p), fy(p), fz(p)) is non-zero. By a replacing the equation f = 0 by a non-zero multiple we can ensure that the vector ( fx(p), fy(p), fz(p)) is a unit vector, which is then called the unit normal to the surface.

As in the case of curves we call the former case the singular surfaces and the latter are called smooth or non-singular surfaces. We will restrict our attention to smooth surfaces.

picture(9565,5259)(556,-4750) (991,-4696)(0,0)[lb]Smooth surface and its plane of tangent lines (6166,-4696)(0,0)[lb]Singular surface and its cone of tangents
We define the linear space of tangent vectors Tp(S) to S at p as the space of all vectors v so that the parametric line p + tv is tangent to S at p.

Exercise 11   Show that the tangent space is precisely,

{v = (x1, y1, z1) | fx(x0, y0, z0)x1 + fy(x0, y0, z0)y1 + fz(x0, y0, z0)z1}

In other words v is tangent if and only if Dv(f )(p) = 0.

The curvature of curves defined in the previous section can be formulated as follows. We have a map from a plane curve C to the unit circle given by sending each point to the unit tangent vector at that point (we need to choose a direction'' along the curve but that can be done locally). As we saw the derivative of this map is the curvature. Now the unit normal direction is a right-angle rotation from the tangent vector, so we could equally well have used the map which takes a point to its unit normal. The latter map also makes sense for a surface S in 3-space; we have a natural map

p n(p) =

which sends S to the (unit) sphere. Curvature should be a measure of the derivative of this map. Let v be any vector in Tp(S) or equivalently v be a vector orthogonal to n(p).

Exercise 12   Show that Dv(n)(p) is orthogonal to n(p) also. (Hint: Use the equation n . n = 1 for all p).

Thus, if v and w are linearly independent vectors in Tp(S) (and hence form a basis of this vector space), then we have

Dv(n)(p) = av + bw andDw(n)(p) = cv + bw

for some constants a, b, c and d.

Definition 4   We define the Gaussian curvature of S at p to be the determinant ad - bc of the above linear transformation.

Exercise 13   Show that for any function f, vectors v, w and constant a we have D(av + w)(f )= aDv(f )+ Dw(f ). Now show that the above definition is independent of the choice of vectors v and w.

We can additionally justify the above definition by noting that the Gaussian curvature of a plane is 0. In order to understand the Gaussian curvature better we must first understand the notion of straight lines'' or geodesics. To begin with let us examine curves lying on the surface. Let p = (x0, y0, z0) be a point of S and (x(t), y(t), z(t)) be a curve on S passing through p at t = 0.

Exercise 14   Show that the tangent vector (xt(0), yt(0), zt(0)) is in the tangent space Tp(S).

We have already seen that a curve has non-trivial curvature if the acceleration required to travel along it at constant speed is non-zero. Thus a curve must be considered straight on the surface S if the projection of this acceleration into Tp(S) is zero. In other words:

Definition 5   Let (x(t), y(t), z(t)) be a parametrised curve (of some order r 2) on the surface S. Moreover, suppose that the speed ((xt)2 + (yt)2 + (zt)2)1/2 is a constant. We say that the curve is geodesic on S if the acceleration at any point of the curve is a multiple of the unit normal to S at that point.

Exercise 15   Consider a parametric solution of order 3 to the equations f (x(t), y(t), z(t)) = 0 and (xt)2 + (yt)2 + (zt)2 = c; morover suppose that the acceleration is a multiple of the normal to S. Show that we obtain the equation for the acceleration vector

(x2, y2, z2) = - (fx(p), fy(p), fz(p))

where, p = (x0, y0, z0) is the starting point'', (x1, y1, z1) is the initial tangent vector and Qf is the quadratic form given by the second derivatives of f, that is

Qf(a, b, c) = fxxa2 + 2fxyab + 2fxzac + fyyb2 + 2fyzbc + fzzc2

The above equation can be written as

(xtt, ytt, ztt)(p) = - (fx(p), fy(p), fz(p))

Given any point (x0, y0, z0) on the surface and a tangent direction (x1, y1, z1) to S at that point, we can inductively solve this equation to obtain the terms upto any required order and obtain the geodesic in parametric form.

Consider the quadratic form Q obtained by restricting the form Qf/(fx2 + fy2 + fz2)1/2 to the tangent space Tp(S).

Theorem 1   Let Q be a quadratic form on a vector space V (over the real numbers) there is an orthonormal basis {e1,..., en} of V so that Q can be written as

Q(u1e1 + ... + unen) = a1u12 + ...anen2

In particular, we have an orthonormal basis {e1, e2} for the tangent space Tp(S) so that Q(ue1 + ve2) = ku2 + lv2. Euler called the numbers k and l the principal curvatures of S.

Exercise 16   Show that the Gaussian curvature of S is kl.

picture(7682,6596)(2092,-7465)
Examples of different curvatures on a tube

We shall now show how one can use geodesics to compute curvature. Let p be any point on the surface S, then for each tangent direction w in Tp(S) we can form the parametric geodesic gp, w(t) of some order r. Let e1, e2 be a basis of Tp(S) as above and w = ue1 + ve2. Consider the map

(u, v) p(u, v) = gp, ue1 + ve2(1)

which gives a parametric form for S upto order r. This parametric form is called the geodesic normal form or geodesic normal co-ordinates and is determined by the geometry of the surface (that is, by understanding the equations of straight lines'' or geodesics on the surface).

To simplify things we take p to be the origin (by translation) and let e3 to be the unit normal to S at p. We use {e1, e2, e3} as a basis in which to express points of space. The above map then becomes (for r = 4),

p(u, v) = (u + A(u, v), v + B(u, v), - frac12Q(u, v) + C(u, v))

where A, B and C are homogeneous of degree 3.

Exercise 17   By using the geodesic equation show that A = Q(u, v)Qu(u, v)/6 and B = Q(u, v)Qv(u, v)/6.

By definition of p(u, v) the lines through the origin in the (u, v) plane go to geodesics in S. What about other lines? Consider the line l = (u0 + su1, v0 + sv1) parametrised by s in the (u, v) plane. It becomes a curve (s) = p(u0 + su1, v0 + sv1) in space. The velocity vector of this curve is (s) = puu1 + pvv1 and the acceleration is

a(s) = (s) = puuu12 + 2puvu1v1 + pvvv12

We need to find the component of this accleration in the tangent plane to S, i. e. is the space of pu and pv.

Exercise 18   Prove the following formulas valid modulo order 2 in u0 and v0
 a(s) . pu = kl /3v1(v0u1 - u0v1) a(s) . pv = kl /3u1(v0u1 - u0v1)

One checks that pu and pv are unit vectors modulo order 2 in u0 and v0. Thus, the magnitude of the acceleration is (kl /3)(v0u1 - u0v1) times the length of the vector (u1, v1) along the line l in the (u, v) plane. Thus the curvature of the image of lines parallel to the origin in the (u, v) plane is also related to the Gaussian curvature of S; which thus also measures the deviation from the parallel postulate.

Next: Manifolds Up: Coordinate Geometry Previous: Curves
Kapil H. Paranjape 2001-01-20