Previous: The Shape of a Surface
2.3. Characterization of the SphereA point P is called an umbilic if the curvatures of all normal sections are equal, so the normal curvature is independent of the direction in the tangent space. Since kN = II(V)/I(V), the condition for P to be an umbilic is equivalent to hij = cgij for all indices i, j. The constant c = c(u1, u2) depends on the point P.
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For a plane, all points are umbilic with normal curvature 0. For a sphere of radius r, all points are umbilic with normal curvature 1/r. We shall now prove the converse.
A surface consisting entirely of umbilics is either a plane or a sphere.
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We have c(u1,u2) XiXj = -Xij so (cXj + j)Xi = 0. Furthermore = 1 implies that j = 0, so (cXj + j) = 0 as well. Therefore cXj + j, being perpendicular to three independent vectors, must be the zero vector, so cXj + j = 0.. By differentiating this expression, we have
(*) cXij + ciXj + ij = 0
and similarly,
(**) cXji + cjXj + cij = 0.
Since we are assuming that the second partial derivatives of X and n are continuous, we have Xij= Xji and ij = ji. Subtracting (**) from (*), we obtain ciXj - cjXi = 0. Since X1 and X2 are linearly independent by hypothesis, it follows that ci = 0 = cj identically, so c is constant.
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If c = 0, then i = 0 = j so is constant, equal to a fixed unit vector A. Then (XA)i = XiA = 0 for i = 1, 2 so XA = d, a constant, and X therefore lies in the plane perpendicular to A at distance d from the origin.
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If c 0, then (X - (1/c)n)i = 0 for i = 1, 2, so X - (1/c) = C, a constant vector, and X(u1,u2) - C = (1/c)(u1,u2) for all u1, u2. Thus the length of X(u1,u2) - C is a constant, 1/c, and the parametrized surface lies on the sphere of radius 1/c about the center C.
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Corollary: A surface consisting entirely of planar points is a plane.
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