d F = (df E 3 ) E 3. (4.1)

4. The Second Fundamental Form In the last section we developed the theory of intrinsic geometry of surfaces by considering the covariant differential d F, that is, the tangential component of df for a vector field F tangent to the given surface S. In the present section we take up the study of the shapes of surfaces by considering the normal component d F of df, which mainly tells us how S is embedded in the ambient space R 3. For example, a cylinder is intrinsically flat like a plane but its shape is different from a plane: for a plane, the normal differential d F is always zero, but this is not the case for a cylinder. Notice that, if E 1, E 2, E 3 for a Darboux frame for a surface S with E 3 normal to S, then d F = (df E 3 ) E 3. (4.1) Since df E 3 is a 1-form, we may consider df E 3, G for given vector fields F and G, which is a scalar function as the result of pairing a covector field with a vector field. Example 4.1. Consider the cylinder C: x 2 + y 2 = 1 with < z <, and the Darboux frame E 1 = ( y, x, 0), E 2 = (0, 0, 1) and E 3 = (x, y, 0) (at the point (x, y, z) on C). A vector field F tangent to C can be expressed as F = f 1 E 1 + f 2 E 2, where f 1 and f 2 are scalars depending on the point at which F is evaluated. Find an expression for the 1-form df E 3. Also, given another tangential vector field G = g 1 E 1 + g 2 E 2, find an expression for the pairing df E 3, G. Solution: We have de 1 = ( dy, dx, 0), de 2 = (0, 0, dz) and E 3 = (x, y, 0). Hence de 1 E 3 = xdy + ydx and de 2 E 3 = 0. Thus df E 3 = (df 1 E 1 + f 1 de 1 + df 2 E 2 + f 2 de 2 ) E 3 = f 1 ( xdy + ydx). Also, df E 3, G = f 1 ( xdy + ydx), G = f 1 ( x dy, G + y dx, G ). For convenience, introduce the standard basis e 1 = (1, 0, 0), e 2 = (0, 1, 0) and e 3 = (0, 0, 1) of R 3. Then dy, G = y G = e 2 G = g 1 e 2 E 1 + g 2 e 2 E 2 = g 1 x. Similarly, dx, G = g 1 y. Thus df E 3, G = f 1 ( x(g 1 x) + y( g 1 y)) = f 1 g 1 (x 2 + y 2 ) = f 1 g 1, in view of the fact that (x, y, z) is a point on C and hence x 2 + y 2 = 1. Example 4.2. Take a parametric curve γ: Y = Y (s), Z = Z(s) in Y Z-plane with Y (s) > 0, assuming the arc-length parametrization: Y (s) 2 + Z (s) 2 = 1. The surface of revolution obtained by revolving γ about the z-axis is, in parametric equations, x = x(u, v) = Y (v) cos u, y = y(u, v) = Y (v) sin u, z = z(u, v) = Z(v). 1

The tangent fields r u and r v along u-curves and v-curves are respectively given by r u = ( Y (v) sin u, Y (v) cos u, 0) and r v = (Y (v) cos u, Y (v) sin u, Z (v)). An easy computation shows r u r v = 0. Normalizing, we have unit tangent fields E 1 and E 2. Taking their cross product, we get a unit normal field E 3 : E 1 = r u 1 r u = ( sin u, cos u, 0), E 2 = r v 1 r v = (Y (v) cos u, Y (v) sin u, Z (v)) E 3 = E 1 E 2 = (Z (v) cos u, Z (v) sin u, Y (v)). Find df E 3, G for given tangent fields F = f 1 E 1 + f 2 E 2 and G = g 1 E 1 + g 2 E 2. Solution: df E 3 = (df 1 E 1 + f 1 de 1 + df 2 E 2 + f 2 de 2 ) E 3 = f 1 de 1 E 3 + f 2 de 2 E 3. Note that de 1 = ( cos u du, sin u du, 0) and hence de 1 E 3 = ( cos u du)(z (v) cos u) + ( sin u du)(z (v) sin u) = Z (v) du. Also, de 2 = ( Y (v) sin u du + Y (v) cos u dv, Y (v) cos u du + Y (v) sin u dv, Z (v)dv). A short computation shows de 2 E 3 = {Z (v)y (v) Y (v)z (v)} dv. Thus df E 3 = f 1 Z (v) du f 2 {Y (v)z (v) Z (v)y (v)} dv. To find df E 3, G, we need to know du, G and dv, G. Recall Rule VF2 from 1.3 that du, Ur u + V r v = U and dv, Ur u + V r v = V. On the other hand, we have G = g 1 E 1 + g 2 E 2 = g 1 r u 1 r u + g 2 r v 1 r v = g 1 Y (v) 1 r u + g 2 r v. Thus du, G = g 1 Y (v) 1 and dv, G = g 2. So we have df E 3, G = {Y (v) 1 Z (v)} f 1 g 1 {Y (v)z (v) Z (v)y (v)} f 2 g 2. (4.2) Exercise 1 in 4.1 tells us that Y Z Z Y is the curvature κ of the generating curve γ of S. Note that the last example is the special case with Y (v) = 1 and Z(v) = v. Identity (4.2) indicates that df E 3, G is a symmetric bilinear form in F and G. As we shall see, this is not accidental! A remarkable thing about the expression df E 3 is the identity d(gf) E 3 = g(df E 3 ), (4.3) 2

that is, the function g can be extracted. This follows immediately from Rule PN1 in the previous section and the obvious identity df E 3 = d F E 3 obtained from (4.1). There is a better way to see (4.3): differentiate the identity F E 3 = 0 (which is a consequence of assumption that F is tangential and E 3 is normal) to get df E 3 + F de 3 = 0, or df E 3 = F de 3. (4.4) Thus (4.3) follows from the obvious identity (gf) de 3 = g(f de 3 ). Identity (4.3) is remarkable because it shows that the 1-form df E 3 at a point p on the surface S only depends on F p, the vector of F at the point p, although df at p also depends on F at points near p. Indeed, suppose that we have another vector field G which coincides with F at the point p: F p = G p. Then F G is zero at p and hence we can write F G = fh for some smooth scalar function f with f(p) = 0 and some vector field H. Thus df E 3 dg E 3 = d(f G) E 3 = d(fh) E 3 = fdh E 3, which vanishes at p. The reader should find out how this argument breaks down when df E 3 is replaced by df in order to enhance his or her understanding. The vector-valued differential form de 3 on S will be called the Weingarten form for S. One can see that de 3 is tangential by differentiating E 3 E 3 = 1, giving us E 3 de 3 = 0. Actually, we know this already: recall de 3 = ω 1 3E 1 + ω 2 3E 2 from (3.6) in 4.3, from which we see that de 3 is tangential because both E 1 and E 2 are. On the LHS of (11.4), E 3 is normal and df in general is neither normal nor tangential. However, both F and de 3 on the RHS are tangential. Before going any further, let us recall the basic setting from the last section: given a Darboux frame E j (j = 1, 2, 3) for the surface S, where E 1, E 2 tangential and E 3 is normal. The fundamental forms θ j are determined by dr = θ 1 E 1 + θ 2 E 2 (θ 3 = 0), and their differentials satisfy the first structural equations dθ 1 = θ 2 ω 1 2, dθ 2 = θ 1 ω 2 1, (dθ 3 = ) 0 = θ 1 ω 3 1 + θ 2 ω 3. (4.5) The connection forms ω j k are determined by de j = 3 k=1 ωk j E k, with ω k j + ωj k = 0 and their differentials satisfy the second structural equations dω 2 1 = ω 3 1 ω 2 3, dω 3 1 = ω 2 1 ω 3 2, dω 3 2 = ω 1 2 ω 2 1. (4.6) Now we claim: 3

Rule VF3. θ 1 and θ 2 form a coframe of S dual to the tangent frame E 1, E 2 in the sense that θ j, E k = δ j k. Here δ j k is Kronecker s delta; in detail, θ1, E 1 = θ 2, E 2 = 1 and θ 1, E 2 = θ 2, E 1 = 0. That θ j (j = 1, 2) is a coframe means that every 1-form ω defined on S can be written as ω = g 1 θ 1 + g 2 θ 2 for some scalar functions g 1 and g 2 on S. This rule, a companion to Rule VF2 in 1.3, is a bit hard to swallow. Its justification needs some elaborate theoretical background and hence is postponed to the end of the section. The key for studying the shape of a surface is the last identity in (4.5), namely θ 1 ω1 3 + θ 2 ω2 3 = 0. Since θ 1, θ 2 form a coframe for S, we may put ω1 3 = h 11 θ 1 + h 12 θ 2 and ω2 3 = h 21 θ 1 + h 22 θ 2. (4.7) Thus 0 = θ 1 ω1 3 + θ 2 ω2 3 = h 12 θ 1 θ 2 + h 21 θ 2 θ 1 = (h 12 h 21 )θ 1 θ 2. Since θ 1 θ 2 is the area form, it is non-vanishing everywhere. So we must have h 12 h 21 = 0, or h 12 = h 21. (4.8) Let F = f 1 E 1 + f 2 E 2 and G = g 1 E 1 + g 2 E 2 be vector fields tangent to S, where f j = F E j and g j = G E j. We compute df E 3, G as follows. From (4.4), we have df E 3 = F de 3 = F (ω3 1 E 1 + ω3 2 E 2 ) = ω3 1 F E 1 ω3 2 F E 2 = f 1 ω3 1 f 2 ω3 2 = f 1 ω1 3 + f 2 ω2 3 Thus df E 3, G = f 1 ω1, 3 G + f 2 ω2, 3 G. Now ω1, 3 G = h 11 θ 1 + h 12 θ 2, g 1 E 1 + g 2 E 2 = h 11 g 1 + h 12 g 2, (4.9) in view of Rule VF3 above. Similarly, we have ω2, 3 G = h 21 g 1 + h 22 g 2. Finally, df E 3, G = f 1 (h 11 g 1 + h 12 g 2 ) + f 2 (h 21 g 1 + h 22 g 2 ) = 2 h jkf j g k. (4.10) j,k=1 Due to (11.8), the last expression is a symmetric bilinear form in (f 1, f 2 ) and (g 1, g 2 ), and, as a consequence, df E 3, G = dg E 3, F. We call II(F, G) df E 3, G de 3 F, G the second fundamental form for the surface. By the way, the first fundamental form is just the inner product: I(F, G) = F G (= f 1 g 1 + f 2 g 2 ). Both fundamental forms are symmetric in the sense that I(F, G) = I(G, F) and II(F, G) = II(G, F). 4

[ ] h11 h The trace and the determinant of the symmetric matrix 12 h 21 h 22 the second fundamental form, namely associated with H = h 11 + h 22 K = h 11 h 22 h 2 12, (4.11) are called the mean curvature and the total curvature of S respectively; (some books define the mean curvature H as h 11 h 22, or ( h 11 h 22 )/2 ). Example 4.3. Find the total curvature of the surface of revolution in Example 4.2. Solution: Comparing the answer to Example 4.2 with (4.10), we have h 11 = Z /Y, h 22 = {Y Z Z Y } and h 12 = h 21 = 0; (Y and Z are functions of v). So the total curvature K is {Y Z Z (Z ) 2 Y }/Y. Differentiating (Y ) 2 + (Z ) 2 = 1, we have Y Y + Z Z = 0. Thus K = {Y ( Y Y ) (Z ) 2 Y }/Y = {(Y ) 2 + (Z ) 2 }Y /Y = Y /Y. (In the next section we give another way to obtain this answer.) A pleasant surprise is: the total curvature K is the same as the Gaussian curvature K described in the last section; (the same letter K used here for the total curvature is not a mistake!) This is surprising because K = h 11 h 22 h 2 12 is an expression involving extrinsic quantities h jk, while the Gaussian curvature given in the last section, determined by dω2 1 = K θ 1 θ 2, is intrinsic. This highly nontrivial fact, known as theorem egregium of Gauss, is one of the most important discoveries in classical differential geometry. Now Cartan s method of moving frame makes it almost transparent. If you play with Cartan s structural equations long enough, you ll get it! Example 4.4. Prove the theorem egregium of Gauss. Solution: The first identity in (4.6) gives dω2 1 = ω2 3 ω3 1 = ω2 3 ( ω1) 3 = (h 21 θ 1 + h 22 θ 2 ) ( h 11 θ 1 h 12 θ 2 ) = (h 11 h 22 h 2 12) θ 1 θ 2. Comparing with dω2 1 = K θ 1 θ 2, we immediately get K = h 11 h 22 h 2 12. Done! Finally we explain why we should believe in Rule VF3. Let us first consider the situation that E j (j = 1, 2, 3) is a frame filled a spatial region U in R 3 ; (technically, U is 5

an open set in R 3 ). Again θ j is determined by dx = 3 j=1 θj E j, where x = (x 1, x 2, x 3 ). We claim: θ j, E k = δ j k. To prove this, we introduce the standard basis vectors for R3 : e 1 = (1, 0, 0), e 2 = (0, 1, 0) and e 3 = (0, 0, 1). Write E k = (E k1, E k2, E k3 ) (k = 1, 2, 3). Then E kl = E k e l (k, l = 1, 2, 3). Notice that x l = ( x1 x l, x 2 x l, ) x 3 = e l ; l = 1, 2, 3. x l Hence dx l, E k = x l E l = e l E k = E kl. On the other hand, from the identity 3 k=1 θk E k = dx and E j E k = δ jk, we have θ j = 3 k=1 θk E j E k = E j dx = (E j1, E j2, E j3 ) (dx 1, dx 2, dx 3 ) = 3 Thus θ j, E k = 3 l=1 E jl dx l, E k = 3 l=1 E jle kl = E j E k = δ j k. Done. l=1 E jldx l. Now we consider a Darboux frame E j (j = 1, 2, 3) for a surface S. This frame is defined only at points on S. We extend each E j to some Ẽj (j = 1, 2, 3) to some spatial region U, which is a neighborhood of S. θ j, Ẽk = δ j k (j, k = 1, 2, 3), where θ j are 1-forms determined by It follows from our previous discussion that dx = 3 j=1 θ j Ẽ j. (4.12) Clearly, the restriction of each Ẽj to S is E j for each j. Along the surface S, the tangential component (dx) of dx is, in view of (4.12) above and the fact that the restriction of Ẽ j to S is E j, given by (dx) = θ 1 E 1 + θ 2 E 2. So the restriction of θ 1 and θ 2 to S are θ 1 and θ 2 respectively. Thus θ j, E k = θ j, Ẽk = δ j k. There is one question remains: why can we extend E j to Ẽj? This is a very technical question and we answer briefly as follows. Anyone with proper background in modern mathematical analysis should have no problem to fill in the details of the proof outlined below. The uninterested reader may skip this technical Mumbo Jumbo. To begin with, we assert that every (smooth) function f on S can be extended to a function f defined on a (spatial) neighborhood of S. This can at least be done in a local level, mainly because at any point p of the surface we can choose a local coordinate (u, v, w) so that in a neighborhood of p, S is described by the equation w = 0. To extend f globally, we may use a partition of unity to piece local extensions together. Once we know that a (smooth) function can be extended, we can extend every (smooth) vector field by extending its components. Now let us agree that each vector 6

field E j of the given moving frame can be extended to a vector field Ēj. The problem here is that {Ēj (j = 1, 2, 3)} is not necessarily an orthonormal frame. All we can say is that, at a point close enough to S, {Ēj } is almost orthonormal and hence is linearly independent. We may allow the frame Ēj (j = 1, 2, 3) to be linearly independent at each point by shrinking its domain appropriately. Now we can use the Gram-Schmidt process to obtain an orthonormal frame out of Ēj (j = 1, 2, 3), which is our required extension of the Darboux frame E j (j = 1, 2, 3). Exercises 1. In each of the following parts, find the second fundamental form, the mean curvature and the total curvature of the given surface of revolution: (a) Cone: x 2 + y 2 = z 2 (z > 0). (b) Sphere: x 2 + y 2 + z 2 = R 2 (R > 0). (c) Paraboloid: x 2 + y 2 = z. (d) Hyperboloid: x 2 + y 2 = 1 + z 2 (e) Catenoid: x = cosh u cos v, y = cosh u sin v, z = u. (f) Psuedosphere: x = sech u cos v, y = sech u sin v, z = tanh u u. (g) Torus: x = (R + r cos u) cos v, y = (R + r cos u) sin v, z = r sin u (0 < r < R). Hint: Apply the solution to Example 4.2 with caution: the generating curve Y = Y (v), Z = Z(v) there has the arc-length parametrization: (Y ) 2 + (Z ) 2 = 1. 2. (a) What happens to the mean curvature H and the total curvature K of a surface if the unit normal field E 3 on S is reversed, that is, replaced by E 3? (b) Same question if S is dilated. In detail, for a constant a > 0, let S a be the surface such that a point p is on S if and only if ap is on S a. Notice that S 1 = S. Denote by H a and K a the mean curvature and the total curvature of S a respectively. What are the relations between H a and H 1, and between K a and K 1? 3. Show that if the mean curvature H and the total curvature K of a surface S are vanishing everywhere, then S is a plane. Hint: Since [h ij ] is a symmetric matrix with vanishing eigenvalues (H, K = the sum and the product of eigenvalues), we must have h ij 0. By (11.8), ω 3 1 = ω 3 2 = 0. So the Weingarten form de 3 also vanishes.) 4. We say that a point p on a surface S is umbilic, if h 11 = h 22 and h 12 = h 21 = 0 at that point, in other words, there is a scalar λ such that II(F, G) p = λ I(F, G) p for all tangential vector fields F and G. Prove that, if every point of a surface S is umbilic, then S is either planar or spherical. 7

Hint: By assumption, there is a scalar function such that ω 1 3 = λ θ 1 and ω 2 3 = λ θ 2. So dn de 3 = ω 1 3E 1 + ω 3 2E 2 = λ(θ 1 E 1 + θ 2 E 2 ) = λ dx. Assume that the surface S is parametrized by u, v so that x, N and λ are functions of u, v. Then dn = λ dx gives N u = λ x u and N v = λx v. So N uv = λ v x u + λx uv and N vu = λ u x v + λx vu. Hence λ v x u = λ u x v. From x u x v 0 we can derive λ u = λ v = 0. Thus λ is a constant. Now dn = λdx can be rewritten as d(n λx) = 0. 5. Let S be a surface given by parametric equation r = r(u, v). Recall that the metric tensor for S is ds 2 = g uu du 2 + 2g uv dudv + g vv dv 2, where g uu = r u r u, g uv = r u r v and g vv = r v r v. Also, N = r u r v is normal to S with N = (g uu g vv g 2 uv) 1/2. We may take E 3 = N/ N. Let L uu = r uu N, L uv = r uv N, L vv = r vv N; (in classical differential geometry, the second fundamental form is defined to be the symmetric tensor L uu du 2 + 2L uv dudv + L vv dv 2, which differs from ours, as we shall see). Let F and G be tangential vector fields. (a) Show that df E 3, G = N 1 dn F, G. get dn F = N df. Recall that E 3 = N/ N. Hint: Again, from N F = 0 we (b) Verify that dn F, G = L uu F u G u + L uv F u G v + L vu F v G u + L vv F v G v, where F u = F, du, F v = F, dv, G u = G, du and G v = G, dv. Hint: By Rule VF2 in 1.3, we have F = F, du r u + F, dv r v. (c) Show that the mean curvature and the total curvature of S are given by H = L uug vv 2L uv g uv + L vv g uu g uu g vv g 2 uv, K = L uul vv L uv g uu g vv guv 2. 6. In each of the following parts, use the result of the last exercise to compute the mean curvature and the total curvature of the given parametric surface: (a) the unit sphere, using the stereographic parametrization. (b) the torus given in part (g) of Exercise 1 above. (c) the helicoid r(u, v) = (u cos v, u sin v, bv) (b > 0); (note that, for fixed v, equation r = r(u, v) with u as a parameter describes a line). 7. Suppose that a surface S is given by an equation of the form z = h(x, y). Show that the mean curvature and the total curvature of S are given by H = (1 + h2 x)h yy 2h x h y h xy + (1 + h 2 y)h xx (1 + h 2 x + h 2 y) 3/2, K = h xxh yy h 2 xy (1 + h 2 x + h 2 y) 2. 8