§4. The Second Fundamental Form In the last section we developed the theory of intrinsic geometry of surfaces by consid- ering 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 R3. 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 E1, E2, E3 for a Darboux frame for a surface S with E3 normal to S, then d⊥F = (dF · E3) E3. (4.1) Since dF · E3 is a 1-form, we may consider hdF · E3, Gi 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: x2 + y2 = 1 with −∞ < z < ∞, and the Darboux frame E1 = (−y, x, 0), E2 = (0, 0, 1) and E3 = (x, y, 0) (at the point (x, y, z) on C). A vector field F tangent to C can be expressed as F = f1E1 + f2E2, where f1 and f2 are scalars depending on the point at which F is evaluated. Find an expression for the 1-form dF · E3. Also, given another tangential vector field G = g1E1 + g2E2, find an expression for the pairing hdF · E3, Gi. Solution: We have dE1 = (−dy, dx, 0), dE2 = (0, 0, dz) and E3 = (x, y, 0). Hence dE1 · E3 = −xdy + ydx and dE2 · E3 = 0. Thus dF · E3 = (df1 E1 + f1dE1 + df2 E2 + f2dE2) · E3 = f1(−xdy + ydx). Also, hdF · E3, Gi = hf1(−xdy + ydx), Gi = f1(−xhdy, Gi + yhdx, Gi). For convenience, 3 introduce the standard basis e1 = (1, 0, 0), e2 = (0, 1, 0) and e3 = (0, 0, 1) of R . Then hdy, Gi = ∇y · G = e2 · G = g1e2 · E1 + g2e2 · E2 = g1x. Similarly, hdx, Gi = −g1y. Thus 2 2 hdF · E3, Gi = f1(−x(g1x) + y(−g1y)) = −f1g1(x + y ) = −f1g1, in view of the fact that (x, y, z) is a point on C and hence x2 + y2 = 1. Example 4.2. Take a parametric curve γ: Y = Y (s), Z = Z(s) in YZ-plane with Y (s) > 0, assuming the arc-length parametrization: Y 0(s)2 + Z0(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 ru and rv along “u-curves” and “v-curves” are respectively given 0 0 0 by ru = (−Y (v) sin u, Y (v) cos u, 0) and rv = (Y (v) cos u, Y (v) sin u, Z (v)). An easy computation shows ru · rv = 0. Normalizing, we have unit tangent fields E1 and E2. Taking their cross product, we get a unit normal field E3: −1 E1 = |ru| ru = (− sin u, cos u, 0), −1 0 0 0 E2 = |rv| rv = (Y (v) cos u, Y (v) sin u, Z (v)) 0 0 0 E3 = E1 × E2 = (Z (v) cos u, Z (v) sin u, −Y (v)). Find hdF · E3, Gi for given tangent fields F = f1E1 + f2E2 and G = g1E1 + g2E2. Solution: dF · E3 = (df1E1 + f1dE1 + df2E2 + f2dE2) · E3 = f1dE1 · E3 + f2dE2 · E3. Note that dE1 = (− cos u du, − sin u du, 0) and hence 0 0 0 dE1 · E3 = (− cos u du)(Z (v) cos u) + (− sin u du)(Z (v) sin u) = −Z (v) du. 0 00 0 00 00 Also, dE2 = (−Y (v) sin u du + Y (v) cos u dv, Y (v) cos u du + Y (v) sin u dv, Z (v)dv). 0 00 0 00 A short computation shows dE2 · E3 = {Z (v)Y (v) − Y (v)Z (v)} dv. Thus 0 0 00 0 00 dF · E3 = −f1Z (v) du − f2{Y (v)Z (v) − Z (v)Y (v)} dv. To find hdF · E3, Gi, we need to know hdu, Gi and hdv, Gi. Recall Rule VF2 from §1.3 that hdu, Uru + V rvi = U and hdv, Uru + V rvi = V . On the other hand, we have −1 −1 −1 G = g1E1 + g2E2 = g1|ru| ru + g2|rv| rv = g1Y (v) ru + g2rv. −1 Thus hdu, Gi = g1Y (v) and hdv, Gi = g2. So we have −1 0 0 00 0 00 hdF · E3, Gi = −{Y (v) Z (v)} f1g1 − {Y (v)Z (v) − Z (v)Y (v)} f2g2. (4.2) Exercise 1 in §4.1 tells us that Y 0Z00 −Z0Y 00 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 hdF · E3, Gi is a symmetric bilinear form in F and G. As we shall see, this is not accidental! A remarkable thing about the expression dF · E3 is the identity d(gF) · E3 = g(dF · E3), (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 · E3 = d⊥F · E3 obtained from (4.1). There is a better way to see (4.3): differentiate the identity F · E3 = 0 (which is a consequence of assumption that F is tangential and E3 is normal) to get dF · E3 + F · dE3 = 0, or dF · E3 = −F · dE3. (4.4) Thus (4.3) follows from the obvious identity (gF) · dE3 = g(F · dE3). Identity (4.3) is remarkable because it shows that the 1-form dF · E3 at a point p on the surface S only depends on Fp, 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: Fp = Gp. 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 · E3 − dG · E3 = d(F − G) · E3 = d(fH) · E3 = fdH · E3, which vanishes at p. The reader should find out how this argument breaks down when dF · E3 is replaced by dF in order to enhance his or her understanding. The vector-valued differential form dE3 on S will be called the Weingarten form for S. One can see that dE3 is tangential by differentiating E3 ·E3 = 1, giving us E3 ·dE3 = 0. 1 2 Actually, we know this already: recall dE3 = ω3E1 + ω3E2 from (3.6) in §4.3, from which we see that dE3 is tangential because both E1 and E2 are. On the LHS of (11.4), E3 is normal and dF in general is neither normal nor tangential. However, both F and dE3 on the RHS are tangential. Before going any further, let us recall the basic setting from the last section: given a Darboux frame Ej (j = 1, 2, 3) for the surface S, where E1, E2 tangential and E3 is j 1 2 3 normal. The fundamental forms θ are determined by dr = θ E1 + θ E2 (θ = 0), and their differentials satisfy the first structural equations 1 2 1 2 1 2 3 1 3 2 3 dθ = θ ∧ ω2, dθ = θ ∧ ω1, (dθ = ) 0 = θ ∧ ω1 + θ ∧ ω . (4.5) j P3 k k j The connection forms ωk are determined by dEj = k=1 ωj Ek, with ωj + ωk = 0 and their differentials satisfy the second structural equations 2 3 2 3 2 3 3 1 2 dω1 = ω1 ∧ ω3, dω1 = ω1 ∧ ω2, dω2 = ω2 ∧ ω1. (4.6) Now we claim: 3 1 2 Rule VF3. θ and θ form a coframe of S dual to the tangent frame E1, E2 in the j j sense that hθ , Eki = δk. j 1 2 1 2 Here δk is Kronecker’s delta; in detail, hθ , E1i = hθ , E2i = 1 and hθ , E2i = hθ , E1i = 0. That θj (j = 1, 2) is a coframe means that every 1-form ω defined on S can be written as 1 2 ω = g1θ + g2θ for some scalar functions g1 and g2 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 1 3 2 3 1 2 θ ∧ ω1 + θ ∧ ω2 = 0. Since θ , θ form a coframe for S, we may put 3 1 2 3 1 2 ω1 = h11θ + h12θ and ω2 = h21θ + h22θ . (4.7) 1 3 2 3 1 2 2 1 1 2 1 2 Thus 0 = θ ∧ ω1 + θ ∧ ω2 = h12θ ∧ θ + h21θ ∧ θ = (h12 − h21)θ ∧ θ . Since θ ∧ θ is the area form, it is non-vanishing everywhere. So we must have h12 − h21 = 0, or h12 = h21. (4.8) 1 2 1 2 Let F = f E1 + f E2 and G = g E1 + g E2 be vector fields tangent to S, where j j f = F · Ej and g = G · Ej.