§4. the Second Fundamental Form in the Last Section We Developed The

§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 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. We compute hdF · E3, Gi as follows. From (4.4), we have

1 2 1 2 dF · E3 = −F · dE3 = −F · (ω3 E1 + ω3 E2) = −ω3 F · E1 − ω3 F · E2 1 1 2 2 1 3 2 3 = −f ω3 − f ω3 = f ω1 + f ω2

1 3 2 3 Thus hdF · E3, Gi = f hω1, Gi + f hω2, Gi. Now

3 1 2 1 2 1 2 hω1, Gi = hh11θ + h12θ , g E1 + g E2i = h11g + h12g , (4.9)

3 1 2 in view of Rule VF3 above. Similarly, we have hω2, Gi = h21g + h22g . Finally,

X2 1 1 2 2 1 2 j k hdF · E3, Gi = f (h11g + h12g ) + f (h21g + h22g ) = hjkf g . (4.10) j,k=1

Due to (11.8), the last expression is a symmetric bilinear form in (f 1, f 2) and (g1, g2), and, as a consequence, hdF · E3, Gi = hdG · E3, Fi. We call

II(F, G) ≡ hdF · E3, Gi ≡ hdE3 · F, Gi 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 1g1 + f 2g2). Both fundamental forms are symmetric in the sense that I(F, G) = I(G, F) and II(F, G) = II(G, F).

4 · ¸ h h The trace and the determinant of the symmetric matrix 11 12 associated with h21 h22 the second fundamental form, namely

2 H = h11 + h22 K = h11h22 − h12, (4.11) are called the mean curvature and the total curvature of S respectively; (some books deﬁne the mean curvature H as −h11 − h22, or (−h11 − h22)/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

0 0 00 0 00 h11 = −Z /Y, h22 = −{Y Z − Z Y } and h12 = h21 = 0;

(Y and Z are functions of v). So the total curvature K is {Y 0Z0Z00 − (Z0)2Y 00}/Y . Diﬀerentiating (Y 0)2 + (Z0)2 = 1, we have Y 0Y 00 + Z0Z00 = 0. Thus

K = {Y 0(−Y 0Y 00) − (Z0)2Y 00}/Y = −{(Y 0)2 + (Z0)2}Y 00/Y = −Y 00/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 2 not a mistake!) This is surprising because K = h11h22 − h12 is an expression involving extrinsic quantities hjk, while the Gaussian curvature given in the last section, determined 1 1 2 by dω2 = K θ ∧θ , 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

1 3 1 3 3 dω2 = ω2 ∧ ω3 = ω2 ∧ (−ω1) 1 2 1 2 2 1 2 = (h21θ + h22θ ) ∧ (−h11θ − h12θ ) = (h11h22 − h12) θ ∧ θ .

1 1 2 2 Comparing with dω2 = K θ ∧ θ , we immediately get K = h11h22 − h12. Done!

Finally we explain why we should believe in Rule VF3. Let us ﬁrst consider the 3 situation that Ej (j = 1, 2, 3) is a frame ﬁlled a spatial region U in R ; (technically, U is

5 3 P3 j an open set in R ). Again θj is determined by dx = j=1 θ Ej, where x = (x1, x2, x3). j j 3 We claim: hθ , Eki = δk. To prove this, we introduce the standard basis vectors for R : e1 = (1, 0, 0), e2 = (0, 1, 0) and e3 = (0, 0, 1). Write Ek = (Ek1,Ek2,Ek3)(k = 1, 2, 3).

Then Ek` = Ek · e` (k, ` = 1, 2, 3). Notice that µ ¶ ∂x1 ∂x2 ∂x3 ∇x` = , , = e` ; ` = 1, 2, 3. ∂x` ∂x` ∂x`

Hence hdx`, Eki = ∇x` · El = e` · Ek = Ek`. On the other hand, from the identity P3 k k=1 θ Ek = dx and Ej · Ek = δjk, we have

X3 X3 j k θ = θ Ej · Ek = Ej · dx = (Ej1,Ej2,Ej3) · (dx1, dx2, dx3) = Ej`dx`. k=1 `=1

j P3 P3 j Thus hθ , Eki = `=1Ej`hdx`, Eki = `=1Ej`Ek` = Ej · Ek = δk. Done.

Now we consider a Darboux frame Ej (j = 1, 2, 3) for a surface S. This frame is

deﬁned only at points on S. We extend each Ej to some E˜ j (j = 1, 2, 3) to some spatial region U, which is a neighborhood of S. It follows from our previous discussion that ˜j ˜ j ˜j hθ , Eki = δk (j, k = 1, 2, 3), where θ are 1-forms determined by

X3 ˜j dx = θ E˜ j. (4.12) j=1

Clearly , the restriction of each E˜ j to S is Ej 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 ˜1 ˜2 ˜1 ˜2 1 E˜ j to S is Ej, given by (dx)> = θ E1 + θ E2. So the restriction of θ and θ to S are θ 2 j ˜j ˜ j and θ respectively. Thus hθ , Eki = hθ , Eki = δk. There is one question remains: why can we extend Ej to E˜ j ? 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 Ej of the given moving frame can be extended to a vector ﬁeld E¯ j. The problem here is that {E¯ j (j = 1, 2, 3)} is not necessarily an orthonormal frame. All we can say is that, at a point close enough to S, {E¯ j} is almost orthonormal and hence is linearly independent. We may allow the frame E¯ j (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 E¯ j (j = 1, 2, 3), which is our required extension of the Darboux frame Ej (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: x2 + y2 = z2 (z > 0). (b) Sphere: x2 + y2 + z2 = R2 (R > 0). (c) Paraboloid: x2 + y2 = z. (d) Hyperboloid: x2 + y2 = 1 + z2 (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 0)2 + (Z0)2 = 1.

2. (a) What happens to the mean curvature H and the total curvature K of a surface

if the unit normal ﬁeld E3 on S is reversed, that is, replaced by −E3 ? (b) Same

question if S is dilated. In detail, for a constant a > 0, let Sa be the surface such

that a point p is on S if and only if ap is on Sa. Notice that S1 = S. Denote by Ha

and Ka the mean curvature and the total curvature of Sa respectively. What are the

relations between Ha and H1, and between Ka and K1? 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 [hij] is a symmetric matrix with vanishing eigenvalues (H,K = the sum and the product of eigenvalues), we must 3 3 have hij ≡ 0. By (11.8), ω1 = ω2 = 0. So the Weingarten form dE3 also vanishes.)

4. We say that a point p on a surface S is umbilic, if h11 = h22 and h12 = h21 = 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 1 1 2 2 Hint: By assumption, there is a scalar function such that ω3 = λ θ and ω3 = λ θ . 1 3 1 2 So dN ≡ dE3 = ω3E1 + ω2E2 = λ(θ E1 + θ E2) = 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 Nu = λ xu and Nv = λxv. So Nuv = λvxu + λxuv and Nvu = λuxv + λxvu.

Hence λvxu = λuxv. From xu × xv 6= 0 we can derive λu = λv = 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 2 2 2 tensor for S is ds = guudu + 2guvdudv + gvvdv , where guu = ru · ru, guv = ru · rv 2 1/2 and gvv = rv · rv. Also, N = ru × rv is normal to S with |N| = (guugvv − guv) . We

may take E3 = N/|N|. Let

Luu = ruu · N,Luv = ruv · N,Lvv = rvv · N;

(in classical differential geometry, the second fundamental form is defined to be the 2 2 symmetric tensor Luudu + 2Luvdudv + Lvvdv , which differs from ours, as we shall see). Let F and G be tangential vector fields.

−1 (a) Show that hdF · E3, Gi = −|N| hdN · F, Gi. Hint: Again, from N · F = 0 we

get dN · F = −N · dF. Recall that E3 = N/|N|. u u u v v u v v (b) Verify that hdN · F, Gi = LuuF G + LuvF G + LvuF G + LvvF G , where F u = hF, dui, F v = hF, dvi, Gu = hG, dui and Gv = hG, dvi.

Hint: By Rule VF2 in §1.3, we have F = hF, duiru + hF, dvirv. (c) Show that the mean curvature and the total curvature of S are given by

Luugvv − 2Luvguv + Lvvguu LuuLvv − Luv H = 2 ,K = 2 . guugvv − guv guugvv − guv 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 2 2 2 (1 + h )hyy − 2hxhyhxy + (1 + h )hxx hxxhyy − h H = x y ,K = xy . 2 2 3/2 2 2 2 (1 + hx + hy) (1 + hx + hy)