1. Vector Differentials and Moving Frames, Darboux Frames in Particular

CHAPTER 4. GEOMETRY OF SURFACES

§1. Vector Diﬀerentials and Moving Frames, Darboux Frames in Particular

The differential of a vector field in a Euclidean space is straightforward to define. If F n is a vector field in R with components F1,F2,...,Fn, that is, F = (F1,F2,...,Fn), then

dF is just the vector-valued diﬀerential form with components dF1, dF2, . . . , dFn, that is,

dF = (dF1, dF2, . . . , dFn). For example, writing r = (x, y, z), we have dr = (dx, dy, dz). Another example: let 2 2 2 2 F = (x − y , 2xy). The components of this vector ﬁeld are F1 = x − y and F2 = 2xy

with dF1 = 2xdx − 2ydy and dF2 = 2xdy + 2ydx and hence dF = 2(xdx − ydy, xdy + ydx),

or dF = 2(xdx, ydx) + 2(−ydy, xdy) = G1dx + G2dy, where the “coeﬃcients” G1 and G2

of dF are themselves vector fields given by G1 = 2(x, y) and G2 = 2(−y, x). In general, a vector-valued differential in the Euclidean space Rn can be written as Pn k=1 Vkdxk, where Vk are vector fields. Another description is that a vector-valued differential is an expression like a vector with differential forms as its components. In actual computation of differentials of vector fields, we often use various versions of the product rule, depending on what kind of products we are dealing with. In what follows, F and G are vector fields and f is a scalar field in Rn:

Rule VD1 (Product Rule). d(fG) = df. G + fdG, d(F · G) = dF · G + F · dG, and when n = 3, d(F × G) = dF × G + F × dG.

(F · G is the dot product and F × G is the cross product of F and G.) Often a vector ﬁeld or a vector-valued diﬀerential spreads over a smaller set called a submanifold, such as a curve or a surface, so that it depends on a smaller number of parameters, as shown in the following example.

Example 1.1. Find the differentials dT and dN of the unit tangent field T and the unit normal field N respectively along the parabola y = x2/2.

Solution: Put the parabola y = x2/2 into parametric equations as x = t, y = t2/2. Then the velocity vector is given by v = (dx/dt, dy/dt) = (1, t). The unit tangent ﬁeld T, which is a ﬁeld of unit vectors tangent to the parabola, can be obtained by normalizing v:

T = |v|−1v = (1 + t2)−1/2(1, t) ≡ (1 + x2)−1/2(1, x).

1 To find the normal direction at each point on the parabola, we treat the parabola as the level curve f = 0 of the scalar field f(x, y) = y − x2/2 and compute the gradient of f: ∇f = (∂f/∂x, ∂f/∂y) = (−x, 1). Normalizing ∇f, we get the unit normal field

N = |∇f|−1∇f = (1 + x2)−1/2(−x, 1).

Alternatively, we can get N by turning T 90o: turning a vector v = (a, b) in 90o anticlockwise, we get v⊥ = (−b, a). By the product rule, we have

dT = d((1 + x2)−1/2). (1, x) + (1 + x2)−1/2 d(1, x) = (−x(1 + x2)−3/2). (1, x) dx + (1 + x2)−1/2 (0, 1) dx = (1 + x2)−3/2 {−x(1, x) + (1 + x2)(0, 1)} dx = (1 + x2)−3/2 (−x, 1) dx.

Similarly, we have dN = (1 + x2)−3/2 (−1, −x) dx. Putting ω = (1 + x2)−1 dx, we can rewrite the answer as dT = ω N, dN = −ω T.

In the above example, the pair of vector ﬁelds T, N form an orthonormal frame ﬁeld, or a moving frame, along the parabola. By an orthonormal frame in the Euclidean space Rn we mean an orthonormal basis in this linear space, which consists of n unit vectors orthogonal to each other, say e1, e2,..., en, with ej · ek = δjk (the Kronecker delta, which is 1 when j = k and is 0 when j 6= k), that is, ej · ej = 1 and ej · ek = 0 for j 6= k, 1 ≤ j, k ≤ n. By a moving frame along a manifold M (such as a curve, or a surface, or a hypersurface) in Rn we mean at each point p of the manifold M, a frame

E1(p), E2(p),..., En(p) attaching to p is assigned. Thus each Ek (1 ≤ k ≤ n) individually is a vector ﬁeld along M, and together they form an orthonormal basis at each point of M. The elegant method of moving frames is credited to Elie Cartan for studying diﬀerential geometry, although Darboux used them before him.

Suppose that a moving frame E1, E2,..., En is given. Then, for each k (1 ≤ k ≤ n),

the differential dEk can be written as a linear combination of E1, E2,..., En with some differential forms as coefficients, say X n j dEk = ω Ej; (1.1) j=1 k

j j The diﬀerential forms ωk are called the connection forms. We can write out ωk explicitly n by using the orthonormality of the frame {Ek}k=1, that is, Ej · Ek = δjk. Indeed, Xn Xn ` ` j dEk · Ej = ωk E` · Ej = ωkδ`j = ω . `=1 `=1 k

2 j and hence we have ωk = dEk · Ej. Applying “d” to both sides of the orthonormal relation Ej · Ek = δjk and using the product rule, we have dEj · Ek + Ej · dEk = 0, that is, k j k k k ωj + ωk = 0. In particular, when j = k, we have ωk + ωk = 0 and hence ωk = 0. Thus we conclude

k j k ωk = 0 and ωk = −ωj for all j, k with 1 ≤ j, k ≤ n. (1.2)

In Example 1.1, we may take E1 = T and E2 = N. Relation (1.2) reduces four connection j 2 forms ωk (1 ≤ j, k ≤ 2) to one. In that example, (1.1) becomes dE1 = ω1E2 = ω E2 and 1 2 −1 dE2 = ω2E1 = −ωE1, where ω = (1 + x ) dx.

As usual, we denote the Cartesian coordinate functions by x1, x2, . . . , xn. (For any n point p = (p1, p2, . . . , pn) in R , xk(p) is the kth component of p, namely, xk(p) = pk. n So xk is indeed a function deﬁned on R .) Put x = (x1, x2, . . . , xn), which is considered as a vector ﬁeld. (When n = 2 or 3, we often use the letter r instead. Thus r = (x, y)

or r = (x, y, z).) The vector-valued diﬀerential form dx = (dx1, dx2, . . . , dxn) is a linear

combination of E1,..., En with some diﬀerential forms as coeﬃcients, say Xn k dx = θ Ek. (1.3) k=1

k k As before, we can write out θ explicitly as: θ = dx · Ek. (In the next chapter we shall Pn k 2 see that the “metric tensor” of a submanifold is given by k=1 (θ ) .)

1 2 Example 1.2. Find the covector ﬁelds θ and θ associated with the frame E1 = T

and E2 = N in Example 1.1. Solution: We have r = (x, y) and dr = (dx, dy) = (dx, d(x2/2)) = (dx, xdx). Since 2 −1/2 2 −1/2 E1 = T = (1 + x ) (1, x) and E2 = N = (1 + x ) (−x, 1),

1 2 −1/2 2 1/2 θ = dr · E1 = (1 + x ) {(dx, xdx) · (1, x)} = (1 + x ) dx,

and similarly θ2 = (1 + x2)−1/2{(dx, xdx) · (−x, 1)} = 0. (As we shall see, θ2 = 0 is

expectable because E1 and E2 form a so-called Darboux frame for the parabola.)

A moving frame E1, E2,..., En along a submanifold M, such as a curve or a surface,

is called a Darboux frame for M if each Ek is either tangent to M or normal to M. The moving frame T, N in Example 1.1 is a Darboux frame along the parabola y = x2/2. For a general curve in R2 described by parametric equations x = x(t), y = y(t), we can get a Darboux frame as follows. First compute the velocity v ≡ (dx/dt, dy/dt). Let p v = |v| ≡ (dx/dt)2 + (dy/dt)2 and ds = v dt; (1.4)

3 (|v| is the magnitude of the velocity v, called the speed, and ds, called a line element of p the curve, is often written as ds = dx2 + dy2, which is rather problematic but good for mnemonic purpose). We normalize v to get a unit tangent ﬁeld T = v/v; (here we tacitly assume that v 6= 0). Turning T 90o anticlockwisely, we get a unit normal vector N. In this way we obtain a Darboux frame along the curve consisting of E1 = T and E2 = N. Now equations (1.1) become dT = ωN and dN = −ωT. Here the connection form ω depends only on one variable, namely the parameter t. Hence it can be written in the form f dt, where f is a function with a single variable t. Recall the relation ds = vdt from (1.4). Let us assume that v 6= 0 so that we may rewrite this relation as dt = v−1ds. Thus we have ω = f dt = κ ds, where κ = v−1f. We call κ = κ(t) the curvature of the parametric curve at point (x(t), y(t)). In Example 1.1, we have v = (1, t) and ω = (1 + t2)−1dt (notice that x = t). Hence v = |v| = (1 + t2)1/2, which gives ds = vdt = (1 + t2)1/2dt. Consequently ω = (1 + t2)−3/2(1 + t2)1/2dt = (1 + t2)−3/2ds. Therefore the curvature of the parabola in that example is given by κ(t) = (1+t2)−3/2 at the point (t, t2/2). For a general formula for computing curvature of a planar parametric curve, see Exercise 1 of the present section. The identity dT = (κ ds)N tells us that the curvature κ is negative if the curve bends in the opposite direction of N.

Next, consider a spatial curve in parametric equations x = x(t), y = y(t), z = z(t). We construct a Darboux frame along this curve as follows. First, we take the velocity vector v ≡ (dx/dt, dy/dt, dz/dt) and normalize it to get a unit tangent ﬁeld T = |v|−1v. Then take the derivative T0 ≡ dT/dt and normalize it to obtain a unit vector ﬁeld

N = |T0|−1T0,

called the principal normal of the curve; (we tacitly assume that v 6= 0 and T0 6= 0). Diﬀerentiating both sides of T · T = |T|2 = 1 and applying the product rule, we obtain 2T · T0 = 0. Thus T · N = 0, telling us that N ⊥ T. Finally, let

B = T × N,

which is called the binormal to the curve. The vector ﬁelds E1 = T, E2 = N and

E3 = B along the curve form a Darboux frame, called the Frenet-Serrat frame. From N = |T0|−1T0 we have 0 0 0 dE1/dt = T = |T |N = |T |E2.

2 0 3 1 3 k So ω1 = |T |dt, ω1 = 0, and ω3 = −ω1 = 0. Thus, the nine connection forms ωj are

4 2 3 2 3 reduced to two, namely ω1 and ω2. Putting ω = ω1 and η = ω2, (1.1) gives: dT = ω N,

dN = −ω T +η B, (1.5)

dB = −η N, which are called Frenet-Serrat’s equations. Again, using the relation ds = vdt, where p v = |v| = x0(t)2 + y0(t)2 + z0(t)2, the connection forms ω and η can be written as ω = κ ds and η = τ ds. The functions κ = κ(t) and τ = τ(t) ] are called the curvature 2 0 and the torsion respectively of the curve at (x(t), y(t), z(t)). Recall that ω = ω1 = |T | dt Thus we have κ vdt = κ ds = ω = |T0| dt and hence κ = v−1|T0| ≥ 0. Thus, unlike the planar curve, the curvature of a spatial curve does not take negative values.

Example 1.3. Find the Frenet-Serrat frame and the associated connection forms, as well as the curvature κ and the torsion τ, for the helix described by the parametric equations x = cos t, y = sin t and z = t.

Solution: A direct computation shows v = (dx/dt, dy/dt.dz/dt) = (− sin t, cos t, 1), √ √ giving v ≡ |v| = ((− sin t)2 + (cos t)2 + 1)1/2 = 2 and T = v/|v| = (− sin t, cos t, 1)/ 2. √ Next, T0 = (− cos t, − sin t, 0)/ 2, which gives N = |T0|−1T0 = (− cos t, − sin t, 0). The binormal B can be obtained by forming T × N: µ¯ ¯ ¯ ¯ ¯ ¯¶ 1 ¯ cos 1 ¯ ¯ 1 − sin t ¯ ¯ − sin t cos t ¯ 1 B = √ ¯ ¯ , ¯ ¯ , ¯ ¯ = √ (sin t, − cos t, 1). 2 ¯ − sin t 0 ¯ ¯ 0 − cos t ¯ ¯ − cos t − sin t ¯ 2 √ √ √ So we have dT = d(− sin t, cos t, 1)/ 2 = (− cos t, − sin t, 0)dt/ 2 = N dt/ 2 and √ √ similarly, dB = (cos t, sin t, 0) dt/ 2 = −N dt/ 2. Thus, according to the notation in √ √ (9.5), the required connection forms are ω = dt/ 2 and η = dt/ 2. On the other hand, √ √ since v = 2, we have ω = κ ds = κ. 2 dt and consequently κ = 1/2. Similarly, τ = 1/2.

3 A Darboux frame for a surface S in R has two tangential ﬁelds, say T1 and T2, and one normal ﬁeld, say N. When S is given by parametric equations x = x(u, v), y = y(u, v), z = z(u, v), we can obtain a Darboux frame T1, T2, N for S as follows. Write

r = r(u, v) ≡ (x(u, v), y(u, v), z(u, v)).

Compute the tangent ﬁelds

ru ≡ ∂r/∂u and rv ≡ ∂r/∂v,

5 which are tangent to “u”-curves and “v”-curves in S respectively. Let n = ru ×rv, which is normal to S. Orthonormalize ru, rv to obtain unit tangent ﬁelds T1 and T2 and normalize n to obtain a unit normal ﬁeld N (= |n|−1n).

Example 1.4. Consider the unit sphere S2 parametrized by the latitude φ and longitude θ (−π < θ < π, −π/2 < φ < π/2): x = cos θ cos φ, y = sin θ cos φ, z = sin φ. j j Find a Darboux frame and the associated connection forms ωk and forms θ . Solution: Let us write r = (x, y, z) = (cos θ cos φ, sin θ cos φ, sin φ). Then we have t1 = rθ = (− sin θ cos φ, cos θ cos φ, 0), t2 = rφ = (− cos θ sin φ, − sin θ sin φ, cos φ) and 2 n = t1 × t2 = (cos θ cos φ, sin θ cos φ, cos φ sin φ) = cos φ. r. It is easy to check t1 · t2 = 0.

Normalizing t1, t2 and n, we get a Darboux frame:

T1 = t1/|t1| = (− sin θ, cos θ, 0), T2 = t2/|t2| = (− cos θ sin φ, − sin θ sin φ, cos φ), and N = n/|n| = r = (cos θ cos φ, sin θ cos φ, sin φ). (Note that |t2| = 1 and |t1| = 1 2 | cos φ| = cos φ because −π/2 < φ < π/2.) Comparing dr = θ T1 + θ T2 with

dr = rθ dθ + rφ dφ ≡ t1dθ + t2dφ = T1. |t1| dθ + T2. |t2| dφ = T1. cos φ dθ + T2. dφ we have θ1 = cos φ dθ, θ2 = dφ. (1.6)

(This will give us the metric tensor (θ1)2 + (θ2)2 = cos2 φ dθ2 + dφ2 and the area form θ1 ∧ θ2 = cos φ dθ ∧ dφ.) Also,

dT1 = (− cos θ, − sin θ, 0) dθ

dT2 = (sin θ sin φ, − cos θ sin φ, 0) dθ − (cos θ cos φ, sin θ cos φ, sin φ) dφ dN = (− sin θ cos φ, cos θ cos φ, 0) dθ + (− cos θ sin φ, − sin θ cos φ, cos φ) dφ.

2 3 3 The connection forms ω1, ω1 and ω2 in

2 3 2 3 3 3 dT1 = ω1T2 + ω1N, dT2 = −ω1T1 + ω2N, dN = −ω1T1 − ω2T2 can be computed as follows:

2 2 2 ω1 = dT1 · T2 = (cos θ sin φ + sin θ sin φ) dθ = sin φ dθ 3 2 2 ω1 = dT1 · N = (− cos θ cos φ − sin θ cos φ) dθ = − cos φ dθ (1.7) 3 ω2 = dT2 · N = (sin θ sin φ cos θ cos φ − cos θ sin φ sin θ cos φ) dθ + (− cos2 θ cos2 φ − sin2 θ cos2 φ − sin2 φ) dφ = −dφ.

6 As you can see, even for a simple example in differential geometry, the amount of work could be quite formidable. Naturally we need to find ideas for “seeing things” (such as reduction by symmetries) in order to avoid heavy and tedious computation. In fact, many general formulas in differential geometry are good for theoretical purposes rather than practical calculation!

Exercises

1. Let γ be a planar curve given by parametric equations x = x(t) and y = y(t). Assume p 0 2 0 2 1 0 0 v = v(t) ≡ x (t) + y (t) 6= 0 for all t. Verify that T = v (x , y ) and N = 1 0 0 v (−y , x ) form a Darboux frame along γ. Verify dT = ωN and dN = −ωT, where

x0y00 − y0x00 ω = dt. (1.8) v2

Thus we obtain the formula κ(t) = (x0y00 − y0x00)/v3 for computing curvature. Also, 1 2 employ (1.8) to ﬁnd the connection 1-form ω for the parabola y = 2 x by using the parametrization x = 2t, y = 2t2. Show that the curvature at the point (x, x2/2) is κ = (1 + x2)−3/2.

2. In each of the following cases, ﬁnd the Darboux frame, T, N, its associated connection form ω and the curvature function κ for the given planar parametric curves. (a) Archimedian spiral: x = eat cos t, y = eat sin t;(a is any constant). (b) catenary: x = t, y = cosh t. 2 1 3 (c) cubic curve x = t , y = t − 3 t . R t 2 Partial answers: (a) κ = 1/as, where s = 0 v; (b) κ = 1/ cosh t. 3. Given a planar curve γ with parametric equations x = x(t), y = y(t). The invo-

lute of γ is deﬁned to be the parametric curve γe : X = X(t),Y = Y (t) given by (X(t),Y (t)) = (x(t), y(t)) + κ(t)−1N, where N and κ are respectively the unit normal

vector ﬁeld and the curvature of γ. (a) Show that the tangents of γe are normal to γ. (b) Find the evolute of the catenary x(t) = t, y(t) = cosh t.

Note: We call κ(t)−1 the radius of curvature. The circle with radius κ(t)−1 centered at (x(t), y(t)) + κ(t)−1N is “kissing” the curve γ at the point P ≡ (x(t), y(t)). It is called the osculating circle of γ at P . So the involute of γ is nothing but the trajectory of the center of the osculating circle.

7 4. In this exercise we describe a way to construct a planar curve with a given curvature R t function κ(t), a ≤ t ≤ b. Fix any c between a and b. Let φ(t) = c κ(u) du. Show that the parametric curve x = x(t), y = y(t)(a < t < b) given by

Z t Z t x(t) = cos φ(t) dt, y(t) = sin φ(t) dt c c has the prescribed curvature κ(t).

5. Consider a spatial parametric curve r(t) = (x(t), y(t), z(t)). As usual, we write r0(t) = (x0(t), y0(t), z0(t)) and r00(t) = (x00(t), y00(t), z00(t)). Verify that Frenet-Serrat’s equations (1.5) hold with ω = κ. vdt and η = τ. vdt (v = |r0|), where

|r0 × r00| [r, r0, r00] κ = κ(t) = , (curvature) τ = τ(t) = − (torsion); (1.9) |r0|3 |r0 × r00|2

([a, b, c] designates the triple product (a × b) · c).

6. (a) Use (1.9) to compute the curvature κ and the torsion τ for the helix x(t) = cos t, y(t) = sin t and z(t) = t. Verify that your answer agrees with the one in Example 1.3 above. (b) Find Frenet-Serrat frame, the curvature and the torsion for the twisted 2 2 3 cubic given by the parametric equations x(t) = t, y(t) = t , y = 3 t . 7. Let γ be a parametric curve γ in the XY -plane given by X = X(t) and Y = Y (t). Assume that X0(t)2 + Y 0(t)2 ≡ 1; that is, t is an arc length parameter. Consider the cylinder given by x(u, v) = X(u), y(u, v) = Y (u), z(u, v) = v. Show that

0 0 0 0 T1(u, v) = (X (u),Y (u), 0), T2 = (0, 0, 1), N(u, v) = (−Y (u),X (u), 0)

form a Darboux frame for the cylinder. Verify dT2 = 0, dT1 = ηN and dN = −ηT1, where η = (X0(u)Y 00(u) − Y 0(u)X00(u)) du.

8. Let Y = Y (t), Z = Z(t) be a parametric curve in the yz-plane with Y (t) > 0 for all t. Let r(u, v) = (Y (u) cos v, Y (u) sin v, Z(u)) be the surface of rotation it generates.

Compute ru, rv and ru ×rv and verify that, after normalization, we obtain a Darboux frame 0 0 0 T1 = h(u)(Y (u) cos v, Y (u) sin v, Z (u)), T2 = (− sin v, cos v, 0), N = h(u)(−Z0(u) cos v, −Z0(u) sin v, Y 0(u)),

0 2 0 2 −1/2 j where h(u) = (Y (u) + Z (u) ) . Find the connection forms ωk and the funda- mental forms θj associated with this Darboux frame in terms of the functions Y (u), Z(u) and their derivatives.