Chapter 6 Some Special Curves on Surfaces

In this chapter we introduce three families of curves on surfaces, namely, principal, asymptotic and geodesic curves. Principal curves follow paths of maximal bending, asymptotic curves follow paths of zero bending and geodesicscurvesalwaysgo"striaghtahead"inthesurface.

6.1 Vector Fields Along Curves

Our discussion begins with an introduction to vector fields along curves.

Definition 135 Let α :(a, b) R3 be a curve given componentwise by α(t)= (x(t),y(t),z(t)).Avectorfield→V along α is a function defined on (a, b) such that V(t) R3 for all t (a, b). Each vector field V along α has the form ∈ α(t) ∈

V(t)=(V1(t),V2(t),V3(t))α(t) , where Vi :(a, b) R. Avectorfield V along α is differentiable if each Vi is differentiable; the→ derivative of a differentiable vector field V along α is the vector field V0 along α defined by

V0(t)=(V10(t),V20(t),V30(t))α(t) . The velocity vector field along α is defined by

α0(t)=(x0(t),y0(t),z0(t))α(t) ; the acceleration vector field along α is defined by

93 94 CHAPTER 6. SOME SPECIAL CURVES ON SURFACES

α00(t)=(x00(t),y00(t),z00(t))α(t) . All vector fields henceforth are assumed to be infinitely differentiable, i.e., smooth.

Definition 136 Let α :(a, b) R3 be a curve, let f :(a, b) R be a differentiable function and let V→, W be vector fields along α. For al→ l t (a, b), define ∈

a. (V + W)(t)=V(t)+W(t) b. (fV)(t)=f(t)V(t) c. (V W)(t)=V(t) W(t). • • The proof of the next Proposition is left for the reader.

Proposition 137 Differentiation of vector fields has the following proper- ties:

a. (V + W)0= V0 + W0

b. (fV)0=f 0V+fV0

c. (V W)0= V0 W + V W0. • • •

6.2 Principal, Asymptotic & Geodesic Curves

Now suppose that the image of a curve α is contained in the image of some regular patch x. Define

Definition 138 A vector vp p is principal if k (vp)=S (vp) ∈ M • vp = k1 or k2; it is asymptotic if k (vp)=S (vp) vp =0. Directions in • p that contain principal tangent vectors are called principal directions;di- M rections in p that contain asymptotic tangent vectors are called asymptotic directions. M

Note that in an asymptotic direction at p, does not bend away from M p (at least instantaneously at p). M 6.2. PRINCIPAL, ASYMPTOTIC & GEODESIC CURVES 95

Definition 139 Acurveα is a principal curve if α0 is a principal tan- ⊂ M gent vector field along α; α is an asymptotic curve if α0 (t) is an asymptotic tangent vector field along α.

Principal curves travel in directions of extreme bending. If p is a nonum- bilic point of , there are exactly two principal curves through p intersecting each other orthogonallyM at p. Asymptotic curves, on the other hand, travel in directions of no bending. Since k measures the com- ponent of acceleration normal to ,normalcurvaturek =0implies that M acceleration α00 along an asymptotic curve α is always tangent to . Principal curves on a cylinder are circular crossections and linesM parallel to the axis; these lines are also asymptotic curves. Principal curves on the hyperbolic paraboloid z = xy are the parabolas z = x2; the asymptotic curves are the xy-coordinate axes xy =0. ±

Lemma 140 Let α be a regular curve in and let U be a surface normal. M a. α is a principal curve if and only if

Dα U (α)=λα0. − 0 b. If α is principal curve, then the principal curvature

α00 U (α) k (α0)= • . α 2 k 0k

Proof. For part a, note that Dα U (α)=λα0 if and only if S (α0)=λα0 − 0 if and only if λ (t) is a principal curvature at α (t) for all t if and only if α0 is a principal tangent vector field along α. For part b, the vector field α0 consists of principal tangent vectors belonging to some principal curvature, say ki. Hence S (α0) α0 α00 U (α) ki = k (α0)= • = • , α 2 α 2 k 0k k 0k where the last equality is given by Lemma 101.

In this lemma, statement (a) gives a simple criterion for a curve to be a principal curve, while statement (b) gives the principal curvature along a curve known to be principal. 96 CHAPTER 6. SOME SPECIAL CURVES ON SURFACES

Definition 141 Let be the image of a regular patch. A geodesic on M M is a curve α :(a, b) whose acceleration vector field α00 is always in the direction normal to→ M, i.e., if U is a surface normal, then M

α00 = λU (α) .

Thusageodesicisacurvethatalwaysgoes“straightahead”inthe surface; its acceleration only serves to keep it in the surface. There is no component of the acceleration tangent to the surface. Let p⊥ denote the 3 M orthogonal complement of p in Rp. M Example 142 Suppose the image of a regular patch contains a straight M segment α(t)=p+tv, where t (a, b). Then α00(t)=0 for all t and ∈ in particular, α00(t) vα(t) =0for all tangent vectors vα(t) α(t). Hence • ∈M α00(t) α⊥(t), for all t and it follows that straight line segments on a regular patch∈ areM geodesics.

Example 143 Let be the cylinder x2 + y2 =1thought of as the level surface for g(x, y, z)=Mx2+y2 1 of height zero. The gradient is g(x, y, z)= (2x, 2y, 0), so a surface normal− U on is given by 5 M

U(x, y, z)=(x, y, 0)p, where p =(x, y, z) . For each a, b, c, d, the curve ∈ M α(t)=(cos(at + b), sin(at + b),ct+ d) is a geodesic in the cylinder because

2 2 2 α00(t)= a cos(at + b), a sin(at + b), 0 = a U(α(t)), − − α(t) − for all t. Furthermore,¡ the geodesic is a ¢

1. (a) Helix if a, c =0: 6 α(t)=(cos(at + b), sin(at + b),ct+ d)

(b) Line if a =0and c =0: 6 α(t)=(cos(b), sin(b),ct+ d) 6.2. PRINCIPAL, ASYMPTOTIC & GEODESIC CURVES 97

(c) Circle if a =0and c =0: 6 α(t)=(cos(at + b), sin(at + b),d)

(d) Point if a, c =0: α(t)=(cos(b), sin(b),d) .

Example 144 Let be the sphere x2 + y2 + z2 = r2.Slice with a Γ through theM origin and obtain a great circle C = ΓM.Choose M ∩ 3 any orthonormal basis e1, e2 for Γ (thought of as a subspace of R ) and parametrize C by { }

α(t)=(r cos at) e1 +(r sin at) e2, with t R and a =0. As with the cylinder, there is a surface normal U on given∈ by 6 M x y z U(x, y, z)= , , , r r r (x,y,z) where (x, y, z) . Then for all t, ³ ´ ∈M 2 2 2 2 α00(t)= a r cos(at)e1 a r sin(at)e2 = a α(t)α(t) = a (U α)(t) , − − α(t) − − ◦ so the great¡ circle C is a geodesic. Since¢ Γ and e1 were arbitrarily chosen, every great circle is a geodesic. Furthermore, if α(t)=re1 is a constant curve, then α00(t)=0 r⊥e1 for all t, and the point re1 is a geodesic. We conclude that points and∈ M great circles on a sphere are geodesics. Proposition 145 Geodesic on the image of a regular patch have constant speed. Proof. Let be the image of a regular patch and let α :(a, b) be M → M a geodesic. Then α0(t) α(t) and α00(t) α⊥(t) so that α0(t) α00(t)=0. Therefore ∈M ∈M •

d 2 d α0(t) = [α0(t) α0(t)] = 2α0(t) α00(t)=0, dt k k dt • • so the speed is constant. Here is a simple geometric way to find examples of geodesics. Let U be a surface normal on .WesaythataplaneΠ is orthogonal to at p if U (p) Π. M M ⊂ 98 CHAPTER 6. SOME SPECIAL CURVES ON SURFACES

Proposition 146 If Π is a plane everywhere orthogonal to and α is a unit speed curve in Π, then α is a geodesic of . M M∩ M

Proof. Since α has constant speed, α0 α0 = c so that α0 α00 =0. • • Furthermore, α0, α00 Π since α Π. By assumption, U (α (t)) Π for all t. ⊂ ⊂ ⊂ Therefore α00 (t)=λ (t) U (α (t)) and is α is a geodesic. We could have used this proposition to find the geodesics in all of the examples above except the helices on the cylinder. Note that meridians on a surface of revolution are geodesics since they are cut from by a plane passing through the axis. M

Exercise 6.2.1 Let α(t)=(x(t),y(t)) be a constant speed curve in the upper half plane y>0 with no self-intersections and let be the surface of revolution obtained by revolving the trace of α about theMx-axis. For each t (a, b) define βt : R by ∈ →M

βt(θ)=(x(t),y(t)cosθ, y(t)sinθ).

The circles βt are called the parallels on . Show that a parallel βt is a y t M geodesic if and only if the slope 0( ) of the line tangent to α at α (t) is zero. x0(t) In summary, let’s compare the essential properties of principal, asymp- totic and geodesic curves.

If a curve α is

principal: k (α0) k1,k2 S (α0)=α0λ ∈ { }

asymptotic: k (α0)=0 S (α0) α0 =0 α00 is tangent •

geodesic: α00 is normal.

6.3 Existence and uniqueness of Geodesics

Intuitively, given a point p on a surface and any tangent vector vp there M should be a geodesic in passing through p with initial velocity vp. After all, a bicycler riding onM the surface and passing through p should be able to continue traveling “straight ahead” with constant speed v and thereby k k 6.3. EXISTENCE AND UNIQUENESS OF GEODESICS 99 follow a geodesic in . Our main theorem asserts that this intuitive idea is indeed correct and thatM the geodesic with these properties is in some sense unique.

Definition 147 Apatchx is orthogonal if F = xu xv =0. • Let be the image of an orthogonal patch x and let α = x (u (t) ,v(t)) M be a curve. Then α0 = u0xu + ν0xv and 2 2 α00 = u00xu +(u0) xuu +2u0ν0xuv + ν00xv +(ν0) xvv.

Let U be a surface normal; we need to write xuu, xuv and xvv in the basis xu, xv, U ,i.e., { } u v xuu = Γuuxu+ Γuuxv+ U

u v xuv = Γuvxu+ Γuvxv+ mU

u v xvv = Γvvxu+ Γvvxv+ nU i for appropriate choice of coefficients Γjk,i,j,k u, v and = xuu U, ∈i { } • m = xuv U and n = xvv U. The coefficients Γjk are known as Christoffel • u • Symbols.TocomputeΓuu, first note that

u v xuu xu = Γuuxu xu + Γuuxv xu + U xu • u • • • = ΓuuE

On the other hand, differentiating E = xu xu with respect to u gives • Eu = xuu xu + xu xuu =2xuu xu. • • • Therefore 1 u xuu xu = Eu = Γ E, • 2 uu which gives E Γu = u . uu 2E Similarly,

1 u xuv xu = Ev = Γ E, • 2 uv 1 v xuv xv = Gu = Γ G, • 2 uv 1 v xvv xv = Gv = Γ G • 2 vv 100 CHAPTER 6. SOME SPECIAL CURVES ON SURFACES give E G G Γu = v , Γv = u , and Γv = v . uv 2E uv 2G vv 2G For the two remaining coefficients, differentiate xu xv =0with respect to u and v. With respect to u we get •

xuu xv + xu xvu =0 • • or 1 v xuu xv = xuv xu = Ev = Γ G, • − • −2 uu which gives E Γv = v uu −2G and similarly with respect to v we obtain G Γu = u . vv −2E Therefore

2 Eu Ev α00 = u00xu +(u0) xu xv + U 2E − 2G µ ¶ E G +2u ν v x + u x + mU 0 0 2E u 2G v µ ¶ 2 Gu Gv +(ν0) xu + xv + nU + ν00xv −2E 2G µ ¶ 2 Eu Ev 2 Gu = u00 +(u0) + u0ν0 (ν0) xu 2E E − 2E · ¸ 2 Ev Gu 2 Gv + ν00 (u0) + u0ν0 +(ν0) xv − 2G G 2G · ¸ 2 2 + (u0) +2u0ν0m +(ν0) n U.

This proves: h i Theorem 148 α is a geodesic if and only if

2 Eu Ev 2 Gu u00 +(u0) + u0ν0 (ν0) =0 2E E − 2E

2 Ev Gu 2 Gv ν00 (u0) + u0ν0 +(ν0) =0. − 2G G 2G 6.3. EXISTENCE AND UNIQUENESS OF GEODESICS 101

The system of second order differential equations above are known as the Geodesic Equations. All geodesics appear as solutions of these equations. Given initial data consisting of a point p on and a tangent vector vp M ∈ p, the theory of ordinary differential equations guarentees the existence M and uniqueness of a geodesic on through p with velocity vp. We state this formally as: M

Theorem 149 Let be the image of a regular patch, let p ,andlet M ∈M vp p. Then there exists an open interval (a, b) containing 0 andageo- ∈ M desic α :(a, b) such that α(0) = p and α0(0) = vp. The geodesic α is unique in the following→ M sense: If β :(c, d) is any other geodesic such → M that β(0) = p and β0(0) = vp,then(c, d) (a, b) and α(t)=β(t), for all t (c, d). ⊂ ∈ Definition 150 Let α :(a, b) be a geodesic with initial conditions → M α(0) = p and α0(0) = vp. We say that α is maximal if every other geodesic β :(c, d) with initial conditions β(0) = p and β0(0) = vp satisfies (c, d) (a,→ b)Mand α(t)=β(t), for all t (c, d). ⊂ ∈ In light Theorem 149, we now see that the geodesics on the sphere are exactly the constant points and great circles; on the cylinder the geodesics are straight lines, circles, helices, and constant points. The connection between distance on a surface and its geodesics is the content of our next theorem, which we state without proof:

Theorem 151 Let α be a curve on the image of a regular patch whose trace joins two points p and q. If length [α] length [β] for everyM other curve β on whose trace joins two points p and≤ q, then α is a geodesic. M This theorem does not imply that a geodesic joining two given points always exists. Here is a simple counterexample:

Example 152 Let be the plane with the origin removed, i.e., = R2 (0, 0) . ThedistancefromM p =(1, 0) to q =( 1, 0) is easily seenM to be −2, however,{ } every curve β whose trace joins p and−q has length strictly greater than 2. Furthermore, the difference 2 length [β] can be made arbitrarily small by appropriately choosing β. Therefore− no distance minimizing path in R2 (0, 0) joining p and q exists and it follows that no geodesic joining p and−q{exists.}