Elementary Differential Geometry: Lecture Notes. Gilbert Weinstein

Transcription

1 Elementry Differentil Geometry: Lecture Notes Gilbert Weinstein

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3 Contents Prefce 5 Chpter 1. Curves 7 1. Preliminries 7 2. Locl Theory for Curves in R Plne Curves Fenchel s Theorem 15 Exercises 16 Chpter 2. Locl Surfce Theory Surfces The First Fundmentl Form The Second Fundmentl Form Exmples Lines of Curvture More Exmples Surfce Are Bernstein s Theorem Theorem Egregium 39 Exercises 41 Chpter 3. Locl Intrinsic Geometry of Surfces Riemnnin Surfces Lie Derivtive Covrint Differentition Geodesics The Riemnn Curvture Tensor The Second Vrition of Arclength 56 Exercises 59 Chpter 4. Globl Theory of Surfces Differentible Mnifolds Some Multi-liner Algebr Integrtion 65 Exercises 65 Index 67 3

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5 Prefce These notes re for beginning grdute level course in differentil geometry. It is ssumed tht this is the students first course in the subject. Thus the choice of subjects nd presenttion hs been mde to fcilitte s much s possible concrete picture. For those interested in deeper study, second course would tke more bstrct point of view, nd in prticulr, could go further into Riemnnin geometry. Much of the mteril is borrowed from the following sources, but hs been dpted ccording to my own tste: [1] M. P. Do Crmo, Differentil geometry of curves nd surfces, Prentice-Hll. [2] L. P. Eisenhrt An introduction to differentil geometry with use of the tensor clculus, Princeton University Press. [3] W. Klingenberg, A course in differentil geometry, Springer-Verlg. [4] B. O Neill Elementry differentil geometry, Acdemic Press. [5] M. Spivk, A comprehensive introduction to Differentil Geometry, Publish or Perish. [6] J. J. Stoker, Differentil Geometry, Wiley & Sons. The prerequisites for this course re: liner lgebr, preferbly with some exposure to multiliner lgebr; clculus up to nd including the inverse nd implicit function theorem; the fundmentl theorem of ordinry differentil equtions concerning existence of solutions, uniqueness, nd continuous dependence on prmeters, nd some knowledge of liner systems of ordinry differentil equtions; liner first order prtil differentil equtions; complex nlysis including Liouville s theorem; nd some elementry topology. It is highly recommended for the students to complete ll the exercises included in these notes. Gilbert Weinstein Birminghm, Albm April

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7 CHAPTER 1 Curves 1. Preliminries Definition 1.1. A prmetrized curve is smooth C function γ : I R n. A curve is regulr if γ 0. When the intervl I is closed, we sy tht γ is C on I if there is n intervl J nd C function β on J which grees with γ on I. Definition 1.2. Let γ : I R n be prmetrized curve, nd let β : J R n be nother prmetrized curve. We sy tht β is reprmetriztion orienttionpreserving reprmetriztion of γ if there is smooth mp τ : J I with τ > 0 such tht β = γ τ. Note tht the reltion β is reprmetriztion of γ is n equivlence reltion. A curve is n equivlence clss of prmetrized curves. Furthermore, if γ is regulr then every reprmetriztion of γ is lso regulr, so we my spek of regulr curves. Definition 1.3. Let γ : I R n be regulr curve. For ny compct intervl [, b] I, the rclength of γ over [, b] is given by: L γ [, b] = b γ dt. Note tht if β is reprmetriztion of γ then γ nd β hve the sme length. More specificlly, if β = γ τ, then L γ [ τc, τd ] = Lβ [c, d]. Definition 1.4. Let γ be regulr curve. We sy tht γ is prmetrized by rclength if γ = 1 Note tht this is equivlent to the condition tht for ll t I = [, b] we hve: L γ [, t] = t. Furthermore, ny regulr curve cn be prmetrized by rclength. Indeed, if γ is regulr curve, then the function st = t γ, is strictly monotone incresing. Thus, st hs n inverse function τs function, stisfying: dτ ds = 1 γ. It is now strightforwrd to check tht β = γ τ is prmetrized by rclength. 7

8 8 1. CURVES 2. Locl Theory for Curves in R 3 We will ssume throughout this section tht γ : I R 3 is regulr curve in R 3 prmetrized by rclength nd tht γ 0. Note tht γ γ = 0. Definition 1.5. Let γ : I R 3 be curve in R 3. The unit vector T = γ is clled the unit tngent of γ. The curvture κ is the sclr κ = γ. The unit vector N = κ 1 T is clled the principl norml. The binorml is the unit vector B = T N. The positively oriented orthonorml frme T, N, B is clled the Frenet frme of γ. It is not difficult to see tht N + κt is perpendiculr to both T nd N, hence we cn define the torsion τ of γ by: N + κt = τb. Note tht the torsion, unlike the curvture, is signed. Finlly, it is esy to check tht B = τn. Let X denote the 3 3 mtrix whose columns re T, N, B. We will cll X lso the Frenet frme of γ. Define the rottion mtrix of γ: 0 κ ω := κ 0 τ 0 τ 0 Proposition 1.1 Frenet frme equtions. The Frenet frme X = T, N, B of curve in R 3 stisfies: 1.2 X = Xω. The Frenet frme equtions, Eqution 1.2, form system of nine liner ordinry differentil equtions. Definition 1.6. A rigid motion of R 3 is function of the form Rx = x 0 +Qx where Q is orthonorml with det Q = 1. Note tht if X is the Frenet frme of γ nd Rx = x 0 + Qx is rigid motion of R 3, then QX is the Frenet frme of R γ. This follows esily from the fct tht Q is preserves the inner product nd orienttion of R 3. Theorem 1.2 Fundmentl Theorem. Let κ > 0 nd τ be smooth sclr functions on the intervl [0, L]. Then there is regulr curve γ prmetrized by rclength, unique up to rigid motion of R 3, whose curvture is κ nd torsion is τ. Proof. Let ω be given by 1.1. The initil vlue problem X = Xω, X0 = I cn be solved uniquely on [0, L]. The solution X is n orthogonl mtrix with det X = 1 on [0, L]. Indeed, since ω is nti-symmetric, the mtrix A = XX t is constnt. Indeed, A = XωX t + Xω t X t = Xω + ω t X t = 0, nd since A0 = I, we conclude tht A I, nd X is orthogonl. Furthermore, det X is continuous, nd det X0 = 1, so det X = 1 on [0, L]. Let T, N, B be the columns of X, nd let γ = T, then T, N, B is orthonorml nd positively oriented on [0, L]. Thus, γ is prmetrized by rclength, γ = T, nd N = κ 1 T is

9 2. LOCAL THEORY FOR CURVES IN R 3 9 the principl norml of γ. Similrly B is the binorml, nd consequently, κ is the curvture of γ nd τ its torsion. Now suppose tht γ is nother curve with curvture κ nd torsion τ, nd let X be its Frenet frme. Then there is rigid motion Rx = Qx + x 0 of R 3 such tht Rγ0 = γ0, nd QX0 = X0. By the remrk preceding the theorem, QX is the Frenet frme of the curve R γ, nd thus both QX nd X stisfy the initil vlue problem: Y = Y ω, Y 0 = QX0. By the uniqueness of solutions of the initil vlue problem, it follows tht QX = X. In prticulr, R γ = γ, nd since R γ0 = γ0 we conclude R γ γ. Assuming γ0 = 0, the Tylor expnsion of γ of order 3 t s = 0 is: γs = γ 0s γ 0s γ 0s 3 + Os 4. Denote T 0 = T 0, N 0 = N0, B 0 = B0, κ 0 = κ0, nd τ 0 = τ0. We hve γ 0 = T 0, γ 0 = κ 0 N 0, nd γ 0 = κ 0N 0 + κ 0 κ 0 T 0 + τ 0 B 0. Substituting these into the eqution bove, decomposing into T, N, nd B components, nd retining only the leding order terms, we get: γs = s + Os 3 κ τ T + 2 s2 + Os 3 N + 6 s3 + Os 4 B The plnes spnned by pirs of vectors in the Frenet frme re given specil nmes: 1 T nd N the osculting plne; 2 N nd B the norml plne; 3 T nd B the rectifying plne. We see tht to second order the curve stys within its osculting plne, where it trces prbol y = κ/2 s 2. The projection onto the norml plne is cusp to third order: x = 3τ/2 y 2/3. The projection onto the rectifying plne is to second order line, whence its nme. Here re few simple pplictions of the Frenet frme. Theorem 1.3. Let γ be regulr curve with κ 0. Then γ is stright line. Proof. Since T = κ = 0, it follows tht T is constnt nd γ is liner. Theorem 1.4. Let γ be regulr curve with κ > 0, nd τ = 0. Then γ is plnr. Proof. Since B = 0, B is constnt. Thus the function ξ = γ γ0 B vnishes identiclly: ξ0 = 0, ξ = T B = 0. It follows tht γ remins in the plne through γ0 perpendiculr to B. Theorem 1.5. Let γ be regulr curve with κ constnt nd τ = 0. Then γ is circle.

10 10 1. CURVES Proof. Let β = γ + κ 1 N. Then β = T + 1 κt + τb = 0. κ Thus β is constnt, nd γ β = κ 1. It follows tht γ lies in the intersection between plne nd sphere, thus γ is circle. 3. Plne Curves 3.1. Locl Theory. Let γ : [, b] R 2 be regulr plne curve prmetrized by rclength, nd let κ be its curvture. Note tht κ is signed, nd in fct chnges sign but not mgnitude when the orienttion of γ is reversed. The Frenet frme equtions re: e 1 = κ e 2, e 2 = κ e 1 Proposition 1.6. Let γ : [, b] R 2 be regulr curve with γ = 1. Then there exists differentible function θ : [, b] R such tht 1.3 e 1 = cos θ, sin θ. Moreover, θ is unique up to constnt integer multiple of 2π, nd in prticulr θb θ is independent of the choice of θ. The derivtive of θ is the curvture: θ = κ. Proof. Let = t 0 < t 1 < < t n = b be prtition of [, b] so tht the dimeter of e 1 [t i 1, t i ] is less thn 2, i.e., e 1 restricted to ech subintervl mps into semi-circle. Such prtition exists since e 1 is uniformly continuous on [, b]. Choose θ so tht 1.3 holds t, nd proceed by induction on i: if θ is defined t t i then there is unique continuous extension so tht 1.3 holds. If ψ is ny other continuous function stisfying 1.3, then k = 1/2πθ ψ is continuous integer-vlued function, hence is constnt. Finlly, e 2 = sin θ, cos θ hence nd we obtin θ = κ Globl Theory. e 1 = κ e 2 = θ sin θ, cos θ, Definition 1.7. A curve γ : [, b] R n is closed if γ k = γ k b. A closed curve γ : [, b] R n is simple if γ,b is one-to-one. The rottion number of smooth closed curve is: 1.4 n γ = 1 θ θb, 2π where θ is the function defined in Proposition 1.6. We note tht the rottion number is lwys n integer. For reference, we lso note tht the rottion number of curve is the winding number of the mp e 1. Finlly, in view of the lst sttement in Proposition 1.6, we hve: n γ = 1 κ ds. 2π Theorem 1.7 Rottion Theorem. Let γ : [0, L] R 2 be smooth, regulr, simple, closed curve. Then n γ = ±1. In prticulr 1 κ ds = ±1. 2π [0,L] [0,L]

11 3. PLANE CURVES 11 For the proof we will need the following technicl lemm. We sy tht set Ω R n is str-shped with respect to x 0 Ω if for every y Ω the line segment x 0 y lies in Ω. Lemm 1.8. Let Ω R n be str-shped with respect to x 0 Ω, nd let e: Ω S 1 be continuous function. Then there exists continuous function θ : Ω R such tht: 1.5 e = cos θ, sin θ. Moreover, if ψ is nother continuous function stisfying 1.5, then θ ψ = 2πk where k is constnt integer. In fct, it is sufficient to ssume tht Ω is simply connected, but we will not prove this more generl result here. Proof. Define θx 0 so tht 1.5 holds t x 0. For ech x Ω define θ continuously long the line segment x 0 x s in the proof of Proposition 1.6. Since Ω is str-shped with respect to x 0, this defines θ everywhere in Ω. It remins to show tht θ is continuous. Let y 0 Ω. Since x 0 y 0 is compct, it is possible to choose δ smll enough tht the following holds: y x 0 y 0 nd y y < δ implies ey ey < 2 or equivlently ey nd ey re not ntipodl. Let 0 < ɛ < π. Then there exists neighborhood U B δ y 0 of y 0 such tht y U implies θy θy 0 = 2πky + ɛ y where ɛ y < ɛ nd ky is integer-vlued. It remin to prove tht k 0. Let y U nd consider the continuous function: Since φs = θ x 0 + sy x 0 θ x 0 + sy 0 x 0, 0 s 1. x 0 + sy x 0 x 0 + sy 0 x 0 = sy y0 < δ, it follows from our choice of δ tht e x 0 + sy x 0 nd e x 0 + sy 0 x 0 re not ntipodl. Thus, φs π for ll 0 s 1, nd since φ0 = 0 we conclude tht φ < π. In prticulr nd it follows tht 2πky + ɛ y = θy θy 0 = φ1 < π, 2πky 2πky + ɛ y + ɛ y < 2π. Since ky is integer-vlued this implies ky = 0. Proof of the Rottion Theorem. Pick line which intersects the curve γ nd pick lst point p on this line, i.e., point with the property tht one ry of the line from p hs no other intersection points with γ. Let h be the unit vector pointing in the direction of tht ry. We ssume without loss of generlity tht γ is prmetrized by rclength, γ0 = γl = p. Now, let Ω = {t 1, t 2 R 2 : 0 t 1 t 2 L}, nd note tht Ω is str-shped. Define the S 1 -vlued function: γ t 1 if t 1 = t 2 ; et 1, t 2 = γ 0 if t 1, t 2 = 0, L; γt 2 γt 1 otherwise. γt 2 γt 1

12 12 1. CURVES We clim tht e is continuous on Ω. Indeed, let = {t 1, t 2 : t 1 = t 2 } denote the digonl in Ω, let Ω = Ω \ {0, L} nd suppose tht t 1, t 2 Ω, t 0 [0, L]. Then we cn write: where γt 2 γt 1 = t 2 t 1 γ t 0 + Qt 1, t 2, 1 Qt 1, t 2 = t 2 t 1 t2 t 1 γ t γ t 0 dt. Thus, if we cn show tht Qt 1, t 2 0 s t 1, t 2 Ω tends to t 0, t 0, we get: γt 2 γt 1 γt 2 γt 1 = γ t 0 + Qt 1, t 2 γ t 0 + Qt 1, t 2 γ t 0. nd we will hve shown continuity t t 0, t 0 on digonl. We now show tht Qt 1, t 2 0 s t 1, t 2 t 0, t 0. Let ɛ > 0, then since γ is continuous t t 0 there δ such tht if t t 0 < δ, then γ t γ t 0 < ɛ. Now, if the distnce between t 1, t 2 nd t 0, t 0 is less thn δ, then t 1 t 0, t 2 t 0 < δ, hence t t 0 < δ for ll t [t 1, t 2 ], nd thus it follows tht γ t γ t 0 < ɛ. Consequently, we find: Qt 1, t 2 = 1 t2 γ t γ t 0 dt t 2 t 1 1 t2 γ t γ t 0 dt ɛ. t 2 t 1 t 1 t 1 A similr rgument, using in ddition the fct tht γ is closed, cn be used to show continuity of e t 0, L. Now by the Lemm, there is continuous function θ : Ω R such tht e = cos θ, sin θ. We clim tht θl, L θ0, 0 = ±2π which proves the theorem, since θt, t is continuous function stisfying 1.3 in Proposition 1.6, nd thus cn be used on the right-hnd side of 1.4 to compute the rottion number. To prove this clim, note tht, for ny 0 < t < L, the unit vector e0, t = γt γ0 γt γ0 is never equl to h. Hence, there is some vlue α such tht θ0, t θ0, 0 α+2πk for ny integer k. Thus, θ0, t θ0, 0 < 2π, nd since e0, L = e0, 0 it follows tht θ0, L θ0, 0 = ±π. Since the curves e0, t nd et, L re relted vi rigid motion, i.e., et, L = Re0, t where R is rottion by π, it follows tht ψt = θt, L θ0, L θ0, t θ0, 0 is constnt. Since clerly ψ0 = 0, we get θ0, L θ0, 0 = θl, L θ0, L, nd we conclude: θl, L θ0, 0 = θt, L θ0, L + θ0, t θ0, 0 = ±2π. Definition 1.8. A piecewise smooth curve is continuous function γ : [, b] R n such tht there is prtition of [, b]: = 0 < 1 < < b n = b such tht for ech 1 j n the curve segment γ j = γ [j 1, j] is smooth. The points γ j re clled the corners of γ. The directed ngle π < ψ j π from γ j to γ j + is clled the exterior ngle t the j-th corner. Define

13 3. PLANE CURVES 13 θ j : [ j 1, j ] R s in Proposition 1.6, i.e., so tht γ j = cos θ j, sin θ j. The rottion number of γ is given by: n γ = 1 n θj j θ j j n ψ j. 2π 2π j=1 Agin, n γ is n integer, nd we hve: n γ = 1 κ ds + 1 2π 2π [,b] n ψ j. The Rottion Theorem cn be generlized to piecewise smooth curves provided corners re tken into ccount. Theorem 1.9. Let γ : [0, L] R 2 be piecewise smooth, regulr, simple, closed curve, nd ssume tht none of the exterior ngles re equl to π. Then n γ = ± Convexity. Definition 1.9. Let γ : [0, L] R 2 be regulr closed plne curve. We sy tht γ is convex if for ech t 0 [0, L] the curve lies on one side only of its tngent t t 0, i.e., if one of the following inequlity holds: γ γt0 e 2 0, γ γt0 e 2 0. Theorem Let γ : [0, L] R 2 be regulr simple closed plne curve, nd let κ be its curvture. Then γ is convex if nd only if either κ 0 or κ 0. We note tht n orienttion reversing reprmetriztion of γ chnges κ 0 into κ 0 nd vice vers. Thus, ignoring orienttion, those two conditions re equivlent. We lso note tht the theorem fils if γ is not ssumed simple. Proof. We my ssume without loss of generlity tht γ = 1. Let θ : [0, L] R be the continuous function given in Proposition 1.6 stisfying: e 1 = cos θ, sin θ, nd θ = κ. Suppose tht γ is convex. We will show tht θ is wekly monotone, i.e., if t 1 < t 2 nd θt 1 = θt 2 then θ is constnt on [t 1, t 2 ]. First, we note tht since γ is simple, we hve n γ = ±1 by the Rottion Theorem, nd it follows tht e 1 is onto S 1, see Exercise 1.5. Thus, there is t 3 [0, L] such tht j=1 e 1 t 3 = e 1 t 1 = e 1 t 2. By convexity, the three prllel tngents t t 1, t 2, nd t 3 cnnot be distinct, hence t lest two must coincide. Let p 1 = γs 1 nd p 2 = γs 2, s 1 < s 2 denote these two points, then the line p 1 p 2 is contined in γ. Otherwise, if q is point on p 1 p 2 not on γ, then the line through q perpendiculr to p 1 p 2 intersects γ in t lest two points r nd s, which by convexity must lie on one side of p 1 p 2. Without loss of generlity, ssume tht r is the closer of the two to p 1 p 2. Then r lies in the interior of the tringle p 1 p 2 s. Regrdless of the inclintion of the tngent t r, the three points p 1, p 2 nd s, ll belonging to γ, cnnot ll lie on one side of the tngent, in contrdiction to convexity. If p 1 p 2 {γs: s 1 s s 2 }, then j=1

14 14 1. CURVES p 1 p 2 = {γs: s 2 s L} {γs: 0 s s 1 }. However, in tht cse, we would hve θs 2 θs 1 = θl θ0 = 2π, contrdiction. Thus, we hve p 1 p 2 = {γs: s 1 s s 2 } = {γt: t 1 t t 2 }. In prticulr θt = θt 1 = θt 2. Conversely, suppose tht γ is not convex. Then, there is t 0 [0, L] such tht the function φ = γ γt 0 e 2 chnges sign. We will show tht θ lso chnges sign. Let t +, t [0, L] be such tht min φ = φt < 0 = φt 0 = φt + = mx φ. [0,L] [0,L] Note tht the three tngents t t, t + nd t 0 re prllel but distinct. Since φ t = φ t + = 0, we hve tht e 1 t nd e 1 t + re both equl to ±e 1 t 0. Thus, t lest two of these vectors re equl. We my ssume, fter reprmetriztion, tht there exists 0 < s < L such tht e 1 0 = e 1 s. This implies tht θs θ0 = 2πk, θl θs = 2πk with k, k Z. By the Rottion Theorem, n γ = k + k = ±1. Since γ0 nd γs do not lie on line prllel to e 1 t 0, it follows tht θ is not constnt on either [0, s] or [0, L]. If k = 0 then θ chnges sign on [0, s], nd similrly if k = 0 then θ chnges sign on [s, L]. If kk 0, then since k + k = ±1, it follows tht kk < 0 nd θ chnges sign on [0, L]. Definition Let γ : [0, L] R 2 be regulr plne curve. A vertex of γ is criticl point of the curvture κ. Theorem 1.11 The Four Vertex Theorem. A regulr simple convex closed curve hs t lest four vertices. Proof. Clerly, κ hs mximum nd minimum on [0, L], hence γ hs t lest two vertices. We will ssume, without loss of generlity, tht γ is prmetrized by rclength, hs its minimum t t = 0, its mximum t t = t 0 where 0 < t 0 < L, tht γ0 nd γt 0 lie on the x-xis, nd tht γ enters the upper-hlf plne in the intervl [0, t 0 ]. All these properties cn be chieved by reprmetrizing nd rotting γ. We now clim tht p = γ0 nd q = γt 0 re the only points of γ on the x-xis. Indeed, suppose tht there is nother point r = γt 1 on the x-xis, then one of these points lies between the other two, nd the tngent t tht point must, by convexity, contin the other two. Thus, by the rgument used in the proof of Theorem 1.10 the segment between the outer two is contined in γ, nd in prticulr pq is contined in γ. If follows tht κ = 0 t p nd q where κ hs its minimum nd mximum, hence κ 0, contrdiction since then γ is line nd cnnot be closed. We conclude tht γ remins in the upper hlf-plne in the intervl [0, t 0 ] nd remins in the lower hlf-plne in the intervl [t 0, L]. Suppose now by contrdiction tht γ0 nd γt 0 re the only vertices of γ. Then it follows tht: κ 0 on [0, t 0 ], κ 0 on [t 0, L]. Thus, if we write γ = x, y, then we hve κ y 0 on [0, L], nd x = κy, hence: 0 = L 0 x ds = L 0 κy ds = L 0 κ y ds.

15 4. FENCHEL S THEOREM 15 Since the integrnd in the lst integrl is non-negtive, we conclude tht κ y 0, hence y 0, gin contrdiction. It follows tht κ hs nother point where κ chnges sign, i.e., n extremum. Since extrem come in pirs, κ hs t lest four extrem. 4. Fenchel s Theorem We will use without proof the fct tht the shortest pth between two points on sphere is lwys n rc of gret circle. We lso use the nottion γ 1 + γ 2 to denote the curve γ 1 followed by the curve γ 2. Definition Let γ : [0, L] R n be regulr curve prmetrized by rclength. The sphericl imge of γ is the curve γ : [0, L] S n 1. The totl curvture of γ : [0, L] R n is: K γ = γ ds. I We note tht the totl curvture is simply the length of the sphericl imge. Theorem Let γ be regulr simple closed curve in R n prmetrized by rclength. Then the totl curvture of γ is t lest 2π: K γ 2π, with equlity if nd only if γ is plnr nd convex. The proof will follow from two lemmt which re interesting in their own right. Lemm Let γ : [0, L] R n be regulr closed curve prmetrized by rclength. Then the sphericl imge of γ cnnot mp into n open hemisphere. If γ mps into closed hemisphere, then γ mps into n equtor. Proof. Suppose, by contrdiction, tht there is v S n 1 such tht γ v > 0. Then 0 = γ v L γ v 0 = L 0 γ v ds > 0. If γ v 0, then the sme inequlity shows tht γ v 0, hence γ lies in the plne perpendiculr to v through γ0. Lemm Let n 3, nd let γ : [0, L] S n 1 be regulr closed curve on the unit sphere prmetrized by rclength. 1 If the rclength of γ is less thn 2π then γ is contined in n open hemisphere. 2 If the rclength of γ is equl to 2π then γ is contined in closed hemisphere. Proof. 1 First observe tht no piecewise smooth curve of rclength less thn 2π contins two ntipodl points. Otherwise the two segments of of the curve between p nd q would ech hve length t lest π, nd hence the length of the curve would hve to be t lest 2π. Now pick point p on γ nd let q on γ be chosen so tht the two segments γ 1 nd γ 2 from p to q long γ hve equl length. Note tht p nd q cnnot be ntipodl. Let v be the midpoint long the shorter of the two segments of the gret circle between p nd q. Suppose tht γ 1 intersects the equtor, the gret circle v x = 0. Let γ 1 be the reflection of γ with respect to v, then the length of γ 1 + γ 1 is the sme s the length of γ hence is less thn

16 16 1. CURVES 2π. But γ 1 + γ 1 contins two ntipodl points, contrdiction. Thus, γ 1 cnnot intersect the equtor. Similrly, γ 2 cnnot intersect the equtor, nd we conclude γ stys in the open hemisphere v x > 0. 2 If the rclength of γ is 2π, we refine the bove rgument. If p nd q re ntipodl, then both γ 1 nd γ 2 re gret semi-circle, thus, γ stys in closed hemisphere. 1 So we cn ssume tht p nd q re not ntipodl nd proceed s before, defining v to be the midpoint on the shorter rc of the gret circle between p nd q. Now, if γ 1 crosses the equtor, then γ 1 + γ 1 contins two ntipodl points on the equtor, nd the two segments joining these points enter both hemispheres. Thus, these segments re not semi-circle, nd consequently both hve rclength strictly greter thn π. Thus the rclength of γ 1 + γ 1 is strictly lrger thn 2π contrdiction. Similrly, γ 2 does not cross the equtor, nd we conclude tht γ stys in the closed hemisphere v x 0. Proof of Fenchel s Theorem. Note tht the totl curvture is simply the rclength of the sphericl imge of γ. By Lemm 1.13 γ is not contined in n open hemisphere, so by Lemm 1.14 K γ = γ ds 2π. I If the rclength of γ is 2π, then by Lemm 1.14, γ is contined in closed hemisphere, nd by Lemm 1.13, γ mps into n equtor. If n > 3, we my proceed by induction until we obtin tht γ is plnr. Once we hve tht γ is plnr, the Rottion Theorem gives n γ = ±1. Without loss of generlity, 2 we my ssume tht n γ = 1. Hence 0 κ κ ds = Kγ 2π = 0, I nd it follows tht κ = κ 0, which by Theorem 1.10 implies tht γ is convex. Exercises Exercise 1.1. A regulr spce curve γ : [, b] R 3 is helix if there is fixed unit vector u R 3 such tht e 1 u is constnt. Let κ nd τ be the curvture nd torsion of regulr spce curve γ, nd suppose tht κ 0. Prove tht γ is helix if nd only if τ = cκ for some constnt c. Exercise 1.2. Let γ : I R 4 be smooth curve prmeterized by rclength such tht γ, γ, γ re linerly independent. Prove the existence of Frenet frme, i.e., positively oriented orthonorml frme X = e 1, e 2, e 3, e 4 stisfying e 1 = γ, nd X = Xω, where ω is nti-symmetric, tri-digonl, nd ω i,i+1 > 0 for i n 2. The curvtures of γ re the three functions κ i = ω i,i+1, i = 1, 2, 3. Note tht κ 1, κ 2 > 0, but κ 3 hs sign. Exercise 1.3. Prove the Fundmentl Theorem for curves in R 4 : Given functions κ 1, κ 2, κ 3 on I with κ 1, κ 2 > 0, there is smooth curve γ prmeterized by rclength on I such tht κ 1, κ 2, κ 3 re the curvtures of γ. Furthermore, γ is unique up to rigid motion of R 4. 1 In fct, since γ is smooth, γ1 nd γ 1 re contined in the sme gret circle, nd hence γ is itself gret circle. 2 Reversing the orienttion of γ if necessry.

17 EXERCISES 17 Exercise 1.4. Let γ : [, b] R 2 be regulr plne curve with non-zero curvture κ 0, nd let β = γ + κ 1 N be the locus of the centers of curvture of γ. 1 Prove tht β is regulr provided tht κ 0. 2 Prove tht ech tngent l of β intersects γ t right ngle. A curve stisfying 1 nd 2 is clled n evolute of γ. 3 Prove tht ech regulr plne curve γ : [, b] R 2 hs t most one evolute. Exercise 1.5. A convex plne curve γ : [, b] R 2 is strictly convex if κ 0. Prove tht if γ : [, b] R 2 is strictly convex simple closed curve, then for every v S 1, there is unique t [, b] such tht e 1 t = v. Exercise 1.6. Let γ : [0, L] R 2 be strictly convex simple closed curve. The width wt of γ t t [0, L] is the distnce between the tngent line t γt nd the tngent line t the unique point γt stisfying e 1 t = e 1 t see Exercise 1.5. A curve hs constnt width if w is independent of t. Prove tht if γ hs constnt width then: 1 The line between γt nd γt is perpendiculr to the tngent lines t those points. 2 The curve γ hs length L = πw. Exercise 1.7. Let γ : [0, L] R 2 be simple closed curve. By the Jordn Curve Theorem, the complement of γ hs two connected components, one of which is bounded. The re enclosed by γ is the re of this component, nd ccording to Green s Theorem, it is given by: A = x dy = xy dt, γ where the orienttion is chosen so tht the norml e 2 points into the bounded component. Let L be the length of γ, nd let β be circle of width 2r equl to some width of γ. Prove: 1 A = 1 2 γ xy yx dt. 2 A + πr 2 Lr. 3 The isoperimetric inequlity: 4πA L 2. 4 If equlity holds in 3 then γ is circle. Exercise 1.8. Prove tht if convex simple closed curve hs four vertices, then it cnnot meet ny circle in more thn four points. γ

18

19 CHAPTER 2 Locl Surfce Theory 1. Surfces Definition 2.1. A prmetric surfce ptch is smooth mpping: X : U R 3, where U R 2 is open, nd the Jcobin dx is non-singulr. Write X = x 1, x 2, x 3, nd ech x i = x i u 1, u 2, then the Jcobin hs the mtrix representtion: x 1 1 x 1 2 dx = x 2 1 x 2 2 x 3 1 x 3 2 where we hve used the nottion f i = f u i = f/ u i. According to the definition, we re requiring tht this mtrix hs rnk 2, or equivlently tht the vectors X 1 = x 1 1, x 2 1, x 3 1 nd X 2 = x 1 2, x 2 2, x 3 2 re linerly independent. Another equivlent requirement is tht dx : R 2 R 3 is injective. Exmple 2.1. Let U R 2 be open, nd suppose tht f : U R is smooth. Define the grph of f s the prmetric surfce Xu 1, u 2 = u 1, u 2, fu 1, u 2. To verify tht X is indeed prmetric surfce, note tht: dx = f 1 f 2 so tht clerly X is non-singulr. A diffeomorphism between open sets U, V R 2 is mp φ: U V which is smooth, one-to-one, nd whose inverse is lso smooth. If detdφ > 0, then we sy tht φ is n orienttion-preserving diffeomorphism. Definition 2.2. Let X : U R 3, nd X : Ũ R3 be prmetric surfces. We sy tht X is reprmetriztion of X if X = X φ, where φ: Ũ U is diffeomorphism. If φ is n orienttion-preserving diffeomorphism, then X is n orienttion-preserving reprmetriztion. Clerly, the inverse of diffeomorphism is diffeomorphism. Thus, if X is reprmetriztion of X, then X is reprmetriztion of X. Definition 2.3. The tngent spce T u X of the prmetric surfce X : U R 3 t u U is the 2-dimensionl liner subspce of R 3 spnned by the two vectors X 1 nd X Note tht the tngent plne to the surfce XU t u is ctully the ffine subspce Xu + T ux. However, it will be very convenient to hve the tngent spce s liner subspce of R 3. 19

20 20 2. LOCAL SURFACE THEORY If Y T u X, then it cn be expressed s liner combintion in X 1 nd X 2 : 2 Y = y 1 X 1 + y 2 X 2 = y i X i, where y i R re the components of the vector Y in the bsis X 1, X 2 of T u X. We will use the Einstein Summtion Convention: every index which ppers twice in ny product, once s subscript covrint nd once s superscript contrvrint, is summed over its rnge. For exmple, the bove eqution will be written Y = y i X i. The next proposition show tht the tngent spce is invrint under reprmetriztion, nd gives the lw of trnsformtion for the components of tngent vector. Note tht covrint nd contrvrint indices hve different trnsformtion lws, cf. 2.1 nd 2.2. Proposition 2.1. Let X : U R 3 be prmetric surfce, nd let X = X φ be reprmetriztion of X. Then T φũ X = Tũ X. Furthermore, if Z Tφũ X, nd Z = z i X i = z j Xj, then: i=1 2.1 z i = z j ui ũ j, where dφ = u i / ũ j. Proof. By the chin rule, we hve: 2.2 Xj = ui ũ j X i. Thus Tũ X Tφũ X, nd since we cn interchnge the roles of X nd conclude tht Tũ X = Tφũ X. Substituting 2.2 in z j Xj, we find: X, we nd 2.1 follows. z i X i = z j ui ũ j X i, Definition 2.4. A vector field long prmetric surfce X : U R 3, is smooth mpping Y : U R 32. A vector field Y is tngent to X if Y u T u X for ll u U. A vector field Y is norml to X if Y u T u X for ll u U. Exmple 2.2. The vector fields X 1 nd X 2 re tngent to the surfce. The vector field X 1 X 2 is norml to the surfce. We cll the unit vector field N = X 1 X 2 X 1 X 2 the unit norml. Note tht the triple X 1, X 2, N, lthough not necessrily orthonorml, is positively oriented. In prticulr, we cn see tht the choice of n orienttion on X, e.g., X 1 X 2, fixes unit norml, nd vice-vers, the choice of unit norml fixes the orienttion. Here we chose to use the orienttion inherited from the orienttion u 1 u 2 on U. Definition 2.5. We cll the mp N : U S 2 the Guss mp. 2 We often visulize Y u s being ttched t Xu, i.e. belonging to the tngent spce of R 3 t Xu; cf. see footnote 1.

21 2. THE FIRST FUNDAMENTAL FORM 21 The Guss mp is invrint under orienttion-preserving reprmetriztion. Proposition 2.2. Let X : U R 3 be prmetric surfce, nd let N : U S 2 be its Guss mp. Let X = X φ be n orienttion-preserving reprmetriztion of X. Then the Guss mp of X is N φ. Proof. Let v V. The unit norml Ñv of X t v is perpendiculr to T v X. By Proposition 2.1, we hve Tφv X = T v X. Thus, Ñv is perpendiculr to T φv X, s is Nφv. It follows tht the two vectors re co-liner, nd hence Ñv = ±Nφv. But since φ is orienttion preserving, the two pirs X 1, X 2 nd X 1, X 2 hve the sme orienttion in the plne T v X. Since lso, the two triples X 1 φv, X 2 φv, Nφv nd X 1 v, X 2 v, Nv hve the sme orienttion in R 3, it follows tht Nφv = Ñv. 2. The First Fundmentl Form Definition 2.6. A symmetric biliner form on vector spce V is function B : V V R stisfying: 1 BX + by, Z = BX, Z + bby, Z, for ll X, Y V nd, b R. 2 BX, Y = BY, X, for ll X, Y V. The symmetric biliner form B is positive definite if BX, X 0, with equlity if nd only if X = 0. With ny symmetric biliner form B on vector spce, there is ssocited qudrtic form QX = BX, X. Let V nd W be vector spces nd let T : V W be liner mp. If B is symmetric biliner form on W, we cn define symmetric biliner form T Q on V by T QX, Y = QT X, T Y. We cll T Q the pull-bck of Q by T. The mp T is then n isometry between the inner-product spces V, T Q nd W, Q. Exmple 2.3. Let V = R 3 nd define BX, Y = X Y, then B is positive definite symmetric biliner form. The ssocited qudrtic form is QX = X 2. Exmple 2.4. Let A be symmetric 2 2 mtrix, nd let BX, Y = AX Y, then B is symmetric biliner form which is positive definite if nd only if the eigenvlues of A re both positive. Definition 2.7. Let X : U R 3 be prmetric surfce. The first fundmentl form is the symmetric biliner form g defined on ech tngent spce T u X by: gy, Z = Y Z, Y, Z T u X. Thus, g is simply the restriction of the Eucliden inner product in Exmple 2.3 to ech tngent spce of X. We sy tht g is induced by the Eucliden inner product. Let g ij = gx i, X j, nd let Y = y i X i nd Z = z i X i be two vectors in T u X, then 2.3 gy, Z = g ij y i z j. Thus, the so-clled coordinte representtion of g is t ech point u 0 U n instnce of Exmple 2.4. In fct, if A = g ij, nd Bξ, η = ξ Aη for ξ, η R 2 s in Exmple 2.4, then B is the pull-bck by dx u : R 2 T u X of the restriction of the Eucliden inner product on T u X.

22 22 2. LOCAL SURFACE THEORY The clssicl Guss nottion for the first fundmentl form is g 11 = E, g 12 = g 21 = F, nd G = g 22, i.e., E F gij = F G Clerly, F 2 < EG, nd nother condition equivlent to the condition tht X 1 nd X 2 re linerly independent is tht det g ij = EG F 2 > 0. The first fundmentl form is lso sometimes written: ds 2 = g ij du i du j = E du F du 1 du 2 + G du 2 2. Note tht the g ij s re functions of u. The reson for the nottion ds 2 is tht the squre root of the first fundmentl form cn be used to compute length of curves on X. Indeed, if γ : [, b] R 3 is curve on X, then γ = X β, where β is curve in U. Let βt = β 1 t, β 2 t, nd denote time derivtives by dot, then L γ [, b] = b γ dt b g ij βi βj dt. Accordingly, ds is lso clled the line element of the surfce X. Note tht g contins ll the intrinsic geometric informtion bout the surfce X. The distnce between ny two points on the surfce is given by: dp, q = inf{l γ : γ is curve on X between p nd q}. Also the ngle θ between two vectors Y, Z T x X is given by: cos θ = gy, Z gy, Y gz, Z, nd the ngle between two curves β nd γ on X is the ngle between their tngents β nd γ. Intrinsic geometry is ll the informtion which cn be obtined from the three functions g ij nd their derivtives. Clerly, the first fundmentl form is invrint under reprmetriztion. The next proposition shows how the g ij s chnge under reprmetriztion. Proposition 2.3. Let X : U R 3 be prmetric surfce, nd let X = X φ be reprmetriztion of X. Let g ij be the coordinte representtion of the first fundmentl form of X, nd let g ij be the coordinte representtion of the first fundmentl form of X. Then, we hve: 2.4 g ij = g kl u k where dφ = u i / ũ j. Proof. In view of 2.2, we hve: g ij = g X i, X u k j = g ũ i X k, ul ũ j X l ũ i u l ũ j, = uk ũ i u l ũ j gx u k k, X l = g kl ũ i u l ũ j.

23 3. THE SECOND FUNDAMENTAL FORM The Second Fundmentl Form We now turn to the second fundmentl form. First, we need to prove technicl proposition. Let Y nd Z be vector fields long X, nd suppose tht Y = y i X i is tngentil. We define the directionl derivtive of Z long Y by: Y Z = y i Z i = y i Z u i. Note tht the vlue of Y Z t u depends only on the vlue of Y t u, but depends on the vlues of Z in neighborhood of u. In ddition, Y Z is reprmetriztion invrint, but even if Z is tngent, it is not necessrily tngent. Indeed, if we write Y = ỹ i Xi, then we see tht: ỹ i i Z = ỹ i Z ũ i = y ũ i Z u k j u j u k ũ i = yj j Z. The commuttor of Y nd Z cn now be defined s the vector field: [Y, Z] = Y Z Z Y. Proposition 2.4. Let X : U R 3 be surfce, nd let N be its unit norml. 1 If Y nd Z re tngentil vector fields then [Y, Z] T u X. 2 If Y, Z T u X then Y N Z = Z N Y. Proof. Note first tht since X is smooth, we hve X ij = X ji, where we hve used the nottion X ij = 2 X/ u i u j. Now, write Y = y i X i nd Z = z j X j, nd compute: Y Z Z Y = y i z j X ji + y i i z j X j y i z j X ij z j j y i X i = y i i z j z i i y j X j. To prove 2, extend Y nd Z to be vector fields in neighborhood of u, nd use 1: Y N Z Z N Y = N Y Z Z Y = 0. Note tht while proving the proposition, we hve estblished the following formul for the commuttor: 2.5 [Y, Z] = y i i z j z i i y j X j Definition 2.8. Let X : U R 3 be surfce, nd let N : U S 2 be its unit norml. The second fundmentl form of X is the symmetric biliner form k defined on ech tngent spce T u X by: 2.6 ky, Z = Y N Z. We remrk tht since N N = 1, we hve Y N N = 0, hence Y N is tngentil. Thus, ccording to 2.6, the second fundmentl form is minus the tngentil directionl derivtive of the unit norml, nd hence mesures the turning of the tngent plne s one moves bout on the surfce. Note tht prt 2 of the proposition gurntees tht k is indeed symmetric biliner form. Note tht it is not necessrily positive definite. Furthermore, if we set k ij = kx i, X j to be the coordinte representtion of the second fundmentl form, then we hve: 2.7 k ij = X ij N.

24 24 2. LOCAL SURFACE THEORY This eqution leds to nother representtion. Consider the Tylor expnsion of X t point, sy 0 U: Xu = X0 + X i 0 u i ijx0 u i u j + O u 3 Thus, the elevtion of X bove its tngent plne t u is given up to second-order terms by: Xu X0 Xi 0u i N = 1 2 k ij0u i u j + O u 3. The prboloid on the right-hnd side of the eqution bove is clled the osculting prboloid. A point u of the surfce is clled elliptic, hyperbolic, prbolic, or plnr, depending on whether this prboloid is elliptic, hyperbolic, cylindricl, or plne. In clssicl nottion the second fundmentl form is: L M kij = M N Clerly, the second fundmentl form is invrint under orienttion-preserving reprmetriztions. Furthermore, the k ij s, the coordinte representtion of k, chnges like the first fundmentl form under orienttion-preserving reprmetriztion: k ij = k X u, X u m u l j = k ml ũ i ũ j, Yet nother interprettion of the second fundmentl form is obtined by considering curves on the surfce. The following theorem is essentilly due to Euler. Theorem 2.5. Let γ = X β : [, b] R 3 be curve on prmetric surfce X : U R 3, where β : [, b] U. Let κ be the curvture of γ, nd let θ be the ngle between the unit norml N of X, nd the principl norml e 2 of γ. Then: 2.8 κ cos θ = k γ, γ. nd Proof. We my ssume tht γ is prmetrized by rclength. We hve: γ = β i X i,. κe 2 = γ = β i X i + β i βj X ij. The theorem now follows by tking inner product with N, nd tking 2.7 into ccount. The quntity κ cos θ is clled the norml curvture of γ. It is prticulrly interesting to consider norml sections, i.e., curves γ on X which lie on the intersection of the surfce with norml plne. We my lwys orient such plne so tht the norml e 2 to γ in the plne coincide with the unit norml N of the surfce. In tht cse, we obtin the simpler result: κ = k γ, γ. Thus, the second fundmentl form mesures the signed curvture of norml sections in the norml plne equipped with the pproprite orienttion.

25 4. EXAMPLES 25 Definition 2.9. Let X : U R 3 be prmetric surfce, nd let k be its second fundmentl form. Denote the unit circle in the tngent spce t u by S u X = {Y T u X : Y = 1}. For u U, define the principl curvtures of X t u by: k 1 = min ky, Y, k 2 = mx ky, Y. Y S ux Y S ux The unit vectors Y S u X long which the principl curvtures re chieved re clled the principl directions. The men curvture H nd the Guss curvture K of X t u re given by: H = 1 2 k 1 + k 2, K = k 1 k 2. If we consider the tngent spce T u X with the inner product g nd the unique liner trnsformtion l: T u X T u X stisfying: 2.9 g ly, Z = ky, Z, Z T u X, then k 1 k 2 re the eigenvlues of l nd the principl directions re the eigenvectors of l. If k 1 = k 2 then k = λg nd every direction is principl direction. A point where this holds is clled n umbilicl point. Otherwise, the principl directions re perpendiculr. We hve tht H is the trce nd K the determinnt of l. Let g ij be the inverse of the 2 2 mtrix g ij : g im g mj = δ i j. Set lx i = l j i X j, then since k ij = glx i, X j = l m i g mj, we find: l j i = k img mj. It is customry to sy tht g rises the index of k nd to write the new object k j i = k im g mj. Here since k ij is symmetric, it is not necessry to keep trck of the position of the indices, nd hence we write: l j i = kj i. In prticulr, we hve: 2.10 H = 1 2 ki i, K = det k ij det. g ij Now, k ij = g im g jl k lm, nd we hve Hence, we conclude k 2 = k ij k ij = tr l 2 = k k 2 2 = 4H 2 2K K = 2H k 2 4. Exmples In this section, we use u 1 = u, nd u 2 = v in order to simplify the nottion Plnes. Let U R 2 be open, nd let X : U R 3 be liner function: Xu, v = Au + Bv, with A, B R 3 linerly independent. Then X is plne. After reprmetriztion, we my ssume tht A nd B re orthonorml. In tht cse, the first fundmentl form is: ds 2 = du 2 + dv 2.

26 26 2. LOCAL SURFACE THEORY Furthermore, A B = 1, nd N = A B is constnt, hence k = 0. In prticulr, ll the points of X re plnr, nd we hve for the men nd Guss curvtures: H = K = 0. It is of interest to note tht if ll the points of prmetric surfce re plnr, then XU is contined in plne. We will lter prove stronger result: X hs reprmetriztion which is liner. Proposition 2.6. Let X : U R 3 be prmetric surfce, nd suppose tht its second fundmentl form k = 0. Then, there is fixed vector A nd constnt b such tht X A = b, i.e., X is contined in plne. Proof. Let A be the unit norml N of X. Let 1 i 2, nd note tht N i is tngentil. Indeed, N N = 1, nd differentiting long u i, we get N N i = 0. However, since k = 0 it follows from 2.6 tht N i X j = k ij = 0. Thus, N i = 0 for i = 1, 2, nd we conclude tht N is constnt. Consequently, X N i = X i N = 0, nd X N is lso constnt, which proves the proposition Spheres. Let U = 0, π 0, 2π R 2, nd let X : U R 2 be given by: Xu, v = sin u cos v, sin u sin v, cos u. The surfce X is prmetric representtion of the unit sphere. A strightforwrd clcultion shows tht the first fundmentl form is: ds 2 = du 2 + sin 2 u dv 2, nd the unit norml is N = X. Thus, N i = X i, nd consequently k ij = N i X j = X i X j = g ij, i.e., k = g. In prticulr, the principl curvtures re both equl to 1 nd ll the points re umbilicl. We hve for the men nd Guss curvtures: H = 1, K = 1 Proposition 2.7. Let X : U R 3 be prmetric surfce nd suppose tht ll the points of X re umbilicl. Then, XU is either contined in plne or sphere. Proof. By hypothesis, we hve 2.12 N i = λx i. We first show tht λ is constnt. Differentiting 2.12, we get N ij = λ j X i +λx ij. Interchnging i nd j, subtrcting these two equtions, nd tking into ccount N ij N ji = X ij X ji = 0, we obtin λ i X j λ j X i = 0, e.g., λ 1 X 2 λ 2 X 1 = 0. Since X 1 nd X 2 re linerly independent, we conclude tht λ 1 = λ 2 = 0 nd it follows tht λ is constnt. Now, if λ = 0 then ll points re plnr, nd by Proposition 2.6, X is contined in plne. Otherwise, let A = X λ 1 N, then A is constnt: A i = X i λ 1 N i = 0, nd X A = λ 1 is lso constnt, hence X is contined in sphere.

27 4. EXAMPLES Ruled Surfces. A ruled surfce is prmetric surfce of the form: Xu, v = γu + vy u for curve γ : [, b] R 3, nd vector field Y : [, b] R 3 long γ. The curve γ is the directrix, nd the lines γu + ty u for u fixed re the genertors of X. We my ssume tht Y is unit vector field. Provided Ẏ 0. We will lso ssume tht Ẏ 0. In this cse, it is possible to rrnge by reprmetriztion tht γ Ẏ = 0, in which cse γ is sid to be line of striction. Indeed, if this is not the cse, then we cn set φ = γ Ẏ / Ẏ 2, nd note tht the curve α = γ + φy lies on the surfce X, nd stisfies α Ẏ = 0. Consequently, the surfce: Xs, t = αs + ty s is reprmetriztion of X. Furthermore, there is only one line of striction on X. Indeed, if β nd γ re two lines of striction, then since both β is curve on X we my write β = γ + φy for some function φ nd consequently: β = γ + φy + φẏ. Tking inner product with Ẏ nd using the fct tht Y is unit vector, we obtin φ Ẏ 2 = 0 which implies tht φ = 0 nd thus, β = γ. We hve X u = γ + vẏ, X v = Y, nd X vv = 0. Thus, the first fundmentl is: 1 + v 2 Ẏ gij = 2 γ Y γ Y 1 nd det g ij = 1 + v 2 Ẏ 2 + γ Y 2 v 2 Ẏ 2. Hence, dx is non-singulr except possibly on the line of striction. Furthermore, k vv = N X vv = 0, hence det k ij = k 2 uv nd if det k ij = 0 then Nv X u = N v X v = 0, i.e., N is constnt long genertors. We hve proved the following proposition. Proposition 2.8. Let X be ruled surfce. Then X hs non-positive Guss curvture K 0, nd Ku = 0 if nd only if N is constnt long the genertor through u Cylinders. Let γ : [, b] R 3 be plnr curve, nd A be unit norml to the plne which contins γ. Define X : [, b] R R 3 by: Xu, v = γu + va. The surfce X is cylinder. The first fundmentl form is: ds 2 = du 2 + dv 2, nd we see tht for cylinder dx is lwys non-singulr. After possibly reversing the orienttion of A, the unit norml is N = e 2. Clerly, N v = 0, nd N u = κe 1. Thus, the second fundmentl form is: κ du 2 The principl curvtures re 0 nd κ. We hve for the men nd Guss curvtures: H = 1 κ, K = 0. 2

28 28 2. LOCAL SURFACE THEORY A surfce on which K = 0 is clled developble Tngent Surfces. Let γ : [, b] R 3 be curve with nonzero curvture κ 0. Its tngent surfce is the ruled surfce: Xu, v = γu + v γu. Since γ γ = 0, the curve γ is the line of striction of its tngent surfce. We hve X u = e 1 + vκ e 2 nd X v = e 1, hence the first fundmentl form is: 1 + v 2 κ 2 1 gij =. 1 1 The unit norml is N = e 3, nd clerly N v = 0. Thus, Hyperboloid. Let γ : 0, 2π R 3 be the unit circle in the x 1 x 2 -plne: γt = cost, sint, 0. Define ruled surfce X : 0, 2π R R 3 by: Xu, v = γu + v γu + e 3 = cosu v sinu, sinu + v cosu, v. Note tht x x 2 2 x 3 3 = 1 so tht XU is hyperboloid of one sheet. A strightforwrd clcultion gives: N = v 2 cosu v sinu, sinu + v cosu, v, nd N v 2 2 = 1 + 4v 2 + 4v 4. It follows from Proposition 2.8 tht X hs Guss curvture K < Lines of Curvture Definition A curve γ on prmetric surfce X is clled line of curvture if γ is principl direction. The following proposition, due to Rodriguez, chrcterizes lines of curvture s those curves whose tngents re prllel to the tngent of their sphericl imge under the Guss mp. Proposition 2.9. Let γ be curve on prmetric surfce X with unit norml N, nd let β = N γ be its sphericl imge under the Guss mp. Then γ is line of curvture if nd only if 2.13 β + λ γ = 0. Proof. Suppose tht 2.13 holds, then we hve: γ N + λ γ = 0. Let l be the liner trnsformtion on T u X ssocited with k s defined by 2.9. Then, we hve for every Y T u X: g l γ, Y = k γ, Y = γ N Y = λg λ γ, Y. Thus, l γ = λ γ, nd γ is principl direction. similr. The proof of the converse is

29 5. LINES OF CURVATURE 29 It is cler from the proof tht λ in 2.13 is the ssocited principl curvture. The coordinte curves of prmetric surfce X re the two fmily of curves γ c t = Xt, c nd β c t = Xc, t. A surfce is prmetrized by lines of curvture if the coordinte curves of X re lines of curvture. We will now show tht ny nonumbilicl point hs neighborhood in which the surfce cn be reprmetrized by lines of curvture. We first prove the following lemm which is lso of independent interest. Lemm Let X : U R 3 be prmetric surfce, nd let Y 1 nd Y 2 be linerly independent vector fields. The following sttements re equivlent: 1 Any point u 0 U hs neighborhood U 0 nd reprmetriztion φ: V 0 U 0 such tht if X = X φ then X i = Y i φ. 2 [Y 1, Y 2 ] = 0. Proof. Suppose tht 1 holds. Then Eqution 2.5 shows tht [ X 1, X 2 ] = 0. However, since the commuttor is invrint under reprmetriztion, it follows tht [Y 1, Y 2 ] = 0. Conversely, suppose tht [Y 1, Y 2 ] = 0. Express X i = j i Y j nd Y i = b j i X j, nd note tht b j i is the inverse of j i. We now clculte: 0 = [X i, X j ] = [ k i Y k, l jy l ] = l i Yl k j l j Yl k i Yk + k i l j[y k, Y l ] = l ib m l m k j l jb m l m k i Yk = i k j j k i Yk. Since Y 1 nd Y 2 re linerly independent, we conclude tht: 2.14 i k j j k i = 0. Now, fix 1 k 2, nd consider the over-determined system: ũ k u i = k i, i = 1, 2. The integrbility condition for this system is exctly 2.14, hence there is solution in neighborhood of u 0. Furthermore, since the Jcobin of the mp ψu 1, u 2 = ũ 1, ũ 2 is dψ = k i, nd det k i 0, it follows from the inverse function theorem, tht perhps on yet smller neighborhood, ψ is diffeomorphism. Let φ = ψ 1, then φ is diffeomorphism on neighborhood V 0 of ψu 0, nd if we set X = X φ, then: X i = X j u j ũ i = X jb j i = Y i. Proposition Let X : U R 3 be prmetric surfce, nd let Y 1 nd Y 2 be linerly independent vector fields. Then for ny point u 0 U there is neighborhood of u 0 nd reprmetriztion X = X φ such tht X i = f i Y i φ for some functions f i.

30 30 2. LOCAL SURFACE THEORY Proof. By Lemm 2.10 is suffices to show tht there re function f i such tht f 1 Y 1 nd f 2 Y 2 commute. Write [Y 1, Y 2 ] = 1 Y 1 2 Y 2, nd compute: [f 1 Y 1, f 2 Y 2 ] = f 1 f 2 1 Y 1 2 Y 2 + f1 Y1 f 2 Y2 f 2 Y2 f 1 Y1. Thus, the commuttor [f 1 Y 1, f 2 Y 2 ] vnishes if nd only if the following two equtions re stisfied: We cn rewrite those s: Y2 f 1 1 f 1 = 0 Y1 f 2 2 f 2 = 0. Y2 log f 1 = 1 Y1 log f 2 = 2. Ech of those eqution is liner first-order prtil differentil eqution, nd cn be solved for positive solution in neighborhood of u 0. In neighborhood of non-umbilicl point, the principl directions define two orthogonl unit vector fields. Thus, we obtin the following Theorem s corollry to the bove proposition. Theorem Let X : U R 3 be prmetric surfce, nd let u 0 be non-umbilicl point. Then there is neighborhood U 0 of u 0 nd diffeomorphism φ: Ũ0 U 0 such tht X = X φ is prmetrized by lines of curvture. If X is prmetrized by lines of curvture, then the second fundmentl form hs the coordinte representtion: k1 g 11 0 kij = 0 k 2 g 22 Definition A curve γ on prmetric surfce X is clled n symptotic line if it hs zero norml curvture, i.e., k γ, γ = 0. The term symptotic stems from the fct tht those curve hve their tngent γ long the symptotes of the Dupin indictrix, the conic section k ij ξ i ξ j = 1 in the tngent spce. Since the Dupin indictrix hs no symptotes when K > 0, we see tht the Guss curvture must be non-positive long ny symptotic line. The following Theorem cn be proved by the sme method s used bove to obtin Theorem Theorem Let X : U R 3 be prmetric surfce, nd let u 0 be hyperbolic point. Then there is neighborhood U 0 of u 0 nd diffeomorphism φ: Ũ0 U 0 such tht X = X φ is prmetrized by symptotic lines. 6. More Exmples A surfce of revolution is prmetric surfce of the form: Xu, v = fu cosv, fu sinv, gu, where ft, gt is regulr curve, clled the genertor, which stisfies ft 0. Without loss of generlity, we my ssume tht ft > 0. The curves γ v t = ft cosv, ft sinv, gt, v fixed.

31 6. MORE EXAMPLES 31 re clled meridins nd the curves β u t = fu cost, fu sint, gu, u fixed. re clled prllels. Note tht every meridin is plnr curve congruent to the genertor nd is furthermore lso norml section, nd every prllel is circle of rdius fu. It is not difficult to see tht prllels nd meridins re lines of curvture. Indeed, let γ v be meridin, then choosing s in the prgrph following Theorem 2.5 the correct orienttion in the plne of γ v, its sphericl imge under the Guss mp is σ v = N γ v = e 2, nd by the Frenet equtions, σ v = κe 1 = κ γ v. Thus, using Proposition 2.9 nd the comment immeditely following it, we see tht γ v is line of curvture with ssocited principl curvture κ. Since the prllels β u re perpendiculr to the meridins γ v, it follows immeditely tht they re lso lines of curvture. We derive this lso follows from Proposition 2.9 nd furthermore obtin the ssocited principl curvture. A strightforwrd computtion gives tht the sphericl imge of β u under the Guss mp is: τ u = N β u = cβ u + B where B R 3 nd c R re constnts. Thus, τ u = c β u nd β u is line of curvture with ssocited principl curvture c. The plne, the sphere, the cylinder, nd the hyperboloid re ll surfces of revolution. We discuss one more exmple. The ctenoid is the prmetric surfce of revolution obtined from the generting curve cosht, t : Xu, v = coshu cosv, coshu sinv, u. The norml N is esily clculted: cosv Nu, v = coshu, sinv coshu, sinhu coshu If γ v t is meridin, then σ v t = Nt, v is its sphericl imge under the Guss mp, nd differentiting with respect to t, we get the principl curvture ssocited with meridins: κu, v = 1/ coshu. Similrly, the principl curvture ssocited with prllels is: 1/ coshu. Thus, we conclude tht 1 H = 0, K = coshu 2. Definition A prmetric surfce X is miniml if it hs vnishing men curvture H = 0. For exmple, the ctenoid is miniml surfce. The justifiction for the terminology will be given in the next section. The following proposition is immedite from Proposition Let X be miniml surfce. Then X hs non-positive Guss curvture K 0, nd Ku = 0 if nd only if u is plnr point. We will set out to construct lrge clss of miniml surfces. We will use the Weierstrss Representtion. Definition A prmetric surfce X is conforml if the first fundmentl form stisfies g 11 = g 22 nd g 12 = 0. A prmetric surfce X is hrmonic if X = X 11 + X 22 = 0.