在二維黎曼流形上的四頂點定理

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(1)國立臺灣師範大學數學系碩士班碩士論文. 指導教授：. 林俊吉. 博士. The Four-vertex Theorem in Riemannian 2-manifolds. 在二維黎曼流形上的四頂點定理. 研究生：黃志強. 中華民國九十五年六月.

(2) Abstract We survey the results on the four-vertex theorems, beginning from 1909. Moreover, we focus on curves in Riemannian 2-manifolds and generalize the Osserman’s proof [6]. Some four-vertex theorems for special curves in Riemannian 2-manifolds are derived in this thesis. Key words: vertex, curvature, circumscribed circle, four-vertex theorem. 摘要我們研究從 1909 年開始的頂點定理之結果。更進一步，我們著重於在二維黎曼流形上的曲線，並且推廣奧瑟曼的證明。在這篇論文，某些二維黎曼流形上特殊曲線的四頂點定理被得到。關鍵字：頂點，曲率，外接圓，四頂點定理. i.

(3) Contents §1 Introduction. 1. §2 Comparisons and discussions on the proofs of the four-vertex theorem 1~2 §3 The four-vertex theorem in Hadamard’s 2-manifolds with negative constant curvature. 2~5. §4 The four-vertex theorem in Hadamard’s 2-manifolds for some special curves. 5~11. §5 The four-vertex theorem in Riemannian 2-manifolds for some special curves. 11~13. §6 Summary. 13~14. Reference. 15. ii.

(4) §1 Introduction Recently, the four-vertex theorem received more attention in global geometry of curves. In 1909, Mukhopadhaya proved that a simple closed convex planar curve has at least four vertices, where a vertex is defined as a critical point (or local extreme) of the curvature. The theorem, without assuming convexity of curves, was proved by Kneser in 1912. After Kneser’s proof, some generalizations or analogies were published. In this thesis, we would like to generalize the four-vertex theorem from plane case to Riemannian 2-manifods, where a vertex is defined as a critical point of the geodesic curvature. Jackson verified the four-vertex theorem for curves in spheres [1], and in surfaces with constant curvature [3], [4]. He also reproved the plane case by using inscribed circles [2]. Additionally, the plane case is also elegantly reproved by Osserman from the approach of circumscribed circles [6]. A new proof for convex curves in the hyperbolic plane was given by Singer in 2001 [7]. Meanwhile, another proof for the plane case was also shown by Ou in 2003 [8]. With regard to the previous studies, it is natural to ask whether the four-vertex theorem still holds in surfaces with variable curvature. However, Jackson gave a counterexample: there exists a sufficiently small distance circle centered at a nondegenerate point of Gauss curvature which has only two vertices [3] (p508). Thus, the four-vertex theorem cannot hold for arbitrary simple closed curves in surfaces with variable curvature. In this thesis, some four-vertex theorems for special curves in Riemannian 2-manifolds with variable curvature are derived.. §2 Comparisons and discussions on the proofs of the four-vertex theorem The proofs of the four-vertex theorem are discussed and compared in this chapter. Here, the most important contributions of references mentioned in chapter 1 are summarized in Table 1.. 1.

(5) Author. Jackson. Osserman. Ou. Singer. Surface. Essence. Statement. Plane. Inscribed circle. Compare the curvature between curves and inscribed circles. Sphere. Inversion. Establish a vertex correspondence between planes and spheres. Surface with constant curvature. Vertex preserving transformation. Establish a vertex correspondence between planes and surfaces with the constant curvature. Plane. Circumscribed circle. Compare the curvature between curves and the Circumscribed circle. Plane. Equality for the derivative of the curvature. Consider a conformal map form the upper plane to the interior of the curve. Ghys’ theorem. Find a diffeomorphism between the curve and a circle so that its hyperbolic trace is relative to the curvature. Hyperbolic plane. Table 1 The verifications of references in Table 1 are examined to generalize the four-vertex theorem for manifolds with the variable curvature. In Ou’s proof, the equality for the curvature function intensely depends on zero Gauss curvature. If the Gauss curvature does not identify zero, the arbitrary choice of a conformal map disappears. Therefore, the proof is not easy to generalize. In Singer’s paper, the relation between the hyperbolic trace and the curvature function was established from the magical existence of the diffeomorphism. It may be not easy to apply these theories for surfaces with the variable curvature. The best choice can be either Jackson's proof or Osserman's because theirs are of the type of construction. However, the smallest circumscribed circle is unique, but the largest inscribed circle is not. Therefore, we want to reprove the four-vertex theorem by following the Osserman’s idea.. §3 The four-vertex theorem in Hadamard’s 2-manifolds with negative constant curvature Let M be a Hadamard’s 2-manifold (complete, simply connected Riemannian manifold with nonpositive curvature) with negative constant Gauss curvature K . The 2.

(6) geometry in M is similar to Euclidean plane. For example, M is diffeomorphic to R 2 ; two points on M precisely determine one geodesic. Moreover, the law of cosines of a triangle ΔABC on M is cosh( − K a ) = cosh( − K b) cosh( − K c ) − sinh( − K b) sinh( − K c ) cos θ. where a = d ( B, C ), b = d (C , A), c = d ( A, B ) and θ is the vertex angle at A . By the law of cosines, one can establish the law of perpendicular bisector: Given a geodesic segment AB , a point P is at the perpendicular bisector of AB if and only if d ( P, A) = d ( P, B ) . For four-vertex theorem in the plane, Osserman mentioned “the essence of the proof may be distilled in a single phrase: consider the circumscribed circle.” Therefore, first of all, we need to check the existence and uniqueness of the circumscribed circle in M . Definition 3.1 Let γ be a simple closed curve in M . The circumscribed circle of γ is the smallest distance circle containing the interior of γ . Denote it by CR , where R is the smallest radius. Lemma 3.1 Let γ be a simple closed curve in M , then there exists the unique circumscribed circle of γ . Proof. Choose a distance circle C containing the interior of γ . For all p ∈ Int(C ) , Rp := sup d (q, p ) = max d (q, p ) because of the continuity of d . Define q∈γ. q∈γ. f : Int(C ) → R such that f ( p ) = R p , then f is continuous and gets the minimum R .. If there are two points p1 , p2 such that f ( p1 ) = f ( p2 ) = R . Let C1 , C2 be two circles with the centers p1 , p2 and radius R , respectively. Let C1 ∩ C2 = { A, B} and p* be the middle point of p1 , p2 . There is a circle with the center p* and the radius d ( p* , A) < R so that it contains the interior of γ . It is contradictory to the definition of the circumscribed circle. □ Lemma 3.2 Any subarc of the circumscribed circle CR greater than a semicircle intersects γ . Proof. If there is a semicircle PQ such that PQ ∩ γ = ∅ . Let δ = d ( PQ, γ ) > 0 , and. Ω=. Int(C R ) − {interior of circles circle with a center ∈ PQ and radius = δ } 3. ..

(7) Along the perpendicular bisector of PQ , slightly move the center of CR so as to detach it far away from PQ . Denote the new point by O* . Let CR* be the circle with the center O* and the radius R . Then CR* ∩ Ω = ∅ . Hence, there is a smaller circle containing γ . It is contradictory to the definition of CR . □ The lemma 3.2 states that the circumscribed circle CR meets γ at least two points. Afterwards, we discuss the relations between the curvature and the position of the curve. The proof of the following lemma can refer to Jackson’s paper [2], [3]. We change the statement of the lemma in this thesis. Lemma 3.3 Let oriented curve γ have the same unit tangent at a point P as a positively oriented distance circle C of radius R in M . Let kg be the geodesic curvature of γ .Then (1) if kg ( P ) > − K coth( − K R ) , a neighborhood of P on γ lies inside of C , (2) while. if. a. neighborhood. of. P. on. γ. lies. inside. of. C , then. kg ( P ) ≥ − K coth( − K R ) .. In order to further promote the proof, we encounter some problems: (1) How do we define ‘the movement to the left of the circumscribed circle’? (2) Three points do not merely make a circle in general. Hence we need to slightly modify the notion of this proof. Lemma 3.4 Let γ be a simple closed, positively oriented curve tangent to the circumscribed circle CR at P1 , P2 . Let γ * be the positively oriented subarc of γ form P1 to P2 . Then there is a point Q in γ * such that kg (Q ) ≤ − K coth( − K R ) . Proof. By Lemma 3.2, we may assume P1 , P2 are contained in a semicircle. Consider the family - of all distance circles of radius R , centered at a point in the perpendicular bisector of the geodesic segment P1 P2 . Since γ is simple and lies. 4.

(8) inside of circular arc P1 P2 , there is a distance circle C * in - satisfying C * lies to left of γ * . Moreover, C * never touches P1 , P2 and never crosses γ * , but C * meets γ * .We know a circle meets a differential arc at an interior point without crossing it, then they are tangent. By Lemma 3.3, there is an interior point Q such that kg (Q ) ≤ − K coth( − K R ) . □. The main theorem describes the relation between the circumscribed circle and vertices. Theorem 3.1: Let γ be a simple closed curve on M and CR be the geodesic circumscribed circle of γ . Then (1) the number of γ ∩ CR is at least 2. (2) if the number of γ ∩ CR is at least n , then γ has at least 2 n vertices. Proof. By Lemma 3.2, the first part is done. Let {P1 , P2 , , Pn } = γ ∩ CR . By Lemma , n . On the other hand, let γ i be the. 3.3, kg ( Pi ) ≥ − K coth( − K R) , i = 1, 2,. positively oriented subarc of γ form Pi to Pi +1 ( Pn +1 = P1 ). By lemma 3.4, there is a Qi such that kg (Qi ) ≤ − K coth( − K R) . Hence, we can conclude there is a local minimum at an interior point of γ i . Let γ i* be the subarc of γ from Qi to Qi +1 ( Qn +1 = Q1 ). Then there exists a local maximum at an interior point of γ i* . Therefore, we complete the proof. □. §4 The four-vertex theorem in Hadamard’s 2-manifolds for some special curves Let N be a Hadamard’s 2-manifold with the variable Gauss curvature K . N is diffeomorphic to R 2 although the space N may be not symmetrical. Moreover, two points precisely determines one geodesic. Some properties can be obtained by comparison theorems or variations [9], [10]. 5.

(9) Property1: If a triangle in N. has three edges a, b, c and the angle θ. corresponding to the edge c , then a 2 + b 2 − 2ab cos θ ≤ c 2 .. Moreover, K ≤ − k 2 < 0 , where k is a constant, then cosh( ka ) cosh( kb) − sinh( ka ) sinh(kb) cos θ ≤ cosh( kc ) . Property 2: Let γ : ( −∞, ∞ ) → N be a geodesic and p be a point off γ , in N . Let d ( s ) = d ( p, γ ( s )) , then there exists exactly one s0 such that d ( s0 ) = min d ( s ) . The minimizer is the perpendicular point. Moreover, d ( s ) is a convex function. Lemma 4.1 Let γ be a simple closed curve in N , then there exists the unique circumscribed circle of γ . Proof. It is similar to Lemma 3.1. We only deal with the uniqueness of the circumscribed circle. Let C1 , C2 be two circumscribed circles of γ with the centers O1 , O2 and radius R , respectively. Let the region Ω be the interior of the intersection of C1 , C2 . Take an arbitrary point P in the geodesic segment O1O2 . For all points Q ∈ ∂Ω , d ( P, Q ) ≤ R ' < R because d (t ) = d (Q, α (t )) is a convex function, where α (t ) is a parameterization of O1O2 . Therefore there is a circle of radius R ' centered at P and containing γ . This is contradictory. □ Lemma 4.2 Any subarc of the circumscribed circle CR greater than a semicircle intersects γ . Proof. Let PQ , δ and Ω be the same notations in Lemma 3.2. Along the perpendicular bisector of PQ , move the center of CR below. δ 2. so as to detach it. far away PQ . By the triangle inequality and Property 2, the new point O* satisfies d (O* , A) ≤ R ' < R for all A∈ ∂Ω . It is contradictory. □ Lemma 4.3 (due to Jackson [3]) Let γ and γ be two positively oriented curves tangent in the same direction at P . Let kg and kg be the geodesic curvatures of γ and γ , respectively. 6.

(10) Then (1) if kg ( P ) > kg ( P ) , then a neighborhood of P in γ lies on or outside γ , (2)while if a neighborhood of P in γ lies outside γ , then kg ( P ) ≥ kg ( P ) . Proof: Choose an orthogonal coordinate in N with γ as the curve u = 0 , and the geodesics orthogonal to γ as the u -curves. Let u denote arc length along these geodesics and v denote arc length along γ measured in the positive direction from P , the first fundamental form is ds 2 = du 2 + Gdv 2 , where G (0, v ) = 1 . Because γ is. tangent to γ at P , its equation can be written as u = u (v ) and u (0) = u '(0) = 0 . Afterwards, compute the geodesic curvature of γ and γ . Then. Gu 2G , where the bar denotes that functions are evaluated for u = 0 . And kg =. kg =. −2Gu ''+ 2Gu u '2 + Gv u '+ GGu . 2 G (u '2 + G )3/ 2. By kg ( P ) > kg ( P ) , in a neighborhood of 0, 2Gu u '2 + Gv u '+ GGu Gu (u '2 + G )3/ 2 u '' ≤ − . 2G 2G G By expanding that functions as a finite Taylor series of two terms, one can get. u '' ≤ K u ' + L u , where K and L are sufficiently large positive constants. R. P. Boas, Jr. has established the following result [11]. If f ( x ) is of class C 2 in (0, a ) , f (0) = f '(0) = 0 , and f ''( x) ≤ K f '( x) + L f ( x) in (0, a ) , where K and L are constants, then f ( x ) ≤ 0 in some interval (0, b) . We understand uncomplicatedly this result by assuming f ( x ) is analytic at 0. Let an be the first nonvanishing coefficient of the power series of f ( x ) at 0. Then x 2− n f ''( x) approaches a nonzero limit, while x 2− n f '( x) and x 2− n f ( x) approach zero, as x → 0 . Hence an < 0 and f ( x ) ≤ 0 , near 0. Continuing the proof of this lemma, we can gain u ≤ 0 , near 0. Hence the lemma can be derived. The second part is derived by contradicting to (1).□ If the Gauss curvature of a manifold is variable, then the curvature of a distance 7.

(11) circle is not a constant. Therefore, we cannot expect the four-vertex theorem on these manifolds. Afterwards, we want to understand how the geodesic curvature of distance circles fluctuates. Lemma 4.4 A complete Riemannian 2-manifold satisfies −b 2 ≤ K ≤ − a 2 < 0 , where a, b are constants, then the geodesic curvature kg R of distance circles of radius R satisfies a coth aR ≤ kg R ≤ b coth bR In another case, −b 2 ≤ K ≤ 0 , 1 ≤ kg R ≤ b coth bR R. Proof: Take the geodesic polar coordinate in N , then the first fundamental form is ds 2 = d ρ 2 + Gdθ 2 . And. ( G ) ρρ ( ρ , θ ) + K ( ρ ,θ ) G ( ρ ,θ ) = 0 G (0, θ ) = 0, ( G ) ρ (0,θ ) = 1. (*).. Compare this equation with ( G ) ρρ ( ρ ,θ ) − a 2 G ( ρ ,θ ) = 0 G (0,θ ) = 0, ( G ) ρ (0,θ ) = 1 G ( ρ ,θ ) =. , whose solution is. 1 sinh( a ρ ) .For convenience, denote u ( ρ ) = a. 1 sinh(a ρ ) . Since the Gauss curvature is a nonpositive, u ( ρ ) is increasing and never equals to zero. Then ( G )( ρ , θ ) is a solution of (*) and v ( ρ ) =. ρ. 0= ∫ [u (t )(v ''(t ) − a 2v(t )) −v(t )(u ''(t ) + K (t )u (t ))]dt 0. ρ. ρ. 0. 0. = ∫ [u (t )v ''(t ) −v(t )u ''(t )]dt + ∫ [(−a 2 − K (t ))u (t )v(t )]dt ≥ u ( ρ )v '( ρ ) − v ( ρ )u '( ρ ). Hence u '( ρ ) v '( ρ ) ≥ u ( ρ ) v( ρ ) One can compute the geodesic curvature of distance circle with the radius R is. kg R (θ ) =. ( G ) ρ ( R, θ ) G ( R, θ ). .. 8.

(12) Therefore, we can conclude this lemma.□ In order to generalize the proof of Lemma 3.4, we search the minimal number x such that x ≥ R > 0 and min kg R (θ ) ≥ max kg x (θ ) . By Lemma 4.4 and computation, 1 1 x = coth −1 ( ) b bR 1 a x = coth −1 ( coth(aR )) b b. if −b 2 ≤ K ≤ 0. and R <. if −b 2 ≤ K ≤ − a 2 < 0 and R <. 1 , b. 1 b coth −1 ( ) . a a. Hence, we can compare the geodesic curvature between the curve and distance circles. Lemma 4.5 Let N be a Hadamard’s 2-manifold satisfying −b 2 ≤ K ≤ − a 2 < 0 [or −b 2 ≤ K ≤ 0 ]. Let γ be a simple closed, positively oriented curve tangent to the circumscribed geodesic circle CR of γ at P1 , P2 , successively. Assume R < R<. 1 b coth −1 ( ) [or a a. 1 ]. Let γ * be the positively oriented subarc of γ form P1 to P2 .Choose the b. circle C ' of the radius. 1 a 1 1 coth −1 ( coth( aR )) [or coth −1 ( ) ] passing through b b b bR. P1 , P2 and positively oriented form P1 to P2 . If there is a point in γ * lying in the interior of C ' . Then there exists a local minimum of geodesic curvature of γ in γ * . Proof. By property 2, there is the unique point P in geodesic segment P1 P2 such that geodesic segment O ' P is perpendicular to P1 P2 , where O ' is the center of. C ' . Consider the family - of all distance circles of radius. 1 a coth −1 ( coth( aR )) b b. 1 1 [or coth −1 ( ) ], centered at a point in the geodesic O ' P .Then there exists a b bR. distance circle C * in - satisfying C * lies to left of γ * . Moreover, C * never touches P1 , P2 and never crosses γ * , but C * meets γ * . Since γ is simple and a point in γ * lies in the interior of C ' , the geodesic curvature of the contact point of γ * and C * is smaller than or equal to a coth aR [or 1/ R ]. Hence there exists a local minimum of geodesic curvature of γ in γ * .□. 9.

(13) Give any point P in a geodesic segment AB . Along the perpendicular geodesic of AB passing through P , move P to P ' so that the distance PP ' is at most L . Then we say the rectangular neighborhood N ( L ) of AB is the collection of all points P ' . Moreover, we say the dilatation of a rectangular neighbor is 2L . length( AB ) Theorem 4.1 Let N be a Hadamard’s 2-manifold satisfying −b 2 ≤ K ≤ 0 . Let γ be a simple closed curve tangent to the circumscribed circle CR of γ only at two points P1 , P2 . Assume R <. 1 .If γ lies in the interior of the rectangular neighborhood of the b. geodesic segment P1 P2 , with the dilatation small than or equal to 1 1 1 1 coth −1 ( ) − (coth −1 ( )) 2 − 1 . 2 bR bR (bR ) bR. Then γ has at least four vertices. Proof: By Lemma 4.2, the length of P1 P2 is 2R . Consider the circle C passing through P1 , P2 , with radius. 1 1 coth −1 ( ) and the center O . We want to find the b bR. maximum of d (O, PP 1 2 ) independent of the choice of manifolds. Then the maximum is. 1 1 (coth −1 ( ))2 − R 2 2 b bR by contradicting to Property 1. Let the positively oriented subarcs γ 1 , γ 2 of γ be from P1 to P2 and from P2 to P1 , respectively. Let C1 , C2 be two distance circles with the centers O1 , O2 and radius. 1 1 coth −1 ( ) , respectively. If the dilatation of a b bR. rectangular neighborhood N ( L ) of P1 P2 is smaller than or equal to 1 1 1 1 coth −1 ( ) − (coth −1 ( )) 2 − 1 2 bR bR (bR ) bR. 10.

(14) Hence. 1 1 1 1 L ≤ coth −1 ( ) − 2 (coth −1 ( ))2 − R 2 . b bR b bR Because γ lies in the interior of N ( L ) , γ 1 and γ 2 both meet the intersection of the interior of C1 , C2 , respectively. By Lemma 4.5, γ 1 and γ 2 both have a local minimum of geodesic curvature of γ . Hence γ has at least four vertices. □ Theorem 4.2 Let N be a Hadamard’s 2-manifold satisfying −b 2 ≤ K ≤ − a 2 < 0 . Let γ be a simple closed curve tangent to the circumscribed circle CR of γ only at two points P1 , P2 . Assume R <. 1 b coth −1 ( ) .If γ a a. lies in the interior of the rectangular. neighborhood of the geodesic segment P1 P2 , with the dilatation small than. a a cosh( coth −1 ( coth(aR ))) a 1 1 b b coth −1 ( coth(aR)) − cosh −1 ( ). bR b aR cosh(aR) Then γ has at least four vertices. Proof: This estimate is the same as Theorem 4.1 except for the maxima of d (O, PP 1 2) .. a a cosh( coth −1 ( coth(aR))) 1 b b cosh −1 ( ) .□ The maximum of d (O, PP 1 2 ) is a cosh(aR). §5 The four-vertex theorem in Riemannian 2-manifolds for some special curves The method in chapter 4 works for manifolds with positive curvature if the distance circles we discussed are convex. In order to simplify the content, we don’t restate the similar lemmas in chapter 4. We search the minimal number x such that x ≥ R > 0 and min kg R (θ ) ≥ max kg x (θ ) .Hence x=. 1 tan(aR ) a. 1 a x = cot −1 ( cot(aR )) b b. if 0 ≤ K ≤ a 2 ≠ 0 if 0 < b 2 ≤ K ≤ a 2. 11. and R <. 1 π tan −1 ( ) a 2. and R <. 1 b bπ cot −1 ( cot( )) a a 2a.

(15) 1 a 1 b bπ x = coth −1 ( cot(aR )) if 0 ≠ −b 2 ≤ K ≤ a 2 ≠ 0 and R < cot −1 ( coth( )) b b a a 2a. Use the comparison theory between manifolds and manifolds with positive constant curvature k 2 . And the law of cosine for manifolds with positive constant curvature is cos(kc ) = cos(ka ) cos( kb) + sin( ka ) sin( kb) cos θ , where three edges lengths are a, b, c and the angle θ corresponding to the edge c . We may get the following property. Property1: If a triangle ΔABC has that three edges lengths are a, b, c and the angle θ corresponding to the edge c .Assume the Gauss curvature K satisfies K ≤ k 2 ≠ 0 , where k is a constant, then cos(kc ) ≤ cos(ka ) cos( kb) + sin( ka ) sin( kb) cos θ .. Afterwards, we estimate the largest length of the rectangular neighborhood N ( L ) of the geodesic segment P1 P2 . We gain 1 cos(tan(aR )) cos −1 ( ) a cos(aR ). when 0 ≤ K ≤ a 2 ≠ 0 ,. a a cos( cot −1 ( cot(aR))) 1 b b cos −1 ( ) a cos(aR). when 0 < b 2 ≤ K ≤ a 2 ,. 1 cos −1 ( a. a a cos( coth −1 ( cot(aR))) b b ) when 0 ≠ −b 2 ≤ K ≤ a 2 ≠ 0 . cos(aR). Hence the following three theorems and one corollary hold. Theorem 5.1 Let N be a simply connected, oriented, complete Riemannian manifold satisfying 0 ≤ K ≤ a 2 ≠ 0 . Let γ be a simple closed curve tangent to the circumscribed circle CR of γ only at two points P1 , P2 . Assume R <. 1 π tan −1 ( ) .If γ lies in the interior a 2. of the rectangular neighborhood of the geodesic segment P1 P2 , with the dilatation small than 1 1 cos(tan(aR)) tan(aR) − cos −1 ( ). aR aR cos(aR ) Then γ has at least four vertices.. 12.

(16) Theorem 5.2 Let N be a simply connected, oriented, complete Riemannian manifold satisfying 0 < b 2 ≤ K ≤ a 2 . Let γ be a simple closed curve tangent to the circumscribed circle CR of γ only at two points P1 , P2 . Assume R <. 1 b bπ cot −1 ( cot( )) .If γ lies in a a 2a. the interior of the rectangular neighborhood of the geodesic segment P1 P2 , with the dilatation small than. a a cos( cot −1 ( cot(aR))) a 1 1 b b cot −1 ( cot(aR)) − cos −1 ( ). bR b aR cos(aR) Then γ has at least four vertices. Corollary 5.1 Let N be a simply connected, oriented, complete Riemannian manifold satisfying K = a 2 > 0 . Let γ be a simple closed curve tangent to the circumscribed circle CR. of γ only at two points P1 , P2 . Assume R <. π 2a. . Then γ has at least four vertices.. Theorem 5.3 Let N be a simply connected, oriented, complete Riemannian manifold satisfying 0 ≠ −b 2 ≤ K ≤ a 2 ≠ 0 . Let γ be a simple closed curve tangent to the circumscribed circle CR of γ only at two points P1 , P2 . Assume R <. 1 b bπ cot −1 ( coth( )) .If γ a a 2a. lies in the interior of the rectangular neighborhood of the geodesic segment P1 P2 , with the dilatation small than. a a cos( coth −1 ( cot(aR ))) a 1 1 b b coth −1 ( cot(aR )) − cos −1 ( ). bR b aR cos(aR) Then γ has at least four vertices.. §6 Summary We summarize the results in chapter 4 and 5.. 13.

(17) Main theorem A simply connected, orientated complete Riemannian 2-manifold satisfies k2 ≤ K ≤ k1 . Let γ be a simple closed curve tangent to the circumscribed circle CR of γ only at two points P1 , P2 . Assume 0 < R < C1 , for some constant C1 = C1 (k1 , k2 ) (Table 2) .If γ lies in the interior of rectangular neighborhood of geodesic segment P1 P2 , with the dilatation small than a constant C2 = C2 (k1 , k2 , R) (Table 2).Then. γ. has at least four vertices.. C1 = C1 (k1 , k2 ). k1 = 0 k2 = −b 2 ≠ 0 k1 = − a 2 < 0 k2 = −b. 2. k1 = a 2 ≠ 0 k2 = 0 k1 = a 2 k2 = b 2 > 0. 1 b. 1 1 1 1 coth −1 ( ) − (coth −1 ( )) 2 − 1 2 bR bR (bR) bR. 1 b coth −1 ( ) a a. a a cosh( coth −1 ( coth(aR ))) a 1 1 b b coth −1 ( coth(aR)) − cosh −1 ( ) bR b aR cosh(aR). 1 π tan −1 ( ) a 2 1 b bπ cot −1 ( cot( )) a a 2a. 1 b bπ cot −1 ( coth( )) a 2a k2 = −b 2 < 0 a k1 = a 2 > 0. C2 = C2 (k1 , k2 , R). 1 1 cos(tan(aR)) tan(aR) − cos −1 ( ) cos(aR) aR aR a 1 1 cot −1 ( cot(aR )) − cos −1 ( bR b aR. a a cos( cot −1 ( cot(aR))) b b ) cos(aR ). a a cos( coth −1 ( cot( aR))) a 1 1 b b coth −1 ( cot(aR )) − cos −1 ( ) bR b aR cos(aR ). Table 2. 14.

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