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I u I | l 0. .. ... lus. .Yl.-..lv. . . . .. .,r‘ . .....‘...¢O.n~w. ...J L“ .u . ..T 1‘ . 1 'I .... . v MUM} r . .Ho’fil‘fl . . .‘r l-1\ \ . | “7‘ w .40 30!; \ I; A . pl 1 t ‘ . ..l 1'9»: .! / ?;’ MS U is an Affirmative Action/Equal Opportunity Institution 0- 12771 LIBRARY Mlchigan State University PLACE ll RETURN BOX to remove this checkout from your mood. TO AVOID FINES Mum on or bofoto dot. duo. DATE DUE DATE DUE DATE DUE MSU Is An Affirmative Adlai/Equal Opportunlty Institution Walnut ELLIPTIC FUNCTIONS, THETA FUNCTION, AND SUBMANIFOLDS IN SPACE FORMS By Jz'e Yang A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1997 ABSTRACT ELLIPTIC FUNCTIONS, THETA FUNCTION, AND SUBMANIFOLDS IN SPACE FORMS By Jz'e Yang In the first part of the thesis(Chapter 1), we study slant surfaces in C2. The complete classification for all proper slant surfaces with constant Gaussian curvature and nonzero constant mean curvature in C2 is obtained in this part. In 1993, B. Y. Chen introduced an important Riemannian invariant 6M for a Riemannian n-manifold M ", namely take the scalar curvature and subtract at each point the smallest sectional curvature. He proved that every submanifold M n in a Riemannian space form Rm(e) satisfies a sharp inequality: n2(n — 2) 6M 3 2(n_1) H2+(n+1)(n-2)e. In the second part of the thesis (Chapter 2 and 3), first we classify hypersurfaces with constant mean curvature in a Riemannian space form which satisfy the equality case of the inequality. Next, by utilizing J acobi’s elliptic functions and Theta function we obtain the complete classification of conformally flat hypersurfaces in Riemannian space forms which satisfy the equality. To my parents, my wife Ying and daughter Jessica iii ACKNOWLEDGMENTS I am very grateful to my thesis advisor, Professor Bang-Yen Chen, from whom I learnt much of differential geometry, for his constant help, advice, and encouragement. I would also like to thank Professors David E. Blair, with whom I took my first differential geometry course, Gerald Ludden, Wei-Eien Kuan, Zhengfan Zhou for their advice in many helpful conversations, academically and socially. iv TABLE OF CONTENTS 0 INTRODUCTION 1 1 SLAN T SURFACES WITH CONSTANT MEAN CUEVATURE IN C2 4 1.1 Introduction ................................ 4 1.2 Preliminaries ............................... 6 1.3 Basic Equations .............................. 7 1.4 Proof of Theorem 1.1 ........................... 10 1.5 Proof of Theorem 1.2 for K a Positive Constant ............ 15 1.6 Proof of Theorem 1.2 for K a Negative Constant ........... 21 2 HYPERSURFACES WITH CONSTANT MEAN CURVATURE SATISFYING CHEN’S EQUALITY 28 2.1 A Remannian Invariant and Chen’s Equality .............. 28 2.2 Main Results ............................... 30 2.3 Prelimilaries ................................ 32 2.4 Proof of Theorems 2.1, 2.2 and 2.3 ................... 36 3 COMFORMALLY FLAT HYPERSURFACES SATISFYING CHEN’S EQUALITY 43 3.1 Main Results ............................... 43 3.2 The Jacobi Elliptic Functions, Theta Function and Zeta Function . . 49 3.3 Two Lemmas ............................... 52 3.4 Proof of Theorem 3.1 ........................... 61 3.5 Exact Solutions of Differential Equations of Picard Type ....... 67 3.6 Proof of Theorem 3.2 ........................... 77 3.7 Proof of Theorem 3.3 ........................... 89 BIBLIOGRAPHY 109 CHAPTER 0 INTRODUCTION Let (N, 9, J) denote an almost Hermitian manifold equipped with an almost complex structure J and almost Hermitian metric g. A submanifold M of N is called slant if its Wirtinger angle is constant. Complex submanifolds and totally real submanifolds are two special classes of slant submanifolds which have Wirtinger angle 0 and 1%, respectively. It is known that there exist ample examples of slant submanifolds other than complex and totally real submanifolds. The first part of the thesis studies slant surfaces in the complex 2-plane (22. It is well—known that helical cylinders in (C2 are flat proper slant surfaces with nonzero constant mean curvature. Conversely, we prove that every flat proper slant surface with nonzero constant mean curvature in C2 is an open portion of a helical cylinder. Although it is known that there exist abundant examples of slant surfaces with constant mean curvature and constant Gaussian curvature in non-fiat complex-space- forms. However, we prove that there do not exist prOper slant surfaces in C2 with constant mean curvature and nonzero constant Gaussian curvature. According to the well-known Nash imbedding theorem, every Riemannian n- manifold can be realized as a submanifold in a Riemannian space form, in particular, in a Euclidean space. For a submanifold in a Riemannian space form, Chen proved in 1993 a general inequality involving sectional curvature, scalar curvature and the squared mean curvature of the submanifold. Chen’s inequality has some important applications. For example, it gives rise to the second Riemannian obstruction for a Riemannian manifold to admit a minimal isometric immersion into a Euclidean space. It also gives rise to an obstruction to Lagrangian isometric immersions of compact Riemannian manifolds with finite fundamental group 7r1 into complex space forms. Since Chen’s inequality is very general and sharp, it is natural and interesting to understand submanifolds which satisfy the equality case of this inequality. Re- cently, there are several interesting papers which investigate submanifolds satisfying Chen’s equality. In this thesis, we investigate the most fundamental case; namely, hypersurfaces satisfying Chen’s equality. First, we give a complete classification of hypersurfaces with constant mean curvatures satisfying the equality. Next, by uti- lizing the Jacobi elliptic functions and the Theta function, we completely classify conformally flat hypersurfaces satisfying the equality. Let M be an n-dimensional (n > 2) hypersurface in a Riemannian space form Rn+1(e), (e = 1, —1 or O) which satisfies the equality. We show that, if e = 0, M is either minimal or an open portion of a spherical hypercylinder; if e = 1, M is either totally geodesic or a tubular hypersurface with radius g about a 2-dimensional minimal surface; and if c = —1, M is either totally geodesic, or an open portion of a tubular hypersurface with radius cosh’1(\/2) about a 2-dimensional totally geodesic surface of R"+1(—1), or a “suitable tubular hypersurface” about a minimal surface in the de Sitter space-time S?+1(1). In order to classify conformally flat hypersurfaces satisfying the equality, we need to define some special families of Riemannian manifolds: P: (a > O), 02((1 > 1), D2 (0 < a < 1), F", L", A: (a > 0), B2 (0 < a < 1), G", H: (a. > 0), W3, (0. > 0) and Ya"(0 < a < 1) via warped products of IR ( or an open interval ) and some Riemannian space form by some warp functions that may involve the Jacobi elliptic functions. For example, Pg, 0:, are the warped productszl X,“ Sn‘1(“4—4‘l), R X”, [In—1“:—1 pa = ak cn(a:r, 4/5?) and 17,, = fidnfi-x, 78%), and cn(a:c, k), dn(%, k) are the Jacobi ), where elliptic functions with modulus k. Topologically, S" is the two point compactification of P}, C]: as well as of B: and the Riemannian metrics defined on Pf, C}: or B: can be extended smoothly to S". Let 13:, C2 and B2 denote the n-sphere together with the Riemannian metrics given by the smooth extensions of the metrics on Pf, C: and B}: to S", respectively. We prove that if M is a conformally flat hypersurface of a Riemannian space form which satisfies the basic equality, then either M is totally geodesic or M is an open portion of one these ten special families of Riemannian manifolds. Furthermore, we are able to determine these immersions explicitly. If the the ambient space is spherical, there exist three families of such hypersurfaces. One of the families is the immersion of P2 into 54(1) and its local expression involves the Jacobi elliptic functions and the Theta function. In order to get the expression, we have to solve a family of second order ODEs of Picard type whose coeffients involve the Jacobi elliptic functions, namely. u” (:r) +2asc(ax)dn(a3:)u’(:1:) — u(:L') = 0. We call such an equation a differential equation of Picard type since a similar equation was studied by E. Picard in 1879. However the method of Picard does not work for our equations, so we need to develop a new approach to obtain the general solutions for this type of ODEs. Our results seem to have independent interest by themselves. If the ambient spaces are hyperbolic, we are able to obtain the complete classification via nine families of immersions from the following Riemannian manifolds Ag, G", H2, Ya", L3, C2, D2, etc, to the ambient space. In order to establish the local expressions of the immersions of C2 and D2, we need to solve two families of ODEs similar to the one mentioned above. CHAPTER 1 SLANT SURFACES WITH CONSTANT MEAN CUEVATURE IN C2 In this chapter, we completely classify proper slant surfaces with constant Gaussian curvature and nonzero constant mean curvature in C2. 1 . 1 Introduction Let M be a Riemannian n-manifold and (M , 9, J) an almost Hermitian manifold with almost complex structure J and almost Hermitian metric 9. Let TPM be the tangent space to M at p. An isometric immersion f : M —> M is called holomorphic if at each point p E M we have J (TPM ) = TPM. The immersion is called totally real if J (TpM ) C TiM for each 1) E M, where TiM is the normal space of M in M at p. For each nonzero vector X tangent to M at p, the angle 7(X) between JX and T pM is called the Wirtinger angle of X. The immersion f : M —> M is said to be slant if 7(X) is a constant (which is independent of the choice of p E M and X E TpM. see [1] for details). The Wirtinger angle 7 Of a Slant immersion is called the slant angle. Homolorphic and totally real immersions are slant immersions with slant angle 0 and 325, respectively. A slant immersion is said to be proper slant if it is neither holomorphic nor totally real. The simpliest and most important examples of slant submanifolds are Slant sur- faces in C2 , where C2 is the Euclidean 4-space 1R4 equipped with its canonical complex structure. In [2], B. Y. Chen constructed ample examples of such surfaces. He also proved that there is no proper slant surface in C2 with parallel mean curvature vector (cf. also [3]). Thus the following open problem proposed in [2] by B. Y. Chen is very interesting: Problem: Classify slant surfaces in (C2 with nonzero constant mean curvature. It is known that helical cylinders in C2 are flat proper Slant surfaces with nonzero constant mean curvature [2]. The first result of this chapter is to prove that the converse of this fact is also true. Namely we prove the following: Theorem 1.1. A flat proper slant surface with nonzero constant mean curvature in C2 is an open portion of a helical cylinder. B. Y. Chen and L. Vrancken Show in [4] that there exist many proper slant surfaces with constant mean curvature or with constant Gaussian curvature in complex-space- forms. However, in this paper we prove the following nonexistence theorem: Theorem 1.2. There do not exist proper slant suraces with nonzero constant mean curvature and nonzero constant Gaussian curvature in C2. 1 .2 Preliminaries Let C2 be R4 equipped with its complex structure J and M be a proper slant surface isometrically immersed in C2 . For any vector X tangent to M, set JX=PX+FX where PX and F X are respectively the tangential and normal components of JX. It is clear that P is an endomorphism of the tangent bundle TM and that F is a normal-bundle—valued 1-form on TMp. Let 61 be an unit local vector field in TM. We choose a canonical orthonormal local frame €1,82,83,€4 such that 62 = (sec 7)Pel, e3 2 (csc ’y)Fel, 64 = (csc ’7)F82. Such an orthonomal frame is called an adapted Slant frame. Let w1,w2,w3, an be the dual frame of 61,82,63,€4. Then the structure equations are given by de = —wAB /\ 013, deB = —wAC /\ was, MAB + wBA = 0- where A, 821, 2, 3, 4. 7‘, 3:3, 4. i,j=1, 2. Restricting to M, w, = 0. Then we have _ 1' _ 1‘ _ T (Uri — hijwj, hij —- hJ‘l' A helical cylinder in 1&4 is defined by (1.2.1) :r(u,v) = (u, kcosv,mv, ksin v) where m and k are nonzero constants. With complex structure Jo in IR", (1.2.1) defines a proper Slant surface in C2 with slant angle cos-1(77—ngn—T—g), where Jo 3 ($1,$2,333,$4) *——> (—173, —$4,$1,$2) If we choose 1 W "2'+ " E 63 = (O, — cos 2}, O, ——- sin n), 61: (0, —ksinv,m,kcos v), 62 = (1,0,0, 0), 1 .3, — _.m( 0, —m sin v, ——k, m cos 2)), then 61, 62, 63, 64 form an adapted slant frame with respect to J0, where 63 is in the direction of mean curvature vector. The connection form of ( 1.2.1) is given by { 0 0 Egg—gull 0 \ 0 0 O 0 (1.2.2) (wAB) = k WW1 0 0 ‘W’i—flwl K O 0 #021 0 ) 1.3 Basic Equations Let M be a proper slant surface with constant mean curvature 5 ¢ 0. We recall from [2] that (1-3-1) hi2 : hi1, hgz = hi2: (1.3.2) w34 — (4112 = cot 7{(trh3)w1 + (trh4)w2}. where "y is the slant angle. We can choose an adapted slant frame such that 63 is parallel to the direction of mean curvature vector. Thus the Shape Operator A3 and A4 take the following forms: c—Aa a /\ a A /\—0z Let a = ccot'y > 0. By (1.3.2), we have (.034 — (4)12 = awl. Using the structure equations, we get dwl = 0. So, locally, there exists a variable a: such that d1: 2 col. Since dwl = —w12 /\ (.02, we Obtain that can = ,8602, and (1134 2 3w + awl. Assume tag 2 f (:r,y)dy, where f (2:,y) is a C"2 function and nowhere zero on a neighborhood U of (0,0). SO, on U, the metric tensor is g = (1172 + f2(:v, y)dy2. Since (S31 2 (c —- A)w1 + am, we have dLU31 = (62A “I" 810 — afl)w1/\ (4)2. On the other hand, dw31 = —w32 /\ W21 — w34 /\ 6041 = (2015 — aA)w1 /\ wg, So, by comparing these two expressions, we deduce that 62A + em: = 3afi — 01A. Similarly, we have 61A — 820 2 —CB + 3A6 + 0.0. Let K denote Gaussian curvature of M, then K011 /\ £02 2 dang —_- (61,6 — B2)w1/\w2. But (10112 = —w13 A 0132 — a114 /\ W42 2 (CA — 2A2 — 2&2)w1 /\ (.02, therefore Kzelfl—fizch—2A2—2a2. Consequentely, we have elA — 620 = —cfi + 3A3 + no, (1-3-3) 62) + 610 = 3016 — a/\, K=elfl—52=c/\—2/\2—2a2. In particular, (1.3.3) implies K S Cg. Since tag 2 f(:1:,y)dy, 10 gidmdy = dwg = —w21 /\ wl 0:2: 2 —fifd:rdy. Therfore fl = _§-f,{, and 0’_f _ _,,2f_ _ a: 83:2 — 8x 01: (1.3.4) = 52f - f(fi2 + K) = —Kf. 1.4 Proof Of Theorem 1.1 Now, we consider the flat case, i.e. K = 0. By (1.3.4), we have f($, y) = p(y)$ + (1(31) on some open neighborhood U of (0, 0). Let a, 6, '7 as in the previous section. First, if a E 0 and then A E 0 on U by (1.3.3) since 0 ¢ 0. In this case, we can compute its connection form as follows: __ ._ _ 3 3 _ _ wlg — fiwg —— 0, (.4131 — huwl + hlgwg — (c — A)w1 + awg — cw], _ _ _ 3 3 _ ._ w34 — ng + awl — awl, c032 — hmwl + h22w2 — awl + Awg — 0, _ h4 h4 _ - A c041 — 11+ 12w; — aw1+ L02, 4 (1242 = h‘]1w1 + h22w2 = Awl — am 2 0. 11 Thus we have ( 0 0 —cw1 0 \ 0 0 0 0 (WAR) = can 0 0 awl K 0 0 —aw1 0 / Therefore, we can chose m and k such that c = fig. Let — COS—1 _____7n_.__ 7 W ‘ By assumption, __ _ m a — CCOL ’Y — —m. Thus we get (1.2.1). i.e., ll! is helical cylinder on U. So we have proven the theorem in this case. If there exists a point z E U such that a(z) aé 0, we can find an Open subset V Of U such that a is nowhere zero on V. Lemma Let V as be chosen above, then p(y) E 0 on V. Proof. Let us assume that p(y) does not vanish identically. Let z’ E V be a point such that p(z') 76 0, we can choose an open subset of V, say W, such that p(y) is nowhere zero on W. Thus on W, — 2 a: M f(1r,y)—p(./)( +p(y))- So, without loss of generality (by replacing p(y)dy by dY if necessary), we can assme f(:v.y) = x + 9(y) on W- 12 21 In this case, ,6 2 —%§- = -%. From (1.3.3), we have ' anew-3A, (9:1: fay—- f aa, 1Q+£9£_ Ba a)‘ a a ‘"__ ’ (1.4.1) ] f y a: go (C—4A)5}-—4aa—$=O, 6A Ba \ (C — 4A)??? — 40—8—37 — 0. Solving (1.4.1) and noticing that CA — 2A2 — 2a2 2 O, we have 8A_4aa (9A_ 40 (42) a_—C_(7+af)v 'a_y—_?(C—A+aaf)a 1. ' 8a c-4/\ 0 6a c—4A —= — —=— —A . ax C (f af)? 6!! C (C +a’af) Let A = % +aa, and B = c— a+aaf. Since 5:53} = 5:23, we have 40 BA BB "flag”:‘aal—O Since a 75 0 on W, Q1. + 9.1.3. — 0 By (91: _ ' Using (1.4.1), this implies (1.4.3) Pa : QA — D, where (*) P:£;-+3av Q:§_a2fa Dz: f f f 13 Therefore P2 2 = 622A2 — 2QDA + D2. By (1.3.3), we have (1.4.4) 2(102 + 622))? — (c1D2 + 4QD)A + 21)2 = 0. From (1.4.3), 619 aa _ 8Q ax an EQ+P5§~8$+ 03: 82: By (1.4.4) and (1.4.1), this implies (1.4.5) (a‘o’f4 — 2a2g’f2 — 2g' — ag’2)/\ — 2acf2 = 0. Thus (1.4.4) and (1.4.5) give (1.4.6) ((ag')2 + 3ag' + 2)(9a2f4 + 6ag’f2 + 9’2) = 0. SO, if there exists a point zo E W such that ((ag’)2 + 3ag’ + 2)(z0) 75 O, we can choose an open neighborhood W’ of zo in W such that (ag’)2 + 3ag’ + 2 is nowhere zero on W’. Thus, by (1.4.6), we have 9a2f4 + 6ag’f2 + g’2 = O on W’, i.e., (:r + g(y))2 =2 f2 = :gggfll on W’. This is a contradiction. In the following, we assume (ag’)2 + 3ag’ + 2 = 0 on W. Thus, 9’ = l or —% on W, then f(a:,y) = x+by+C on W, where b = —% or —Z. a a 14 Now, from (1.4.3), we have Qfa+p§2_ 99- Qé_§2_0 3y 6y 0y 0y 0y _ By doing similarly computations as before, we have (—4afDP + 3cP2 — 4DQ + ac fPQ + cpyo + 4cQ2 — ciao,» (1.4.7) +2D2 + achP + chP — c2P2 — cDPy — 4cDQ = 0 Notice that, in this case P—i+3a, Q= —f2 —a2fa D: i E f f' Substituting the above and (1.4.7) back in (1.4.4) and simplifying, we have (1.4.8) 3a6c4f8 + (—8a"c4 — 38a5c4b — 1241661112”6 + (—276a?c4 — 706a3c4b — 532a4c4b2 — I20c15c“123)f4 + (36c4 — 144ac4b — 43201254112 — 338a2c4b3 — 74a4c4b4)f2 — 2064b2 — 66ac”b3 — 55a264b4 — 12a3c4b5 = 0. The leading term of (1.4.8) is 3a6c4 75 0, and the other coefficients are constants. Thus we get f E constant on W. This contradicts f = :1: + by + C. So, we have completed the proof of the lemma. Returning to the proof of Theorem 1, From the above lemma, we know f (51:, y) = 91 q(y) on W. By (at), we can see that P, Q and D are functions of y. Also ,6 2 ~13;— = 0. Moreover, from (*), we see that P , Q and D can not be simultaneously zero at any point on W, otherwise we have c = 0 at this point. Thus A is a fuction of y, and so I5 is a. By using (1.3.3), we have Id_a_ fdy 1dA_ —aa, and SO, a 2 CIA, where 01 is some constant. By (1.4.1), we get 2(1+ of»? — CA = 0 then C A—O, OI” A—m. If A = 0, we have a = 0. This contradicts our assumption that (1 ¢ 0 on W. If A = W???) # 0, by (1.4.9) and g(y) 76 0, we get a = 0. This contradicts our assumption. Thus we have completed the proof of Theorem 1.1. E] 1.5 Proof Of Theorem 1.2 for K a Positive Con- stant From (1.3.4), we have g} = —12f. For simplicity, we assume I = 1, i.e. K = 1. Thus, in a neighborhood U of (0, 0), f(:r,y) = 91(31) sin$+g2(y) COS-”II 1 = —— sin(-r + 51(30)- 9? + 9% where g(y) = cos‘1g1(y). Since 92(0) # 0 (otherwise f(0, 0) = 0), therefore, without loss of generality, we may assume f(:1:, y) = sin(:z: + g(y)). From (1.3.3) we have c2 2 8. if c2 = 8, we have 01 = 0 and A = i, and then a = 0. 16 This contradicts the assumption of a > 0. Thus we assume c2 — 8 > 0. Let 6 = a: + g(y), since f = sin 0, so 5 = _§-ff_ = —cot 6. Thus, from (1.3.3), we have ’ (9A 1 0a 53; — sinOUy- — (c— 3A) cot0+aa, '._1_§"A' + 6_a = —3acot0 — (IA, (1 5 1) ] SInQBy 0:1: . . 0A BO: (0 — 4A)a—x — 4015; _ 0, 8A Ba c—4A ——4a—=O. . ( lay 6y Solving (1.5.1) and taking (1.3.3) into account, we obtain f g; 2 —4Aa(a cot 0 + (M — Ba), 8A - .8— : —4Aa((c — A — 3B) c056 + aas1n6), (1.5.2) 4 8y .9. = (c — 4A)A(a cot6 + aA — Ba), (91: Ba - 8—3; = —(c — 4A)A((c — A — 3B) cosl9 + aa sm 6), \ whereA= Eff—3 and B: E. If a E 0 on U, we have A E g on U, then a = 0 by (1.3.3). This contradicts our assumption. If there exists a point z E U such that (1(2) 75 O, we can find an open subset V C U such that oz nowhere zero on V. Thus, on V, by using 8133 = 8222, (1.4.2) and (1.3 .3), we have (1.5.3) Pa 2 QA — D 17 where 3a sin2 6 cos 6 + g’ _ —-3B sin2 6 + 0 sin2 6 _ P=3acos6+g’csc26= , sm6 , D ’ (1 — a2)sin26+3cosz6 _ 3 — (2+a2)sin26 Q 2 sin26 sin6 Also, by (1.3.3), we have (1.5.4) 2(P2 + Q2)/\2 + (P2c + 4DQ)A — 2D2 — P2. From (1.5.3), 6P 6a 6Q 6A BD ‘53”5; — 53*+Qa;"a7 Consider (1.3.3), (1.5.3) and (1.5.4). This implies (1.5.5) KA+F=0, where K 2 OP, — AP2ca + 4ABaQ2 + 4AP2Ba — PQ’ + 4APD cot 0 — 4AQDa —AQPc cot 6, F = —P,D — APDc cot 6 — 4ABDQa + 2AQP cot 6 + DxP + 4AaD2. Putting (1.5.5) back into (1.5.4) gives (1.5.6) 2(P2 + Q2)F2 + (P2c + 4DQ)FK + (21)2 + P2)K2 = 0. Let 18 , and t = g’(y). 6 ._ - 2_ U—SlIl 2 Then (1.5.6) becomes the following polynomial equation in a, (1.5.7) blg’ulB + (21711.17 + ‘ ° ' + blu + ()0 = 0, where b, are functions of a, c and t. That is to say that they are functions of y. The followings are some coefficients we will use later: he bu bra b16 18874368(a — 1)2a2(1 + a)2(a2 + c2 — 9) . -(576 — 64c2 — 88a2c2 + 8a"‘c2 + 9a2c4), 169869312(a — 1)2a2(1+ a)2(a2 + c2 — 9) - -(576 — 64c2 — 88a2c2 + 8a4c2 + 9a204), 49152a(192471552a - 492549120a3 + 331043328a5 - 30965760a7 —14684544ac” + 38083584a3c2 + 23807616a5c2 — 56238336a7c2 +9461760a902 — 430080allc2 — 3647616ac" + 5267976a63c4 —1O946376a50" + 10212216a7c4 — 886200a9c" + 298368ac6 —17913a3c6 + 138138a5c” — 418593a7c6 + 2688a308 + 37527715738 —3939840t + 8607744a2t — 4953600a4t451584a6t + 83404802t 45539271281 —- 4953600a4t + 45158465621 — 195008638: +6272a’0c2t — 39424511: -— 133576a2c" + 469344a4c4t —7293432a6c4t + 20416a8c4t - 51206t + 10902a2c6t — 26721a4c6t +14063a6cfit + 162a2c8 — 81a408t, 1179648(a — 1)a2(a + I)(a.2 + c2 — 9)(—32716 + 334080a2 + 34048c2 19 13536a262 — 55680a462 + 464067.602 + 256a4 — 4916a2c4 + 52006464 -270.2C6. We will Show that, for any a and c, the coefficients of (1.5.7) cannot be identically zero simultaneously on W. If this is true, let bk be the biggest 2', (0 S i g 18), such that bk is not identically zero on W . Thus there exists an open subset W’ of W such that bk is nowhere zero on W’, So, on W’ we have with all coefficients as function Of y and leading coefficient 1. Thus, we can write u = F(y) on W’, i.e., sin2 x—‘Lg—w = F(y) on W’, where F(y) is some function Of y. This is a contradiction. Thus, to prove the theorem, it is sufficient to prove that, for any b and c, the coeficients b,, i = 0, 1, 2, . - - , 18, can not be identically zero Simultaneously on W. Case 1. a 76 1 and a2+c2 # 9. In this case, U18 and b17 can not be zero simultaneously. Otherwise, we have (1.5.8) 576 — 64c2 — 88a2c2 + 80.402 + 96%“ = 0, —327168 + 334080a2 + 3404862 + 13536a”c2 — 55680a"c2 (1.5.9) +4640a602 + 256C4 — 4916a2c4 + 5200a4c" — 27a206 = O. Solve (1.5.8) for 02 gives 64 + 88a2 — 864 :t @07360? + (—64 — 88a2 + 8a")2 1.5.10 2 ( ’ C 18612 20 Putting the above back into (1.5.8), we get I43327232(a2 — 4)(a.2 — N466 = 0. Thus a2 = 4 since a2 76 1 and a # 0. Now put a2 = 4 back into (1.5.10), we get 62 = 4. This contradicts the assumption C2 > 8. Case 2. a = 1. In this case, we have D18 = b17 : b15 : b15 = b14 : 0. But b13 = 49152(165888t — 10368062t + 23328c4t — 2268c2t + 81c8t = 3981312(c — 2)(c + 2)(c2 —- 8)3t. So if b13 E 0 on W, we have t E 0 on W since 02 > 8. Thus ( 5.7) becomes (-6115295232 + 4586471424c2 — 1242169344c4 + 14332723206 — 59719682652112 +... + (47775744c2 — 2583705664 + 4064256c6 — 198144c8)u4 +(373248c4 — 82944c6 + 4608c8)u3 = 0. where b12 = —5971968(c2 _ 4)2(c2 _ 8)2, 2 which is not zero since 0 > 8. Therefore b12 and b13 can not be identically zero simultaneously on W. Case 3. a2 + c2 = 9. In this case, C2 = 9 — a2. Substituting into (1.5.7), it becomes (—1179648a6 + 2359296718 — 117964871104112 + +(—3276288a6 +1317888a.8 — 105984a’0)u4 +(373248a6 — 82994408 + 4608a’0)u3 = 0. 21 But by) = —1179648(a2—1)2a6. Thus if b12 = 0, a = 1 since a > 0. So 02 = 9—a2 = 8. This contradicts the assumption c2 > 8. Thus b12 79 0. [:1 1.6 Proof Of Theorem 1.2 for K a Negative Con- stant Now we consider K = ——12 76 0. For Simplicity, we assume I = 1, i.e. K = ——1. AS we did in the last section, we have 3.5 = —6 f and 3?; = f, and thus f (at, y) = 91(9)ear + 92(0)” in a neighborhood U of (0,0) and f 75 0 on U. Since gl and 92 are not simutaneously vanished at 0, otherwise f (0, O) = 0. Thus without loss of generality, we assume 9(0) ¢ 0. Therefore there exists an Open subset V of U such that 91 (y) is nowhere zero on V, so on V SO, without loss of generality, by replacing gl(y)dy by dY if necessary, we can assumef(a:, y) 2 ex + g(y)e‘$. Therefore From (1.3.3) and (1.6.1) cA — 2A2 -— 2012 = —1, 22 we have V 8A 2 6—11: ... 4Aa(2a + acA - [3001): 6A 2 .6— : —4Aoz(c6A + aca — 66 — 5C )f, (1.6.2) l 3” a 6—2: _ A(c — 4A)(2a + acA — 650‘), 80 2 6—3] = —A(c — 4A)(c)6A + aca — 66 — 36 )f- where A = 271;? Ifa E 0 on V, by (1.6.1) we have ,/ 2 A __-—_ C :l: 4C "l” 4 on V, which is a nozero constant. So by (1.3.3), we have a E 0. This contraducts with our assumption. If there exists a point z E V such that 07(2) ;£ 0, we can find an open subset W Of V such that or is never zero on W. Thus, on W, 03:6,, 2 £26: and (1.3.3) gives (1.6.3) Pa : QA — D, where _ 66 6f P — (acf6+3acf6 Cay +acax), = (3cf62—a2cf+-a—fflc+fc§g, 6:1: (91: = 672? + (6 + 853,024). From (1.3.3) and (1.6.3), we have (1.6.4) 2(1D2 + Q2)A2 - (ch2 — 4QD)A + 21)2 — P2 = 0. 23 On the other hand, from (1.6.3), BP 80 6Q 8A 6D By (1.6.1), (1.6.4), this implies (1.6.5) GA + F = 0, where G = PIQ — PQAfic2 — 8aQ2A — 4DPA66 + 4DQAac + I’zAac2 —8P2Aa — PQI + 2QA,662 F = —PID + PDAflc2 + 8aQDA + 2QPA6C + 2P2Aac + PQI. Combining (1.6.4) and (1.6.6), we have (1.6.6) 2(P2 + (22)}?2 + (P2c + 4DQ)FG + (2D2 — P"’)G2 = 0. Let Then (1.6.6) becomes (1.67) 0351136 + 6352135 + + blu + b() = 0, where 60 = 18a2A2c4(5184 + 1555262 + 8640a" + 576a6 + 64(1202 24 —1872a4c2 — 2016a6C2 — 2400.862 — 80.1062 — 810.264 — 567a“c4 —459a6c4 — 29a8c4 — 27a"c6 — 27a6c”)g18 9a3A2c5(576 + 576a2 — 72a2c2 — 128a”c2 — 8a6c2 — 9azc4 — 15a4c4)g’7, 536 = 962426400368 + 21888a2 + 1267264 + 1152a6 —— 1152c2 + 1424a2c2 800a"c2 —— 2112a”c2 — 352a’3c2 — 16awc2 + 32c4 — 34a2c4 —— 6a4c" —646a6c4 — 58a8c4 + 27a"c6 — 54a6c6), 535 = 96(62 — 2)A2c5(—576 — 576a2 + 32c2 -— 240.262 + 960462 +8a6c2 — 3a2c4 + 15a4c”). By the same argument as in the last section, it is sufficient to prove that, for any a and c, the coefficients of (1.6.7) can not simultaneously be identically zero on V. Case 1. g E 0 on V. In this case, t = g’ E 0 on V . and (1.6.7) becomes (1.6.8) 130 +816 + 132212 = 0. where b0 2 (a2 — 2)2c2(8 — a2c2), b1 Also it is easy to compute P = 3acu Q = (2 — a2)cu, and D = —6u. We will prove b,, 2' = 1, 2, 3, can not be zero simultaneously. 25 Subcase 1.1. a2 = 2. By Pa 2 QA — D, we have 3acua = 611 , i.e., 2 (1.6.8) a = — = by (1.6.1), we have A _=. constant on V. But 6 = —i = —1. Fro m(1.3.3), we get 61' Ba + aA = 0. By using (1.6.8), this implies (1.6.9) A = ——. Putting (1.6.8) into (1.6.1) gives c2 + 11 = 0. This is a contradiction. Subcase 1.2. 0202 = 8 but a2 # 2. In this case, put a2 = 385 in bl We have _ 8 51 = —,(512 +192c2 + 24c4 + 06) C which is never zero. Case 2. There exists 2 E V such that g(z) ¢ 0. Thus we can find an Open subset V’ of V such that g(y) is nowhere zero on V’. We will prove that b0, b1, b35 and b36 can not be zero Simultaneously on V’. To this end, we assme b0 2 bl = b35 = b36 = 0. Then by b35 = 0, we have (1.6.10) a2 = 2, or (1.6.11) —-576 — 576a2 + 32c2 — 246%2 + 96a4C2 + 80.662 -— 36%“ + 15647:“ = 0. 26 Subcase 2.1. If (1.6.10) is true, we put a2 = 2 back to b36 = 0, it gives 324(11+ c4)(32 — 8c2 — c4) = 0. Thus c2 = ‘—8—+§8—@ z 2.9282. On the other hand, put a2 = 2 in bl = 0 gives 6(—288 +12062 +136) 2 0, so c2 = :Li’bggifl x 1.9767. This is a contradiction. Subcase 2.2. if (1.6.11) is true. From bl = 0, we have (1.6.12) 576 + 576a2 — 720.262 — 1280.462 — 86%2 — 90.264 — 156%“ = 0. Then (1.6.11)+(1.6.12) gives 4c2(—8 + 24a2 + 8a4 + 3a2c2) = 0, then c2 = 8(1- 3a2 — a4). 3a2 Put the above into (1.6.12) and b35 = 0, we have (1.6.13) 3 — 34a2 — 1364 +1066 + 468 = 0, and (1.6.14) 16 — 309a2 +1768a4 + 520(16 -— 992a8 — 293a10 + 144a12 + 46a14 = 0. Solving (1.6.13), we have (12 3 0.08523 or a2 x 1.8009 27 Putting the above back into the left hand side of (1.6.14) to LHS of (1.6.14)z2.7704 or —15.3325 respectively. Both give contradictions. Cl CHAPTER 2 HYPERSURFACES WITH CONSTANT MEAN CURVATURE SATISFYING CHEN’S EQUALITY In this chapter, we will completely classify hypersurfaces in real space forms with constant mean curvature satisfying Chen’s equality. 2.1 A Remannian Invariant and Chen’s Equality According to the well-known Nash imbedding theorem, every Riemannian n-manifold admits an isometric immersion into the Euclidean space IE"("+1)(3"+“)/2. In general, there exist enormously many isometric immersions from a Riemannian manifold into Euclidean spaces if no restriction on the codimension is made. Associated to sub- manifold Of a Riemannian manifold there are several extrinsic invariants beside its intrinsic invariants. Among intrinsic invariants, sectional curvature and scalar curva- ture are the most fundamental ones. On the other hand, among extrinsic invariants, 28 29 the mean curvature function and shape operator are most fundamental. One of the most fundamental problems in submanifold theory is to obtain simple relationships between the main extrinsic invariants and the main intrinsic invariants of a submanifold and to find applications. Many famous results in differential geometry such as the isoperimetric inequality and Gauss-Bonnet’s theorem, among others, are results in this direction. Let M n be an n—dimensional Riemannian manifold. In 1993, Chen introduced an important Riemannian invariant 6M of M n by 6M (p) = 7(p) — inf K (p), where inf K is the function which assigs to each p E M ” the infimum of K (7r), where 7r runs over all 2-planes in TPM and 7' is defined by 7' = 2K,- K(e,~ A 6,), where {e1,.. .,en} is an orthonormal basis of TPM". For n z: 2, this invariant vanishes trivially. If M " is any submanifold immsersed in an m—dimensional Riemannian space form R’"(e) of constant sectional curvature 6, Chen proved in [7] a sharp inequality involving Chen invariant 6M and the squared mean curvature H 2, namely n2 (”—2) 2 Inequality (2.1.1) is known as Chen’s inequality and has some important appli- cations, for example, it gives rise to the second Riemannian obstruction for a Rie- mannian manifold to admit a minimal isometric immersion into a Euclidean space. It also gives rise to an Obstruction to Lagrangian isometric immersions from compact Riemannian manifolds with finite fundamental group 771 into complex space forms (see [9] for details). Since (2.1.1) is a very general and sharp inequality, it is natural and important to investigate and to understand submanifolds in a Riemannian space form which satisfy 30 the equality case Of Chen’s inequality, which is known as Chen’s equality: - _ n2(n " 2) 2 (2.12) 6M — m}! + (H +1)(n — 2)€. Submanifolds satisfying this basic equality were studied recently in many papers (cf. for instant, [7],[8],[9],[11],[12],[14],[19],[22],[23]). In this respect, we would like to point out in particular that 3-dimensional totally real submanifolds satisfying Chen’s equality in the nearly Kahler 6-sphere 86(1) have been completely classified in [23] by F. Dillen and L. Vrancken; and minimal hypersurfaces in non-flat Riemannian space forms satisfying Chen’s equality were classified completely in [14] by B. Y. Chen and L. Vrancken; roughly speaking they proved that a non-totally geodesic minimal hypersurface Of S"+1(1) satisfies equality (2.1.2) if and only if it is a tubular hypersurface with radius % about a 2-dimensional minimal surface in 5"“(1) and a non-totally geodesic minimal hypersurface of the hyperbolic (n + 1)-space H "+’( —1) satisfies equality (2.1.2) if and only if it is a “suitable tubular hypersurface” about a minimal surface in the de Sitter space-time S?+’(1) (cf. [14] for details). We will investigate the most fundamental case; namely hypersurfaces satisfying Chen’s equality. We will deal with hypersurfaces with constant mean curvature in this chapter and conformally fiat hypersurfaces in the next chapter. Since Chen’s equality is trivial when n = 2, we will consider n-dimensional submanifolds for n > 2. 2.2 Main Results Theorem 2.1 A hypersurface 1V1" (n > 2) of a Euclidean (n + 1)-space 1E"+’ with constant mean curvature satisfies equality (2.1.2) if and only if either M " is minimal or M n is an open portion of a spherical hypercylinder R x Sn’1(r). Theorem 2.2 Let M n (n > 2) be a hypersurface with constant mean curvature in the 31 sphere S”+’(1). Then M" satisfies equality (2.1.2) if and only if one of the following two cases occurs. 1. M n is a totally geodesic hypersurface. 2. There is an open dense subset U of M n and a non-totally geodesic, isometric, minimal immersion ¢ : B2 —> S”+1(1) from a surface B2 into 5"“(1) such that U is an open subset of the unit normal bundle NB2 defined by NpB2 : {g E T¢(,,)S"+1(1)| < {,5 >= 1 and < 5,6.(Tsz) = 0}. Let IE]1+2 denote the (n + 2)-dimensional Minkowski space-time with the Lorenzian metric g = —da:f +dzr§ + - - 1+ (trim. Recall that the unit hyperbolic space H "+1(—1) and the unit de Sitter Space-time SI’+1(1) are isometrically imbedded in IE?+2 respec- tively in the following standard ways: H”+I(—1)={x = ($1,...,:rn+2) E IE?+2| < x,x >= —1}, Sf+1(1) : {x E IE’I’+2| < x,x >= 1}. Theorem 2.3 Let M ” (n > 2) be a hypersurface with constant mean curvature in the hyperbolic space H"+1(—1). Then M" satisfies equality (2.1.2) if and only if one of the following three cases occurs. 1. M ” is a totally geodesic hypersurface. 2. M " is a tubular hypersurface with radius r = coth_’(\/2) about a 2-dimensional totally geodesic surface of H"+’(—1). 3. There is an open dense subset U of M" and a non—totally geodesic, isometric, minimal immersion (b : B2 —> S?“ from a surface B2 into the de Sitter space- 32 time Si’+’(1) such that U is an open subset of the unit normal bundle NB2 of B2 defined by N1082 = {E E T¢(p)S[’+l(1)|<€,€ >= —1 and < §,¢*(TpB2) >= 0}. 2.3 Prelimilaries Assume that M " is a hypersurface in real-Space-form R"+1(e), e = 1, 0, or -1. We shall make use of the following convention on the ranges of indices: 15A,B,CSH+1; 152393113371; 33046.73”- Denote by A = A); the Shape Operator of M n in R"+1(c) with respect to a unit normal vector 5 and by h the second fundamental form M n in R"+1(6). Let A1, A2, ..., An be the eigenvalues of A with respect to orthonormal eigenvector field e1,e2, ..., en, i.e., we have (2.31) Ac, 2 /\,‘6,‘. If M" satisfies the Chen’s equality (2.1.2), then, by rearranging e1, 62, ..., en if neces- sary, we have ([7]) (2.3.2) A1 = 0., A2 = ,u — a, An = p, 33 Let 021, 022, ..., w" denote the dual frame of e1, e2, ..., en. Then Cartan’s structure equa- tions give (2.3.3) dw’ = —w; /\ 02’, d6);- = w}, /\ a): + (e + A,A,-)w’ A to], where (623) are the connection forms. From (2.3.3) and Codazzi equation, we have e,-A-= A,—,\. of e (2.34) J ( J) ( 2) (A1 ‘ Ak)w;-°(e,-) = (A1" Ak)w,”(ej) for distinct i, j, k. Let F]; be defined by Ve,ej 2 22:1 I‘fjek, where V is the Levi-Civita connection on M ". Then we have k_ k j k _ k , k __ i w, -— XII-,0), I}, — (1),-(e3), I}, — PM. I: In this way, (2.3.4) becomes e,A-= Ai—AI‘} (2.3.5) ’ ( ” ’ (A7 - 29113;: (A.- - A9113- for distinct i, j,k. Let H denote the mean curvature of M" in R"+’(e). Then it — 1 H: p. n In the following, Let V denote the Open subset of M " on which M has exactly three distinct principle curvatures, i.e. (2.3.6) V = {x E M|a(;t) 75 0,a(:1:) 96 u,a(:c) 74 g}, where a and a are given by (2.3.2). From (2.3.2) and (2.3.5) we obtain the following 34 lemma. Lemma 2.1 Let M n be a hypersurface in a Riemannian space form Rn+1(e). If M" satisfies (2.1.2), then on V we have Ia 261‘: Ann F312,” Am F60 — F 0 a = 6 82ft 32H 2 610‘" a) 2.3.7 F“ = , F0 = ———, F = ———‘—_, ( ) “a a —)u 20 CE ) 12 H- 2a ega ea u— a eaa P1 = , P2 _ P1 _ 21 2a 02 a 1 011 ,u _ a 0, AG, , lug: e wk+fl——2+ 9” u 11—0 u-(a ) ”—0 n 820. 1 61 u—a 2 a a 2.3.8 4 1: ———-— + —— + —— A. . ( ) L02 ,u—2aw p—2aw fi—2a(],z::3 w (412‘— _ A (4)1 800‘ _ a) 2 BQHWO \ a a a 1 where A0, 2 F21 Proof. It is clear by (2.3.5). C] We also need the following lemma. Lemma 2.2 Let M" be a hypersurface in R"+’(6). If .M" satisfies equation (2.1.2), then on V we have el/i eaa (eaa)2 2aA§ (elu)2 €1( ) — 6’1( )_ 2 — _ 2 (23) 14-0 14—0 (u- a) 14-20 (1.4-a) . .9 _(e 11 2a) (62/1)_ _Z( —— efla)F 0+ an + e, (1“ _ 2a)": ——/‘ “3;“, 2(u-a’ (14-0) (14 2(14-20) u-a (2.3.10) _€a(fl - a’Aa _ (flfllkzzfll = 81(0 — H)(€2#) __ 1 Z aAgl‘ga, (4 - 2a In - CW (14 - 20)a u — a 0,1), 35 61( aAa ) _ e2( 80a ) _ (eaa)(e2a) _ “A0610, _ a) (2311) ”—a’ ”'a (I‘ ’“W 2a) (u-a)(u—2a)’ ’ l (M _ 20) A061” (”Gleam ‘0) " esaa aAs B (fl—a)? — 0’01- 20.) —C§(I«L" Flt—01““). Proof. By applying the second Cartan’s structure equation and by using Lemma 2.1, we may compute did}, in two different ways. Formulas (2.3.9), (2.3.10) and (2.3.11) are Obtained by comparing the coefficients of w1 A 020,612 /\ w“, and 011 A w2 of d013,, respectively. [:1 Similarly, by computing do)? in two different ways and by comparing the coeffi- cients of wl A a)", 022 A 020,021 A w2 of dwg, respectively, we obtain Lemma 2.3 Let A!" be a hypersurface in a Riemainnian space form R"+I(€). If M" satisfies equality (2.1.2), then on V we have e 62_/4 e ..(a — 74) 2am: _ (4.07 — ..))2 _ (e217)? (2312) 2( )+ a( ) (fl—(1.)(ll—2a) 0.2 0.2 ) ' ' = 810‘ " a)(€1#) _ n 660‘ _ a) 60 6 (waxy—2a) :6. a P 1"“ l“ 8 Egg _ e _ Aaeaa _ 14-6 _ a _ (82H)(€1#) 1( a) “A“ 44-0 a-(u 2a)A a( ) dbl-a) (2'3'13) (820.)(81/1.) aAaega = (ll—aXu—Za) _ (ll-aXu-Qa) _;Afiraai 36 e (M) __ e A _ ”A032“ _ (€101 — a))(ea(;i — a)) (2 314) 1 a 2 a (’1’ _ a)(l‘l’ — 2a) a(/1. — 2a) . . (f‘ — 2a)Aa(e2fll _ _(€1(H - a))(eaa) _ ” fl _ 660‘ _ a) fi + am — a) ‘ (I. — am. _ 2..) @9402 ———-,, 1“...)- 2.4 Proof Of Theorems 2.1, 2.2 and 2.3 Let M n be a hypersurface with constant mean curvature H in a Riemannian space form R"+’(e) (n > 2, e = 1, -1, or 0 ) which satisfies Chen’s equality (2.1.2). If M" is not minimal, then a = fill is a nonzero constant. In this case, (2.3.9) and (2.3.12) reduce respectively to 2 _ ZaAfl/J - “)2 (a _ “Xeaeaal _ 2(eaa) — I14 _ 2a + (5 + ”(1)0" " a)2 (2.4.1) n - Z (Baalfgam - a), 16¢a a(eaeaa) — 2(eaa)2 = — (M £12: 2a) + (e + ()1 — 6))62 (2.4.2) 5 +a 2 (@6650. (#0 From (2.4.1) and (2.4.2) we obtain on V that 2aA2",,(u2 — 36a + 3a?) (14 - a) (M - 2a) eaeaa = — + (u — 2a)(a2 — an — e) n (2.4.3) + 2(640)F€a, 0¢fl (......)2 = -5263, + 01 — .52) — gum - a). 37 Since ,u is constant, (2.3.13) reduces to A a 2 - 3 2 2 " (2.4.4) 60A,, = “(e “X” a“ + a ) + Z Aflrga. Now, by applying Lemma 2.1 and Cartan’s structure equations, we can compute do); and drugI in two dofferent ways. Afetr that, by comparing the coefficients of w1 /\ wfl and w2 A w5 with fl 79 a in the formulas of dad}, and of to: so obtained, we find respectively the following two formulas: 2(eaa)(ega) 2a(u — a)AaAfl " a = P7 a we a + H _ a + l1 _ 2a 72::3(870) afi 2(eaa)(ega) 20,3A0A3 " e eaa — + = e a I“; . fl a (u—a)(u-2a) .24.“) " By taking the difference of these two equations, we obtain on V that (2.4.5) (eaa)(ega) = —a2AOA5, a # fl. Now, by applying (2.4.3), we may also obtain —(e..a)2 = «HA: + (u - an?) + gum — a), 4m)? -—- aw + (u — am + 3am — a). By taking the difference of these two equations, we get (2.4.6) (eaa)2 — (830.)2 = a2(A?3 -— A3,). 38 Combining (2.4.5) and (2.4.6), we obtain, on V, (2.4.7) (eaa)2 = a2A2, a ¢ 6. Case (i): e = 1, R"+1(1) = 5"“(1). If M” is non-minimal, follow the exact arguments as in Case (iii) by replacing e = —1 by c = 1, we know that a is constant on each component of V. Therefore (2.4.7) implies A0 = 0. Then the second identity of (2.4.3) implies that V is an empty set. Thus, M n is an isoparametric hypersurface of R"+1( 1) with at most two distinct principle curvatures given either by A1 = 0, A2 = = An = ,u or by A1 = A2 = 211', /\3 = = An = p. Both cases are impossible according to a well-known result of E. Cartan [5]. Consequently, M" is minimal in 5"“(1). Let U denote the open subset of M " consisting of non-totally geodesic points. Then U is an open dense subset of M ". Now, by a result of [7], U has relative nullity n — 2. Thus, by applying a result of Dajczer and Gromoll (Lemma 2.2 of [18]), the Gauss image 82 of U is a minimal surface in the unit shpere. Consequently, U is an open subset of unit bundle N B2 defined in Theorem 2.2. Case (ii): 6 = 0, R"+1(0) 2 En“. If M n is non-minimal, then the second equation in (2.4.3) implies a = 0 or a = p on V which contradicts the definition of V. Thus V is an empty set. Hence, M is a non-minimal isoparametric hypersurface with at most two distinct principle curvatures given either by A1 = 0, A2 = = An = ,u or by A1 = A2 = %u,)\3 = = An = n. It is well-known that the first case occurs if and only if M " is an open portion of a spherical hypercylinder IR x Sn‘1(r) for some 1' > 0. It is also known that the latter case cannot occur. Case (iii): 6 = —1,R"+1(—1) = H"+1(—1). First we assume that M" is non- minimal. We claim that the function a is constant on each component of V. We divide the proof of this claim into three cases. 39 Case (iii-1): n 2 5. From (2.4.7) we have (2.4.8) (63(1)2 2 (640.)2 = = (ena)2 = a2Ag, a = 3, ..., 71.. Without loss of generality, we may assume 630. = aA3. By using (2.4.5), we have 640. = —aA4 and 65a 2 —aA5 which imply (64a)(e5a) = a2A4A5. Thus eaa = A0, = 0 on V by (2.4.5) and (2.4.8). Hence, by applying the first equation in (2.4.3), we obtain a2 — up — c = 0 on V. Since a is constant, this implies that a is constant on each component of V. Case (iii-2): n = 4. from (2.4.5) and (2.4.7) we may assume, without loss of generality, that 63a 2 (LA, and 64a 2 —aA3. By differentiating the second equation of (2.4.3) with respect to 63, we get (2.49) —2(€30)(6363a) = 2(esa)a(A§ + (u — (2)2) + a2(2A3(e3A3) —2()u — a)(e3a) + §(,u - 2a)e3a. On the other hand, since A4 = e3(ln a), (2.4.4) yields (63a)A3(/l2 —3au+4a2) 63a 4 A = —I‘ . e3 3 (u-ZaXu-ala + a 33 Substituting this into (2.4.9), we find _. e a e e a = a 20A§(fl2 - 3au+3a2) (2.4.10) ( 3 )( 3 3 ) (e3 )( (u-a)(#‘2“) +201 — 2a) — (€4G)F§3)° + aw - 20W - a) If 63a does not vanish identically on V, there exists an Open set W C V such that 40 63a is nowhere zero on W. Thus, on W, (2.4.8) becomes —e3e3a = 2c1.A§(p2 — 3au + 3a?) (2.4.11) (11 - a)(# - 2a) 6 +101 — 2a) — (64a)F§3. + aw - (1)04 — 20) Combining (2.4.3) and (2.4.11), we obtain ,u = 2a. which implies that a is ocnstant on W. This contradicts the assumption that 63a 7E O on W. So 83a E O on V. A similar argument yields 64a 5 0 on V. Thus by applying (2.4.5), we know that A3 vanishes identically on V. Hence, by applying (2.4.3) again, we have a2 —— an + g = 0 on V. Consequently, a is constant on each component of V. 8 Case (iii-3): n = 3. From (2.4.3), we have —2(e3a)(e363a) = a2(2A3(63A3) — 2(63a)(u — a)) +2a 2, every component of V is an isoparametric hypersurface of the hyperpolic space. On the other hand, according to a well-known result of Cartan, every isoparametric hypersurface of H "+1(—1) has at most two distinct principle curvatures. T herfore we conclude that each component of V has at most two distinct principle curvatures. This contradicts the definition of V. Thus, V must be the empty set. Consequently, M " is an isoparametric hypersurface of H "+1(—1) with exactly two distinct principle curvatures. Therefore, by applying Cartan’s classfication theorem of isoparametric hypersurfaces in hyperbolic spaces, M " is an open set of the Riemannian product of H 2(—%) and Sn’2(1) isometrically imbeded in the hyperbolic (n. + 1)-space in the standard way. Such a hypersurface is a tubular hypersurface with radius r = coth‘1(\/2) about a 2-dimensional totally geodesic surface (cf. [8]). Now, assume M n is a non-totally geodesic minimal hypersurface. Let U denote the open subset of M " consisting of non-totally geodesic points. Then U is an open dense subset of M ". Now, by a result of [7], U has relative nullity n — 2. In this case, by applying an argument similar to-spherical case, we may conclude that the Gauss image B2 of U is a minimal surface in the unit de Sitter space-time S?+1(1). 42 Consequently, U is an open subset of the unit bundle N B2 defined in Theorem 2.3. The converses are easy to-verify. El CHAPTER 3 COMFORMALLY FLAT HYPERSURFACES SATISFYING CHEN’S EQUALITY In this chapter, we will study conformally flat hypersurfaces satisfying Chen’s equality (2.1.2) in Riemannian space forms. By utilizing the Jacobi elliptic functions and the Theta function we obtain the complete classification of such hypersurfaces. 3. 1 Main Results In order to state our results, we recall three families of Riemannian manifolds, P:(a > 1), Cg(a > 1), D2(O < a < 1) and the two exceptional spaces F ", L", first introduced by Chen in [10]. Let cn(u, k), dn(u, k) and sn(u, k) denote the three main Jacobi’s elliptic functions with modulus k. The nine other elliptic functions nd(u, k), nc(u, k), ns(u, k), sc(u, k), cd(u,k), ds(u,k), cs(u,k), dc(u,k), sd(u,k) are defined by taking reciprocals and quotients. For example, sd(u, k): sn(u, k)/dn(u, k), nd(u, k)=1/dn(u, k) (cf. [24] and the next section for details). 43 44 We define 2 _ (3.1.1) pa = akcn(a:r, k), k = —:l/_2—1, a > 0, a a 0. 2a (31.2) 770 = Edn(;x,k), k = w, 0 < G <1, 2 (2.1.3) pa = akcn(a:1:,k), k = a +1 (1 >1. fia’ Let S"+1(c) and H "+1(—c) denote the n-sphere with constant sectional curvature c and the hyperbolic n-space with constant sectional curvature —c, respectively. For n > 2,Pg‘, D2 and C: are the Riemannian n-manifolds given by the warped prod- uct manifolds I x“, Sn‘1(¢;—1),IR xna H""1(9ff—l) and I xpa S”“1(94—4=i) with warp functions pa, 77a and pa, respectively, where I denote the open interval on which the corresponding warp function is positive. The two exceptional spaces F n and L" are ) and IR x 111"“, respectively. the warped product manifolds IR Xi/fi H "‘1(—% S€Ch(;r:) D2, F" and L" are complete Riemannian n-manifolds, but P: and C: are not com- plete. Topologically, S" is the two point compactfication of both P: and C2. From [13] we know that the Riemannian metrics defined on P: and C]: can be extended smoothly to their two point compactifications S ". We denote by 15: and (53 the sphere S " together with the Riemannian metrics given by the smooth extensions of the met- rics on P: and C: to 3", respectively. We remark that PI, D2, C: are indeed isometric to the wraped products n-manifolds I X), Sn'1(1), I xi, H"‘1(—1), I x), S"’1(1) with warped functions 2pa/(a2 + 1), 2/1a/ (a2 — 1), 2110/ (a2 + 1), respectively. Let A2(a > 1),B;‘(0 < a < 1),G”,H;‘(a > 0) and Ya"(0 < a < 1) denote 45 respectively the following warped product manifolds: n— 1 1 n- 1R)<\fla—7——lcosh:z:S 1(1), (_§I§) X\/1———O7COS$S 1(1)? R XCOShfl? [En—1’ R XWCoshx Hn_1(—1)1 (07 00) XWsinhx Sn—1(1)' When n = 2, the second factor S "’1 or H "‘1 in each of the wraped product manifolds will be replaced either by 81(1) or by IR. The geometry of A3112 and H a2 is similar in the sense that one can be obtained from the others by applying some suitable scalings on the first factor IR. Clearly, Ag, G", H I: are complete Riemannian manifolds. Topologically, S" is the two point compactification of Bf. As for P: and CZ, the warped metric on B",1 can be extended smoothly to its two point compactification, a fact that follows from (3.1.6). We denote by B]: the shere S" together with the Riemannian metrics on 8" extended from the metric on Bf. For n 2 2 and any real number a > 0, there is a well-known Lagrangian immersion from the unit n-sphere S" into a complex Euclidean n-space C" defined by a 3.1.4 to , n =——— n. 1 , ( ) ..(yo 211 y) 1+yg(yl y 310.11 you.) where yg 41 y? + + yf, = 1. The immersiom 1110, due to Whitney, has a unique self- intersection point wa(—1,0, ...0) = wa(1,0....,0). The S" together with the metric induced from Whitney’s immersion wa, denoted by Wg‘, is called a Whitney n-sphere. In this chapter, we first sharpen a result of [22] (Propositon 2) to the following Theorem 3.1. Let x : M" —> IE"+1(n > 2) be an isometric immersion of a confor- mally fiat n-manifold into a Euclidean (n+1)-space. Then it satisfies Chen’s equality (2.1.2) if and only if one of the following four cases occurs: 1. M" is totally geodesic. 46 2. M " is an open portion of a spherical hypercylinder 5'"1 x IR. 3. M " is an open portion of a round hypercone. 4. n = 3 and M3 is an open portion of a Whitney 3—sphere Wu3 for some a > 0 and, up to rigid motions, the immersion x : W: —-> IE4 is given by (3.1.5) 1 V2 V2 x \f2— x/2 x(:c,y1,y2,y3) = (2/0 sd2(—a—:r)d:z:,aylsd(—a—:r),aygsd(—7‘—:r),aygsd(—E—x)), where yf+y§+y§ = % and k = f is the modulus of the Jacobi elliptic functions. Our main results in this chapter are the following. Theorem 3.2. Let x : M" —) S"+1(1) C 15”” (n > 2) be an isometric immersion of a conformally flat n-manifold. Then it satisfies Chen’s equality (2.1.2) if and only if one of the following three cases occurs. 1. M" is an open portion of S"(1) and the immersion x : M" —-> Sn+1(1) is totally geodesic. 2. M" is an open portion of B: —> S"+1(1) C 15"” given by (3.1.6) x(:1:,y1, ..., yn) = (sin :r,acos:1:, \/1 — agyl cos :5, ..., \/1 — a231,, cos 1:) with gfi+y§+...+y,21 = 1. 3. n = 3, M3 is an open portion of 15: for some a > 1 and, up to rigid motions, the immersion x : 132 ——> 54(1) C If3 is given by 1 (31-7) Mai/1,112,313) = J(y1cn(a$),y2cn(a$), yacn(a$),Jcos X,Jsin X), where J = (/a2k’2—cn2(ax),yf + y; + 31% = 1,7 = sn“l(§), and k = x/a2 — 1/(\/2a),k’ = \/a2 +1/(\/§a) are the modulus and the complementary 47 modulus of Jacobi ’s elliptic functions respectively, and X = %x+ -—2——-1—(ln 9W - 7) 9(aa: + 7) + 2mm». where 9(a) = O(u,k) is the Theta function and Z(u) = Z(u,k) is the Zeta function. Theorem 3.3. Let x : M” -—) H”+I(—1) C IET‘2 (n > 2) be an isometric immersion of a conformally flat n-manifold. Then it satisfies Chen’s equality (2.1.2) if and only if one of the following nine cases occurs. 1. M" is an open portion of H"(—1) and the immersion x : M" —> H"+1(—1) is totally geodesic. 2. M" is an open portion of A: for some a > 1 and, up to rigid motions, the immersion x : A: -—> H"+l(—1) C IE]1+2 is given by (3.1.8) x(a:,y1, ...,yn) = (acosh 2:,sinh 2:, Wm cosh :r, ..., My" cosh so) with yf+y§+m+yfi = 1. 3. M" is an open portion of G" and, up to rigid motions, the immersion x : G" —+ H"+1(—1) C IE]l+2 is given by ug+... +u3, h , 2 )cos a: x(a:,u2,...,u,,) = ((1+ (3.1.9) 2 2 u + + u , 2 2 " cosh :1:,smh :r,u2 cosh :1:, ...,un cosh :13). 4. M" is an open portion of H: for some a > 0, and, up to rigid motions, the 48 immersion x : H: —+ H"+1(——1) C IE]1+2 is given by (3.1.10) x(:r,y1, ...,yn) = (\/a2 +1y1 coshx, ..., \/a2 + 1y" cosh 2:, acosh 11:,sinhx) with yf—yg—u. —-y,2, = 1. . M" is an open portion of Y0" for some 0 < a < 1, and, up to rigid motions, the immersion x : Ya" ——> H "+1(—1) C IE]l+2 is given by (3.1.11) x(:r,y1, ...,yn) = (cosh 3:, asinh 2, Wm sinh :13, ..., \/1———c12y,, sinh :r) with yf+y§+...+y3, = 1. . n = 3, M" is an open portion of F3 and, up to rigid motions, the immersion x : F3 —> H4(—1) C IE? is given by (3.1.12) x(:1:,u,v) = (\/2coshucosh v, \/2coshusinh v, \/23inh u,cos (E, sin :12). . n = 3, M" is an open portion of L3 and, up to rigid motions, the immersion x : L3 —> H4(—1) C IE? is given by (3.1.13) 1 1 x(:c, u, v) = secha:(:z:2 + u2 + v2 + cosh2 a: + 1,3732 + u2 + v2 + cosh2 a: — 1,213, a, v). . n = 3, M" is an open portion of C: for some a > 1 and, up to rigid motions, the immersion x : C2 —> H4(—1) C IE," is given by (3.1.14) x(:r;,y1,y2, milk—[(6 cosh )7, E sinh y, ylcn(a:t), yzcn(aa:), y3cn(a:c)) 49 with y? + yg + y§ = 1, where l = \/a"’lc'2 + cn2(aat), where 7 = sn‘1(1/(ak2)),k = \,/a2 +1/(x/2a) and k’ = x/a2 —1/(\/2a). 9. n = 3, M" is an open portion of D2 and, up to rigid motions, the immersion x : D: —> H4(—1) C IE? is given by x(:2:,u,v) 2 (3.1.15) 1 acfikdnéx) cosh u cosh v, kdn(%:r) cosh u sinh v, kdn(%x) sinh u, go cos Z, (9 sin Z), where k = Via/«Ha? and k' = f—l—a2/\/_1+a2, 7 = sn’1(k/a), p = \/lc2dn2(%x) — a2k’2, and 3.2 The Jacobi Elliptic Functions, Theta Function and Zeta Function We review very briefly some known facts on Jacobi’s elliptic functions, Theta function and Zeta function for later use (see [1], [24] or [26] for details). Let 0 be the temperature at time t at any point in a solid material whose con- ducting properties are uniform and isotropic. If p is the material’s density, 3 its specific heat, and k its thermal conductivity, 0 satisfies the heat conduction equation: nV20 = BO/Bt, where n = k/sp is the diffusivity. In the special case where there is no variation of temperation in the 2:- and y-directions, the heat flow is everywhere 50 parallel to the z-axis and the heat equation reduced to the form: 029 _ as 0 = 0(z,t). Consider the boundary conditions: 0(0, t) = 0(7r, t) = O and 0(2, 0) = 7ré(z — 7r/2) for O < z < 7r, where 6(2) is Dirac’s unit impulse function. Then the solution of the boundary value problem is given by (3.2.2) 0(2, t) = 2 2(—1)"e"(2"+1)2'“ sin(2n +1)z. 11:0 By writing e“"‘ = q, the solution of (3.2.2) assume the form (3.2.3) 91(2, q) = 2 Z(—1)"q<"+1/2)2 sin(2n +1)z, n=0 which is the first of the four theta functions. For simplicity, we shall often suppress the dependance on q. If one changes the boundary conditions to 80/02 = O on 7. = 0 and z = 7r with 6(2, 0) = n6(z — 7r/ 2) for 0 < z < 7r, then the corresponding solution of the boundary value problem of the heat equation (3.2.1) is given by (3.2.4) 04(2) = 64(2, q) = 1+ 2 Z(—1)"q"2 cos 2712. 71:1 The theta function 01(2) of (3.2.3) is periodic with period 27r. Incrementing 2 by %7r yields the second theta function: (3.2.5) 62(2) = 02(2, q) = 91(2 + Q, q) = 2 Z (WW/2)? cos(2n +1)z. n20 51 Similarly, incrementing 2 by g for 64 yields the third theta function: 71' 2,q) =1+2an2cos2nz. n=l (3.26) 63(2) 2 63(2, Q) 2 04(2 + The four theta functions 01, 02, 63, 04 can be extended to complex values for z and q such that [q] < 1. The elliptic functions snu, cnu and dnu are defined as ratios of theta functions: _ 03(0)01(z) _ 64(0)02(z) _ 64(0)03(z) (3.2.7) snu — W, cnu — W, dnu —- W, where z = u/6§(0). Define parameters It and k' by k = 92(0)/9§(0), k' = 03(0)/0§(0) which are called the modulus and the complementary modulus of the elliptic function. It and k’ satisfy k2 + k’2 = 1. When it is required to state the modulus explicitly, the elliptic functions of Jacobi will be written sn(u, k), cn(u, k), dn(u, k). The elliptic functions snu, cnu and dnu satisfy the following relations: (3.2.8) sn2u + cn2u = 1, dn2u + kgsn2u = 1, k2cn2u + k'2 : dn2u, 2snucnvdnv (3.2.9) sn(u + v) + 3110‘ — v) = 1 _ k28n2usn2v’ (3.2.10) sn'(u) = cn(u)dn(u), cn'(u) = —sn(u)dn(u), dn'(u) = —kzsn(u)cn(u). 52 The Theta function, 9(a), and the Zeta function, Z(u), are defined by (3.2.11) e(u) = 04%), Z(u) = 8%(1n04), K = 913(0), and satisfy the following identities: (3.2.12) Z(u + v) = Z(u) + Z(v) - k2snusnvsn(u + v), (3.2.13) Z(u) = 38)) , From (3.2.12) we have (3.2.14) k2snusnv[sn(u + v) + sn(u -— v)] 2 Z(u — v) — Z(u + v) + 22(v). 3.3 Two Lemmas Lemma 3.1. Let W: denote the warped product manifold I x a Sd Sn‘l(1), 2 7‘ I («23715) with the warped product metric given by 2 fl 1 3..31 —_ d 2 + _a d2 _. ,_ , ( ) g I S ( (1. I fi)90 2 where I denotes the largest open interval containing 0 such that sd(€$) is nowhere zero on I and go is the standard metric on the unit (n — 1)-sphere. Then the Whitney n—sphere W: is topologically the two point compactification of 14’: Moreover, the metric on W: is the smooth extension of the warped product metric on W: to its two point compactification. 53 Proof. Let S" denote the unit n-sphere with the north and south poles, {N , S }, being removed and let {u1, ug, ..., un} denote the spherical coordinate system on S" given by 310 = COS U1, 311 = Sin ul cos U2, (3.3.2) yn_1 = sin ul... sin un_1 cos un, y, = sin v.1... sin un_1 sin an. From (3.1.4) and (3.3.2) we know that the metric induced from Whitney’s immer- sion wa on S" is given by a2 a2 sin2 u1 . . = ___._____._ d 2 — (333) g (1+cos2u1) ful4h(1+cos2u1)g0 Put (3.3.4) m(ul) = [In ——-2——dt. o M??? Then a "I a a 1 3.3.5 :1: u = —/ dt = —sn_l sinu ,—. Thus . x/i 1 (3.3.6) S111 ul — sn(7x, :5). From (3.2.8), (3.3.3), (3.3.4) and (3.3.6), we obtain (3.3.1). This shows that W: is the S" endowed with the metric induced from Whitney’s immersion wa. Hence, topologically, the Whitney n-sphere W: is two point compactification of W: and, moreover, the metric on W: is the smooth extension of the warped product metric on W: to its two point compactification. E] 54 The following Lemma is important in this chapter. Lemma 3.2. Let M 3 C 124(6) be a conformally fiat hypersurface of a Riemannian space form R4(e) satisfying Chen’s equality (2.1.2), then M 3 has at most two distinct principle curvatures. Proof. Assume M 3 is a conformally flat hypersurface in a 4-dimensional Riemannian space form 124(6) satisfying equality (2.1.2). Let L be the symmetric 2-tensor defined by (3.3.7) L = —Ric + £9, where Ric, r and 9 denote respectively the Rice tensor, the scalar curvature and the metric tensor of M 3 . Then by a result of H. Weyl we have for vectors Y, Z, W tangent to M 3. Let /\1 = a, A2 = u — a, A3 = u be the principle curvatures of M 3 with their corresponding principle directions e1,e2, e3 given as in Section 2.3. Thus, from the equation of Gauss and (3.3.8), we have (339) (A: — ADI—€1- = 3(e,-H)/\J — %e,-T — eiAg, (A? - Almi- = (A? - Ain‘t-3. (3.3.10) r = 36 + ,u2 + all — a2, for distinct i, j, k(i, j, k = 1, 2, 3), where H = gu is the mean curvature function. 55 Equation (3.3.9) and (3.3.10) imply 1 . 2(e,u)Aj — 261.012 + an — a2) — e,/\§ = (A: —- A?)I‘{j = —(A,- + Aj)e,/\j. Thus 1 1 (AI: — Aj)e,-/\j = (*2/\j + [.l + 5a)e,u + (ill - (1)610“ By taking (i, j) equal to (1, 2), (2, 3) and (3, 1) respectively, we obtain (3.3.11) (u -— 2a)e1a = aelu, (3.3.12) (u — 2a)e2a = (2a — 3a)e2u, (3.3.13) new 2 (2p — 3a)e3u. Let V denote the open subset of M 3 on which M 3 has exactly three distinct principle curvatures, i.e., V is given by (2.3.6). Suppose that V is not empty. In this remaining part of the proof, we shall work on this non-empty subset to obtain a contradiction. From (2.3.5) and (3.3.9) we obtain (3.3.14) r31, = 0, for distinct i, j, k. Equation (3.3.14) implies (3.3.15) a); = e2aaw1 + we); 56 By taking the exterior derivative of (3.3.15) and by applying Cartan’s structure equa- tions, we obtain the following formulas on V by comparing the corresponding coeffi- cients in the resulting formula of do); 61(61(u-a))_e( e20 )_ (624)2 (8104-0))2 (3.3.16) (53% — a) u - 2a ([1 — 2‘02 (,1 _ M2 : a(u-a) +a(u—a)+e, €20 (€2a)(€30) _ (€30)(€2#) (3'3”) 63(0 - 2a) (u — c004 — 24) 404 — a) ’ €10“ — a) €10I — a)(€3(fl * a) _ (elflleslu - 0) (3.3.18) e3(———# _ 2a )+ a(#_ 20) — 04/1 _ a) . By (3.3.13) we have A3 = F31 = 0. Thus, by applying (2.3.10), (2.3.13), (3.3.11), (3.3.12) and (3.3.13), we find (3.3.19) em = Hewett), (3.3.20) em = (jgggfiemxewl On the other hand, by Lemma 2.1, (3.3.11) and (3.3.12), we have [61162]” :: _P%181# + Ff2e2’u : _# Combining this with (3.3.19) and (3.3.20) we find (3-3-21) (at/Maw) = 0- ' 57 Therefore, (elp)(e2u) :2 0 on V. If e211 i 0 on V, there exists an open subset W C V on which 62p 72E 0. On W, we have 6111 = 0 identically. Moreover, from (2.3.11) and (3.3.17) we have 3a2 — 3au + 2p2 (3.3.22) (213 — 3a)ege3u = — ap. (egu)(e3,a), (3.3.23) (2p — 3a)e3e2p = 6a2 _ 3:: - 2’12 (eou)(e3u). Combining the above two equations we obtain (3.3.24) (2p — 3a)[eg, e3],u = 3(ii;flll(eou)(e3u). On the other hand, we also have 8301 — 0) 62# 3 [62.63114 = 1336214 - P321230 = — 63/4 - 71—6314 = -;(62#)(63u)- Therefore, we find (3-3-25) (2/1 - 3a)(62u)(63u) = 0- If (egu)(e3u) aé 0 on W, then on an open subset W’ of W on which (egu)(egp) 75 0, we have (3.3.26) 2p — 3a 2 0. Since 2p — 3a 2 0 on W’, (3.3.12) and (3.3.13) imply ega = e3a = 0 on W’. On the other hand, since e1 p = 0 on W, (3.3.11) yields ela = 0, Therefore, we obtain em = 62” = e3u = 0 which is a contradiction. Consequently, we must have e3/i E 0 58 on W, from which we obtain (ela)(e3a) E 0 on W by virtue of (3.3.13). Thus, by (3.3.15) and (3.3.12), we obtain (3.3.27) (2h — 3o.)(egh)2 = 0.01. — 2a)2(a,a + 6). On the other hand, from (3.3.11), we know that 2p — 3a # 0 on W. Thus, there is an open subset W1 C W on which 2p - 3a # 0. On W1, we have (3.3.27) yields 4: am + om - 20)? (3.3.28) (62/!) 2]}. _ 3a By using (3.3.12) and (3.3.28) we obtain 1 egegu = ( [—6a5 — 15a4u + 38a3u2 — 23a2p3 + 4ap4 2p — 3a)2 +e(—16a3 + 27a2u — Map2 + 2p3)]. (3.3.29) On the other hand, (3.3.12) and (3.3.16) imply that on W1 we have 2p -— 3a 12,112 —- 33au + 23a2 (ll _ 2a)2e2€2/1 (,u _ 2(1)? (82/1) = —a(,u - a) — e. By using (3.3.28), the above equation becomes _ 1 5 4 3 2 2 3 4 —e2e2p—W[12a —9ap—2a u —au +2au -I~e(11a3 — 13a2p + an2 + 2113)]. (3.3.30) Combining (3.3.30) and (3.3.29) we get (3.3.31) 6a3 — 12a2u + 6a,a2 + 6(4/1 — 5a) 2 0. 59 Differentiating (3.3.31) and applying (3.3.12) we find (3.3.32) -—30a3 + 72a2p — 54a,u2 + 12”3 + 6(70. —— 6,.) = 0. Also, by combining (3.3.31) and (3.3.32) we get (3.3.33) 6a2u + 6p3 — 12cm2 + 6(7u — 9a) 2 0. The above implies a _ 96 +12,u2 :l: \/81 + 486p2 3.3.34 ( ) 12” By substituting (3.3.34) into (3.3.32) we conclude that )u must satisfy the following polynormial equation with constant coefficients: (3.3.35) 162 + 120,.2 + 886,14 + 32,16 = d:\/3(9 + 85/1 + BMW/27 + 165p? which is impossible since otherwise it is locally a constant function. Thus we must have 62;; E 0. Similarly, we may prove that 81/}, E 0, too. Hence, by applying (3.3.11) - (3.3.13), we obtain (3.3.36) ela = 620. = 61p 2 62/1. 2 0. Now we claim that 83/1. E O on V, also. In fact, otherwise there exists an open subset 01 C V on which 63/; ¢ 0. If 2;; — 3a = O on V, then (3.3.13) implies 63a = 0. Thus, by (3.3.16), we get a(u — a) + 6 = 0. Thus 2112 + 96 = 0 which is impossible. 60 If ,u = 3a, then (3.3.13) yields [1.83/1 = 0, which is also a contradiction. Hence, we have 2p 75 3a and 41 ;£ 3a on V. Now, from (3.3.16) we get (244 - 3a)(u - 3a) flaw — 0) (3.3.37) (63402 = (1(44 — a) + 6. Thus there exists an open subset 02 C 01 on which 2 _ (am - a) + 6)n2a(u - a) (3.3.38) (6344) —- (ZH- 30“” _ 3a) . On the other hand, (2.3.9) and (3.3.13) imply (214 - 3a)63u _ (€3#)2(2H - 3a)2 (3339) _€3( Mu - a) (M - 602442 =a/J.+C. Equations (3.3.38) and (3.3.39) yield 6a5 — 15a3 2 + 11oz);3 — Zap4 y. u e e p = (3.3.40) 3 3 (u — 3a)(2u — 3a)2 +c(3a.3 — 15oz); + 11cm2 — 2p3)]. Similarly, by applying (2.3.12), (3.3.13) and (3.3.38), we have 6363/2 2 gap — 3a)(u — 3a)2[6a5 - 30a4p + 45a3/L2 + 26a2p3 (3.3.41) 14 +5a/44 + 6(3a3 + 6a2p — 10a);2 + 3443)]. Summing the above two equations yield (3.3.42) #201 — 2a)[—24a4 + 48a3u —- 30a2u2 + 6a,.3 + €(15a2 — 15cm + 42.2)] = 0. On the other hand, from (3.3.14), we know that the distribution fl spanned 61 by 81,632 is integrable. Also, the distribution .7: spanned by 63 is clearly integrable. Therefore, there exists a local coordinate system {x1,x2,x3} such that 63 = B/Bt, where t = 2:3. Hence, by applying (3.3.36), we know that both a and p depend only on t. Therefore, (3.3.13) yields da .3.43 —-— = 2 — . (3 ) Md“ .11 3a By solving (3.3.43) we obtain _ l1 —3 (3.3.44) a _ 5 + Ca for some constant C. By substituting the above into (3.3.42), we know that u must satisfies a polynomial equation with constant coefficients. Therefore, 63;; = 0 on 02 which is a contradiction. Consequently, both ,u and a. are constants on each component of V. Hence, by (2.3.9), (2.3.12) and (3.3.16), we get up = (u — (3)41 = —e which is clearly impossible. Therefore, we know that V is an empty set. Hence, M 3 has at most two distinct principle cuevatures. This complete the proof of the Lemma. D 3.4 Proof of Theorem 3.1 If x : M " —-) IE"+1 is an isometric immersion of a conformally fiat n-manifold with n > 2 which satisfies Chen’s equality (2.1.2). Then, by Lemma 3.2 and a well-known result of E. Cartan and J. A. Schouten on confromally flat hypersurfaces (cf. p154 of [6]), we know that M " has a principle curvature with multiplicity at least n -— 1 for n = 3 as well as for n 2 4. Thus, by applying (2.3.2), we know that either (i) the principle curvatures of M n are given by A1 = 0, A2 = = An = p, or (ii) n = 3 and 62 then the principle curvatures of M 3 are given by A1 = A2 = p/ 2, A3 = )1 7A 0. If Case (i) occurs, M " is either totally geodesic, or an open portion of spherical hypercylinder, or an open portion of a round hypercone (cf. [22]). Now, we assume case (ii) occurs. Denote by U the open subset of M 3 on which the mean curvature function is nonzero. Then U is a non-empty open subset of M 3. We shall work on U unless mentioned otherwise. From (2.3.5) we obtain (3.4.1) 1‘31 2 A3 = 0, 61/1. 2 82/1. 2 0, F§3 2 P33 2 0. Denote by .7: and .751 the distributions spanned by {61, 62} and {63}, respectively. By (3.4.1) we know that the integral curves of fl are geodesics and the distribution 7: is integrable. Consequently, there exist local coordinate systems {2311,21} such that f is spanned by {63/821, (9/81)} and e3 = (9/022. From (3.4.1) we know that ,u depends only on 2:, i.e., 11 = 11(2). Also, from (2.3.8) and (3.4.1), we have I I (3.4.2) to; = u (37)..)1, a); = “ (”412. Using the above we obtain (3.4.3) we3 = {figep j = 1, 2. Therefore, each integral submanifold of .7 is an extrinsic sphere of E4. Hence, by applying a result of Hiepko [21] (cf. also Remark 2.1 of [20]), we know that M 3 is locally the warped product I x “3) 32(1), where f (:5) is a suitable warped function. 63 So, the metric of M 3 is given by (3.4.4) 9 = (1352 + f2(:1:)go, where go is the standard metric of 82(1). In particular, if we choose the spherical coordinate system {0, (15} for 52(1), we have (3.4.5) 9 2 d2:2 + f2(:13)(al(f>2 + cos2 ¢d02). Applying the above equation, we obtain a _ a -43 2 1'2: V%ax—O’ Vfi86_f66’ V§30¢=fa¢ 8 , 8 8 __ 8 (3.4.6) Vgflz—ff 7:, VEQBF—_tan65—6’ 8 a V955-2 —ff' cos 256% $+sin¢cos¢—¢. By computing dwé and by using (3.4.2) and Cartan’s structure equations, we find 3 (3.4.7) ,u"(:1:) + ,1 g”) = 0. Integrating once (3.4.7) yields 4 (3.4.8) 41/2 + 114 = 217’ for some real number a > 0. Now, we claim that U is dense in M 3. If it is not, then M 3 — U has nonempty interior. From the definition of U, we know that the squared mean curvature function, u and 11’ vanish identically on the interior of M 3 —— U. On the other hand, (3.4.8) says 64 that this is impossible due to the continuity of the squared mean curvature function. Thus, U must be an open dense subset of M 3. Solving (3.4.8) yields 11 = 1:42stan + b, -\}3) for some constant b. By applying a translation in a: if necessary, we obtain \/2 «2 1 3.4.9 =—-———d—-— k 192—. ( ) #(zv) aS(ax.), 2 From (3.4.5) and from our assumption on the the principle curvatures, we know that the second fundamental form h of M 3 in IE4 satisfies 60_ 23_12 23_122 ' ' 3 a a a (9 8 M29}, 5(5) = (1982,55) " Mag, 66) " 0. where E is a unit normal vector field of M 3 in LB4 and p is given in (3.4.9). Applying (3.4.6), (3.4.10) and the equation of Codazzi, we may obtain [1’ f = p f’ . Therefore, (3-4-11) f($) = 014113), for some nonzero constant c. On the other hand, using (3.4.6) we can compute the sectional curvature K23 of the plane section spanned by {8/845, 6/60}. We also can compute K23 by using the equation of Gauss. By comparing these two different expressions of K23, we find u4f2 = 4(1 — f’2). Thus by using (3.4.9) and (3.4.11), we find (3.4.12) f(x) = {333({33 k), k = [\DIH From (3.4.5) and (3.4.12) and Lemma 3.2 we know that M 3 is an open portion of 65 the Whitney 3—sphere W: for some a > 0. (3.4.5), (3.4.6), (3.4.9), (3.4.12) and the formula of Gauss imply that the immersion x satisfies the following system of partial differential equations: (3.4.13) (3.4.14) (3.4.15) (3.4.16) (3.4.17) (3.4.13) 62x a fi fi \/2- 0x a 3 fl W _fiCd(—;$)Sd(7$)nd(7 )5— de (_a 113K, 32x 2 (92x . 3X _862 _ cos (lb—6452 +sm¢cos¢5fiy Solving (3.4.18) yields (3.4.19) x(a:,9, (b) = B(:r, l9) c0343 + 0(1), (15), for some function B(:z:, 0), C(zr, (b) of two variables. Substituting (3.4.19) into (3.4.16) 66 yields 63 _ v2 «2 v2 52:— — Tcs(—a—x)nd(—x)B, (3.4.20) 0C \/2 \/2 \/2 0C 830$ = 7cs(7$)nd(—a—x)-a—¢. Solving the above system yields (3.4.21) B(:r, 6) = G(0)sd(—\—c/l——gx), C(x, (1’)) = F(¢)sd(—\—/d—2x) + H(:1:). Combining (3.4.19) and (3.4.21) gives (3.4.22) x(a:, <15, 0) = G(6)sd(—\:—§z) cos ¢ + F(¢)sd(—\:—§-$) + H(:c). By taking the partial derivatives of (3.4.14) and (3.4.15) with respect to ()5 and 0 respectively, we find (3.4.23) F"'(q§) + F'(¢) = 0, G"'(6) + (7(0) 2 0. From (3.4.22) and (3.4.23) we find (3.4.24) x(:z:, 65, 9) = sd(¥$)(c1 cos qb cos 0+c2 cos (6 sin 0+03 cos ¢+c4 sin (b) +A(:1:), where c1, ..., c4 are constant vectores. Substituting (3.4.24) into (3.4.14) and using (3.4.13) yields (3.4.26) «2 2 x(:c, (b, 0) = sd(—‘:::r)(cl cos ()5 cos 0+C2 cos cf) sin 0+c3 cos 6+c4 Sin ¢)+c5 fsd3(—(;—a:)d:1:, where 01,...,c5 are constant vectors in IE4. 67 By choosing suitable initial conditions, we obtain (3.1.5) from (3.4.26). Conse- quently, up to rigid motion of E4, the immersion x is given by (3.1.5). The converse can be verified by straight-forward computation. E] 3.5 Exact Solutions of Differential Equations of Pi- card Type For the proof of Theorem 3.2 and Theorem 3.3, we need the exact solutions of some differential equations with Jacobi’s elliptic functions in their coefficients. The results obtained in this section seem to be of independent interest in themselves. Proposition 5.1. For any real number a > 0, the general solution of the second order differential equation: (3.5.1) y"($) + 2asc(aa:)nd(ax)y'(a:) — y(:1:) = 0 is given by y(x) = c1y1(a:) + egg/20:) with 311(33) = \/a2kl2_ cn2(a$) cos (lg—Ia: +2———1—————:;_;k(:_k’_ 111% + 2aZ(’7)23)), and 312(3) = (/a2kr2_ cn2(a:r) sin (kin: +2——1——1\/\/C;::_k’(1n w + 2aZ(7)a:)), where k = Va? — 1/(\/2a) and k' = \/a2 + 1/(x/2a) are the modulus and the comple- mentary modulus of the Jocobi elliptic functions, 9 the Theta function, Z the Zeta function, and ”y = sn‘1(\/—1/(ak2)). Proof. The trick to solve (3.5.1) is to make two key transformations. First we make 68 the transformation: (3-5-2) 31(27) = f($) exp(\/-_19(IL‘)). where f (:13) and g(:1:) are real-valued functions. Then (3-5-3) 21’ = (f’ + \/——1fg’)exp(\/-_lg). (36.4) 1": (f” - f (9’)2 + \/-_1(2f'9’ + fg")) ex136/719)- substituting (3.5.3) and (3.5.4) into (3.5.1) we get by taking the imaginary part (3.5-5) f (2)9”(2) + 2(f'(1?) + aSC(0413)<1n(a£l?)f (17))9'($) = 0. and (3-5-6) f"($) - f(3:)(g'(:1:))2 + 2060(afvldn(antlf'(i1?) — f($) = 0- Equation (3.5.5) can be written as (3.5.7) (ln g'(:r))' = -2(1n f(:z:))' + 2(ln cn(az))' which yields (3.5.8) g'(:2:) — for some constant a. 69 Substituting (3.5.8) into (3.5.6) we obtain a second order nonlinear equation: (3.5.9) f"(:I:) + 2asc(a:c)dn(a:r)f'(:1:) — f(:1:) = M. f3($) We make the second transformation by putting (3.5.10) f(:1:) = h(sc), u = cn2(a:r). From (3.5.9) and (3.5.10) we obtain another nonlinear equation: (3.5.1,) (aadn (ax)sn2(2h(u)h"(u) — (6(a)?) +a2(k2cn2(ax)sn2(az) — dn2(aa:))h(u)h’(u) = h2(u) + azcn4(a:c), which, by (3.2.8), is equivalent to the following nonlinear equation: 209141 - 1006214 + l6’2)h(U)h"(U) - 02062142 + k'2)h(U)h'(U) —a2(k2u + k'2)(u — u2)h’2(u) = h2(u) + 012112. (3.5.12) If h = b — cu is a linear function in u with constant coefficients, then (3.5.12) becomes (3.5.13) b(b — azck'z) + 2c(a2clc'2 — b)u + (a2 + c2 — a2bcl~c2 + a262(k2 — 16'2))u2 = 0. It is straight-forward to verify that (3.5.13) holds if and only if a = ca2kk’ and b = azck’z. This shows taht h(u) : C(a2k’2 — u) is a solution of (3.5.12). If we choose 0 = 1, we obtain a = azkk’ and (3.5.14) f(:1:) = \/a2k’2 — cn2(a:r). 70 By computing (3.5.8) and (3.5.14) we obtain (3.5.15) g(z)= [0 62366112612) dz, a2 k’2 — cn2(aa:) On the other hand, from (3.2.8) and (3.2.9) we have azkk’cn2(aa:) _ k’ _ azk’3sn2(a$) azlrc’2 — cn2(a:1:) _ k k(a2k2 + sn2(a$)) _ _k_’ _ ak’2 cn(7)dn(7)sn2(a$) ) _ k (/a2k4 + 1 1 — k2sn2(7)sn2(ax) (3.5.16) I ’2 — k— —- ——aL—-(sn(a:1: + )+ sn(aa: — ))sn(a:1:) k 2\/a2k4 + 1 7 7 \/—1a2k'2 — I + W(W(sn(am + ’7) + sn(aa: — 7))sn(7)sn(a:r)), where sn(7) = \/—1/(ak2),dn(7) = k’/k,cn(7) = k’2/k2. Combining, (3.2.14) and (3.5.16), we obtain ank’cn2(a$) _ k’ \/—1a2k’2 3. .17 — — —— ( 5 ) azli"2 — cn2(a:1:) k + 2W (Z(aa: — 7) — Z(aa: + ’7) + 2Z('y)). From (3.5.15) - (3.5.17), we find go): 24+ +,——‘/\/_k—'c ”126 —7)Z(aa: + 11+ 2Z(’7))dx (3.5.18) +_\/:_;_ak’2 9(ax— ) _ _+2\/a2k4+—_1119(1W(+7) + 26‘2””), where we applied (3.2.13) and the fact that O is an even function. Therefore, by (3.5.2), (3.3.14) and (3.5.18), we conclude that the functions y1, y2 defined in Propo- sition 5.1 are independent solutions of (3.5.1). Consequently, the general solution of (3.5.1) is given by the linear combination of y1,y2. El 71 Proposition 5.2. For any real number a > 1, the general solution of (3.5.19) y"(:r) + 2asc(a2:)dn(a2:)y'(a:) + y(:c) 2: 0 is given by y(:c) = c1y1(2:) + 02y2(:1:) with k' 1 y1($) : \/a2k’2 + CD2(CL.’L‘) COSh (E — 5 n W and y2(3:) = \/a2k’2 + cn2(a:1:) sinh (g — éln w -— aZ(7)a:) where k 2 V0.2 + 1/(\/2a) and k’ = x/a2 - 1/(\/§a) are the modulus and the comple- mentary modulus of the Jacobi elliptic functions and 7 = sn‘1(1/ak2)). Proof. This can be proved by using the same trick given in the proof of Proposition 5.1. After making the key transformation (3.5.2) for (3.5.19), we obtain (3.5.5) and (3.5.20) f"(:2:) — f(:t)(g'(:z:))2 + 2asc(a:r)dn(a:z:)f'(:c) + f(:c) = 0. Solving (3.5.5) yields , _ acn2(a2;) (3.5.21) 9 (2:) — —f2($) . for some constant a. By substituting (3.5.21) into (3.5.20) we obtain (3.5.22) f"(a:) + 2asc(ax)dn(aa:)f'(:1:) + f(:1:) = W. After making the second key transformation f (2:) : (/h(u) with u = cn2(a.r) for 72 (3.5.22), we find (3 5 23) 2a2u(1 — u)(k2u + k'2)h(u)h”(u) — a2(162u2 + k’2)h(u)h’(u) —a2(k2u + k'2)(u — u2)h’2(u) = -—h2(u) + azuz. It is then straight-forward to verify that a linear function h = b — cu is a solution of (3.5.23) if and only ifb = -—a2ck’2 and a2 = 62(1—(14)/4. In particular, if we choose 0 = —-1, we obtain _ 2kkl 2 (3.5.24) 1(4) = (£2142 +cn2, g’(:v) = QZ+C§§(§;‘§’- On the other hand, from (3.2.8), (3.2.9) we have azkk’cn2(a$) __ k_’ _ azk’zsmasc) azk’2 + cn2(a:r — k 13(an2 — sn2(a.’r)) _ k_’ _ cn(7)dn(1)sn2(a$) — k 1 - k2sn2(ax)sn2(a:r) (3.5.25) k, 1 = h — §(sn(a:1: + 'y) + sn(aa: — 7))sn(a:r) = % _ g(k2(sn(ax + 7) + sn(aa: — 7))sn(’y)sn(a:r)). where sn(y) = 1/(ak2),dn(’y) = k’/k,cn(’y) = (k’/k)2. Combining (3.2.14) and (3.4.25), we obtain aZkk’cn2(a:c) _ k’ a (3.5.26) a216,? + cn2(a$) — I — §(Z(aa: —- 'y) — Z(aa: + 7) + 2Z(7)). 73 From (3.5.24) and (3.5.26), we have go) — % —§ imam — 7) — Z(ax + "1) + 2261112: 3.5.27 1 ( ) = k—x — ll ____O(a:c _ 7) — aZ('y):1:, where we applied (3.2.13) and the fact that O is an even function. Therefore, by (3.5.2), (3.5.24) and (3.5.27) we conclude that 21 (:17) : (/a2k’2 + cn2(a.1:) exp ( — [122:1 + 21D g——————E:::+ — Z; + aZ(7):1:) is a solution of (3.5.19). By applying the method of reduction of order, we know that 22(37): \/a2k’2 + cn2(aa:) exp (k—g; _ lln @(cm: _ 7) k 2 O(a:1: + ’y) — aZ(’7):1:) is a second independent solution of (3.5.19). Consequently, the functions y1,y2 defined in Proposition 5.2 are two independent solutions of (3.5.19). Hence, the general solution of (3.5.19) is given by the linear combination of y1,y2. [2] Proposition 5.3. For any real number 0 < a < 1, the general solution of (3.5.28) y ".(1: )+ 2akcn(::z: )sd(::1:y :1:)y ':1:() + y(:1:) = 0 is given by y(:1:) = clyl(a:) + 02y2(:1:) with y1(:1:) = \/k2dn2(%:r) — a2k’2 cos (k'x— a} H 99 FIDR‘IQ B H :i I fi t-J IQ B4 :9: H v and 312(37) = \/k2dn2(%x) - a2k’2 Sin (k'a: — ' —1 1n Egg—ii; ._ ,/_1 E 74 where k = x/2a/sqrt1 + a2 and k’ 2 v1 — a2/\/1 + a2 are the modulus and the com- plementary modulud of the Jocobi’s elliptic functions and 7 = sn‘1(k/a). Proof. This can also be proved by using the same trick. After making the key transformation (3.5.2) for (3.5.28), we obtain (3.5.29) f(rc)g”(:v) + 2012) + akcn(%x)sd(%x)f(m))g’(rv) = o, and (3.5.30) f”(:z:) — f(:1:)(g'(:1:))2 + 2akcn(%zz:)sd(%:r)f’(a:) + f(x) = 0. Solving (3.5.29) yields (3.5.31) 9'01?) _ _, for some constant 01. By substituting (3.5.31) into (3.5.30) we obtain ,, a a , _ azdn4(9:r) (3.5.32) f (2:) + 2akcn(;:1:)sd(;$)f (2:) + f(:1:) — W. After making the second key transformation f (1:) = (/h(u) with u = dn2(a:r/ k) for (3.5.30), we find 2a2u(1 — u)(u -— k'2)h(u)h"(u) + (12(16’2 —— u2)h(u)h’(u) —a2(u — k’2)(u — u2)h’2(u) = —k2h2(u) + a2k2u2. (3.5.33) It is then straight-forward to verify that a linear function h = b + cu is a solution of (3.5.33) if and only if b = —a2ak'2/lc2 and a2 = a2k’2. In particular, if we choose 75 c 2 162,0 = azk”, we obtain _ azk’dn2(%:1:) ’ 14261.2(?) — a2k'2' (3.5.34) f(:1:) = \/kzdn2(%x) — 62km, g'(:c) On the other hand, from (3.2.8), (3.2.9) we have a22k'dn2(%:1:) : k’ _ \/_—1k2cn(7)dn(7)sn2(%x) kzdn (fix) — a2k’2 1 — k2sn2(%x) (3.5.35) a a a a = k’ _ ,/_. ._ 2 _ .. _ _ 12k(k (sn(kx+:y) +sn(k:1: 7)sn(7)sn(k:1:)), where sn(y) = k/a,dn(7) = k'2, cn(y) = \/-1k’. Combining (3.2.14) and (3.5.33), we obtain azk’dn2(%:1:) 3.5.36 ( ) k2dn2(7‘:-:1:) — 6211'? = k' — flfimfiz — 7) — 26:4 + 7) + 2Z6». From (3.5.21), (3.5.24) and (3.5.34), we have 9(2) = k'x — «:15; f0 $12122: — 7) — min + 1) + 22(1)2=)drr 16 (3.5.37) \/:1‘ 9G3; _ ,7) a 2 ln — \/——1 = k :1: — O(%x + 7) ;Z('y)x. Therefore, by (3.5.2), (3.5.34) and (3.5.37), we conclude that the functions y1,y2 defined in Proposition 5.3 are independent solutions of (3.5.28). Hence, the general solution of (3.5.28) is given by the linear combinatin of y1, yr). 1:] Corollary 1. For any real number a > 0, the general solution of a2—1 Via (3.5.38) z"(:r) = (a2 + 1)nc2(a:1:, k)z, k = is given by z(:1:) = clcn(aa:, k)y1(:1:) + cgcn(aa:,k)y2(:1:) where y1(:1:),y2(:1:) are defined 76 in Proposition 5.1. Proof. This follows from Proposition 5.1 and the fact that equation (3.5.38) can be obtained from (3.5.1) by making the transformation y(1:) = cn(ax, k)z(:1:). [:1 Corollary 2. For any real number a > 1, the general solution of Va2+1 (3.5.39) z"(:1:) = (a2 — 1)nc2(az, k)z, k = x/2a is given by z(:1:) = clcn(a:1:, k)y1(:1:) + cgcn(a:1:,k)y2(:1:) where y1(:r), y2(:1:) are defined in Proposition 5.2. Proof. This follows from Proposition 5.2 and the fact that equation (3.5.39) can be obtained from (3.5.19) by making the transformation y(2:) = cn(ax, k)z(:1:). E] Corollary 3. For any real number 0 < a < 1, the general solution of (3.5.40) z"(::)=(az—1)nd2(fix,k)z, 1.: «2... k.____./r——62 k \/1+a2 x/l-i-a2 is given by 2(32) 2 cldn(a:1:/k, k)y1(:1:) + cgdn(a:1:/k, k)y2(:r) where y1, y2 are defined in Proposition 5. 3. Proof. This follows from Proposition 5.3 and the fact that equation (3.5.40) can be obtained from (3.5.28) by making the transformation y(:1:) = dn(ax/k, k)z(:1:). E] Remark 1. Conversely, since d (-)(u-’y) _ @(In W - Z(u — 7) — Z(au — ’7) (3.5.41) = k28n(U)Sn(1)(sn(u - 7) + sn(u + 7)) - 2Z(or) _ 2k2cn(i)dn(i)sn(7)sn2(u) _ 1 — kzsn2(7)sn2(u) - 2Z(7), it is straight-forward to verify that two functions y1,y2 defined in Proposition 5.1 (respectively, in Proposition 5.2 and 5.3) are indeed independent solutions of (3.5.1) 77 (respectively, of (3.5.19) and of (3.5.28)). Remark 2. In 1879, E. Picard discovered in [25] a method for solving the differential equation: (3.5.42) y”(:1:) + nk2cn(:1:)sd(:r)y’(:1:) + erg/(2:) = 0, where n is a positive integer and 01 a constant. Although equation (3.5.42) is quite similar to the equations (3.5.1), (3.5.19) and (3.5.28), unfortunely, Picard’s method does not apply to these equations. 3.6 Proof of Theorem 3.2 Let x : M" —> 5"“(1) C 1E"+2 be an isometric immersion of a conformally flat n-manifold with n > 2 which satisfies Chen’s equality (2.1.2). Then, by Lemma 3.2 and a well-known result of E. Cartan and J. A. Schouton on conformally flat hypersurfaces(cf. p.154 of [6]), we know that M n has a. principle curvature with multiplicity at least n — 1 for n = 3 as well as for n 2 4. Thus, by applying (2.3.2), we know that either (i) the principle curvatures of M n are given by A1 = 0,/\2 = = An = 11, or (ii) 11 = 3 and the principle curvature of M3 are given by A1 = A2 = 11/ 2, A3 = ,u # 0. We treat these two cases separately. Case (i): A1 = 0 and A2 = : An = 11. If u = 0 identically, then M" is totally geodesic in 8"“(1). Now, suppose that M " is not totally in 8"“(1). Denote by U the open subset of M n on which the mean curvature function is nonzero. Then U is non-empty. We will prove in the following that U is the whole manifold M n in this case. We denote by ’D and ’Di the distributions on the open subset U spanned by {el} and {e2, ..., en}, respectively. 78 Now, by puttingj = Li = 2,3, ...,n in (2.3.5), we get w](e1) = 0 which implies that integral curves of ’D are geodesics. Also by using the first equation of (2.3.5), we find (3.6.1) e211 2 = emu = 0. It is easy to see from (2.3.4) that (3.6-2) 411(6)): 0. 1 s z‘ aéj s n, which implies that Di is integrable. Consequently, there exist local coordinate sys- tems{:1:1,a:2,...,:1:n} such that (9/8362, ...,6/8xn span Di and e1: (9/33 with :1: = 3:1. From (3.6.1), we know that )1 depends only on 0:, i.e., u = 11(2). Choosing i = 1 for the first equation in (2.3.4), we get u’(:1:) = —u(x)w{(ej) for anyj Z 2. Thus (3.6.3) Vejel = :wf(ej)ek = w{(ej)ej = —(ln ,u)'ej. k=2 Using (3.6.3) we know that each integral submanifold of ’D‘L is an extrinsic sphere of 8"“(1), i.e., it is a totally umbilical submanifold with nonzero parallel mean curvature vector in 5"“(1). Hence, the distribution Di is a spherical distribution. Therefore, by applying a result of Hiepko [21] (cf. Remark 2.1 of [20]), we know that U is locally the warped product I Xftr) Sn‘1(1), where f(:1:) is a suitable warped function. Therefore, the metric on U is given by (3.6.4) g 2 do:2 + f2(:1:)g0, where go is the metric of ism—1(1). In particular, if we choose the spherical coordinate 79 system {u2, ...,un} on S"“1(1), then we have (3.6.5) g = d2:2 + f2(:1:)(du§ + cos2 uzdug + + cos2 112... cos2 un_1du?,). By (3.6.5) we obtain (3.6.6) o a f’ a a ,6 V555”, V253... 7‘5; V-u—a_ ‘ffif’ a a ___=_ — <' Vfiauj (tanu,)auj, 2 z 0 on U. Hence, by the continuity of the squared mean curvature function H 2 = (n — 1)2u2/n2, we know that the mean curvature function is nowhere zero and therefore U is the whole manifold M ". Consequently, 81 (3.6.4) and (3.6.12) imply that M " is isometric to an open portion of the Riemannian manifold B: which was defined in Section 3.1. By applying (3.6.5), (3.6.6), (3.6.7), (3.6.12) and the formula of Gauss, we conclude that the isometric immersion x : M n —> Sn'1(1) C EH2 satisfies the following system of partial differential equations: 82x (3.613) "a? Z “X, Bzx Bx . (3.6.14) 828112- — -—(tan :1:)8—uj, ] — 2,3, ...,n, 62x (9x . .1 = — t i —a 2 S . .1 (3 6 5) 8122an (anu )Buj 1, < J 2 (3.6.16 9—3;— = 1(1— a2) sin 22:9E + a\/1— a2 cos 2:6 —- ((1 — a2) cos2 2:)x, 6112 2 62: 82x 2 32x 1 , ax . (3.617) 671?; : (COS Uj)'a—UJ§ + §(Sll’l 2Uj)5;;, ] = 2, ...,n — 1. Solving (3.6.13) yields (3.6.18) x = P(u2, .., un) sin2: + Q(u2, .., un) cos :3, foe some IE"+1-valued functions P = P(u2, ..., un) and Q = Q(u2, ..., 11"). Substituting (3.6.18) into (3.6.14), we know that P is a constant vector, we denote it by c1. Thus (3.6.19) x = 01 sin2: + Q(u2, ..., 11") cos 2:, 82 Substituting (3.6.19) into (3.6.15) with 2' = 2, we obtain 6%) 86) _ ._ (3.620) aujauz + (tan U2)a—uj — 0, j —- 3, ..., n which implies a .. (3.621) ‘82 + (tan U2)Q = $2013), “2 for some function $2 = 62(212). Therefore, by solving (3.6.21), we have (3.6.22) Q = ¢2(u2) + Q3(U3, ..., un) cos U2 for some function 452 = ¢2(u2) and Q3 2 Q3(u3, ..., un). Similarly, by substituting (3.6.19) and (3.6.22) into (3.6.15) with z' = 3 andj > 3, we find (3-6-23) Q3 = $3013) + Q4(U4. ..., Un) 005713, for some function (253 = (153(212) and Q4 2 Q4(u4, ...,un). Repeating such procedure n — 2 times, we obtain Q = (152%) + Q3(U3. un) cos u2, Q3 = ¢3(U3) + Q4014, "'aun) COS U3, (3'624) Q4 : $4014) + Q5016, "'a un) cos “’4: Qn—l : ¢n—1(un—1) + ¢n(un) COS un—la 83 with ¢n(un) = Qn(u,,). Substituting (3.6.24) into (3.6.19) we get (3.6.25) x = Cl sin :1: + (15204.2) 008$ + ¢3(U3) COS U2 Cosx + +¢n_1(un_1) cos U2... cos un_2 cos a: + (15,,(un) cos U2... cos un_1 cos :1:. Substituting (3.5.25) into (3.5.17) with j = n — 1, we obtain (5.6.26) qfiflun) + ¢n(un) = cos un_1¢2_1 + sin un_1¢;,_1(un_1), which implies that (3.6.27) (bflun) + ¢n(un) 2 kn, (3.6.28) cos un_1¢x_1(un_1) + sin un_1q5;,_1(un_1) = k,,, for some constant vector kn. Solving (3.6.27) yields (3.5.29) (15,, = c,,+1 sin un + cn+2 cos u" + kn, for some constant vectors cn+1,,cn+2. Combining (3.6.25) and (3.6.29) we obtain x = 01 sinx + (152(u2) cosa: + (153(u3) cos U2 cosrr (3-6-30) +¢n_1(un_1) cos U2... cos un_1 cos 33+ +cn+1 cos 1L2... cos un_1 sin u” cos a: + cn+2 cos U2... cos un cos :1:, for some constant vectors cl, ..., an”. 84 Now, we choose the initial condition at 0=(0, ..., 0) as follows: x(0) = (162 + V1 -— a2en+2, 23(0) 2: 61, (3.6.32) 3x 3x 672(0) = \/1 — a262, au = \/1 — a2en+1, where {61, ..., en”) is the natural coordinate basis of 1153"”. Then, by applying (3.6.31) and (3.6.32), we obtain (3.6.33) 01 = 61, 82 = (162, C3 = \/1— a263, ..., cn+2 = \/1 — a26n+2. Consequently, by (3.6.31) and (3.6.33) we conclude that, up to rigid motions of 1173"”, the immersion x is given by (3.1.5) in Theorem 3.1. It is clear from (3.1.5) that the immersion x can be extended to the two point compactification 132 of B2. Case (ii): A1 = A2 = p/2 and A3 = u 75 O. In this case (2.3.5) yields (3.6.34) r3, = A3 = 0, 61/1 = e2” = 0, 111,, = r33 = 0. Denote by .7: and fl the distributions spanned by {81, 82} and {63}, respectively. By (3.6.34) we know that the integral curves of fl are geodesics and the distribution .7 is integrable. Consequently, there exist a local coordinate system{a:, u, 22} such that f is spanned by {B/Bu, 0/0v} and 83 = 0/033. From (3.6.34) we know that )1 depends only on :1:, i.e., p = h(zr). Also, from (2.3.8) and (3.6.34), we have (3.6.35) w; = w , to; = w . 85 Using (3.6.35) we obtain (3.6.36) we3 = figej, j = 1,2. Therefore, each integral submanifold of .7: is an extrisic sphere of 84(1). Hence, by applying a result of Heipko, we know that M 3 is locally the warped product I x fix) 82(1), where f (x) is a suitable warp function. In particular, if we choose the spherical coordinate system {0, (25} for 82(1), we have (3.6.37) 9 = dx2 + f2(1:)(d¢2 + cos.2 6662). By computing do); and by using (3.6.35) and Cartan’s structure equations, we find W17) (3.6.38) u"(x) + 2 + p(a:) = 0. Let 2/2 = 2u(x). Then (3.6.38) becomes 1/2”(a:)+21/J3(1:)+2p(:r) = 0. Hence, by applying Lemma 5.3 of [10], we obtain 2 _ a2 — 1 (3.6.39) 13(3) 2 (/2(a — 1)cn(a;z:,k), k — x/2a , where a > 1 is real number. Now, by applying (3.6.37), (3.6.39) and the equation of Codazzi, we obtain ,u’ f =- ,u f ’. Therefore (3.6.40) 16:) = cm), for some IlOIl-ZGI‘O constant C. On the other hand, using (3.6.37) we can compute the sectional curvature K23 of 86 the plane section spanned by {ii/8(1), {9/30}. On the other hand, we may also compute K23 by using the equation of Gauss. Comparing the two different expressions of K23 so obtained, yields 1 (3.6.41) 0260 = 1 — c2p2 -— 102434. Substituting (3.6.39) into (3.6.41) we have (:2 = 1/(a4 — 1). Therefore (3.6.42) f = figijcnmx, k), u = 2(a2 — 1)cn(a:r;, k), k = $2.; 1. (5.6.37) and (5.6.42) imply that M 3 is an open portion of the warped product manifold I Xfi/mcmax) 52(1) which is isometric to P3, first introduced by B. Y. Chen in [10] (also see [13]). (3.6.37), (3.6.39), (3.6.42) and the formula of Gauss imply that the immersion x satisfies the following system of partial differential equations: (3.6.43) % = (/2(a2 — 1)cn(a.:z:, k)€ — x, (92x 2a 5255 = Clz—H-cn(azr)dn(a$)sn(am)5; (3.6.44) 2(02 _ 1) 3 2 —a2—+—1—cn (ax)€ — a2 +1cn (ax)x, 2 (9379; = a223- 1 cos2 ¢cn(ax)dn(a:r)sn(a$)% + sin ¢cos¢g—: (3.6.45) 2(a2 — 1) +.__—___ a2 +1 cos2 ¢cn3(a:1:)£ —— cos2 q§cn2(a:1:)x, a2+1 62x (3.6.46) 638$ = —asc 82x (3.6.47) 82:80 — —asc 82x . .4 (3 6 8) (908(1) (ax)dn(a:1:) g, (ax)dn(a:z:)gx—, 6% 60 = —- tan (fig—3. By taking the partial derivative of (3.6.44) with respect to d) and by applying the equation of Weingarten, we obtain 3 (3.6.49) 6 x by virtue of (3.2.8), (3.2.10) and (12—1 2a2 ’ (3.6.50) k2 = Solving (3.6.49) yields (3.6.51) 0?— 8x _%, x(:r, (15, 6) = A(:1:,0) sin (3 + B(:1:, 6) cos (b + C(x, 6), for some [fi-valued functions A,B,C of two variables. Substituting (3.6.51) into (3.6.46) yields 8A (3.6.52) E Solve the above, we have (3.6.53) 2 —asc(a:r)dn(a:13)A, A(a:, 0) = E(6)cn(a:z:), 8—15: 33: = —asc(a2:)dn(a:z:)B. B(:r, 0) = D(0)cn(a:1:) 88 for some IEfi-valued functions E, D. Thus, (3.6.54) x(2:, ()5, 0) = E(0)cn(a:r) sin (15 — D(6)cn(a:c) cos 45 + 0(2), 6). Substituting (3.6.54) into (3.6.47) yields 2 8 C - —asc(aa:)dn(a$)a—C— 611369 — 86 ' (3.6.55) By solving (3.6.55) we find C(cc,6) = G(0)cn(a:z:) + K(:2:) for some Ei-valued functions G(0) and K (3:) Thus, (3.5.54) gives (3.6.56) x(x, (b, 6’) = E(6)cn(aa:) sin (b + D(0)cn(a:z:) cos <15 + G(6)cn(a2:) + K(:r). Substituting (3.6.56) into (3.6.48) yields E’ (6’) = 0(0) = 0. Thus, E and G are constant vectors in E5. Consequently, from (3.6.56) we know that x takes the form: (3.6.57) x(a:, <15, 9) = clcn(a2:) sin <15 + D(6)cn(a:c) cos Q5 + F(;r). where c1 ia a constant vector. From (3.6.43) we have (3.6.58) 6: 1 (82x + x). ,/2(..2 — 1)cn(a$) 6102 By (3.6.45), (3.6.57), (3.6.58) and a long computation, we obtain (5.6.59) F”(:I:) + 2asc(ax)dn(a:r)F'(:z:) — F(:1:) = 0, (3.6.60) D”(0) + 19(9) = 0, 89 by virtue of (3.2.8), (3.2.10) and (3.6.50). Solving (3.6.60) yields (3.6.61) D(6) = c2 cos 6 + c3 sin 6, for some constant vectors 02, c3 in E5. Therefore, by applying Proposition 5.1, (3.6.57) and (3.6.61), we have x(:1:, d), 6) 2 c1 sin ¢cn(a:z:) + C2 cos ()5 cos 6cn(a:1:) + 03 cos ((5 sin 6cn(a2:) F1k12 +c.4\/a2k’2 — cn2(a$) COS(-k-$ + g—m—TZ—(ln 9(62: + 7 ) \/—1k’2 9 _ +65 \/a2kl2 ._ cn2(a:z:) sin(7c-:r + 2 ;———21———ij (1n —O(::- + Z; (3.6.62) where O and Z are the Theta function and Zeta functions. Now, if we choose the initial conditions at 0=(0, 0, 0) as follows: 6x 1 6x 1 (3-6-63) X(0)= am) = gig-Kl, 55(0) = $63, 6x a(€2—k—, + 64), Egan = 65, where {61, ..., 65} is the standard basis of E5, then we obtain 1 3H4 az—aa (66) c ak’e a =1,...,5. Therefore, up to rigid motions, the immersion x is given by (2.1.7) in Theorem 3.2. The converse can be verified by straigh-forward but long computations. D 3.7 Proof of Theorem 3.3 Let x : M n —> H "+1(—1) C 1E}1+2 be an isometric immersion of a conformally flat n-manifold with n > 2 which satisfies Chen’s equality (2.1.2). By Lemma 3.2 and a 90 result of Cartan and Shouten, either (a) the principle curvatures of M " are given by A1 = 0, A2 = = An = p, or (b) n = 3 and the principle cuevatures of M 3 are given by A1 = A2 = p/ 2, A3 = ,u aé 0. We treat these two cases separately. Case (a): A1: 0 and A2 = = An = u. If p = 0 identically, then M " is totally geodesic. This yields Case 1 of Theorem 3.3. Now, suppose that M " is not totally geodesic in H "+1(-—1). Denote by U the open subset of M n on which the mean curvature function is nonzero. We shall work on this open subset of M " unless mentioned otherwise. Clearly, ,u aé 0 on U. We denote by D and Di the distributions on the open subset U apanned by {81} and {82, ...,en}, respectively. Then, as in the proof of Theorem 3.2, we can prove that integral curves of D are geodesics and Di is integrable. Thus, there exist a local coordinate system {3131,22, ...,xn} such that 6/622, ..., 6/6rrn span Di and 61 = 6/63: with :1: = :61. Also, using (2.3.5) we can prove that ,u depends only on :1:, i.e., ,u = h(ay). If we choose 2’ = 1 for the first equation of (2.3.4), we get p’(:c) = —/3($)w{(ej) for anyj 2 2. Thus 11 (3.7.1) V8181 = wa(ej)ek = w{(ej)ej = (ln p)'eJ-. k=2 Hence, each integral submanifold of DL is an extrinsic sphere of H "+1(—1), i.e., the distribution Dl is spherical. Therefore, U is locally a warped product manifold I x [(x) N"‘1(5), where f (:r)(> 0) is the warp function and N "'1(E) ia a Riemannian space form of constant sectional curvature 6. Since N "’1(E) is of constant curvature, it is conformally flat. Thus, there exists a local coordinate system {132, ..., un} such that metric tensor of N "”1(E) is given by (3.7.2) go = E2(du§ + dug + + d213,). 91 With respect to the coordinate system {23, U2, .. gun} on I x f(:v) N ” 1(c ), we have (3.7.3) 9 = dx2 + f2E2(du§ + + dug). From the above, we have 8 _ f’ 6 E.- a E,- 6 V%6x_0’ V6 6 _af6 a..au,-E'a—u;+73‘au; (3.7.4) 6 6 +E E 6 2 _ _ __’:___ V8162“: ..,-ffE 62: +E 13:,é; E Buk’ for distinct z',j, k(2 S i,j, k S n), where E,- = 6E/6u,. Codazzi’s equation, (3.7.4) and our assumption on principle curvatures imply (3.7.5) 7. = ; for some constant a > 0. Also, from (3.7.4), it follows that the sectional curvatures K12 and K23 of M " associated with the plane section spanned by 6/623, 8/6132 and B/Bug, (9/6233 are given by — f” / f and (E — f’2)/f2, respectively. On the other hand, Gauss equation yields K12 = —1 and K23 = —1 + 13- Comparing these facts with (3.7.5), we get (3.7.6) f” — f = 0, f2 — f’2- _ a2 — 6. Solving the first equation of (3.7.6) yields f (:r) = 01 cosh a: + c2 sinh :1:, where c1, 62 are constants, which we can write as either f (2:) = a sinh(:r+b) or f (:1:) = a cosh(a:+b), for some constant a, b. Thus , by applying a translation in x if necessary, we have f (:1:) = asinha: or f (2:) = acosh :1:. We consider these two cases separately. Case (1): f (:1:) = acosh :1:. In this case, the second equation of (3.7.6) yields 92 a2 = a2 — 6. Thus, after applying a scaling on E if necessary, we have (3.7.7) 5 = 1, f(x) 2 Va? -—1cosha:, u = sechx, a > 1, a _ or (3.7.8) 5 = 0, f(:1:) = cosh :1:, u = sechx, or (3.7.9) 6 = —1, f(:z:) = Mcosh :1:, ,u z a2 +188Ch$, a > 0. Case (Li): 6 = 1,f(;r) 2 Va? — 1cosh 3:,” = fisechxfi > 1. In this case, the open subset U is the whole manifold M ". Therefore, M " is an open portion of the warped product manifold A: = R XJE’TT Sn‘1(1). By choosing spherical cosh :1: coordinates {71.2, ..., an} on Sn'1(1), we obtain (3.7.3) and (3.7.4) with sechcr. (3.7.10) f 2 Va2 - 1cosh 3:, u = fin Therefore, the equation of Gauss implies that the isometric immersion x satisfies the following system of partial differential equations: (3.7.11) ————- = x, (3.7.12) 93 62x 3.7.13 —-—— = —t ,-—, 2 <' ', ( ) 611,611,- anu Bu]- — Z < J 62 1 — 2 6x (3.7.14) —-—§ 2 ___a sinh 223— + aVa2 - 1cosh 336 + ((a2 — 1) cosh2 :1:)x, 8112 2 8:1: 62x 2 82x 1 . . (3.115) au§+1 = COS 117—6713 + g Sln 2Uja—u;, ] = 2, ...,n — 1. where f is a unit normal vector field of M " in H "+1(—1). Solving equation (3.7.11) yields (3.7.16) x = P(u2, ..., un) sinha: + Q(u2, ..., un) cosh :1:, for some IE’1'+1-valued functions P = P(u2, ..., an) and Q : Q(u2, ...,un). By substi- tuting (3.7.16) into (3.7.12), we know that P is a constant vector, denoted by c1. Thus (3.7.17) x 2 c1 sinh :1: + Q(u2, ..., un) cosh :6. Substituting (3.7 .17) into (3.7.13) with 2' = 2 yields (3.7.18) 3% + (tan 62m = (132072), for some function (52 = (252012). Therefore, after solving (3.7.18), we obtain (3.“9) Q = 652(712) + Q3(U3, ..., U") COS Hg for some functions (1)2 = ¢2(u2) and Q3 2 Q3(u3, ...,un). Repeating this procedure 94 n — 2 times, we obtain Q = (152012) + Q3(u3, ..., un) cos 112, (3.120) 'Q3 : ¢3(U3) + Q4014, "'7 un) COS U3, Qn—l : ¢n—l(un—1) + ¢n(un) cosun_1, where 45,,(un) = Qn(un). Substituting (3.7.20) into (3.7.17), we find (3.7.21) x = Cl Sinh 3” + Mu?) C0511 1’ + 853%) cos U2 cosh :1: + +¢n_1(un_1) cos 71.2... cos u,,_2 cosh :1: + ¢n(un) COS 112... cos un_1 cosh 2:. Now, by applying (3.7.21) and (3.7.15), we may obtain in the same way as given in the proof of Theorem 3.1 that x = (:1 sinhx + c2 coshx + 03 sin U2 coshx + (3.7.22) +cn+1 cos U2... cos un_1 sin 21,, cosh at + Cn+2 cos 212... cos un cosh :1:, for some constant vectors c1, ..., on”. If we choose suitable initial conditions for x, fix/62, Bat/Bug, ..., fix/Bun at 0=(0, ..., 0), we will obtain (3.1.8) from (3.7.22). Consequently, up to rigid motions, the immersion x is given by (3.1.8). Case (l-ii): E = 0, f (:13) = cosh :1:, p = sechzr. Again, the open subset U is the whole manifold M ". Thus, M " ia an open portion of the warped product manifold G" 2 IR xcoshx En‘l. Hence, the metric tensor of M " is given by (3.7.23) 9 = dx2 + cosh? m(dug + + d233,). 95 In this case, the equation of Gauss implies that the isometric immersion x satisfies the following system: (3.7.24) —— = x, 82x 6x 3.7.25 — = t h — '= 2 (92x 3.7.26 2 0 2 < ' < ' ( ) 8%an , _ z .7, 2 (3.7.27) 59—); = sinhxcosh 1123(- + cosh 175 + cosh"2 xx, j : 2, ..., n. Bu]- 8:2: After solving this system, we obtain (3.7.28) x(x, ug, ..., U”) = 01 sinh :1: + (02213 + + £13,113, + @212 + + flnun +7) cosh 1:, for some constant vectors cl, 012, ..., an, fig, ..., fin, ’7. If we choose suitable initial conditions for x, Bat/6.7:, ant/Bug, ..., Bx/aun, we may obtain (3.1.9) from (3.7.28). Thus, up to rigid motions, the immersion is given by (3.1.9). Case (l-iii): a = —1, f(:2:) = m cosh x, u = asechx/m,a > 0. Again, the open subset U is the whole manofold M " and M " is an open portion of the warped product manifold H: = 1R choshx H "‘1(—1). Thus, the metric tensor of M " is given by (3.7.29) 9 = das2 + (a2 + 1)(cosh2 :1:)go, 96 where (3.7.30) go = dug + sinh2 u2(du§ + cos2 u3dui + + cos2 u3... cos2 un_1du,2,). If we apply (3.7.29), (3.7.30), and our assumption on the second fundamental form and the equation of Gauss, we know that the isometric immersion x satisfies the following system: (3.7.31) 2:3:- = x, (3.7.32) 6:27:1- = tanhxat? j = 2, ..., 72., (3.7.33) 85321,- = coth 212-517;, j = 3, ...,n, (3.7.34) 0:25;,- : __ tan ujgxm, 3 g z‘ < j g n, (3.7.35) gin); = —92—;-1 sinh 22:22 + ax/Ez2—+_1 cosh x6 + (a2 + 1) cosh2 xx, (3.7.36) 2:73; 2 sinh2 27222535 —- sinhsrcosh 23583;, (3.7.37) 83:2: 2 cos2 mfg—:3;- — sin n, cos (1,-~53, j = 2, ..., n — 1. 97 After solving this system in the same way as in Case (1-i) we obtain x = Cl sinh :1: + c2 cosh :I: + 03 cosh 11,2 cosh a: + c4 sinh 11.2 cos u3 sinh a: (3-7-38) +... + cn+1 sinh U2 cos u3... cos un_1 sin un cosh x +cn+2 sinh U2 cos U3... cos un cosh :13, for some constant vectors c1, ..., on”. If we choose suitable initial condition for x,6x/8:v,8x/8u2, ...,ax/Oun, we will obtain (3.1.10) from (3.7.38). Case (2): f(:1:) = asinh x. In this case, (3.7.6) yields 5 = 012 + a2. Thus, 5 > 0. By applying a scaling on E if necessary, we have 5 = 1 and a = m. In summary, we have (3.7.39) 5 = 1, f(x) = v1 — a2 sinhm, cschx, 0 < a < 1. a ”771:7 From (3.7.39) and continuity, we know that U is a dense open subset of M ". Moreover, locally, U is an open subset of the warped product manifold Ya". Thus the metric tensor of M n is given by 11—] (3.7.40) g = (12:2 + (1 — (12)sinh2 m(dug + cos2 u2du§ + + (H cos2 uj)du,2,). 1:2 Therefore, by applying the equation of Gauss, we know that the isometric immersion x satisfies the following system: (3.7.41) — = x, 62x . . 2 (3 7 4 ) Basauj au,’ 98 82x 6x 3.7.43 2 -t h — < ' ' ( ) 8u26uj an ugauj, 2 _ z < j 2 2 - 1 (3.7.44) 9—);- = 9—— sinh ZxQX— + aVl — a2 sinh xf + (1 — a2) sinh2 xx, 811.2 2 (9x (3 7 45) 32" 62x _ 2 , ° . . '— _ 5—5——cos 14,6 2+smchoqua , 1—2,...,n 1. Uj+1 Uj Uj After solving this system in the same way as in Case (1-i) we obtain x = c1 cosx + Cg sinhx + 0;; sin u2 sinhx + (3.7.46) +cn+1 cos u2... cos un_1 sin u" sinh x + cn+2 cos ug... cos un sinh x, for some constant vectores cl, ..., cu”. If we choose suitable initial conditions for x, 6x/ 8x, 8x/ 011.2, ..., 6x/(9un at 02(0, ..., 0), we will obtain (3.1.11) from (3.7.46). Therefore, up to rigid motions, the immersion is given by (3.1.11). Case (b): n = 3, A1 = A2 = p/2,)\3 = u 75 0. Let f and fl denote the distributions spanned by {81,62} and {63}, respectively. Then as Case (ii) in the proof of Theorem 3.1, we can prove that the integral curves of fl are geodesics and the distribution .7: is a spherical integrable distribution. Thus, there exist a local coordinate system {x, u, U} such that .7: is spanned by {a/Bu, (9/61)} and e3 = B/Bx. As in the proof of Theorem 3.1, we may also prove that u = h(x) depends only on x. Again, according to a result of Hiepko, M 3 is locally a warped product manifold I x f(x) N 2(E), where f (x) > 0 is the warp function and N 2(E) is a surface of constant curvature 6. So, the metric of M 3 can be written as (3.7.47) 9 = dx2 + f2(x)E2(du2 + dv2), 99 where {u,v} is an isothermal coordinate system on N2(E). Applying (3.7.47) we obtain 8 _ 8 _ f’ ('3 0 _ f’ 0 V%ax_0’ V%6u_ fau’ Va‘irav‘ fav’ V 2-133 3.: £627 E822 Eav’ (3.7.48) v 3 —ff’E2_a_+§l—6__ 5:13 831.- 32; Eau Eav’ 6 , 28 E, 8 ”6 Vszx— ”Ea—4‘23; ‘57; From (3.7.47) and the hypothesis on the principle curvatures, we know that the second fundamental form h of M 3 in H 4(-1) satisfies 8 ('3 _ _ __1_ 2 2 22 _l 2 2 “3;" a) = “as." a) ‘ “an, a.) = ’ where g is a unit normal vector field of M 3 in H 4(-1). From (3.7.48), (3.7.49) and equations of Gauss and Codazzi, we find (37-50) f (a?) = OWE), (3.7.51) #”(:1:) + 14351?) _ ”(33) = 0, 2 II L2 _ _ ,2 (3.7.52) K13=_1+%:_f7’ K12:_1+%:Cf2f ’ where oz is a nonzero real number and K12,K13 are sectional curvatures of M 3 of plane sections spanned by {81, 62}, {61, e3}, respectively. 100 Put 112(x) = 2p(x). Then (3.7.51) becomes z/J"(x) + 21/23(x) - 212(x) = 0. Hence, by applying Lemma 5.4 of [10], we know that p = p(x) is one of the following functions: (1) u = fl, (2) p = 2sech(x), (3) h(x) = Mcnmx, T), a > 1, 7—471 «2.. «2 x’ M” We consider these four cases separately. 0 1. In this case, (3.7.52) yields 5 = 02(04 — 1) > 0. By choosing 'c' = 1, we have _\/a2+1 k,_\/a2—1 1 , . = 2 k k = — — __ — —__° (3768) ,u(x) a cn(alT, ), f ak,cn(ax,k), k \/2a ’ \/2a Thus, locally, U is an open portion of the warped product manifold 03. If we choose the spherical coordinate system {(15, 6} on 52(1), we have cn2(ax, 1:) (3.7.69) 9 : dx2 + a216,, (dd)2 + cos2 ¢d62). By (3.7.68), (3.7.69) and formula of Gauss, we know that the immersion satisfies the following system: 02x (3.7.70) 5;; = 2akcn(ax)§ + x, (3.7.71) 3:;- = 01:, cn(ax)dn(ax)sn(ax)g-x; + Egg—€23 + C::(I:f::)x, (3.7.72) :27): = cos2 gig—g; + sin aficos 45%, (3.7.73) 6:2qu 2 ‘GSC(a$)dn(ax)gX—¢, (3.7.74) 68:; = —asc(ax)dn(ax)-870-, (3.7.75) By taking the partial derivative of (3.7.71) with respect to ¢, by applying (3.2.8), (3.7.73) and Weingarten equation, we obtain 63x/8653 + 3x/qu = 0, from which we get (3.7.76) x(x, 65, 6) : A(x, 6) sin (15 + B(x, 6) cos ¢ + C(x, 6), for some E?-valued functions A, B, C. Substituting (3.7.76) into (3.7.73) and (3.7.74) yields A(x, 6) = cn(ax)D(6), B(x, 6) = cn(ax)E(6), C(x, 6) = cn(ax)F(6) + C(x) for some IE?-valued functions D, E, F and G. Thus, we obtain (3.7.77) x = D(6)cn(ax) sin (I) + E(6)cn(ax) cos <2 + F(6)cn(ax) + G(x). Substituting (3.7.77) into (3.7.75) yields D’(6) = F’(6) = 0. Thus, (3.7.77) implies (3.7.78) x = clcn(ax) sin cf) + E(6)cn(ax) cos d) + K(x) for some constant vector c1 and IE?-valued function K (x) By substituting (3.7.78) into (3.7.72) and by applying (3.2.8) and (3.7.70), we find E”(6) + E(6) = 0. Thus, E(6) = 62 cos6 + 6;, sin 6, for some constant vectors Cg, C3. Substituting this into (3.7.77) yields (3.7.79) x = clcn(ax) sin (15 + cgcn(ax) cos gbcos 6+ c3cn(ax) sin 65 cos 6 + K(x). 105 Substituting (3.7.79) into (3.7.70), we have (3.7.80) K"(x) + 2asc(ax)dn(ax)K'(x) + K(x) = 0. Therefore, by applying (3.7.79), (3.7.80) and Proposition 5.2, we know that x takes the form: x(x, 65, 6) = clcn(ax) sin 63 + cgcn(ax) cos 42605 6 + cgcn(ax) sin 66603 6 k’ 1 8 — (3.7.81) +C4 3216’2 + cn2(ax) cosh(;x — §ln neg—3% + —————::—;—- aZ(’y)x) +c5\/a2k’2 + cn2(ax) sinh(%x — én g————E:: _: Z; — aZ(’y)x), where k = W/(fia) and k’ = W/(fia) are the modulus and the com- plementary modulus of the Jacobi elliptic functions and 7 = sn’1(1/(ak2)). Now, by choosing suitable initial conditions, we obtain (3.1.14) from (3.7.81). Therefore, up to rigid motions, the immersion is given by (3.1.14). Case (b-4): h(x) = WCMWx/fi, \/2a/\/1—+717),0 < a < 1. In this case, (3.7.52) yields 6 = —4cr.2672k’2/k2 < 0. By choosing a = —1, we obtain k ,/ _ 2 (3.7.82) u(x) = —k—dn(: x, k), f— - ?dn(kx, k), k = fi, 16’: V1725. Thus, U = M 3 which is an open portion of the warped product manifold D2 = 2 o c X(Ic/(ak’))dIl(ax/k) H (-—1). If we choose the hyperbollc coordinate system {65, 6} on H2(—1), we get 162an (7‘: x) (3.7.83) 9 = dx2 + a2k’2 (462 + cosh2 6662). By (3.7.88) and (3.7.89) and formula of Gauss, we know that the immersionsatisfies 106 the following system: (3.7.34) (3.7.85) (3.7.86) (3.7.37) (3.7.88) (3.7.89) 62X 261 a 8—1172- — —k—dn(;x)§ + X, 62x 163 a a a 3x kdn3(%x) k2dn2(%x) $2— : aklzcn(gx)dn(gx)sn(-Ex)5; + _——ak’2 ——_—a2k’2 x, 82x __ 2 6 6 cosh (1)93— , 0x 3452—81111) ()5 cosh (ta—é, 62x 8x 6x665 - ——ak‘cn(:x )sd(kx )5—¢, 62x 8x86 $)Sd(%$)::, = —akcn(— (92x 6x .6766 — tanh (6'55. Solving (3.7.89) yields (3.7.90) x,(x 65, 6:) B(,x 6)coshq§+C(x ,¢). for some [BE-valued functions B (x, 6), C (x, 65). Substituting (3.7.90) into (3.7.87) yields (3.7.91) BB a a a; — —akcn(Ex)sd(-l;x)B, 107 02C a a 8x (3.7.92) 3$3¢ — -akcn(;x)sd(;x)-ég, Solving the above two equations, we get (3.7.93) B(x, 6) = G(6)dn(%x), C(x, 6) = K(¢)dn(%x) + W(x). for some Ei-valued functions G (6), K (65), W(x). Thus, we obtain a (3.7.94) x(x, 6, 6) = G(6)dn(%x) cosh 6 + K(¢)dn( k By subsituting (3.7.94) into (3.7.85) and (3.7.86), by applying (3.2.8) and (3.7.84), we have (3.7.95) G”(6) — 0(6) = 0, K”(¢) -— 13(6) = 0, (3.7.96) W”(x) + 2akcn(36)sd(36)w'(x) + W(x) = 0. k k Therefore, by applying (3.7.94) - (3.7.96) and Proposition 5.2, we know that x takes the form: x = cldn(%x) cosh 63 cosh 6 + cgdn(;:—x) cosh 65 sinh 6 +c3dn(%x) cosh ¢ + c4dn(-:—x) sinh 65 (3.7.97) +Cs\/k2dn2(%x) — a2k’2 cos(k’x — T ln SEE: ; Z; — \/——1%Z(’y)x) +Cs\/k2dn2(%$) — a2k’2 sin(k'x — \/2__11n SEE: ; Z; — \/——1_%Z('y)x), where k = \/2a/\/1 + a2 and k’ = \/1 - a2/\/1 + a2 are the modulus and the com- 108 plementary modulus of the Jacobi’s elliptic functions and 'y = sn‘1(k/a). Now, by choosing suitable initial conditions, we obtain (3.1.15) from (3.7.97). Therefore, up to rigid motions, the immersion is given by (3.1.15). The converse can be verified by straight-forward long computation. 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