Nov 20, 1996 - Submanifolds satisfying this basic equality were studied recently in .... the hyperbolic n-space with constant sectional curvature kc, respectively.
463
Math. Proc. Camb. Phil. Soc. (1999), 125, 463 Printed in the United Kingdom # 1999 Cambridge Philosophical Society
Elliptic functions, theta function and hypersurfaces satisfying a basic equality B BANG-YEN CHEN JIE YANG Department of Mathematics, Michigan State University, East Lansing, MI 48824, U.S.A. e-mail : bychen!math.msu.edu & jyang!math.msu.edu (Received 20 November 1996 ; revised 21 July 1997) Abstract In previous papers [4, 6], B.-Y. Chen introduced a Riemannian invariant δM for a Riemannian n-manifold M n, 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(ε) satisfies : δM [n#(nk2)]\2(nk1) H #j"(nj1)(nk2) ε. # In this paper, first we classify constant mean curvature hypersurfaces in a Riemannian space form which satisfy the equality case of the inequality. Next, by utilizing Jacobi’s elliptic functions and theta function we obtain the complete classification of conformally flat hypersurfaces in Riemannian space forms which satisfy the equality.
1. Introduction For a submanifold Mn in an m-dimensional Riemannian space form Rm(ε) of constant sectional curvature ε, the first author proves in [4] a sharp inequality involving the sectional curvature K of Mn, the scalar curvature τ of Mn and the squared mean curvature H#. It is convenient to introduce a Riemannian invariant δM of Mn by δM(p) l τ(p)kinf K(p), where inf K is the function assigning to each p ? Mn the infimum of K(π), where π runs over all 2-planes in Tp M and τ is defined by τ B i j K(eiFej), and oe , … , enq is an orthonormal basis of Tp Mn. The inequality " proved in [4] can be written as follows : δM
n#(nk2) H#j"(nj1) (nk2) ε. # 2(nk1)
(1n1)
The equality case of (1n1) holds identically for n l 2 since both sides of (1n1) vanish trivially. Inequality (1n1) 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 from compact Riemannian manifolds with finite fundamental group π into complex space forms (see [6] for details). " Since (1n1) is a very general and sharp inequality, it is natural and interesting to
464
B.-Y. C J. Y
investigate and to understand submanifolds in a Riemannian space form which verify the equality of this inequality, i.e. the equality δM l
n#(nk2) H#j"(nj1)(nk2) ε. # 2(nk1)
(1n2)
When M is a hypersurface with dimension 3, equality (1n2) holds at a point p ? M if and only if, up to permutations, the principal curvatures λ , … , λn of M at p satisfy " the condition λ jλ l λ l ( l λn (cf. [4]). Therefore, if M satisfies (1n2) " # $ identically, M has a principal curvature µ with multiplicity nk2. When n 4, the multiplicity of µ is at least 2. In particular, if the principal curvature µ has constant multiplicity then dµ vanishes along the corresponding principal directions and, moreover, in this case the leaves of the corresponding foliation of M are known to be extrinsic spheres of the ambient space Rm(ε) (cf. [18, 21–23]). In Section 5 we also prove that a similar phenomenon holds for conformally flat hypersurfaces in 4dimensional Riemannian space forms which satisfy equality (1n2). Submanifolds satisfying this basic equality were studied recently in several papers (see for instance [4–6, 8, 9, 12, 15, 16]). In this respect we point out that 3-dimensional totally real submanifolds satisfying this basic equality in the nearly Ka$ hler 6-sphere S' (1) have been completely classified in [16] by F. Dillen and L. Vrancken. In this paper we investigate the most fundamental case ; namely hypersurfaces satisfying the basic equality. In the first part of this article we completely classify CMC hypersurfaces satisfying the basic equality. More precisely, we prove the following theorems. T 1. A hypersurface Mn (n 2) of a Euclidean (nj1)-space n+" with constant mean curvature satisfies equality (1n2) if and only if either Mn is a minimal hypersurface with relative nullity nk2 or Mn is an open portion of a spherical hypercylinder iSn−"(r). T 2. Let Mn (n 2) be a hypersurface with constant mean curvature in the sphere Sn+" (1). Then Mn satisfies equality (1n2) if and only if one of the following two cases occurs : (1) Mn is a totally geodesic hypersurface ; (2) there is an open dense subset U of Mn and a non-totally geodesic, isometric, minimal immersion φ : B# 4 Sn+"(1) from a surface B# into Sn+"(1) such that U is an open subset of NB# 9 Sn+"(1) defined by Np B# l oξ ? Tφ(p) Sn+"(1) Q f ξ, ξg l 1 and fξ, φJ(Tp B#)g l 0q. Let n+# denote the (nj2)-dimensional Minkowski space-time with the Lorentzian " metric g lkdx#jdx#j(jdx#n+ . Recall that the unit hyperbolic space Hn+"(k1) # " # and the unit de Sitter space-time Sn+"(1) are isometrically embedded in n+# " " respectively in the following standard ways : Hn+"(k1) l ox l (x , … , xn+ ) ? n+# Q fx, xg lk1q, # " " Sn+"(1) l ox ? n+# Q fx, xg l 1q. " " T 3. Let Mn (n 2) be a hypersurface with constant mean curvature in the hyperbolic space Hn+"(k1). Then Mn satisfies equality (1n2) if and only if one of the following three cases occurs :
Elliptic functions, theta function and hypersurfaces
465
(1) Mn is a totally geodesic hypersurface ; (2) Mn is a tubular hypersurface with radius r l coth−" (N2) about a 2-dimensional totally geodesic surface of Hn+"(k1) ; (3) there is an open dense subset U of Mn and a non-totally geodesic, isometric, minimal immersion φ : B# 4 Sn+" from a surface B# into the de Sitter space-time " Sn+"(1) such that U is an open subset of the unit normal bundle NB# of B# defined " by Np B# l o ξ ? Tφ(p) Sn+"(1) Q f ξ, ξg lk1 and f ξ, φJ(Tp B#)g l 0q. " In order to state our next results we recall the three families of Riemannian manifolds Pna (a 1), Cna(a 1), Dna(0 a 1) and the two exceptional spaces Fn, Ln, first introduced in [7]. 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) l sn (u, k)\dn (u, k), nd (u, k) l 1\dn (u, k) (cf. [17, 19] and Section 2 for details). We put Na#k1 µa l ak cn (ax, k), k l , a 1, (1n3) N2a a a ηa l dn x, k , k k
0 1
kl
N2a , Na#j1
0
ρa l ak cn (ax, k),
kl
Na#j1 , N2a
a 1.
a
1,
(1n4) (1n5)
Let Sn(c) and Hn(kc) denote the n-sphere with constant sectional curvature c and the hyperbolic n-space with constant sectional curvature kc, respectively. For n 2, Pna, Dna and Cna are the Riemannian n-manifolds given by the warped product manifolds Iiµα Sn−"[(a%k1)\4], iη Hn−"[(a%k1)\4] and Iiρ Sn−"[(a%k1)\4] with a a warped functions µa, ηa and ρa, respectively, where I are the open intervals on which the corresponding warped functions are positive. The two exceptional spaces Fn and Ln are the warped product manifolds i /N Hn−"(k") and isech(x) n−", " # % respectively. Dna, Fn, Ln are complete Riemannian n-manifolds but Pna and Cna are not complete. Topologically, Sn is the two point compactification of both Pna and Cna. From [10] we know that the Riemannian metrics defined on Pna and Cna can be extended smoothly to their two point compactifications Sn. We denote by PV na and CV na the Sn together with the Riemannian metrics given by the (smooth) extensions of the metrics on Pna and Cna to Sn, respectively. We remark that Pna, Dna, Cna are indeed isometric to the warped products n-manifolds Iif Sn−"(1), Iif Hn−"(k1), Iif Sn−"(1) with warped func" # $ tions 2ρa\(a#j1), 2µa\(a#k1), 2ηa\(a#j1), respectively. Let Ana(a 1), Bna(0 a 1), Gn, Hna(a 0) and Yna(0 a 1) denote the warped product manifolds iN
a#−"coshx
icoshx n−",
iN
0
1
π π Sn−"(1), k , iN # Sn−"(1), "−a cosx 2 2
a#+" coshx
Hn−"(k1), (0, _)iN # Sn−"(1), "−a sinhx
466
B.-Y. C J. Y
respectively. When n l 2, the second factor Sn−" or Hn−" in each of the warped product manifolds shall be replaced either by S"(1) or by . The geometry of A#a, G# and H#a are similar in the sense that one can be obtained from the others by applying some suitable scalings on the first factor . Clearly, Ana, Gn, Hna are complete Riemannian manifolds. Topologically, Sn is the two point compactification of Bna. As for Pna and Cna, the warped metric on Ban can be extended smoothly to its two point compactification ; a fact that follows from (1n8). We denote by BV na the Sn together with the Riemannian metrics on Sn extended from the metric on Bna. For n 2 and any real number a 0, there is a well-known Lagrangian immersion from the unit n-sphere Sn into a complex Euclidean n-space n defined by a (1n6) ( y , … , yn, y y , … , y yn), wa( y , y , … , yn) l ! " ! " ! 1jy# " ! where y#jy#j(jy#n l 1. This Whitney’s immersion wa has a unique self! " intersection point wa(k1, 0, … , 0) l wa(1, 0, … , 0). The Sn together with the metric induced from Whitney’s immersion wa, denoted by Wna, is called a Whitney n-sphere. In the second part of this article we completely classify conformally flat hypersurfaces satisfying the basic equality. In this respect first we sharpen proposition 2 of [15] to the following. T 4. Let x : Mn 4 n+"(n 2) be an isometric immersion of a conformally flat n-manifold into a Euclidean (nj1)-space. Then it satisfies equality (1n2) if and only if one of the following four cases occurs : (1) Mn is totally geodesic ; (2) Mn is an open portion of a spherical hypercylinder Sn−"i ; (3) Mn is an open portion of a round hypercone ; (4) n l 3 and M$ is an open portion of a Whitney 3-sphere W$a for some a 0 and, up to rigid motions, the immersion x : W$a 4 % is given by
0& 0 1
0 1
0 1
0 11 ,
N2 N2 N2 N2 1 x sd# x(x, y , y , y ) l x dx, ay sd x , ay sd x , ay sd x " # $ " # $ 2 a a a a !
(1n7)
where y#jy#jy# l " and k l 1\N2 is the modulus of the Jacobi’s elliptic " # $ # functions. Our next main results are those following. T 5. Let x : Mn 4 Sn+"(1) 9 n+#(n 2) be an isometric immersion of a conformally flat n-manifold. Then it satisfies equality (1n2) if and only if one of the following three cases occurs : (i) Mn is an open portion of Sn(1) and the immersion x : Mn 4 Sn+"(1) is totally geodesic ; (ii) Mn is an open portion of BV na for some a ? (0, 1) and, up to rigid motions, the immersion x : BV na 4 Sn+"(1) 9 n+# is given by x(x, y , … , yn) l (sin x, a cos x, N1ka# y cos x, … , N1ka# yn cos x) " " with y#jy#j(jy#n l 1 ; " #
(1n8)
Elliptic functions, theta function and hypersurfaces
467
(iii) n l 3, M$ is an open portion of PV $a for some a 1 and, up to rigid motions, the immersion x : PV a$ 4 S%(1) 9 & is given by
0
5
1 x(x, y , y , y ) l y cn (ax), y cn (ax), y cn (ax), " # $ # $ akh " Θ(axkγ) j2aZ(γ) x11 0khk xj2i 0ln Θ(axjγ) kh i Θ(axkγ) Na#kh#kcn#(ax) sin 0 xj 0ln j2aZ(γ) x111 , k 2 Θ(axjγ) i y#jy#jy# l 1, γ l sn−" 0 1 , i l Nk1, " # $ ak# Na#kh#kcn# (ax) cos
(1n9) 7
6
8
where k l Na#k1\(N2a), kh l Na#j1\(N2a) are the modulus and the complementary modulus of Jacobi’s elliptic functions respectively, Θ(u) l Θ(u, k) is the theta function and Z(u) l Z(u, k) the zeta function. T 6. Let x : Mn 4 Hn+"(k1) 9 n+# (n 2) be an isometric immersion of a " conformally flat n-manifold. Then it satisfies equality (1n2) if and only if one of the following nine cases occurs. (i) Mn is an open portion of Hn(k1) and the immersion x : Mn 4 Hn+"(k1) is totally geodesic. (ii) Mn is an open portion of Ana for some a 1 and, up to rigid motions, the immersion x : Ana 4 Hn+"(k1) 9 n+# is given by " x(x, y , … , yn) l (a cosh x, sinh x, Na#k1 y cosh x, … , Na#k1 yn cosh x) (1n10) " " with y#jy#j(jy#n l 1. " # (iii) Mn is an open portion of Gn and, up to rigid motions, the immersion x : Gn 4 Hn+"(k1) 9 n+# is given by " x(x, u , … , un) # u#j(ju#n u#j(ju#n cosh x, # cosh x, sinh x, u cosh x, … , un cosh x . l 1j # # 2 2
00
1
1
(1n11) (iv) Mn is an open portion of Hna for some a 0 and, up to rigid motions, the immersion x : Hna 4 Hn+"(k1) 9 n+# is given by " x(x, y , … , yn) l (Na#j1 y cosh x, … , Na#j1 yn cosh x, a cosh x, sinh x) (1n12) " " with y#ky#j(ky#n l 1. " # (v) Mn is an open portion of Yna for some a, 0 a 1 and, up to rigid motions, the immersion x : Yna 4 Hn+"(k1) 9 n+# is given by " N x(x, y , … , yn) l (cosh x, a sinh x, 1ka# y sinh x, … , N1ka# yn sinh x) (1n13) " " with y#jy#j(jy#n l 1. " # (vi) n l 3, M$ is an open portion of F$ and, up to rigid motions, the immersion x : F$ 4 H%(k1) 9 & is given by " x(x, u, v) l (N2 cosh u cosh v, N2 cosh u sinh v, N2 sinh u, cos x, sin x). (1n14)
B.-Y. C J. Y
468
(vii) n l 3, M$ is an open portion of L$ and, up to rigid motions, the immersion x : L$ 4 H%(k1) 9 & is given by " x(x, u, v) l sech x i(x#ju#jv#jcosh# xj", x#ju#jv#jcosh# xk", x, u, v) (1n15) % % (viii) n l 3, M$ is an open portion of CV $a for some a 1 and, up to rigid motions, the immersion x : CV a$ 4 H%(k1) 9 & is given by " 5 x(x, y , y , y ) " # $ 1 kh 1 Θ(axkγ) Na#kh#jcn#(ax) cosh l xk ln kaZ(γ) x , akh k 2 Θ(axjγ) 6 (1n16) 7 kh 1 Θ(axkγ) Na#kh#jcn#(ax) sinh xk ln kaZ(γ) x , k 2 Θ(axjγ)
0
0
0
1
1
y cn (ax), y cn (ax), y cn (ax) , " # $
1
y#jy#jy# l 1, " # $ 8
where γ l sn−"(1\(ak#)), k l Na#j1\(N2a) and kh l Na#k1\(N2a). (ix) n l 3, M$ is an open portion of D$a for some 0 a 1 and, up to rigid motions, the immersion x : D$a 4 H%(k1) 9 & is given by " 5 x(x, u, v) l
0 0 1
1 a k dn x cosh u cosh v, akh k
0ak x1 cosh u sinh v, k dn 0ak x1 sinh u, xkγ) a ki Z(γ) x1 ’ k# dn# 0ak x1ka#kh# cos 0khxk2i ln Θ((a\k) Θ((a\k) xjγ) k xkγ) a ki Z(γ) x11 ’ k# dn# 0ak x1ka#kh# sin 0khxk2i ln Θ((a\k) Θ((a\k) xjγ) k k dn
(1n17) 6 7
8
where γ l sn−" (k\a), k l N2a\N1ja# and kh l N1ka#\N1ja#. 2. Jacobi’s 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 (for details see, e.g. [1, 17, 20]). Let θ be the temperature at time t at any point in a solid material whose conducting properties are uniform and isotropic. If ρ is the material’s density, s its specific heat and k its thermal conductivity, θ satisfies the heat conduction equation : κ]# θ l cθ\ct, where κ l k\sρ is the diffusivity. In the special case where there is no variation of temperature in the x- and y-directions the heat flow is everywhere parallel to the z-axis and the heat equation reduced to the form : κ
c# θ cθ l , cs# ct
θ l θ(z, t).
(2n1)
Consider the boundary conditions : θ(0, t) l θ(π, t) l 0 and θ(z, 0) l πδ(zkπ\2) for
Elliptic functions, theta function and hypersurfaces
469
0 z π, where δ(z) is Dirac’s unit impulse function. Then the solution of the boundary value problem is given by _
θ(z, t) l 2 (k1)n e−(#n+")#κt sin (2nj1) z n=! κt − By writing e % l q, the solution (2n2) assumes the form :
(2n2)
_
(2n3) θ (z, q) l 2 (k1)n q(n+"#)# sin (2nj1) z, " n=! which is the first of the four theta functions. When the precise value of q is not important we shall suppress the dependence upon q. If one changes the boundary conditions to cθ\cz l 0 on z l 0 and z l π with θ(z, 0) l πδ(zkπ\2) for 0 z π, then the corresponding solution of the boundary value problem of the heat equation (2n1) is given by _
(2n4) θ (z) l θ (z, q) l 1j2 (k1)n qn# cos 2nz. % % n=" The theta function θ (z) of (2n3) is periodic with period 2π. Incrementing z by "π " # yields the second theta function :
0
1
_ π (2n5) θ (z) l θ (z, q) l θ zj , q l 2 q(n+"#)# cos (2nj1) z. # # " 2 n=! Similarly, incrementing z by "π for θ yields the third theta function : # % _ π (2n6) θ (z) l θ (z, q) l θ zj , q l 1j2 qn# cos 2nz. $ $ % 2 n=" The four theta functions θ , θ , θ , θ can be extended to complex values for z and " # $ % q such that QqQ 1. The elliptic functions sn u, cn u and dn u are defined as ratios of theta functions :
0
1
θ (0) θ (z) θ (0) θ (z) θ (0) θ (z) " , cn u l % # , dn u l % $ , sn u l $ (2n7) θ (0) θ (z) θ (0) θ (z) θ (0) θ (z) # % # % $ % where z l u\θ#(0). Define parameters k and kh by $ k l θ#(0)\θ#(0), kh l θ#(0)\θ#(0) # $ % $ which are called the modulus and the complementary modulus of the elliptic functions ; k and kh satisfy k#jkh# l 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 sn u, cn u and dn u satisfy the following relations : sn# ujcn# u l 1, dn# ujk# sn# u l 1, k# cn# ujkh# l dn# u, sn(ujv)jsn(ukv) l
2 sn u cn v dn v , 1kk# sn# u sn# v
(2n8) (2n9)
snh(u) l cn (u) dn (u), cnh(u) lksn (u) dn (u), dnh(u) lkk# sn (u) cn (u). (2n10) The theta function Θ(u) and the zeta function Z(u) are defined by
0 1
πu , Θ(u) l θ % 2K
Z(u) l
d (ln θ ), % du
π K l θ#(0), 2 $
(2n11)
B.-Y. C J. Y
470
which satisfy the following identities : Z(ujv) l Z(u)jZ(v)kk# sn u sn v sn (ujv), Z(u) l
(2n12)
Θh(u) , Θ(u)
(2n13)
where Θ(u) is an even function. From (2n12) we have k# sn u sn v [sn (ujv)jsn (ukv)] l Z(ukv)kZ(ujv)j2Z(v).
(2n14)
3. Some lemmas Assume that Mn is a hypersurface in a real-space-form Rn+"(ε), ε l 1, 0 or k1. We shall make use of the following convention on the ranges of indices : 1 A, B, C nj1 ;
1 i, j, k n ;
3 α, β, γ n.
Denote by A l Aξ the shape operator of Mn in Rn+"(ε) with respect to a unit normal vector ξ and by h the second fundamental form of Mn in Rn+"(ε). Let λ , … , λn be the " eigenvalues of A with associated orthonormal eigenvectors e , … , en. Then we have " Aei l λi ei. (3n1) If Mn satisfies the basic equality (1n2) then, by rearranging e , … , en if necessary, " we have [4, 5] (3n2) λ l a, λ l µka, λα l µ. " # In the following, let V denote the open subset of Mn on which M has exactly three distinct principal curvatures, i.e.
(
V l x ? U Q a(x) 0, a(x) µ,
and
a(x)
*
µ , 2
(3n3)
where a and µ are given by (3n2). Then there exists a frame (e , … , en) of orthonormal " eigenvector fields on V. Let ω", ω#, … , ωn denote the dual frame of e , e , … , en. Then " # Cartan’s structure equations give dωi lkωijFωj,
dωij lkωikFωkjj(εjλi λj) ωiFωj,
(3n4)
where (ωA ) are the connection forms. From (3n4) and Codazzi’s equation, we have B ei λj l (λikλj) ωij(ej) 5 6 7
(λjkλk) ωkj(ei) l (λikλk) ωki(ej) 8
(3n5)
for distinct i, j, k. Let Γkij be defined by Γkij B f]e ei, ekg where ] is the Levi–Civita connection on Mn. j Then we have ωki l Γkij ωj, k
Γkij l ωki(ej),
Γkij lkΓikj
In this way, (3.5) becomes ei λj l (λikλj) Γjij for distinct i, j, k.
5
(3n6) 6
(λjkλk) Γkji l (λikλk) Γkij 7 8
471
Elliptic functions, theta function and hypersurfaces Let H denote the mean curvature of Mn in Rn+"(ε). Then H l ((nk1)\n) µ. From (3n2) and (3n6) we obtain
L 1. Let Mn be a hypersurface in a Riemannian space form Rn+"(ε). If Mn satisfies equality (1n2) then on V we have a Λ, Γ#α l " 2akµ α e µ Γαα l " , " akµ e a Γ" l # , #" µk2a
a Γ"α l Λ, # µka α
e µ Γαα lk # , # a Γ#α l #
5
Γβα l Γβα l 0, α β " #
e ( µka) Γ# lk " , "# µk2a
eα( µka) , a
(3n7) 6
e a Γ"α l α . " µka
7
8 5
e a aΛ e µ ω"α l α ω"j α ω#j " ωα, µka µka µka n e a e ( µka) a Λ ωα, 67 ω" l # ω"j " ω#j # µk2a µk2a µk2a α= α $ eα( µka) e µ α ω#α l Λα ω"j ω#j # ω , a a
(3n8)
8
where Λα l Γ#α . " We also need
L 2. Let Mn be a hypersurface in Rn+"(ε). If Mn satisfies equality (1n2) then on V we have
0 1 0 1
e µ e a (e a)# 2aΛ#α (e µ)# " keα α k α k k " e " µka µka ( µka)# µk2a ( µka)# (e a)(e µ) 1 n # k lk # (eβ a) Γβααjaµjε, ( µk2a) a µka βα
(3n9)
0 1 0 1
e µ aΛα a#Λα(eαa) e (akµ)Λα " keα e j jα # µka µka ( µka)#( µk2a) µka k
eα( µka) (e µ)(e µ) e (akµ)(e µ) 1 n # l " # k Λαk " aΛβ Γβαα , µk2a ( µka) a ( µk2a) a µka βα
(3n10)
0 1 0 1
aΛα eα a (eα a)(e a) µΛα e ( µka) # " e ke k k " µka # µka ( µka)( µk2a) ( µka)( µk2a) j
0
1
( µk2a)Λα e µ (e a)e ( µka) n eβ a β aΛ " lk # α k Γα k β Γβα . # " ( µka)# a( µk2a) µka µka βα
(3n11)
Proof. By applying the second Cartan’s structure equation and by using Lemma 1 we may compute dω"α in two different ways. Formulas (3n9), (3n10) and (3n11) are obtained by comparing the coefficients of ω"Fωα, ω#Fωα, ω"Fω# of dω"α, respectively. Similarly, by computing dωα in two different ways and by comparing the # coefficients of ω"Fωα, ω#Fωα, ω"Fω# of dω#α, respectively, we obtain
B.-Y. C J. Y
472
L 3. Let Mn be a hypersurface in a Riemannian space form Rn+"(ε). If Mn satisfies equality (1n2) then on V we have
0 1 0
1
e µ e (akµ) 2a#Λ#α (e ( µka))# (e µ)# e # jeα α j k # k α # a a a# a# ( µka)( µk2a) n e ( µka) Β e ( µka)(e µ) " k β l " Γααj( µka) µjε, ( µka)( µk2a) βα a
(3n12)
0 1
e µ Λ e a µka (e µ)(e µ) " Λ e ( µka)k # e # keα Λαk α α k " a µka a( µk2a) α α a( µka) l
(e a)(e µ) aΛα eα a # " k k Λβ Γβαα , ( µka)( µk2a) ( µka)( µk2a) βα
µΛ e a (e ( µka))(e ( µka)) # ke Λ k k " 0a ( µka) 1 # a ( µka)( µk2a) a( µk2a) ( µk2a)Λ (e µ) (e ( µka))(e a) n e ( µkα) # lk " j Λ Γ k k Γ 1. # " a( µka) ( µka)( µk2a) 0 α
e "
α
α
α
(3n13)
α
α
α
β
βα
β α
β
β α
(3n14)
We also need L 4. Let Wg na denote the warped product manifold : Ii(a/N Sn−" (1), #)sd(N#x/a, "/N#) with the warped product metric given by
0
1
N2 a# 1 g, x, g l dx#j sd# 2 a Ns !
(3n15)
where I denotes the largest open interval containing 0 such that sd ((N2\a) x) is nowhere zero on I and g is the standard metric on the unit (nk1)-sphere. Then the Whitney n! sphere Wna is topologically the two point compactification of Wna ; moreover, the metric on Wna is the smooth extension of the warped product metric on Wg na to its two point compactification. Proof. Let Sg n denote the unit n-sphere with the north and south poles, oN, Sq, being removed and let ou , … , unq denote the spherical coordinate system on Sg n given by " 5 y l cos u , y l sin u cos u , … , ! " " " # 6 (3n16) 7 yn− l sin u ( sin un− cos un, yn l sin u ( sin un− sin un. 8 " " " " " From (1n6) and (3n16) we know that the metric induced from Whitney’s immersion wa on Sg n is given by a# a# sin# u " g. du#j (3n17) gl " 1jcos# u ! 1jcos# u " " Put u" a x(u ) l dt. (3.18) " N1jcos# t ! Then u a a a 1 " . (3n19) x(u ) l sn−" sin u , dt l " N2 " N2 N1k" sin# t N2 ! # Thus N2 1 sin u l sn . (3.20) x, " a N2
0
1 &
0
1
&
0
0
1
1
473
Elliptic functions, theta function and hypersurfaces
From (2n8), (3n17), (3n18) and (3n20) we obtain (3n15). This shows that Wg na is the Sg n endowed with the metric induced from Whitney’s immersion wa. Hence, topologically, the Whitney n-sphere Wna is the two point compactification of Wg na and, moreover, the metric on Wna is the smooth extension of the warped product metric on Wg na to its two point compactification. 4. Proofs of Theorems 1, 2 and 3 Let Mn be a hypersurface with constant mean curvature H in a Riemannian space form Rn+"(ε) (n 2, ε l 1, 0, or k1) which satisfies equality (1n2). If Mn is not minimal, then µ l (n\(nk1)) H is a nonzero constant. In this case, (3n9) and (3n12) reduce respectively to (akµ)(eα eα a)k2(eα a)# l
n 2aΛ#α( µka)# j(εjµa)( µka)#k (eβ a) Γβαα( µka), µk2a βα
(4n1)
n 2a% Λ#α j(εj( µka)µ) a#ja (eβ a) Γβαα . ( µka)( µk2a) βα
(4n2)
a(eα eα a)k2(eα a)# lk
From (4n1) and (4n2) we obtain eα eα a lk
5
n 2aΛ#α( µ#k3µaj3a#) j( µk2a)(a#kaµkε)j (eβ a) Γβαα , ( µk2a)( µka) βα
ε (eα a)# lka#(Λ#αj( µka)#)k a( µka), 2
(4n3) 7
6
on V. 8
Since µ is constant, (3n13) reduces to eα Λα l
Λα(eα a)( µ#k3µaj4a#) n j Λβ Γβαα . ( µka)( µk2a) βα
(4n4)
Now, by applying Lemma 1 and Cartan’s structure equations, we can compute dω"α and dω#α in two different ways. After that, by comparing the coefficients of ω"Fω β and of ω#Fω β with β α in the formulas of dω"α and of dω#α so obtained, we find respectively the following two formulas : n 2(eα a)(eβ a) 2a( µka)Λα Λβ j l (eγ a) Γγαβ , µka µk2a γ=$ n 2(e a)(eβ a) 2a$ Λα Λβ eβ eα ak α j l (eγ a) Γγαβ . a ( µka)( µk2a) γ =$ By taking the difference of these two equations, we obtain
eβ eα aj
(eα a)(eβ a) lka# Λα Λβ, on V. Now, by applying (4n3), we may also obtain
α β,
ε k(eα a)# l a#(Λ#αj( µka)#)j a( µka), 2 ε k(eβ a)# l a#(Λ#βj( µka)#)j a( µka). 2
(4n5)
474
B.-Y. C J. Y
By taking the difference of these two equations we get (eα a)#k(eβ a)# l a#(Λ#βkΛ#α).
(4n6)
Combining (4n5) and (4n6) we obtain, on V, (eα a)# l a# Λ#β ,
α β.
(4n7)
Case (i). ε l 1, Rn+"(1) l Sn+"(1). If Mn is non-minimal, then the second equation of (4n3) implies that V is an empty set. Thus, Mn is an isoparametric hypersurface of Rn+"(1) with at most two distinct principal curvatures given either by λ l 0, λ l ( l λn l µ or by λ l λ l " µ, " # # " # λ l ( l λn l µ. Both cases are impossible according to a well-known result of $ Cartan [2]. Consequently, Mn is minimal in Sn+"(1). Let U denote the open subset of Mn consisting of non-totally geodesic points. Then U is an open dense subset of Mn. Now, by a result of [4], U has relative nullity nk2. Thus, by applying a result of Dajczer and Gromoll (lemma 2n2 of [11]), the Gauss image B# of U is a minimal surface in the unit sphere. Consequently, U is an open subset of NB# defined in Theorem 2. Case (ii). ε l 0, Rn+"(0) l n+". If Mn is non-minimal, then the second equation in (4n3) implies a l 0 or a l µ on V which contradicts the definition of V. Thus V is an empty set. Hence, M is a nonminimal isoparametric hypersurface with at most two distinct principal curvatures given either by λ l 0, λ l ( l λn l µ or by λ l λ l "µ, λ l ( l λn l µ. It is " # # " # $ well-known that the first case occurs if and only if Mn is an open portion of a spherical hypercylinder iSn−"(r) for some r 0. It is also known that the later case cannot occur. Case (iii). ε lk1, Rn+"(k1) l Hn+"(k1). First we assume that Mn 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. Case (iii.1). n 5 From (4n7) we have (e a)# l (e a)# l ( l (ena)# l a# Λ#α, α l 3, … , n. (4n8) $ % Without loss of generality, we may assume e a l aΛ . By using (4n5), we have $ $ e alkaΛ and e a lkaΛ which imply % % & & (e a)(e a) l a# Λ Λ . % & % & Thus ea a l Λα l 0 on V by (4n5) and (4n8). Hence, by applying the first equation in (4n3), we obtain a#kaµkε l 0 on V. Since µ is constant, this implies that a is constant on each component of V. Case (iii.2). n l 4 From (4n5) and (4n7) we may assume, without loss of generality, that e a l aΛ and $ % e a lkaΛ . By differentiating the second equation of (4n3) with respect to e , we get % $ $ k2(e a)(e e a) $ $$ ε l 2(e a) a(Λ#j( µka)#)ja#(2Λ (e Λ )k2( µka) e a)j ( µk2a) e a. (4n9) $ $ $ $ $ $ $ 2
475
Elliptic functions, theta function and hypersurfaces On the other hand, since Λ l e (ln a), (4n4) yields % $ (e a)Λ ( µ#k3µaj4a#) e a $ j $ Γ% . eΛ l $ $ $ ( µk2a)( µka)a a $$ Substituting this into (4.9), we find k(e a)(e e a) $ $$ 2aΛ#( µ#k3µaj3a#) ε $ ja( µka)( µk2a)j ( µk2a)k(e a) Γ% . l (e a) % $$ $ ( µka)( µk2a) 4
0
1
(4n10)
If e a does not vanish identically on V, there exists an open set W 9 V such that e a $ $ is nowhere zero on W. Thus, on W, (4n8) becomes ke e a l $$
2aΛ#( µ#k3µaj3a#) ε $ ja( µk2a)( µka)j ( µk2a)k(e a)Γ% . (4n11) % $$ ( µka)( µk2a) 4
Combining (4n3) and (4n11) we obtain µ l 2a which implies that a is constant on W. This contradicts the assumption that e a 0 on W. So e a 0 on V. A similar $ $ argument yields e a 0 on V. Thus, by applying (4n5), we know that Λ vanishes % $ identically on V. Hence, by applying (4n3) again, we have a#kaµjε\2 l 0 on V. Consequently, a is constant on each component of V. Case (iii.3). n l 3 From (4n3) we have ε k2(e a)(e e a) l a#(2Λ (e Λ )k2(e a)( µka))j2a(e a)(Λ#j( µka)#)j (e a)( µk2a). $ $$ $ $ $ $ $ $ 2 $ By using (4n4) this equation becomes k2(e a)(e e a) l (e a) $ $$ $ 4aΛ#( µ#k3µaj3a#) ε $ j2a( µka)( µk2a)j ( µk2a) . i ( µka)( µk2a) 2
(
*
(4n12)
We claim that e a 0 on V. Since, otherwise, there exists an open subset W of V $ such that e a is nowhere vanished on W. Then, on W, (4n12) becomes $ 4aΛ#( µ#k3µaj3a#) ε $ j2a( µka)( µk2a)j ( µk2a). (4n13) k2(e e a) l $$ ( µka)( µk2a) 2 Combining this with the first equation in (4n3) we conclude that a is constant on W which is a contradiction. Therefore, e a vanishes identically on W. Consequently, $ from the second equation of (4n3), we obtain ε a# Λ# lka#( µka)#k a( µka). $ 2
(4n14)
On the other hand, from the first equation in (4n3), we have k2aΛ#( µ#k3µaj3a#) $ j( µk2a)(a#kaµkε) l 0. ( µk2a)( µka)
(4n15)
(4.14) and (4n15) imply that a satisfies a polynomial equation of degree 4 with constant coefficients on V. Hence, a is constant on each component of V.
476
B.-Y. C J. Y
Therefore, we know that, for any dimension n 2, every component of V is an isoparametric hypersurface of the hyperbolic space. On the other hand, according to a well-known result of Cartan, every isoparametric hypersurface of Hn+"(k1) has at most two distinct principal curvatures. Therefore, we conclude that each component of V has at most two distinct principal curvatures. This contradicts the definition of V. Thus, V must be an empty set. Consequently, Mn is an isoparametric hypersurface of Hn+"(k1) with exactly two distinct principal curvatures. Therefore, by applying Cartan’s classification theorem of isoparametric hypersurfaces in hyperbolic spaces, Mn is an open set of the Riemannian product of H#(k") and # Sn−#(1) isometric embedded in the hyperbolic (nj1)-space in a standard way. Such a hypersurface is a tubular hypersurface with radius r l coth−"(N2) about a 2-dimensional totally geodesic surface (cf. [5]). Now, assume that Mn is a non-totally geodesic minimal hypersurface. Let U denote the open subset of Mn consisting of non-totally geodesic points. Then U is an open dense subset of Mn. Now, by a result of [4], U has relative nullity nk2. In this case, by applying an argument similar to spherical case, we may conclude that the Gauss image B# of U is a minimal surface in the unit de Sitter space-time Sn+"(1). " Consequently, U is an open subset of the unit bundle NB# defined in Theorem 3. The converses are easy to verify. Remark 1. Given a non-totally geodesic minimal hypersurface Mn of Sn+"(1) satisfying equality (1n2), let U be the open dense subset consisting of non-totally geodesic points as before. Then the distribution given by the eigenvectors with eigenvalue 0 is integrable and its leaves are totally geodesic in Sn+"(1). Hence, there are local coordinates (u, v, x , … , xn− ) such that c\cx , … , c\cxn− span . The surface # " # " B# of Theorem 2 can be chosen to be B# l oq ? U Q x l 0, … , xn− l 0q and # " φ : B# 4 Sn+" 9 n+# is given by (u, v) j ξ(u, v, 0, … , 0), where ξ is the unit normal vector field of M in Sn+"(1). The surface φ : B# 4 Sn+"(1) can also be considered as the set of corresponding focal points on U in the sense of [22], which in Theorem 3 is no longer possible. However, in the hyperbolic case, φ(B#) can be interpreted as the focal points beyond the horizon of the hyperbolic space. 5. Proof of Theorem 4 First we prove L 5. Let M$ 9 R%(ε) be a conformally flat hypersurface of a Riemannian space form R%(ε) satisfying equality (1n2), then M$ has at most two distinct principal curvatures. This Lemma can be proved as follows. Assume M$ is a conformally flat hypersurface in a 4-dimensional Riemannian space form R%(ε) satisfying equality (1n2). Let L be the symmetric 2-tensor defined by τ L lkRicj g, 2
(5n1)
where Ric, τ and g denote respectively the Ricci tensor, the scalar curvature and the metric tensor of M$. Then by a result of H. Weyl we have (]Y L)(Z, W) l (]Z L)(Y, W),
(5n2)
Elliptic functions, theta function and hypersurfaces
477
for vectors Y, Z, W tangent to M$. Let λ l a, λ l µka, λ l µ be the principal " # $ curvatures of M$ with their corresponding principal directions e , e , e given as " # $ before. Then, from the equations of Gauss and (5n2), we have (λ#j kλ#i ) Γjij l 3(ei H)λjk" ei τkei λ#j , 5 # 6 7 (λ#j kλ#k)Γkji l (λ#i kλ#k)Γkij , 8 τ l 3εjµ#jaµka#,
(5n3) (5n4)
for distinct i, j, k (i, j, k l 1, 2, 3), where H l # µ is the mean curvature function. $ Equations (3n4) and (5n3) imply 2(ei µ) λjk"ei( µ#jaµka#)kei λ#j l (λjjλi)(λjkλi) Γijj # lk(λijλj) ei λj . Thus (λikλj)ei λj l (k2λjjµj" a) ei µj(" µka) ei a. # # By taking (i, j) being (1, 2), (2, 3) and (3, 1) respectively, we obtain ( µk2a) e a l ae µ, (5n5) " " ( µk2a)e a l (2µk3a) e µ, (5n6) # # µe a l (2µk3a) e µ. (5n7) $ $ Let V denote the open subset of M$ on which M$ has exactly three distinct principal curvatures, i.e., V is given by (3n6). Suppose that V is not empty. In the remaining part of this section we shall work on this non-empty subset to obtain a contradiction. From (3n5) and (5n3) we obtain (5n8) Γijk l 0, for distinct i, j, k. Equation (5n8) (or Lemma 1) implies ea e ( µka) ω" l # ω j " ω . # µk2a " µk2a #
(5n9)
By taking the exterior derivative of (5n9) and by applying Cartan’s structure equations, we obtain the following formulas on V by comparing the corresponding coefficients in the resulting formula of dω". # e ( µka) ea (e a)# (e ( µka))# # ke k # k " e " # µk2a ( µk2a)# ( µk2a)# " µk2a
0
1 0
1
(e a) e ( µka) ja( µka)jε, l $ $ a( µka)
0
1
ea (e a)(e a) (e a)(e µ) # # $ # , j lk $ e $ µk2a ( µka)( µk2a) a( µka)
0
1
e ( µka) e ( µka) e ( µka) (e µ) e ( µka) $ $ j" l " . e " $ µk2a a( µk2a) a( µka)
(5n10) (5n11) (5n12)
By (5n7) we have Λ l Γ# l 0. Thus, by applying (3n10), (3n13), (5n5), (5n6) and $ $" (5n7), we find 2µk3a (e µ)(e µ), (5n13) e e µl # "# ( µk2a)# " 3akµ (e µ)(e µ). e e µl # #" ( µk2a)# "
(5n14)
B.-Y. C J. Y
478
On the other hand, by Lemma 1, (5n5) and (5n6), we have ea e ( µka) 3 e µ lk (e µ)(e µ). [e , e ] µ lkΓ" e µjΓ# e µ lk # e µk " # " # #" " "# # µk2a " µk2a # µk2a " Combining this with (5n13) and (5n14) we find (e µ)(e µ) l 0. (5n15) " # Therefore (e µ)(e µ) l 0 on V. " # If e µ ) 0 on V, there exists an open subset W 9 V on which e µ 0. On W, we " # have e µ l 0 identically. Moreover, from (3n11) and (5n11) we have " 3a#k3aµj2µ# (e µ)(e µ), (5n16) (2µk3a) e e µ lk # $ #$ aµ (2µk3a) e e µ l $#
6a#k3aµk2µ# (e µ)(e µ). # $ aµ
(5n17)
Combining (5n16) and (5n17) we obtain (2µk3a) [e , e ] µ l # $
3(2µk3a) (e µ)(e µ). # $ µ
(5n18)
On the other hand, we also have e ( µka) e µ e µk # e µ [e , e ] µ l Γ# e µkΓ$ e µ lk $ # # $ #$ # $# $ a a $ 3 lk (e µ)(e µ). $ µ # Therefore, we find
(2µk3a)(e µ)(e µ) l 0. (5n19) # $ If (e µ)(e µ) ) 0 on W, then on the open subset Wh of W on which (e µ)(e µ) 0 # $ # $ we have 2µk3a l 0. (5n20) Since 2µk3a l 0 on Wh, (5n6) and (5n7) imply e a l e a l 0 on Wh. On the other # $ hand, since e µ l 0 on W, (5n5) yields e a l 0. Therefore, we obtain e µ l e µ l " " " # e µ l 0 which is a contradiction. Consequently, we must have e µ 0 on W, from $ $ which we obtain e a l e a 0 on W by virtue of (5n7). Thus, by (3n9) and (5n6), we " $ have (2µk3a)(e µ)# l a( µk2a)#(aµjε). (5n21) # On the other hand, from (5n6), we know that 2µk3a 0 on W. Thus, there is an open subset W 9 W on which 2µk3a 0. On W , we have (5n21) which yields " " a(aµjε)( µk2a)# . (5n22) (e µ)# l # 2µk3a By using (5n6) and (5n22) we obtain 1 e e µl ok6a&k15a%µj38a$µ#k23a#µ$j4aµ% ## (2µk3a)# jε(k16a$j27a#µk14aµ#j2µ$)q.
(5n23)
Elliptic functions, theta function and hypersurfaces
479
On the other hand, (5n6) and (5n10) imply on W we have " 2µk3a 12µ#k33µaj23a# e e µj (e µ)# lka( µka)kε. # ( µk2a)# # # ( µk2a)% By using (5n22), the above equation becomes 1 ke e µ l o12a&k9a%µk2a$µ#ka#µ$j2aµ% ## (2µk3a)# Combining (5n23) and (5n24) we get
jε(11a$k13a#µjaµ#j2µ$)q.
6a$k12a#µj6aµ#jε(4µk5a) l 0.
(5n24) (5n25)
Differentiating (5n25) and applying (5n6) we find k30a$j72a#µk54aµ#j12µ$jε(7ak6µ) l 0.
(5n26)
Also, by combining (5n25) and (5n26) we get 6a#µj6µ$k12aµ#jε(7µk9a) l 0. Equation (5n27) implies al
9εj12µ#XN81j48εµ# . 12µ
(5n27)
(5n28)
By substituting (5n28) into (5n26) we conclude that µ must satisfy the following polynomial equation with constant coefficients 162j120µ#j88εµ%j32µ' lpN3(9j8εµj8µ%) N27j16εµ#.
(5n29)
which is impossible since otherwise µ is locally a constant function. Thus we must have e µ 0. # Similarly, we may prove that e µ 0, too. Hence, by applying (5n5), (5n6) and (5n7), " we obtain e a l e a l e µ l e µ l 0. (5n30) " # " # Now we claim that e µ 0 on V, too. In fact, otherwise there exists an open subset $ O 9 V on which e µ 0. " $ If 2µk3a l 0 on V, then (5n7) implies e a l 0. Therefore, by (5n10), we get $ a( µka)jε l 0. Thus, 2µ#j9ε l 0 which is impossible. If µ l 3a, then (5n7) yields µe µ l 0, which is also a contradiction. Hence, we have $ 2µ 3a and µ 3a on V. Now, from (5n10) we get (2µk3a)( µk3a) (e µ)# l a( µka)jε. $ µ#a( µka) Thus, there exists an open subset O 9 O on which # " (a( µka)jε) µ#a( µka) (e µ)# l . $ (2µk3a)( µk3a)
(5n31)
(5n32)
On the other hand, (3n9) and (5n7) imply e µ (e µ)#(2µk3a)# $ k $ l aµjε. 0(2µk3a) µ( µka) 1 ( µka)# µ#
ke $
(5n33)
480
B.-Y. C J. Y
Equations (5n32) and (5n33) yield µ e e µl $$ ( µk3a)(2µk3a)# io6a&k15a$ µ#j11a# µ$k2aµ%jε(3a$k15a#µj11aµ#k2µ$)q.
(5n34)
Similarly, by applying (3n12), (5n7) and (5n32), we have µ e e µl $$ (2µk3a)( µk3a)# io6a&k30a%µj45a$µ#j26a#µ$j5aµ%jε(3a$j6a#µk10aµ#j3µ$)q.
(5n35)
Summing up (5n34) and (5n35) yield µ#( µk2a) ok24a%j48a$µk30a#µ#j6aµ$jε(15a#k15aµj4µ#)q l 0.
(5n36)
On the other hand, from (5n8), we know that the distribution U spanned by e , " e is integrable. Also, the distribution spanned by e is clearly integrable. # $ Therefore, there exists a local coordinate system ox , x , x q such that e l c\ct, where " # $ $ t l x . Hence, by applying (5n30), we know that both a and µ depend only on t. $ Therefore, (5n7) yields da µ l 2µk3a. (5n37) dµ By solving (5n37) we obtain µ a l jCµ−$, (5n38) 2 for some constant C. By substituting (5n38) into (5n36), we know that µ must satisfy a polynomial equation with constant coefficients. Therefore, e µ l 0 on O which is $ # a contradiction. Consequently, both µ and a are constant on each component of V. Hence, by (3n9), (3n12) and (5n10), we get aµ l ( µka)µ lkε which is clearly impossible. Consequently, we know that V is an empty set. Hence, M$ has at most two distinct principal curvatures. This completes the proof of Lemma 5. Now, we go back to the proof of Theorem 4. If x : Mn 4 n+" is an isometric immersion of a conformally flat n-manifold with n 2 which satisfies equality (1n2). Then, by Lemma 5 and a well-known result of Cartan and Schouten on conformally flat hypersurfaces (cf. p. 154 of [3]), we know that Mn has a principal curvature with multiplicity at least nk1 for n l 3 as well as for n 4. Thus, by applying (3n2), we know that either (i) the principal curvatures of Mn are given by λ l 0, λ l ( l " # λn l µ, or (ii) n l 3 and the principal curvatures of M$ are given by λ l λ l " µ, " # # λ l µ 0. $ If (i) occurs, Mn is either totally geodesic, or an open portion of spherical hypercylinder, or an open portion of a round hypercone (cf. [15]). Now, we assume (ii) occurs. Denote by U the open subset of M$ on which the mean curvature function is nonzero. Then U is a non-empty open subset of M$. We shall work on U unless mentioned otherwise. From (3n5) we obtain Γ# l Λ l 0, e µ l u µ l 0, Γ" l Γ# l 0. $" $ " # $$ $$
(5n39)
Elliptic functions, theta function and hypersurfaces
481
Denote by and the distributions spanned by oe , e q and oe q, respectively. " # $ By (5n39) we know that the integral curves of U are geodesics and the distribution is integrable. Consequently, there exist local coordinate systems ox, u, vq such that is spanned by oc\cu, c\cvq and e l c\cx. $ From (5n39) we know that µ depends only on x, i.e., µ l µ(x). Also, from (3n8) and (5n39), we have µh(x) µh(x) ω", ω# l ω#. (5n40) ω" l $ µ(x) $ µ(x) Using (5n40) we obtain µh(x) ]e e l (5n41) e , j l 1, 2. j $ µ(x) j U
Therefore, each integral submanifold of is an extrinsic sphere of %. Hence, by applying a result of Hiepko [14] (cf. also [13, remark 2n1]), we know that M$ is locally the warped product Iif(x) S#(1), where f(x) is a suitable warped function. So, the metric of M$ is given by g l dx#jf(x)# g , (5n42) ! where g is the standard metric of S#(1). In particular, if we choose the spherical ! coordinate system oθ, φq for S#(1), we have g l dx#jf #(x) (dφ#jcos# φdθ#).
(5n43)
Applying (5n43) we obtain c c fh c l 0, ]cc l , x cx x cθ f cθ
]cc
]cφc
c c lkf f h , cφ cx
]cθc
c c c lkf f h cos# φ jsin φ cos φ . cθ cx cφ
]cφc
5
c fh c l x cφ f cφ
]cc
c c lktan φ cθ cθ
(5n44) 7
6
8
By computing dω" and by using (5n40) and Cartan’s structure equations, we find $ µ$(x) µd(x)j l 0. (5n45) 2 Solving (5n45) yields 4 4µh#jµ% l , (5n46) a% for some real number a 0. Now, we claim that U is dense in M$. If it is not, then M$kU has non-empty interior. From the definition of U, we know that the squared mean curvature function, µ and µh vanish identically on the interior of M$kU. On the other hand, (5n46) says that this is impossible due to the continuity of the squared mean curvature function. Thus, U must be an open dense subset of M$. Solving (5n46) yields µ(x) l (N2\a) sd ((N2\a) xjb, 1\N2) for some constant b. By applying a translation in x if necessary, we obtain µ(x) l
0
1
N2 N2 sd x, k , a a
kl
1 . N2
(5n47)
B.-Y. C J. Y
482
From (5n43) and from our assumption on the principal curvatures, we know that the second fundamental form h of M$ on % satisfies
0cxc , cxc 1 l µξ, h 0cφc , cφc 1 l "# µf # ξ, h 0cθc , cθc 1 l "# µf # cos# φξ c c c c c c h 0 , 1 l h 0 , 1 l h 0 , 1 l 0, cx cφ cx cθ cφ cθ 5
h
(5n48) 6 7
8
where ξ is a unit normal vector field of M$ in % and µ is given by (5n47). Applying (5n44), (5n48) and the equation of Codazzi, we may obtain µhf l µf h. Therefore, f(x) l cµ(x), (5n49) for some nonzero constant c. On the other hand, using (5n44) we can compute the sectional curvature K of the #$ plane section spanned by oc\cφ, c\cθq. On the other hand, we may also compute K #$ by using the equation of Gauss. By comparing these two different expressions of K , #$ we find µ% f # l 4(1kf h#). Thus, by using (5n47) and (5n49), we find f(x) l
0
1
Na N2 1 sd x, k , k l . 2 a N2
(5n50)
From (5n43), (5n50) and Lemma 4 we know that M$ is an open portion of the Whitney 3-sphere W$a for some a 0. (5n43), (5n44), (5n47), (5n50) and the formula of Gauss imply that the immersion x satisfies the following system of partial differential equations :
0 1 N2 N2 N2 cx N2 c#x a a lk x sd x nd x j x ξ, cd sd$ cφ# N2 0 a 1 0 a 1 0 a 1 cx 2N2 0 a 1 N2 c#x N2 l sd x ξ, cx# a a
c#x c#x cx l cos# φ jsin φ cos φ , cθ# cφ# cφ
(5n52) (5n53)
0 1 0 1 N2 N2 N2 cx c#x l cs 0 x nd x , cx cθ a a 1 0 a 1 cθ
N2 N2 N2 cx c#x l cs x nd x , cx cφ a a a cφ
Solving (5n56) yields
(5n51)
(5n54) (5n55)
c#x cx lktan φ . cθ cφ cθ
(5n56)
x(x, φ, θ) l B(x, θ) cos φjC(x, φ),
(5n57)
for some functions B(x, θ), C(x, φ) of two variables. Substituting (5n57) into (5n54) yields 5 N2 N2 cB N2 l cs x nd x B, cx a a a 6 (5n58) 7 N N N c# C 2 2 2 cC l cs x nd x . cx cφ a a a cφ
0 1 0 1 0 1 0 1
8
483
Elliptic functions, theta function and hypersurfaces Solving (5n58) yields B(x, θ) l G(θ) sd
0Na2 x1 ,
C(x, φ) l F(φ) sd
0Na2 x1jH(x).
(5n59)
Combining (5n57) and (5n59) yields x(x, φ, θ) l G(θ) sd
0Na2 x1 cos φjF(φ) sd 0Na2 x1jH(x).
(5n60)
By taking the partial derivatives of (5n52) and (5n53) with respect to φ and θ respectively, we find Fe(φ)jF h(φ) l 0, Ge(θ)jGh(θ) l 0.
(5n61)
From (5n60) and (5n61) we find x(x, φ, θ) l sd
0Na2 x1 (c" cos φ cos θjc# cos φ sin θjc$ cos φjc% sin φ)jA(x). (5n62)
where c , … , c are constant vectors. " % Substituting (5n62) into (5n52) and using (5n51) yields Ad(x) l
0 1 0 1
N2 N2 2N2 cs x nd x Ah(x) a a a
(5n63)
After solving (5n63) we obtain from (5n62) that x(x, φ, θ) l sd
0Na2 x1 & 0Na2 x1 dx.
i(c cos φ cos θjc cos φ sin θjc cos φjc sin φ)jc sd$ " # $ % &
(5n64)
where c , … , c are constant vectors in %. " & By choosing suitable initial conditions we will obtain (1n7) from (5n64). Consequently, up to rigid motions of %, the immersion x is given by (1n7). The converse can be verified by straightforward computations. 6. Exact solutions of differential equations of Picard type For the proof of Theorems 5 and 6, 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 by themselves. P 1. For any real numbers a 1 and β 0, the general solution of the second order differential equation : yd(x)j2a sc (ax) dn (ax) yh(x)ka# β# y(x) l 0
(6n1)
is given by y(x) l c y (x)jc y (x) with " " # # aβ Nk#jβ# i Θ(axkγ) xj ln jiaZ(γ) x y l Nkh#kβ# cn#(ax) cos " Nkh#kβ# 2 Θ(axjγ)
00
1
1
B.-Y. C J. Y
484 and
y l Nkh#kβ# cn#(ax) sin #
k#jβ# Θ(axkγ) jiaZ(γ) x1 , 00aβNNkh#kβ# 1 xj2i ln Θ(axjγ)
where k l Na#k1\(N2a) and kh l Na#j1\(N2a) are the modulus and the complementary modules of the elliptic functions, and γ l sn−"(iβ\(k Nkh#kβ#)). Proof. The trick to solve (6n1) is to make two key transformations. First we make the transformation : y(x) l f(x) exp (i g(x)), (6n2) where f(x) and g(x) are real-valued functions. Then yh l ( f hji fgh) exp (i g)
(6n3)
yd l ( f dkf(gh)#ji(2f hghjfgd)) exp (i g).
(6n4)
Substituting (6n3) and (6n4) into (6n1) we get f(x) gd(x)j2( f h(x)ja sc (ax) dn (ax) f(x)) gh(x) l 0,
(6n5)
f d(x)kf(x)(gh(x))#j2a sc (ax) dn (ax) f h(x)ka# β# f(x) l 0.
(6n6)
Equation (6n5) can be written as (ln gh(x))h lk2 (ln f(x))hj2 (ln cn (ax))h
(6n7)
which yields gh(x) l
α cn (ax)# , f(x)#
(6n8)
for some constant α. Substituting (6n8) into (6n6) we obtain a second order nonlinear equation : f d(x)j2a sc (ax) dn (ax) f h(x)ka# β# f(x) l
α# cn% (ax) . f(x)$
(6n9)
We make the second transformation by putting f(x) l Nh(u),
u l cn# (ax).
(6n10)
From (6n9) and (6n10) we obtain another nonlinear equation : a# cn# (ax) dn# (ax) sn# (ax) (2h(u) hd(u)k(hh(u))#)ja#(k# cn# (ax) sn# (ax) kdn# (ax)) h(u) hh(u) l a# β# h(u)#jα# cn% (ax),
(6n11)
which, by (2n8), is equivalent to the following nonlinear equation : 2a#u(1ku) (k#ujkh#) h(u) hd(u)ka#(k#u#jkh#) h(u) hh(u) ka#(k#ujkh#) (uku#) hh(u)# l a# β# h(u)#jα#u#.
(6n12)
If h l bkcu is a linear function in u with constant coefficients, then (6n12) becomes a#b( β#bkckh#)j2a#c(ckh#kβ#b) uj(α#ja# β#c#ka#bck#ja#c#(k#kkh#)) u# l 0. (6n13) It is straightforward to verify that (6n13) holds if and only if b l ckh#\β# and α# l c#a#(k#jβ#)(kh#kβ#)\β#. This shows that h(u) l c(kh#\β#ku) is a solution of (6n12).
485
Elliptic functions, theta function and hypersurfaces If we choose c l 1 and α l a N(k#jβ#)(kh#kβ#)\β, then we obtain 1 f(x) l Nkh#kβ# cn# (ax). β
(6n14)
By combining (6n8) and (6n14) we obtain g(x) l
cn# (ax) dx. &!x αβ N(k#jβ#)(kh#kβ#) kh#kβ# cn# (ax)
(6n15)
On the other hand, from (2n8) and (2n9) we have αβ N(k#jβ#)(kh#kβ#) cn# (ax) kh#kβ# cn# (ax) 5
l
αβ Nk#jβ# aβkh# Nk#jβ# sn# (ax) k Nkh#kβ# Nkh#kβ# (kh#kβ#jβ# sn# (ax))
l
aβ Nk#jβ# aβk cn (γ) dn (γ) sn# (ax) k Nkh#kβ# Nkh#kβ# 1kk# sn# (γ) sn# (ax)
l
aβ Nk#jβ# aβk k (sn (axjγ)jsn (axkγ)) sn (ax) Nkh#kβ# 2 Nkh#kβ#
l
aβ Nk#jβ# ai j (k# (sn (axjγ)jsn (axkγ)) sn (γ) sn (ax)), Nkh#kβ# 2 8
0
1 7
where sn (γ) l
iβ kh , dn (γ) l , Nkh#kβ# k Nkh#kβ#
(6n16) 6
cn (γ) l
kh Nk#jβ# . k Nkh#kβ#
Combining (2n14) and (6n16), we obtain aβ N(k#jβ#)(kh#kβ#) cn# (ax) kh#kβ# cn# (ax) l
aβ Nk#jβ# ai j (Z (axkγ)kZ (axjγ)j2Z(γ)). Nkh#kβ# 2
(6n17)
From (6n15), (6n16) and (6n17) we find
& 0
5 aβ Nk#jβ# ai x xj (Z (axkγ)kZ (axjγ)j2Z(γ)) dx Nkh#kβ# 2 ! 6 7 N aβ k#jβ# i Θ (axkγ) xj ln l j2aZ(γ) x , Nkh#kβ# 2 Θ (axjγ)
g(x) l
1
(6n18) 8
where we applied (2n13) and the fact that Θ is an even function. Therefore, by (6n2), (6n14) and (6n18), we conclude that the functions y , y defined in Proposition 1 are " # independent solutions of (6n1). Consequently, the general solution of (6n1) is given by the linear combination of y , y . " # P 2. For any real numbers a 1 and β 0, the general solution of yd(x)j2a sc (ax) dn (ax) yh(x)ja# β#y(x) l 0
(6n19)
B.-Y. C J. Y
486
is given by y(x) l c y (x)jc y (x) with " " # # aβ Nk#kβ# Θ (axkγ) xj" ln y l Nkh#jβ# cn# (ax) cosh k jaZ(γ) x " # Θ (axjγ) Nkh#jβ# and aβ Nk#kβ# Θ (axkγ) xk" ln y l Nkh#jβ# cn# (ax) sinh kaZ(γ) x , # # Θ (axjγ) Nkh#jβ#
0
1
0
1
where k l Na#j1\(N2a) and kh l Na#k1\(N2a) are the modulus and the complementary modules of the elliptic functions and γ l sn−"( β\(k Nkh#jβ#)). Proof. This can be proved by using the same trick given in the proof of Proposition 1. After making the key transformation (6n2) for (6n19), we obtain (6n5) and f d(x)kf(x)(gh(x))#j2a sc (ax) dn (ax) f h(x)ja# β# f(x) l 0.
(6n20)
Solving (6n5) yields gh(x) l
α cn (ax)# , f(x)#
(6n21)
for some constant α. By substituting (6n21) into (6n20) we obtain f d(x)j2a sc (ax) dn (ax) f h(x)ja# β# f(x) l
α# cn% (ax) . f(x)$
(6n22)
After making the second key transformation f(x) l Nh(u) with u l cn# (ax) for (6n22), we find 2a#u(1ku)(k#ujkh#) h(u) hd(u)ka#(k#u#jkh#) h(u) hh(u) ka#(k#ujkh#)(uku#) hh(u)#ja#β#h(u)# l α#u#.
(6n23)
It is then straightforward to verify that a linear function h l bjcu is a solution of (6n23) if and only if b l ckh#\β# and α# l a#c#( β#kk#)( β#jkh#)\β#. In particular, if we choose c l 1 and α l a N( β#kk#)( β#jkh#)\β, then 1 f(x) l Nkh#jβ# cn# (ax), β
gh(x) l
aβ N( β#kk#)( β#jkh#) cn# (ax) . kh#jβ# cn# (ax)
(6n24)
On the other hand, from (2n8), (2n9) we have aβ N( β#kk#)( β#jkh#) cn# (ax) kh#jβ# cn# (ax) l
aβ Nβ#kk# aβkh# Nβ#kk# sn# (ax) k Nβ#jkh# Nβ#jkh# (kh#jβ#kβ# sn# (ax))
l
aβ Nβ#kk# i aβk cn (γ) dn (γ) sn# (ax) k Nβ#jkh# Nβ#jkh# (1kk# sn#(γ) sn# (ax))
l
aβ Nβ#kk# i aβk k (sn (axjγ)jsn (axkγ)) sn (ax) Nβ#jkh# 2 Nβ#jkh#
l
aβ Nβ#kk# ia k (k# (sn (axjγ)jsn (axkγ)) sn (γ) sn (ax)), Nβ#jkh# 2 8
5
7
6
(6n25)
487
Elliptic functions, theta function and hypersurfaces where sn (γ) l
β kh , dn (γ) l , Nkh#jβ# k Nβ#jkh#
cn (γ) l
kh Nk#kβ# . k Nkh#jβ#
Combining (2n14) and (6n25), we obtain aβ N( β#kk#)( β#jkh#) cn# (ax) kh#jβ# cn# (ax) l
aβ Nβ#kk# ia k (Z (axkγ)kZ (axjγ)j2Z(γ)). Nβ#jkh# 2
(6n26)
From (6n24) and (6n26), we have
&
5 aβ Nβ#kk# ia x xk (Z (axkγ)kZ (axjγ)j2Z (γ)) dx Nβ#jkh# 2 ! 6 7 N aβ k#kβ# i Θ (axkγ) xk ln li ki aZ(γ) x, Nβ#jkh# 2 Θ (axjγ)
g(x) l
(6n27) 8
where we applied (2n13) and the fact that Θ is an even function. Therefore, by (6n2), (6n24) and (6n27) we conclude that
0
1
aβ Nk#kβ# Θ (axkγ) xj" ln jaZ(γ) x z (x) l Nkh#jβ# cn# (ax) exp k # Θ (axjγ) " Nβ#jkh#
is a solution of (6n19). By applying the method of reduction of order, we know that z (x) l Nkh#jβ# cn# (ax) exp #
k#kβ# Θ (axkγ) xk" ln kaZ(γ) x1 0aβNNβ#jkh# # Θ (axjγ)
is a second independent solution of (6n19). Consequently, the functions y , y defined " # in Proposition 2 are two independent solutions of (6n19). Hence, the general solution of (6n19) is given by the linear combination of y , y . " # P 3. For any real numbers 0 a 1 and β 0, the general solution of yd(x)j2ak cn
0ak x1 sd 0ak x1 yh(x)ja#k#β# y(x) l 0
(6n28)
is given by y(x) l c y (x)jc y (x) with " " # #
’
y l "
β# dn#
β#k1 i Θ((a\k) xkγ) a xk ln ki Z(γ) x1 0ak x1kkh# cos 0aβk NNβ#kkh# 2 Θ((a\k) xjγ) k
β# dn#
β#k1 i Θ((a\k) xkγ) a xk ln ki Z(γ) x1 , 0ak x1kkh# sin 0aβk NNβ#kkh# 2 Θ((a\k) xjγ) k
and y l #
’
where k l N2a\N1ja# and kh l N1ka#\N1ja# are the modulus and the complementary modules of the elliptic functions and γ l sn−"( β\Nβ#kkh#).
B.-Y. C J. Y
488
Proof. This can also be proved by using the same trick. After making the key transformation (6n2) for (6n28), we obtain
0
0ak x1 sd 0ak x1 f(x)1 gh(x) l 0, a a a# β# f d(x)kf(x)(gh(x))#j2ak cn 0 x1 sd 0 x1 f h(x)j f(x) l 0. k k k# f(x) gd(x)j2 f h (x)jak cn
(6n29) (6n30)
Solving (6n29) yields gh(x) l
α dn# ((a\k) x) , f#
(6n31)
for some constant α. By substituting (6n31) into (6n30) we obtain f d(x)j2ak cn
(ax) . 0ak x1 sd 0ak x1 f h(x)ja#k#β# f(x) l α# dn% f(x)$
(6n32)
After making the second key transformation f(x) l Nh(u) with u l dn# (ax\k) for (6n30), we find 2a#u(1ku)(ukkh#) h(u) hd(u)ka#(u#kkh#) h(u) hh(u) ka#(ukkh#)(uku#) hh(u)#ja# β# h(u)# l α#k#u#.
(6n33)
It is then straightforward to verify that a linear function h l bjcu is a solution of (6n33) if and only if b lkckh#\β# and α# l a#c#( β#k1)( β#kkh#)\( β#k#). In particular, if we choose c l 1, α l a N( β#k1)( β#kkh#)\( βk), we obtain f(x) l
1 β
’
β# dn#
0ak x1kkh#,
gh(x) l
aβ N( β#k1)( β#kkh#) dn#((a\k) x) . (6n34) k( β# dn#((a\k) x)kkh#)
On the other hand, from (2n8), (2n9) we have aβ N( β#k1)( β#kkh#) dn#((a\k) x) k( β# dn#((a\k) x)kkh# 5
l
aβ Nβ#k1 ia βk cn (γ) dn (γ) sn#((a\k) x) k k Nβ#kkh# Nβ#kkh# (1kk# sn# (γ) sn#((a\k) x)
l
aβ Nβ#k1 a a a ki k# sn xjγ jsn xkγ N 2k k k k β#kkh#
0 0 0
(6n35) 6 7
1 0
11 sn (γ) sn 0ak x11 ,
ikh , Nβ#kkh#
dn (γ) l
8
where sn (γ) l
β , Nβ#kkh#
cn (γ) l
kh Nβ#k1 . Nβ#kkh#
From (2n14), (6n34), (6n35), we obtain
&00
1 0
1
1
5 αβ Nβ#k1 a x a a xki Z xkγ kZ xjγ j2Z(γ) x dx 2k k k k Nβ#kkh# ! 6 7 N aβ β#k1 i Θ((a\k) xkγ) a xk ln ki Z(γ) x. l 2 Θ((a\k) xjγ) k k Nβ#kkh#
g(x) l
8
(6n36)
Elliptic functions, theta function and hypersurfaces
489
Therefore, the functions y , y defined in Proposition 3 are independent solutions of " # (6n28). C 1. For any real numbers a 1 and β 0, the general solution of zd(x) l a#(1jβ#) nc# (ax, k) z,
kl
Na#k1 N2a
(6n37)
is given by z(x) l c cn (ax, k) y (x)jc cn (ax, k) y (x) where y (x), y (x) are defined in " " # # " # Proposition 1. Proof. Follows from Proposition 1 and the fact that equation (6n37) can be obtained from (6n1) by making the transformation : y(x) l cn (ax, k) z(x). C 2. For any real numbers a 1 and β 0, the general solution of zd(x) l a#(1kβ#) nc# (ax, k) z,
kl
Na#j1 N2a
(6n38)
is given by z(x) l c cn (ax, k) y (x)jc cn (ax, k) y (x) where y (x), y (x) are defined in " " # # " # Proposition 2. Proof. Follows from Proposition 2 and the fact that equation (6n38) can be obtained from (6n19) by making the transformation : y(x) l cn (ax, k) z(x). C 3. For any real numbers 0
0
zd(x) l a# 1k
1 0 1
β# a nd# x, k z, k# k
a
1 and β 0, the general solution of
kl
N2a , N1ja#
kh l
N1ka# N1ja#
(6n39)
is given by z(x) l c dn (ax\k, k) y (x)jc dn (ax\k, k) y (x) where y (x), y (x) are " " # # " # defined in Proposition 3. Proof. Follows from Proposition 3 and the fact that equation (6n39) can be obtained from (6n28) by making the transformation : y(x) l dn (ax\k, k) z(x). Remark 2. Conversely, since
0
1
5
d Θ(ukγ) ln l Z(ukγ)kZ (ujγ) du Θ(ujγ) l k# sn (u) sn (γ) (sn (ukγ)jsn (ujγ))k2Z(γ) 67 l
2k# cn (γ) dn (γ) sn (γ) sn# (u) k2Z(γ), 1kk# sn# (γ) sn# (u)
(6n40)
8
it is straightforward to verify that the two functions y , y defined in Proposition 1 " # (respectively, in Propositions 2 and 3) are indeed independent solutions of (6n1) (respectively, of (6n19) and of (6n28)). Remark 3. In 1879, Picard [19] discovered a method for solving the differential equation yd(x)jnk# cn (x) sd (x) yh(x)jαy(x) l 0, (6n41) where n is a positive integer and α a constant. Although (6n41) is similar to our differential equations (6n1) and (6n19). Picard’s method does not apply to them.
490
B.-Y. C J. Y 7. Proof of Theorem 5
Let x : Mn 4 Sn+"(1) 9 n+# be an isometric immersion of a conformally flat n-
manifold with n 2 which satisfies equality (1n2). Then, by Lemma 5 and a wellknown result of Cartan and Schouten on conformally flat hypersurfaces (cf. p. 154 of [3]), we know that Mn has a principal curvature with multiplicity at least nk1 for n l 3 as well as for n 4. Thus, by applying (3n2), we know that either (i) the principal curvatures of Mn are given by λ l 0, λ l ( l λn l µ, or (ii) n l 3 and the " # principal curvatures of M$ are given by λ l λ l " µ, λ l µ 0. We treat these two " # # $ cases separately. Case (i). λ l 0 and λ l ( l λn l µ. " # If µ l 0 identically, then Mn is totally geodesic in Sn+"(1). Now, suppose that Mn is not totally geodesic in Sn+"(1). Denote by U the open subset of Mn 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 Mn in this case. We denote by and U the distributions on the open subset U spanned by oe q and " oe , e , … , enq, respectively. # $ Now, by putting j l 1, i l 2, … , n in (3n6), we get ωi (e ) l 0 which implies that " " integral curves of are geodesics. Also by using the first equation in (3n6), we find e µ l ( l en µ l 0. #
(7n1)
It is easy to see from (3n5) that (7n2) ωi (ej) l 0, 2 i j n, " which implies that U is integrable. Consequently, there exist local coordinate systems ox , x , … , xnq such that c\cx , … , c\cxn span U and e l c\cx with x l x . # " " " # From (7n1), we know that µ depends only on x, i.e. µ l µ(x). If we choose i l 1 for the first equation in (3n5), we get µh(x) lkµ(x) ω j(ej) for any " j 2. Thus n
]e e l ωk(ej) ek l ω j(ej) ej lk(ln µ)h ej . " " j " k=#
(7n3)
Using (7n3) we know that each integral submanifold of U is an extrinsic sphere of Sn+"(1), i.e., it is a totally umbilical submanifold with nonzero parallel mean curvature vector in Sn+"(1). Hence, the distribution U is a spherical distribution. Therefore, by applying a result of Hiepko [14] (cf. also [13, remark 2n1]), we know that U is locally the warped product Iif(x) Sn−"(1), where f(x) is a suitable warped function. Therefore, the metric of U is given by g l dx#jf(x)# g , (7n4) ! where g is the metric of Sn−"(1). In particular, if we choose the spherical coordinate ! system ou , … , unq on Sn−"(1), then we have # (7n5) g l dx#jf#(x) (du#jcos# u du#j(jcos# u ( cos# un− du#n). " # # $ # By (7n5) we obtain
491
Elliptic functions, theta function and hypersurfaces c c fh c l , l 0, ]cc x cx x cu f cuk k
]cc
]c c
c
ui cu
j
lk(tan ui)
0
c , cuj
5
c c lkf f h , u# cu cx #
]c c
2i
j,
(7n6) 6 7
1
0
1
j−" c j−" sin 2uk j−" c lk f f h cos# ul j cos# ul cu , cx 2 j k l=# k=# l=k+" 2 i, j, k n. 8
]c c
c
uj cu
From (7n5) and the assumption on the principal curvatures, we know that the second fundamental form h of Mn in Sn+"(1) satisfies
0 0 0
1
0
1
5
c c c c l 0, h l µf # ξ, , , cx cx cu cu # # n−" c c c c h l µf # cos# u ξ, … , h , l µf # cos# uk ξ, 67 , # cu cu cun cun k=# $ $ c c c c l 0, h , l 0, 2 i j n, h , cx cuj cui cuj h
1
0
1
0
1
1
(7n7)
8
where ξ is a unit normal vector field of Mn in Sn+"(1). Let ]- h denote the covariant derivative of the second fundamental form h. Then by the equation of Codazzi, we have (]` cc h) x
0cuc # , cuc #1 l (]` h) 0cxc , cuc #1 . c
c
u#
(7n8)
From (7n6), (7n7) and (7n8) we obtain µh f lkµf h. Therefore µ(x) l
a , f(x)
(7n9)
for some nonzero constant a. Now, by applying (7n6), we know that the sectional curvatures K and K of the "# #$ plane sections spanned by c\cx, c\cu and c\cu , c\cu are equal to kf d\f and # # $ (1kf h#)\f #, respectively. On the other hand, from the equation of Gauss and our assumption on principal curvatures, we have K l 1 and K l 1jµ#. Therefore, by "# #$ combining these facts with (7n9), we obtain f djf l 0, f h#jf # l 1ka#,
(7n10)
which implies in particular that 0 a 1. Solving the first differential equation in (7n10) yields f(x) l C cos (xjb) for some constant b, C. By applying a translation in x if necessary, we have f(x) l C cos x.
(7n11)
Substituting (7n11) into the second differential equation in (7n10), we obtain C l N1ka#. Consequently, we obtain f(x) l N1ka# cos x,
µ(x) l
a sec x, N1ka#
(7n12)
492
B.-Y. C J. Y
where a is a constant satisfying 0 a 1. (7n12) implies µ a\N1ka# 0 on U. Hence, by the continuity of the squared mean curvature function H# l (nk1)# µ#\n#, we know that the mean curvature function is nowhere zero. Hence, U is the whole manifold Mn. Consequently, (7n4) and (7n12) imply that Mn is isometric to an open portion of Riemannian manifold BV na which was defined in the introduction. By applying (7n5), (7n6), (7n7), (7n12) and the formula of Gauss, we conclude that the isometric immersion x : Mn 4 Sn−"(1) 9 n+# satisfies the following system of partial differential equations : c#x lkx, (7n13) cx# c#x cx lk(tan x) , cx cuj cuj c#x cx lk(tan ui) , cui cuj cuj
j l 2, … , n , 2i
j,
c#x cx l "(1ka#) sin 2x ja N1ka# cos xξk((1ka#) cos# x) x, cu# # cx # c#x c#x cx j(" sin 2uj) , j l 2, … , nk1. l (cos# uj) cu#j+ cu#j # cuj " Solving equation (7n13) yields
(7n14) (7n15) (7n16) (7n17)
(7n18) x l P(u , … , un) sin xjQ(u , … , un) cos x, # # for some n+"-valued functions P l P(u , … , un) and Q l Q(u , … , un). Substituting # # (7n18) into (7n14) we know that P is a constant vector, we denote it by c . Thus " (7n19) x l c sin xjQ(u , … , un) cos x. " # Substituting (7n19) into (7n15) with i l 2, we obtain c# Q cQ l 0, j l 3, … , n (7n20) j(tan u ) # cuj cu cuj # which implies cQ j(tan u ) Q l φg (u ), (7n21) # # # cu # for some function φg l φg (u ). Therefore, by solving (7n21), we obtain # # # (7n22) Q l φ (u )jQ (u , … , un) cos u # # # $ $ for some functions φ l φ (u ) and Q l Q (u , … , un). # # # $ $ $ Similarly, by substituting (7n19) and (7n22) into (7n15) with i l 3 and j 3, we find (7n23) Q l φ (u )jQ (u , … , un) cos u , $ $ $ $ % % for some function φ l φ (u ) and Q l Q (u , … , un). Repeating such procedure nk2 $ $ $ % % % times, we obtain Q l φ (u )jQ (u , … , un) cos u 5 # # # $ $ Q l φ (u )jQ (u , … , un) cos u , $ $ $ % % $ (7n24) Q l φ (u )jQ (u , … , un) cos u , 67 % % % % & & < Qn− l φn− (un− )jφn(un) cos un− , 8 " " " "
Elliptic functions, theta function and hypersurfaces
493
with φn(un) l Qn(un). Substituting (7n24) into (7n19) we get x l c sin xjφ (u ) cos xjφ (u ) cos u cos xj( " # # $ $ # jφn− (un− ) cos u ( cos un− cos xjφn(un) cos u ( cos un− cos x. # " " # # "
(7n25)
Substituting (7n25) into (7n17) with j l nk1, we obtain φn(un)jφn(un) l cos un− φn− (un− )jsin un− φn− (un− ), " " " " " "
(7n26)
which implies that φn(un)jφn(un) l kn,
(7n27)
cos un− φn− (un− )jsin un− φn− (un− ) l kn, " " " " " "
(7n28)
for some constant vector kn. Solving (7n27) yields φn l cn+ sin unjcn+ cos unjkn, " #
(7n29)
for some constant vectors cn+ , cn+ . Combining (7n25) and (7n29) we obtain " # x l c sin xjφ (u ) cos xjφ (u ) cos u cos xj( " # # $ $ # jφn− (un− ) cos u ( cos un− cos x " " # # jcn+ cos u ( cos un− sin un cos xjcn+ cos u ( cos un cos x, " # " # #
(7n30)
where we rename φn− (un− )jkn cos un− as φn− (un− ). By applying (7n17) and by " " " " " repeating such procedure nk2 times, we may obtain x l c sin xjc cos xjc sin u cos xjc cos u sin u cos xj( " # $ # % # $ jcn+ cos u ( cos un− sin un cos xjcn+ cos u ( cos un cos x, " # " # #
(7n31)
for some constant vectors c , … , cn+ . " # Now, we choose the initial conditions at 0 l (0, … , 0) as follows : x(0) l aε jN1ka# εn+ , # #
5
cx (0) l ε , " cx 6 7
cx cx (0) l N1ka# εn+ , (0) l N1ka# ε , … , $ " cu cun 8 #
(7n32)
where oε , … , εn+ q is the natural coordinate basis of n+#. Then, by applying (7n31) # " and (7n32), we obtain c lε , " "
c l aε , # #
c l N1ka# ε , … , cn+ l N1ka# εn+ . $ # # $
(7n33)
Consequently, by (7n31) and (7n33) we conclude that, up to rigid motions of n+#, the immersion x is given by (1n8). It is clear from (1n8) that the immersion x can be extended to the two point compactification BV na of Bna. Case (ii). λ l λ l " µ and λ l µ 0. " # # $ In this case, (3n6) yields Γ# l Λ l 0, e µ l e µ l 0, Γ" l Γ# l 0. $" $ " # $$ $$
(7n34)
B.-Y. C J. Y
494
Denote by and the distributions spanned by oe , e q and oe q, respectively. " # $ By (7n34) we know that the integral curves of U are geodesics and the distribution is integrable. Consequently, there exist local coordinate systems ox, u, vq such that is spanned by oc\cu, c\cvq and e l c\cx. $ From (7n34) we know that µ depends only on x, i.e. µ l µ(x). Also, from (3n8) and (7n34), we have µh(x) µh(x) ω" l ω", ω# l ω#. (7n35) $ µ(x) $ µ(x) Using (7n35) we obtain µh(x) ]e e l (7n36) e , j l 1, 2. j $ µ(x) j U
Therefore, each integral submanifold of is an extrinsic sphere of S%(1). Hence, by applying a result of Hiepko, we know that M$ is locally the warped product Iif(x) S#(1), where f(x) is a suitable warped function. In particular, if we choose the spherical coordinate system oθ, φq for S#(1), we have g l dx#jf #(x) (dφ#jcos# φdθ#).
(7n37)
By computing dω" and by using (7n35) and Cartan’s structure equations, we find $ µd(x)j
µ$(x) jµ(x) l 0. 2
(7n38)
Put ψ(x) l 2f(x). Then (7n38) becomes ψd(x)j2ψ$(x)jψ(x) l 0. Hence, by applying lemma 5n3 of [7], we obtain µ(x) l N2(a#k1) cn (ax, k),
kl
Na#k1 N2a
(7n39)
where a is real number 1. Now, by applying (7n37), (7n39) and the equation of Codazzi, we may obtain µh f l µ f h. Therefore, f(x) l cµ(x), (7n40) for some nonzero constant c. On the other hand, using (7n37) we can compute the sectional curvature K of the #$ plane section spanned by oc\cφ, c\cθq. On the other hand, we may also compute K #$ by using the equation of Gauss. Comparing these two different expressions of K so #$ obtained yields (7n41) c#µh# l 1kc# µ#k"c# µ%. % Substituting (7n39) into (7n41) we have c# l 1\(a%k1). Therefore fl
N2 cn (ax, k), Na#j1
µ l N2(a#k1) cn (ax, k),
kl
Na#k1 . N2a
(7n42)
(7n37) and (7n42) imply that M$ is an open portion of the warped product manifold Ii(N2\Na#j1) cn (ax) S#(1) which is isometric to P$a ; first introduced in [7] (see also Introduction and [10]).
Elliptic functions, theta function and hypersurfaces
495
(7n37), (7n39), (7n42) and the formula of Gauss imply that the immersion x satisfies the following system of partial differential equations : c#x l N2(a#k1) cn (ax) ξkx, cx#
(7n43)
c#x 2a cx l cn (ax) dn (ax) sn (ax) cφ# a#j1 cx j
N2(a#k1) 2 cn$ (ax) ξk cn# (ax) x, a#j1 a#j1
(7n44)
c#x 2a cx cx l cos# φ cn (ax) dn (ax) sn (ax) jsin φ cos φ cθ# a#j1 cx cφ j
N2(a#k1) 2 cos# φ cn$ (ax) ξk cos# φ cn# (ax) x, a#j1 a#j1
(7n45)
c#x cx lka sc (ax) dn (ax) , cx cφ cφ
(7n46)
c#x cx lka sc (ax) dn (ax) , cx cθ cθ
(7n47)
c#x cx lktan φ . cθ cφ cθ
(7n48)
By taking the partial derivative of (7n44) with respect to φ and by applying the equation of Weingarten, we obtain c$x cx lk , cφ$ cφ
(7n49)
by virtue of (2n8), (2n10) and k# l
a#k1 , 2a#
kh# l
a#j1 . 2a#
(7n50)
Solving (7n49) yields x(x, φ, θ) l A(x, θ) sin φjB(x, θ) cos φjC(x, θ),
(7n51)
for some functions A, B, C of two variables. Substituting (7n51) into (7n46) yields cA cB lka sc (ax) dn (ax) A, lka sc (ax) dn (ax) B. cx cx
(7n52)
Solving (7n52) yields A(x, θ) l E(θ) cn (ax),
B(x, θ) l D(θ) cn (ax)
(7n53)
for some functions E, D. Thus, x(x, φ, θ) l E(θ) cn (ax) sin φkD(θ) cn (ax) cos φjC(x, θ).
(7n54)
Substituting (7n54) into (7n47) yields c#C cC lka sc (ax) dn (ax) . cx cθ cθ
(7n55)
B.-Y. C J. Y
496
By solving (7n55) we find C(x, θ) l G(θ) cn (ax)jK(x) for some functions G(θ) and K(x). Thus, (7n54) gives x(x, φ, θ) l E(θ) cn (ax) sin φjD(θ) cn (ax) cos φjG(θ) cn (ax)jK(x).
(7n56)
Substituting (7n56) into (7n48) yields Eh(θ) l Gh(θ) l 0. Thus, E and G are constant vectors in &. Consequently, from (7n56) we know that x takes the form : x(x, φ, θ) l c cn (ax) sin φjD(θ) cn (ax) cos φjF(x). " where c is a constant vector. From (7n43) we have " 1 c#x jx . ξl N2(a#k1) cn (ax) cx#
0
1
(7n57)
(7n58)
By (7n45), (7n57), (7n58) and a long computation, we obtain Fd(x)j2a sc (ax) dn (ax) Fh(x)kF(x) l 0,
(7n59)
Dd(θ)jD(θ) l 0,
(7n60)
by virtue of (2n8), (2n10) and (7n50). Solving (7n60) yields D(θ) l c cos θjc sin θ, (7n61) # $ for some constant vectors c , c in &. Therefore, by applying Proposition 1, (7n57) and # $ (7n61), we obtain x(x, φ, θ) l c sin φ cn (ax)jc cos φ cos θ cn (ax)jc cos φ sin θ cn (ax) " # $ kh i akh# Θ(axkγ) ln xj j2a Z(γ) x jc Na#kh#kcn# (ax) cos % k Θ(axjγ) 2 Na#k%j1
0 0 11 kh i akh# Θ (axkγ) jc Na#kh#kcn# (ax) sin 0 xj ln j2a Z(γ) x11 , 0 & k N θ(axjγ) 2 a#k%j1
5
(7n62) 7
6
8
where Θ and Z are the theta and zeta functions. Now, if we choose the initial conditions at 0 l (0, 0, 0) as follows : x(0) l
1 (ε jε ), akh # %
cx cx 1 cx 1 (0) l ε , (0) l ε, (0) l ε, & cφ cx akh " cθ akh $
where oε , … , ε q is the standard basis of &, then we obtain " & 1 cα l ε , α l 1, … , 5. akh α
(7n63)
(7n64)
Therefore, up to rigid motions, the immersion x is given by (1n9) in Theorem 5. The converse can be verified by straightforward but long computations. 8. Proof of Theorem 6 Let x : Mn 4 Hn+"(k1) 9 n+# be an isometric immersion of a conformally flat " n-manifold with n 2 which satisfies equality (1n2). By Lemma 5 and a result of Cartan and Schouten, either (a) the principal curvatures of Mn are given by λ l 0, " λ l ( l λn l µ, or (b) n l 3 and the principal curvatures of M$ are given by λ l " # λ l " µ, λ l µ 0. We treat these two cases separately. # # $
Elliptic functions, theta function and hypersurfaces
497
Case (a). λ l 0 and λ l ( l λn l µ. " # If µ l 0 identically, then Mn is totally geodesic. This yields Case (i) of Theorem 6. Now, suppose that Mn is not totally geodesic in Hn+"(k1). Denote by U the open subset of Mn on which the mean curvature function is nonzero. We shall work on this open subset U unless mentioned otherwise. Clearly, µ 0 on U. We denote by and U the distributions on the open subset U spanned by oe q " and oe , e , … , enq, respectively. Then, as in the proof of Theorem 5, we can prove that # $ integral curves of are geodesics and U is integrable. Thus, there exist local coordinate systems ox , x , … , xnq such that c\cx , … , c\cxn span U and e l c\cx # " " # with x l x . Also, using (3n6), we can prove that µ depend, only on x, i.e., µ l µ(x). " If we choose i l 1 for the first equation in (3n5), we get µh(x) lkµ(x) ω j(ej) for any " j 2. Thus n
]e e l ωk(ej) ek l ω j(ej) ej lk(ln µ)h ej. (8n1) " " j " k=# Hence, each integral submanifold of U is an extrinsic sphere of Hn+"(k1), i.e., the distribution U is spherical. Therefore, U is locally a warped product manifold Iif(x) N n−"(c` ), where f(x), ( f(x) 0), is the warped function and N n−"(c` ) is a Riemannian space form of constant sectional curvature c` . Since N n−"(c` ) is of constant curvature, it is conformally flat. Thus, there exists a local coordinate system ou , … , unq such that metric tensor of N n−"(c` ) is given by # (8n2) g l E#(du#jdu#j(jdu#n). ! # $ With respect to coordinate system ox, u , … , unq on Iif N n−"(c` ), we have # (8n3) g l dx#jf # E#(du#j(jdu#n). # From (8n3) we have c c fh c l , l 0, ]cc x cx x cu f cui i
]cc
5
c E c E c l i j j , ui cu E cuj E cui j
]c c
c c E c E c lkf f h E# j i k k , ]c c ui cu cx E cu E cuk i i ki
(8n4) 6 7
8
for distinct i, j, k (2 i, j, k n), where Ei l cE\cui. Codazzi’s equation, (8n4) and our assumption on principal curvatures imply µl
a f
(8n5)
for some constant a 0. Also, from (8n4), it follows that the sectional curvatures K "# and K of Mn associated with the plane sections spanned by c\cx, c\cu and c\cu , # # #$ c\cu are given by kf d\f and (c` kf h#)\f #, respectively. On the other hand, the equation $ of Gauss yields K lk1 and K lk1jµ#. Combining these facts with (8n5), we get "# #$ f dkf l 0, f #kf h# l a#kc` . (8n6) Solving the first equation of (8n6) yields f(x) l c cosh xjc sinh x, where c , c are " # " # constants. From (8n6) we have either f(x) l α sinh (xjb) or f(x) l α cosh (xjb), for some constants b, α. Thus, by applying a translation in x if necessary, we have f(x) l α sinh x or f(x) l α cosh x. We consider these two cases separately.
498
B.-Y. C J. Y
Case 1. f(x) l α cosh x. In this case, the second equation of (8n6) yields α# l a#kc` . Thus, after applying a scaling on E if necessary, we have c` l 1, f(x) l Na#k1 cosh x,
a sech x, Na#k1
µl
c` l 0, f(x)l cosh x, c` lk1, f(x) l Na#j1 cosh x,
µ l sech x, µl
a 1, or
or
a sech x, Na#j1
(8n7) (8n8)
a 0.
(8n9)
Case 1 a. c` l 1, f(x) l Na#k1 cosh x, µ l (a\Na#k1) sech x, a 1. In this case the open subset U is the whole manifold Mn. Therefore, Mn is an open portion of the warped product manifold Ana l iN # cosh x Sn−"(1). By choosing the a −" spherical coordinates ou , … , unq on Sn−"(1), we obtain (7n5) and (7n6) with # a f l Na#k1 cosh x, µ l sech x. (8n10) Na#k1 Therefore, the equation of Gauss implies that the isometric immersion x satisfies the following system of partial differential equations : c#x l x, cx# c#x cx l tanh x , cx cuj cuj c#x cx lktan ui , cui cuj cuj
(8n11) j l 2, … , n, 2i
j,
c#x 1ka# cx l sinh 2x ja Na#k1 cosh xξj((a#k1) cosh# x) x, cu# 2 cx # c#x c#x cx j" sin 2uj , j l 2, … , nk1. l cos# uj cu#j+ cu#j # cuj " where ξ is a unit normal vector field of Mn in Hn+"(k1). Solving equation (8n11) yields
(8n12) (8n13) (8n14) (8n15)
(8n16) x l P(u , … , un) sinh xjQ(u , … , un) cosh x, # # for some n+"-valued functions P l P(u , … , un) and Q l Q(u , … , un). By sub# " # stituting (8n16) into (8n12), we know that P is a constant vector, denoted by c . Thus " (8n17) x l c sinh xjQ(u , … , un) cosh x, " # Substituting (8n17) into (8n13) with i l 2 yields cQ j(tan u ) Q l φg (u ), # # # cu # for some function φg l φg (u ). Therefore, after solving (8n18), we obtain # # # Q l φ (u )jQ (u , … , un) cos u # # # $ $
(8n18)
(8n19)
Elliptic functions, theta function and hypersurfaces
499
for some functions φ l φ (u ) and Q l Q (u , … , un). Repeating such procedure # # # $ $ $ nk2 times, we obtain Q l φ (u )jQ (u , … , un) cos u # # $ $ # Q l φ (u )jQ (u , … , un) cos u , … , $ $ $ $ % % Qn− l φn− (un− )jφn(un) cos un− , " " " " where φn(un) l Qn(un). Substituting (8n20) into (8n17) yields 5
(8n20) 6 7 8
x l c sinh xjφ (u ) cosh xjφ (u ) cos u cosh xj(jφn− (un− ) " # # $ $ # " " icos u ( cos un− cosh xjφn(un) cos u ( cos un− cosh x. (8n21) # # " # Now, by applying (8n21) and (8n15), we may obtain in the same way as given in the proof of Theorem 4 that x l c sinh xjc cosh xjc sin u cosh xj(jcn+ cos u ( " # " # $ # icos un− sin un cosh xjcn+ cos u ( cos un cosh x, (8n22) " # # for some constant vectors c , … , cn+ . " # If we choose suitable initial conditions for x, cx\cx, cx\cu , … , cx\cun at # 0 l (0, … , 0), we will obtain (1n10) from (8n22). Consequently, up to rigid motions, the immersion x is given by (1n10). Case 1 b. c` l 0, f(x) l cosh x, µ l sech x. Again, the open subset U is the whole manifold Mn. Thus, Mn is an open portion of the warped product manifold Gn l icoshx n−". Hence, the metric tensor of Mn is given by g l dx#jcosh# x (du#j(jdu#n). (8n23) # In this case, the equation of Gauss implies that the isometric immersion x satisfies the following system : c#x l x, (8n24) cx# c#x cx l tanh x , cx cuj cuj
j l 2, … , n,
c#x l 0, 2 i cui cuj
(8n25) (8n26)
j,
c#x cx l sinh x cosh x jcosh xξj(cosh# x) x, cu#j cx
j l 2, … , n.
(8n27)
After solving this system, we obtain x(x, u , … , un) l c sinh x " # j(α u#j(jα#n u#njβ u j(jβn unjγ) cosh x, (8n28) # # # # for some constant vectors c , α , … , αn, β , … , βn, γ. " # # If we choose suitable initial conditions for x, cx\cx, cx\cu , … , cx\cun, we may # obtain (1n11) from (8n28). Thus, up to rigid motions, the immersion is given by (1n11).
500
B.-Y. C J. Y
Case 1 c. c` lk1, f(x) l Na#j1 cosh x, µ l (a\Na#j1) sech x, a 0. Again, the open subset U is the whole manifold Mn and Mn is an open portion of the warped product manifold Hna l iNa#+ coshx Hn−"(k1). Thus, the metric tensor " of Mn is given by g l dx#j(a#j1)(cosh# x) g , (8n29) ! where g l du#jsinh# u (du#jcos# u du#j(jcos# u ( cos# un− du#n). (8n30) ! # # $ $ % $ " If we apply (8n29), (8n30), our assumption on the second fundamental form and the equation of Gauss, we know that the isometric immersion x satisfies the following system : c#x l x, (8n31) cx# c#x cx l tanh x , cx cuj cuj
j l 2, … , n,
(8n32)
c#x cx l coth u , j l 3, … , n, # cu cu cuj j # c#x cx lktan uj , 3 i j n, cui cuj cuk
(8n33) (8n34)
c#x a#j1 cx lk sinh 2x ja Na#j1 cosh xξj(a#j1) cosh# x x, cu# 2 cx # c#x c#x cx l sinh# u ksinh x cosh x , # cu# cu# cu $ # # c#x c#x cx ksin uj cos uj , j l 2, … , nk1. l cos# uj cu#j+ cu#j cuj " After solving this system in the same way as Case 1 a we obtain
(8n35) (8n36) (8n37)
x l c sinh xjc cosh xjc cosh u cosh xjc sinh u cos u sinh x " # $ # % # $ j(jcn+ sinh u cos u ( cos un− sin un cosh x " # $ " jcn+ sinh u cos u ( cos un cosh x, (8n38) # # $ for some constant vectors c , … , cn+ . " # If we choose suitable initial conditions for x, cx\cx, cx\cu , … , cx\cun, we will # obtain (1n12) from (8n38). Case 2. f(x) l α sinh x In this case, (8n6) yields c` l α#ja#. Thus, c` 0. By applying a scaling on E if necessary, we have c` l 1 and α l N1ka#. In summary, we have c` l 1, f(x) l N1ka# sinh x,
µl
a csch x, N1ka#
0
a
1.
(8n39)
From (8n39) and continuity, we know that U is a dense open subset of Mn. Moreover, locally, U is an open subset of the warped product manifold Yna. Thus, the metric tensor of Mn is given by
0
0
1 1
n−" g l dx#j(1ka#) sinh# x du#jcos# u du#j(j cos# uj du#n . # # $ j=#
(8n40)
Elliptic functions, theta function and hypersurfaces
501
Therefore, by applying the equation of Gauss, we know that the isometric immersion x satisfies the following system : c#x l x, (8n41) cx# c#x cx l coth x , cx cuj cuj
j l 2, … , n,
c#x cx lktan ui , cui cuj cuj
2i
(8n42) (8n43)
j,
c#x a#k1 cx l sinh 2x ja N1ka# sinh xξj(1ka#) sinh# x x, cu# 2 cx # c#x c#x cx j" sin 2uj , j l 2, … , nk1. l (cos# uj) cu#j+ cu#j # cuj " After solving this system in the same way as Case 1 a we obtain
(8n44) (8n45)
x l c cos xjc sinh xjc sin u sinh xj(jcn+ cos u " # " # $ # i( cos un− sin un sinh xjcn+ cos u ( cos un sinh x, (8n46) " # # for some constant vectors c , … , cn+ . " # If we choose suitable initial conditions for x, cx\cx, cx\cu , … , cx\cun at # 0 l (0, … , 0), we will obtain (1n13) from (8n46). Therefore, up to rigid motions, the immersion is given by (1n13). Case (b). n l 3, λ l λ l " µ, λ l µ 0. " # # $ Let and U denote the distributions spanned by oe , e q and oe q, respectively. " # $ Then, as Case (ii) in the proof of Theorem 4, we can prove that the integral curves of U are geodesics and the distribution is a spherical integral distribution. Thus, there exist a local coordinate system ox, u, vq such that is spanned by oc\cu, c\cvq and e l c\cx. As in the proof of Theorem 4, we may also prove that µ l µ(x) $ depends only on x. Again, according to a result of Hiepko, M$ is locally a warped product manifold Iif(x) N #(c` ), where f(x) 0 is the warped function and N #(c` ) is a surface of constant curvature c` . So, the metric of M$ can be written as g l dx#jf(x)# E#(du#jdv#),
(8n47)
where ou, vq is an isothermal coordinate system on N #(c` ). Applying (8n47) we obtain c c fh c l 0, ]cc l , x cx x cu f cu
]cv
]cc
c
u cv
5
c fh c l x cv f cv
]cc
E c E c l v j u E cu E cv
(8n48) 7
6
c c E c E c ]cc lk f f h E# j u k v , u cu cx E cu E cv ]cc
c
v cv
lk f f h E#
c Eu c Ev c k j . cx E cu E bv 8
B.-Y. C J. Y
502
From (8n47) and the hypothesis on the principal curvatures, we know that the second fundamental form h of M$ in H%(k1) satisfies h
0
1
1 0 1 c c c c c c h0 , 1 l h0 , 1 l h0 , 1 l 0, cx cu cx cv cu cv
c c l µξ, , cx cx
0
h
c c l " µ f # E# ξ, , # cu cu
h
5
c c l " µ f # E# ξ, , # cv cv
(8n49) 7
6
8
where ξ is a unit normal field of M$ in H%(k1). From (8n48), (8n49) and the equations of Gauss and Codazzi, we find f(x) l αµ(x) µd(x)j
(8n50)
µ$(x) kµ(x) l 0, 2
fd K lk1j" µ# lk , # "$ f
(8n51)
c` kf h# K lk1j" µ# l , "# % f#
(8n52)
where α is a nonzero real number and K , K are the sectional curvatures of M$ of "# "$ the plane sections spanned by oe , e q, oe , e q, respectively. " # " $ Put ψ(x) l 2f(x). Then (8n51) becomes ψd(x)j2ψ$(x)kψ(x) l 0. Hence, by applying lemma 5n4 of [7], we know that µ l µ(x) is one of the following functions : (1) µ l N2 ; (2) µ l 2 sech (x) ; , a 1; 0 NNa#j1 2a 1 Na#j1 N2a (4) µ(x) l N2(a#j1) dn 0 N2 x, Na#j11 , 0
(3) µ(x) l N2(a#j1) cn ax,
a
1.
We consider these four cases separately. Case b 1. µ l N2, f l N2α. In this case, (8n52) yields c` lkα# 0. By choosing c` lk1, we obtain α l 1, f l N2. Therefore, U l M$ and M$ is an open portion of the warped product manifold iN H#(k1) which is isometric to F$. Therefore, if we choose the hyperbolic # coordinate system oφ, θq on H#(k1), we obtain g l dx#j2(dφ#jcosh# φdθ#).
(8n53)
By (8n53) µ l N2 and the formula of Gauss, we know that the immersion satisfies the differential system : c#x l N2 ξjx, cx#
5
c#x l N2 ξj2x ; cφ#
c#x cx lksinh φ cosh φ jN2 cosh# φξj2 cosh# φ x ; 67 cθ# cφ c#x c#x c#x cx l l 0, l tanh φ . cx cθ cx cφ cφ cθ cθ 8
(8n54)
Elliptic functions, theta function and hypersurfaces
503
After solving (8n54) we obtain x(u, φ, θ) l c cos xjc sin xjc sinh φjc cosh φ cosh θjc cosh φ sinh θ " # $ % & for some constants c , … , c . Thus, by choosing suitable initial conditions, we will " & obtain (1n14). Case b 2. µ l 2 sech (x), f l 2α sech (x). In this case, (8n52) yields c` l 0. By choosing α l " we obtain f l sech x. # Therefore, M# is an open portion of the warped product manifold isechx # which is isometric to F$. Thus g l dx#jsech# x(du#jdv#). (8n55) By (8n55), µ l 2 sech x and the formula of Gauss, we know that the immersion satisfies the following system : c#x l 2 sech x ξjx ; cx#
(8n56)
c#x cx l sech# x tanh x jsech$ x ξjsech# x x ; cu# cx
(8n57)
c#x cx l sech# x tanh x jsech$ x ξjsech# x x ; cv# cx
(8n58)
c#x cx c#x cx lktanh x , lktanh x ; cx cu cu cx cv cv
(8n59)
c#x l 0. cu cv
(8n60)
x(x, u, v) l A(x)jB(u, v) sech x,
(8n61)
Solving (8n59) yields
for some &-valued functions A(x), B(u, v). Substituting (8n61) into (8n60) we obtain " B(u, v) l C(u)jD(v) for some functions C(u), D(v). Thus x(x, u, v) l A(x)j(C(u)jD(v)) sech x.
(8n62)
Substituting (8n62) into (8n57) and (8n58) yields 2Cd(u) l sech x(Ad(x)j2(tanh x) Ah(x)jA(x)),
(8n63)
2Dd(v) l sech x(Ad(x)j2(tanh x) Ah(x)jA(x)).
(8n64)
(8n63) and (8n64) imply that there is a constant vector c such that " 5 Cd(u) l Dd(v) l 2c , " 6 7 Ad(x)j2(tanh x) Ah(x)jA(x) l 4c cosh x. 8 " Solving (8n65) yields C(u) l c u#jc ujb , D(v) l c v#jc vjb , 5 " # " " $ # 6 7 A(x) l sech x(c (x#k" cosh 2x)jc xjb ) 8 " # % $ for some constants c , c , c , b , b , b . Combining (8n62) and (8n66) yields # $ % " # $ x(u, φ, θ) l sech x oc (x#ju#jv#kcosh# x)jc ujc vjc xjc q " # $ % &
(8n65)
(8n66)
(8n67)
504
B.-Y. C J. Y
for some constant c . Thus, by choosing suitable initial conditions, we will obtain & (1n15) from (8n67). Consequently, up to rigid motions, the immersion x is given by (1n15). Case b 3. µ(x) l N2(a#j1) cn (ax, (Na#j1)\N2a), a 1. In this case, (8n52) yields c` l α#(α%k1) 0. By choosing c` l 1, we obtain µ(x) l 2ak cn (ax, k),
fl
1 cn (ax, k), akh
kl
Na#j1 Na#k1 . (8n68) , kh l N2a N2a
Thus, locally, U is an open portion of the warped product manifold C$a. If we choose the spherical coordinate system o φ, θq on S#(1), we have cn# (ax, k) (dφ#jcos# φdθ#). g l dx#j a# kh#
(8n69)
By (8n68), (8n69) and the formula of Gauss, we know that the immersion satisfies the following differential system : c#x l 2ak cn (ax) ξjx ; cx#
(8n70)
c#x 1 cx k cn$ (ax) cn# (ax) l cn (ax) dn (ax) sn (ax) j ξj x; cφ# akh# cx akh# a# kh#
(8n71)
c#x c#x cx l cos# φ jsin φ cos φ ; cθ# cφ# cφ
(8n72)
c#x cx lka sc (ax) dn (ax) ; cx cφ cφ
(8n73)
c#x cx lka sc (ax) dn (ax) ; cx cθ cθ
(8n74)
c#x cx lktan φ . cφ cθ cθ
(8n75)
By taking the partial derivative of (8n71) with respect to φ by applying (2n8), (8n73) and the Weingarten equation, we obtain c$x\cφ$jcx\cφ l 0, from which we get x(x, φ, θ) l A(x, θ) sin φjB(x, θ) cos φjC(x, θ), (8n76) for some &-valued functions A, B, C. Substituting (8n76) into (8n73) and (8n74) yields " A(x, θ) l cn (ax) D(θ), B(x, θ) l cn (ax) E(θ), C(x, θ) l cn (ax) F(θ)jG(x) for some functions D, E, F and G. Thus, we obtain x l D(θ) cn (ax) sin φjE(θ) cn (ax) cos φjF(θ) cn (ax)jG(x).
(8n77)
Substituting (8n77) into (8n75) yields Dh(θ) l Fh(θ) l 0. Thus, (8n77) implies x l c cn (ax) sin φjE(θ) cn (ax) cos φjK(x) (8n78) " for some constant vector c and function K(x). " By substituting (8n78) into (8n72) and by applying (2n8) and (8n70), we find
505
Elliptic functions, theta function and hypersurfaces
Ed(θ)jE(θ) l 0. Thus, E(θ) l c cos θjc sin θ for some constants c , c . Substituting # $ # $ this into (8n77) yields x l c cn (ax) sin φjc cn (ax) cos φ cos θjc cn (ax) sin φ cos θjK(x). " # $
(8n79)
Substituting (8n79) into (8n70) yields Kd(x)j2a sc (ax) dn (ax) Kh(x)jK(x) l 0.
(8n80)
Therefore, by applying (8n79), (8n80) and Proposition 2, we know that x takes the form : x(x, φ, θ) l c cn (ax) sin φjc cn (ax) cos φ cos θjc cn (ax) sin φ cos θ " # $ kh Θ (axkγ) jc Na# kh#jcn# (ax) cosh xk" ln kaZ(γ) x % # Θ (axjγ) k
0 1 kh Θ (axkγ) jc Na# kh#jcn# (ax) sinh 0 xk" ln kaZ(γ)x1 , & # Θ (axjγ) k
5
(8n81) 6 7
8
where k l Na#j1\(N2a) and kh l Na#k1\(N2a) are the modulus and the complementary modules of the Jacobi’s elliptic functions and γ l sn−"(1\(ak#)). Now, by choosing suitable initial conditions, we will obtain (1n16) from (8n81). Therefore, up to rigid motions, the immersion is given by (1n16). Case b 4. µ(x) l N2(1ja#) cn ((N1ja#\N2) x, N2a\N1ja#), 0 a 1 In this case, (8n52) yields c` lk4a#α# kh#\k# 0. By choosing c` lk1, we obtain µ(x) l
0 1
2a a dn x, k , k k
fl
0 1
k a dn x, k , akh k
kl
N2a N1ka# , kh l . N1ja# N1ja#
(8n82)
Thus, U l M$ which is an open portion of the warped product manifold D$a l i(k/(akh))dn(ax/k) H#(k1). If we choose the hyperbolic coordinate system oφ, θq on H#(k1), we get k# dn# ((a\k) x) g l dx#j (dφ#jcosh# φdθ#). (8n83) a# kh# By (8n68), (8n69) and the formula of Gauss, we know that the immersion satisfies the following differential system :
0 1
c#x 2a a l dn x ξjx ; cx# k k
0 1 0 1 0 1
c#x k$ a a a cx k dn$ ((a\k) x) k# dn# ((a\k) x) l cn x dn x sn x j ξj x; cφ# akh# k k k cx akh# a# kh# c#x c#x cx l cosh# φ ksinh φ cosh φ ; cθ# cφ# cφ
(8n84) (8n85) (8n86)
B.-Y. C J. Y
506
0 1 0 1 c#x a a cx lkak cn 0 x1 sd 0 x1 ; cx cθ k k cθ
c#x a a cx lkak cn x sd x ; cx cφ k k cφ
(8n87) (8n88)
c#x cx l tanh φ . cφ cθ cθ
(8n89)
x(x, φ, θ) l B(x, θ) cosh φjC(x, φ),
(8n90)
Solving (8n89) yields
for some &-valued functions B(x, θ), C(x, φ). Substituting (8n90) into (8n87) yields "
0 1 0 1 c# C a a cx lkak cn 0 x1 sd 0 x1 . cx cφ k k cφ cB a a lkak cn x sd x B, cx k k
(8n91) (8n92)
Solving (8n91) and (8n92) yields B(x, θ) l G(θ) dn
0ak x1 ,
C(x, φ) l K(φ) dn
0ak x1jW(x),
(8n93)
for some functions G(θ), K(φ), W(x). Thus, we obtain x(x, φ, θ) l G(θ) dn
0ak x1 cosh φjK(φ) dn 0ak x1jW(x).
(8n94)
By substituting (8n94) into (8n85) and (8n86) and by applying (2n8) and (8n84), we obtain Gd(θ)kG(θ) l 0, Kd(φ)kK(φ) l 0, (8n95) Wd(x)j2ak cn
0ak x1 sd 0ak x1 Wh(x)jW(x) l 0.
(8n96)
Therefore, by applying (8n94), (8n95), (8n96) and Proposition 3, we know that x takes the form :
0 1 0 1 a a jc dn 0 x1 cosh φjc dn 0 x1 sinh φ $ % k k a i Θ((a\k) xkγ) a jc ’ k# dn#0 x1ka# kh# cos 0khxk ln ki Z(γ) x1 & k 2 Θ((a\k) xjγ) k a i Θ((a\k) xkγ) a jc ’ k# dn#0 x1ka# kh# sin 0khxk ln ki Z(γ) x1 , ' k 2 Θ((a\k) xjγ) k 5
a a x l c dn x cosh φ cosh θjc dn x cosh φ sinh θ # " k k
(8n97)
8
7
6
Elliptic functions, theta function and hypersurfaces
507
where k l N2a\N1ja# and kh l N1ka#\N1ja# are the modulus and the complementary modules of Jacobi’s elliptic functions and γ l sn−"(k\a). Now, by choosing suitable initial conditions, we will obtain (1n17) from (8n97). Therefore, up to rigid motions, the immersion is given by (1n17). The converse can be verified by straightforward long computations.
9. Some remarks Remark 4. The surfaces A#a, B#a, C#a, D#a, G#, H#a, L#, P#a, W#a and Y#a can be realized as the Riemannian universal covering spaces of some surfaces of revolution in Euclidean 3-space. For examples, surfaces A#N / , B#N , C#N , D#" , G#, H#", L#, P#N , W#N &# & # # # # and Y#N are the universal covering spaces of the surfaces of revolution pictured in / $# Figs 1–10, respectively. Remark 5. Although the shapes of Ana, Gn and Hna are similar for n l 2, their shapes are quite different for n 3. This can be seen from their definitions given in the introduction.
2 Fig. 1. A√5/2 .
2 Fig. 3. C √2 .
Fig. 2. B √52 .
Fig. 4. D12. 2
B.-Y. C J. Y
508
Fig. 5. G 2.
Fig. 7. L 2.
Fig. 9. W √22 .
Fig. 6. H 12.
2 Fig. 8. P √2 .
2 Fig. 10. Y √3/2 .
REFERENCES [1] R. B. A brief introduction to theta function (Holt, Rinehart and Winston, 1961). [2] E. C. Familles de surfaces isoparame! triques dans les espaces a' courbure constante. Ann. Mat. Pura Appl. 17 (1938), 177–191. [3] B. Y. C. Geometry of submanifolds (M. Dekker, 1973). [4] B. Y. C. Some pinching and classification theorems for minimal submanifolds. Archiv Math. 60 (1993), 568–578.
Elliptic functions, theta function and hypersurfaces
509
[5] B. Y. C. Tubular hypersurfaces satisfying a basic equality. Soochow J. Math 20 (1994), 569–586. [6] B. Y. C. A Riemannian invariant and its applications to submanifold theory. Results in Math. 27 (1995), 17–26. [7] B. Y. C. Jacobi’s elliptic functions and Lagrangian immersions. Proc. Royal Soc. Edinburgh Ser. A 126 (1996), 687–704. [8] B. Y. C, F. D, L. V and L. V. Two equivariant totally real immersions into the nearly Ka$ hler 6-sphere and their characterization. Japan J. Math. 21 (1995), 207–222. [9] B. Y. C, F. D, L. V and L. V. Characterization a class of totally real submanifolds of S' by their sectional curvatures. ToV hoku Math. J. 47 (1995), 185–198. [10] B. Y. C and L. V. Lagrangian submanifolds satisfying a basic equality. Math. Proc. Cambridge Phil. Soc. 120 (1996), 291–307. [11] M. D and D. G. Gauss parametrizations and rigidity aspects of submanifolds. J. Differential Geometry 22 (1985), 1–12. [12] R. D, L. V and S. Y. Hypersurfaces with pseudosymmetric Weyl tensor in conformally flat manifolds. Dept. Math. Agricultural Univ., Wroclaw, Preprint Ser. No. 23 (1995). [13] F. D and S. N$ . Semi-parallelity, multi-rotation surfaces and the helix-property. J. Reine Angew. Math. 435 (1993), 33–63. [14] S. H. Eine innere Kennzeichung der verzerrten Produkte. Math. Ann. 241 (1979), 209–215. [15] F. D, M. P and L. V. Einstein, conformally flat and semisymmetric submanifolds satisfying Chen’s equality. Israel J. Math. 100 (1997), 163–169. [16] F. D and L. V. Totally real submanifolds in S'(1) satisfying Chen’s equality. Trans. Amer. Math. Soc. 348 (1996), 1633–1646. [17] D. F. L. Elliptic functions and applications (Springer-Verlag, 1989). [18] T. O. On principal normal vector fields of submanifolds in a Riemannian manifold of constant curvature. J. Math. Soc. Japan 22 (1970), 35–46. [19] E. P. Sur une application de la the! orie des fonctions elliptiques. C. R. Acad. Sc. Paris 89 (1879), 74–76. [20] H. E. R and A. L. Elliptic functions, theta functions and Riemann surfaces (Williams & Wilkins, 1973). [21] H. R. Krummungsfla$ chen von isometrischen Immersionen in Raume konstanter Krummung. Math. Ann. 223 (1976), 169–181. [22] H. R. On the eigenvalues of the shape operator of an isometric immersion into a space of constant curvature. Math. Ann. 243 (1979), 71–82. [23] P. J. R. Homogeneity and some curvature conditions for hypersurfaces. ToV hoku Math. J. 21 (1969), 363–388.
