art jlmbarbosa ruled

Ruled Submanifolds of Space Forms with Mean Curvature of Nonzero Constant Length Author(s): J. L. M. Barbosa and J. A. Delgado

Reviewed work(s):

Source: American Journal of Mathematics, Vol. 106, No. 4 (Aug., 1984), pp. 763-780 Published by: The Johns Hopkins University Press

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CURVATURE OF NONZERO CONSTANT LENGTH

By J. L. M. BARBOSA and J. A. DELGADO

1. Introduction. Let Nm(c) be a m-dimensional Riemannian mani- fold of constant sectional curvature c and Mn+l be a (n + 1)-dimensional submanifold of Nm(c). We say that Mn+I is ruled if there is a foliation of Mn+1 by n-dimensional totally geodesic submanifolds of Nm(c). If Mn+I is ruled we call each leaf of the foliation a ruling of Mn+ . A mapping X: Mn+l Nm(c) is a ruled immersion if, for each p e M there is a neighbor- hood U of p such that X(U) is ruled submanifold. For the case c < 0 we obtain the following non-existence result.

THEOREM 1.1 For c < 0 there is no ruled submanifold of Nm(c) with mean curvature of nonzero constant length.

This is a surprising result if we observe that there are many examples of minimal ruled submanifolds of Nm(c), when c < 0. (See [1]).

If c > 0 the situation is quite different: there are many examples even if their existence is subjected to some restrictions of the codimension and the type of the metric of Mn+'.

THEOREM 1.2 If c > 0 and Mn+l is a ruled submanifold of Nm(c)

with mean curvature of nonzero constant length then m 2 2n +1 and Mn+I is locally isometric to the Riemannian product R X Nn(c).

When the codimension is n and Nm(c) is the Euclidean sphere we give a full description of the set of ruled submanifolds of Nm(c) with mean cur- vature of nonzero constant length. In fact we prove:

THEOREM 1.3 If X: Mn+l -+ S2n+l is a ruled submanifold with mean curvature of nonzero constant length then n + 1 is even. Given a > 0, then for any complete differentiable curve a in the symmetric space 0(2k)/U(k) there exists an isometric ruled immersion Xa,a: R X S2k-1 I

S4k-1 with mean curvature of constant length a, which is well determined

up to a rigid motion of S4k-1. Furthermore if a and 13 differ by a rigid

Manuscript received June 2, 1982.

764 J. L. M. BARBOSA AND J. A. DELGADO

motion of 0(2k)/U(k) then Xae,a _{and X9,a are the same up to a rigid motion} of 54k-1

Recently Barbosa, Dajczer and Jorge [1] have completely described all ruled minimal submanifolds of Nm(c). They also show that the only ruled submanifolds of R'm with mean curvature of nonzero constant length are generalized cylinders, that is, products of the type Rn

X

Now it is natural to ask what happens if we assume Mn+l to satisfy some extra condition such as: to have parallel mean curvature or to have trivial normal bundle. In fact, for ruled submanifolds of Sm, we prove the two conditions are equivalent. (See Propositions (4.7) and (4.10)).

THEOREM 1.4 Let Mn+l be a complete ruled submanifold of Sm with mean curvature of nonzero constant length. If Mn+1 has parallel mean curvature or trivial normal bundle then n = _{1 and M2 is a product of two} circles in S3 C sm.

2. Preliminary results. Let Nm(c), c * 0, be a space form of sec- tional curvature c and Mn+1 C Nm(c) be an (n + 1)-dimensional ruled submanifold of Nm(c). Mn+I is then foliated by n-dimensional totally geo- desic submanifolds of Nm(c). Locally we can always assume that Nm(c) is either the Euclidean sphere Sm or the hyperbolic space Hm according to c being positive or negative. We will consider Scm and Hcmrespectively, as hypersurfaces of the ambient spaces Rm+l and Lm+l, where Lm+l means the Lorentz (m + 1)-dimensional space with the metric

(2.1) ds2 = dX +

E

i= 1

Since the properties of Mn+l being ruled and Mn+l having mean curvature of constant length are preserved by homothety of the ambient spaces there is no lost of generality if we consider only the cases when c = 1 or c = -1 that is, the cases when Npm is either the unit- sphere Sm or the standard hyperbolic space Hm. From now on we assume this.

assume to be parametrized by arc length. Choose n orthonormal vector fields e1(s), . . ., _en(s) along a which generates the tangent space of the leaf through a(s) for each s. According to Lemma (2.3) of [1] we can assume this choice to be made in such way that those vector fields are not moving with respect to each other. If we represent by <, > the metric of the ambient space this amounts to say that, for all 1 c i, j 5 n

(2.2) _{(ef, ej> =} 0

where ' means derivation with respect to s. The manifold Mn?l can then be locally parametrized by

(2.3) X(s , t . . ., tn) = expa(s) I E _Ij(t2, ...* tn)ei(S))

where EjL- 4']2 = 1. Since Nm is either Sm or Hm then we know that

(2.4) expp(rv) = f(r)p + g(r)v

for each p eNm and v E _Tp Nm with

l

f(r) = cos r, g(r) = sin r in Sm C Rm+l

and by

f(r) = cosh r, g(r) = sinh r in Hm c Lm+l.

If we now set eo(s) = a(s) then we may rewrite (2.3) as

n

(2.5) x(s, tl,., tn) =E (P s(tl, _{tn* t)ej} (S

j=O

where so = (, ..., _vPn) is a parametrization of a neighborhood of the

point (1, 0, . . ., 0) of either the unit sphere Sn of R"+l or the standard hyperbolic space Hn of Ln+1. For future references let's put together all the information we have so far about the ei, 0 < i s n.

766 J. L. M. BARBOSA AND J. A. DELGADO

(2.6) (1 if c = 1

= -1 if c = -1,

(e1,ej> =0 if 0 < i,j < n.

Derivation of (2.5) with respect s and tj yields:

n

XS = E (P i(t I ...,9 tn )el!(S)

(2.7)

Xti = iO ,3t (tl, ...*, tn) ei (s).

Then it follows from (2.6) that the metric of Mn+I is given by

(2.8) da2 = Xs 1 2ds2D0 d02,

where dO2 is the metric of N n. The prametrization X of Mn +1 is such that the vectors Xtj, 1 C _{j c n, are tangent to a totally geodesic submanifold of}

NC. Hence, if B denotes the second fundamental form of Mn + 1, we must

have:

(2.9) B(Xt,,Xtk) = 0 1 C j, k < n.

Since (X., Xtj> = 0, 1 c j c n this implies that the mean curvature vector H of Mn+I is a multiple of B(X,, X.). In fact we obtain

(2.10) (Xss)' = (n + 1) IXS 12H,

where

(

(2.11) eoA...Aen AXs AX., = (n + 1) IXsI2eoA...Aen AXs AH

LEMMA 2.12 Given polynomials P(x,y) = _2O Aixly4i and Q(x,y)

= 0 axiy2i

(a) if P(x,y)(X2 + y2) =(Q(x,y))3 then (i) aO = a2 and a1 = ₀

(ii) A0 = 1/2A2 =A4 = a3 andA1 =A3 = 0

(b) if P(x,y) (-x2 + y2) = _{(Q(X,y))3 then}

(i) aO = -a2 and a1 = ₀

(ii) A0 = -1/2A2 =A4 =-aO andA1 =A3 = 0.

Proof: Just observe that x2 + y2, -x + y, x + y are irreducible.

3. Metric and codimension.

(3.1) Proof of Theorem (1.1). We start by observing that from (2.11), the fact that eo, e1, . . ., _{en, X,, H are orthogonal and (2.6) it follows that}

<eO A .. *A en AXs A Xs, eO A. . .A enA A Xss> =

(3.2) =(n + 1)2 <eoA ... Aen A Xs A H, eoA ... Aen A Xs A H> IXs14 -(n + 1)2 IXSI6 IH12

where the minus sign comes from the fact that <eo, eo > = _{(, a,> =-1.}

Since X. = Fin=o (pje; then X., = tEi=O _{spje[. Set}

by= <e ej >

(3.3)

B1kf = <eOA ... Ae A ei A ej, eoA.. .Ae A ek A e'>, i < i,j, k,l < n

Developing both sides of (3.2) we obtain

(304) ik Bykl PiPjPkPf = -(n + 1)2 IH12( ( bij (p,(j)3.

Let's recall that in the hypothesis of Theorem (1.1), sp = ((po, pI, ... , (Pn) iS a parametrization of a neighborhood ,4 of the point (1, 0, ..., 0) of Hn c Ln+ . Since

I

768 J. L. M. BARBOSA AND J. A. DELGADO

r _{BikeXiXJXkXe) Xi-} =X i

(3.5)

-(n + _{1)2 jH12} ( _bij3xi).

Hence it extends to whole Ln+t. Restriting our attention to each 2-plane xo,

xj and using Lemma (2.12) (b) we obtain

for all 1 c j c n. That is, (ej, e> > =-<e6, e6>. On the other hand by our choice of ei, 0 < i < n we have (e, eo> = 0, that is, e_, 0 < i < n are tangent to hyperbolic space at eo. And so (e;, eil> must be positive. Since

(e6, e> = 1 we obtain (e, eie > -1, 1 i c n, which is an absurd. This

proves the Theorem (1.1).

(3.6) Proof of Theorem (1.2) We follow the same path, the differ- ence being that we do not get a minus sign in (3.2), since Mn+l is now a submanifold of the sphere Sm. Hence (3.4) becomes

O Bijke pjppkspQ =

(3.7)

- (n + 1)2 IH 12

(

bj

where

b

( s SBk,XiXjXkX)( 2 Xj') =

( <i,j,k,f<nJ/\ = /

(3.8)

and from (2.12) it follows that

i) b= 0 and bt, = _{bij, i ? j,} 0 s i,j C n,

ii) Biii + ? Biiji = 0 and Biiii = BW= (n + 1)2 IHI2 i j, O < i, j < n.

From this we obtain

(3.9) (<eJ, ej> =

and

(3.10) _(e[,el'>=0 if i*j and

(et, el'> = 1 + (n + 1)2 HI2

for all 0 c i,j c n. It follows from (3.9) that B(XS, _Xti) 1 c i S n form a linearly independent set of vectors. Hence the first normal space of X has at least dimension n. And so Mn+l lies at least in 52n+1 c Sm. Thus m 2 2n + 1. It also follows from (3.9) that

n n

(3.11)

IX

Hence, the metric of Mn+l, computed in (2.8), is given simply by

(3.12) da2 = ds2 dO2,

where dO2 is the metric of Sn. Therefore Mn?l is locally isometric to the product R X Sn. This completes the proof of Theorem (1.2).

4. The cases of "flat normal bundle" and "parallel mean curvature". In this study of ruled submanifolds Mn+I of Sm the hypothe- sis

I

1

770 J. L. M. BARBOSA AND J. A. DELGADO

PROPOSITION (4.1) Let Mn+l be a ruled submanifold of Sm. If there

is a point p EHsuch that D,H = Ofor all v tangent to the ruling of Mn+I at p then n = 1.

Proof: Start by observing that, from (2.10), (3.9), (3.10) and (3.11) one obtains

n

(4.2) (n + 1)H = X + X = i poi(e[ + ei).

Taking v = Xtk _{one can compute DvH by subtrating from (X,,} + X)tk its tangent projection.

Then it comes out

(4.3) (n + 1)DxtH = *o d, (e1" + ei) - _{bk i(p} Pe

0= _atk i=O

where bk is defined by

(4.4) bk = i oP p (e[", ej>

i,j=O atk

Set

(4.5) Vk = -Ed (ei" + ei), k = 1, 29 ... ., n.

i=0 a3tk

It is now a consequence of (4.3) that if the hypothesis of the proposition is satisfied then all Vk are colinear at the point p. On the other hand

(4.6) (Vk, Vf> = (n + 1)2 HI2 i '3, '3e

1=0 atk _atf

= (n + 1)2 _HI12 (Xtk, Xtf>.

Hence the vectors Vk are linearly independent at each point of Mn + 1. Thus it can exist only one Vk and therefore n = _1.

PROPOSITION (4.7) Let Mn+1 be a ruled submanifold of Sm. If the normal curvature tensor of Mn +1 _{is zero at one point then n = 1.}

Ricci equation tell us that

(4.8) R'(Ei, Ej)Ea = <B(Ei, Ek), Ea>B(Ej, Ek) k

- <B(Ej, Ek),Ea,>B(Ei, Ek),

where 1 < _i,j < n + 1, n + 2 < a < m.

Since Mn+l is ruled we can assume further that E1, . . ., En are tan- gent to its rulings. Consequently we will have B(E1, Ek) = 0, 1 < i, k < n. Hence (4.8) reduces to

(4.9) RI(Ei, Ej); = (B(Ei, En+ 1), >>B(Ej, En+1)

- <B(Ej, En + 1), t >B(Ei, En + 1)

for all 1 < _{i, j c n + 1, and any normal vector t. Set Wk = B(Ek, En+1),} 1 c k < n + 1. The hypothesis of the proposition together with (4.9) imply that all Wk are colinear at the point p. On the other hand the set of Wk spans the first normal space of Mn + 1 which has at least dimension n, as we have proved in the previous section. Therefore n = 1.

PROPOSITION (4.10) Let M2 be a ruled submanifold of Sm with mean

curvature of constant length. Thefollowing conditions are then equivalent.

(a) H is parallel on Mn+ 1. (b) H is parallel along the rulings. (c) Mn+ 1 has flat normal bundle.

Proof: Since n =1 then M2 is parametrized byX(s,t) = cost eo(s) + sint e1(s), where eo(s) and e1(s) satisfy (3.9) and (3.10). Now H parallel is equivalent to DXtH = 0 and Dx3H = 0 on M2. From (4.3) it comes out

(4.11) _{2 DxtH} =-sint

(ed" + eo + bej) + cost(e ' + e1 -be'),

where b = (ej', eO> =-(ed, ej). Hence DxtH = 0 is equivalent to the following two equations:

(4.12) _{eA + eo + <e, ei>ei} = _0,

772 J. L. M. BARBOSA AND J. A. DELGADO

Scalar multiplication of (4.12) by e6 yields

(4.14) 4

₁

In particular (e6, ej> is constant on M2.

The computation of DxsH can be done as indicated by the following chain of equalities

2

+ 4 IHI2Xs + (e4, ei>Xt =

cost (e6- + (1 + 4 1HI2)e6 + (e6, ei>ei) +

? sint (ej + (1 + 4 1HI2)ei - _{(e6, ej>eo).}

Hence, DxsH = 0 is equivalent to the following two equations:

(4.15) _{e6 + ( 1 + 4 IH12)e6 + (eW, ei>ei} = ₀

(4.16) e; + (1 + 4 1H1 2)ei - _{(e6, ei>eo = 0}

It is easy to verify that (4.15) can be obtained by taking the derivative of (4.12) and adding this to (4.13) multiplied by (e6, ej' >. Similarly (4.16) is obtained by taking derivative of (4.13) and adding to (4.12) multiplied by (e6, ei' >. Hence the condition "H parallel on M2" is equivalent to have the validity of (4.12) and (4.13) on M2. Therefore (a) and (b) are equivalent.

To show that (b) and (c) are also equivalent we start by observing that Xt and X. are orthonormal. Taking E1 = Xt and E2 = Xs (4.9) becomes:

(4.17)

R'(Xt, Xs)t = _{<B(Xt, XS), >B(Xs, XS)} - _{(B(Xs, XS), >B(Xt, Xs).}

The condition R' = 0 means essentially that B(XS, _{X.) and B(Xt, X.) are} colinear. An easy computation yields

B(Xt, Xs) = (Xst)' = Xst =-sint e6 + cost ej

Since B(Xt, Xt) = 0 the condition R' = 0 is equivalent to the validity of the following two equations which are obtained from (4.17) by taking respec- tively =B(Xt, X3) and t = B(XS, Xe).

(4.18)

cost(e6 + eo) + sint(ef + e1) - (e6, ej> (-sint e6 + cost ei) = 0

(4.19)

(ei, ei> (cost(ed' + eo) + sint(e' + el))

- 4 1H12(-sint e6 + cost ej) = 0

It is clear that (4.18) is equivalent to (4.12) and (4.13). Since (4.14) is also true then (4.19) yields the same equations as (4.18). Hence R' = 0 is equiv- alent to the validity of (4.12) and (4.13) and therefore (c) and (b) are also equivalent. This completes the proof of Proposition (4.10).

(4.20) Proof of Theorem 1.4 From (3.12) and Propositions (4.1), (4.7) and (4.10), X is an isometric immersion of R2 into Sm with the first normal space N of dimension one. Since H is parallel and nonzero then N is parallel and so we can reduce codimension and find 53 such that M2 C

53 C Sm. By observing that X has also mean curvature of constant length as an isometric immersion of R2 into R4 we get from Colares [2] that X is a product of two curves with the constant first curvature. Since X lies in a sphere it is a product of two circles.

5. Examples

(5.1) Let's examine the lowest codimension case, that is, consider a ruled submanifold Mn+l of 52n+1 with mean curvature H of nonzero constant length. Then Mn+l is described parametrically by (2.5) where eo, e1,

en, eo, ej, ..., en form an orthonormal bases of R2n+2. It follows that e0 , O < j 5 n is a linear combination of those vectors. It is easy to see that

(ej', ek > =-a'jk and that (ej', ek> = -(ej ek > and so we conclude that

r

(5.2) a1k + a kj = 0.

Substitution of (5.2) into (3.10) yields

774 J. L. M. BARBOSA AND J. A. DELGADO

Let A represent the matrix (aij). From (5.2) and (5.3) we obtain

(5.4) A + tA = 0, AtA = aI,

where tA stands for the transposed matrix of A, a = (n + 1)2

I

HI

(5.5) A2 =-aI

and since a > 0, the eigenvalues of A are complex numbers with zero real part. Hence A has only invariant subspaces of dimension two. Therefore (n + 1) is even. This proves the following lemma

LEMMA (5.6) If Mn+l is a ruled submanifold of S2n+l with nonzero mean curvature of constant length. Then (n + 1) is even.

Let M(n) represent the space of n X n real matrices and define

(5.7) Hk(a)= {As EA(2k); A +tA =0 and AtA = aI}.

PROPOSITION (5.8) For a > 0, Hk(a) is diffeomorphic to the homo- geneous space 0(2k)/U(k), where 0(2k) represents the orthogonal group of R2k and U(k) the complex unitary group of Ck.

Proof: It is obvious that it is sufficient to prove the result for a = 1. Observe that 0(2k) acts on H2k(i) by

(5.9) (A, B) -BAtB, AEH2k(1), BE 0(2k)

and this action is transitive, that is, given A1, A2 E H2k(l), there exists B E 0(2k) such that A1 = BA2 tB. Set

(5.10) G = {B e 0(2k); BAotB =AO)

where

AO = ( _{Ej-k) E} H2k

Then G is clearly a subgroup of 0(2k). Furthermore, if B E G then

where Bi Ec M(k), i = 1, 2 and

(5.12) B1tBI + B2tB2 = I and B2tBI - _BItB2 = 0.

It is easy to verify that B1 + iB2 E U(k) and this correspondence is an isomorphism. The map from U(k) to 0(2k) given by this correspondence is also a lie group embedding. And so the proposition follows.

(5.13) Proof of the Theorem (1.3) To construct examples of ruled submanifolds M2k C S4k-1 with mean curvature H of constant length we start with a differentiable curve A(s) = (aij(s)) in Hk(a), where a 4k2

_I

(5.14) X= (iX,

where

l0 lIX

a

(__1{L).

I is the 2k

X

(5.15) k(so) = Eo E 0(4k).

Set eo, e1, ..., e2k-1 to be first 2k line of k. Then el, ..., e'k 1 are the

remain 2k lines and furthermore

2k-i

(5.16) e[ =-ei + _aijej O i < 2k-1.

j=O

The only thing that remain to be shown is that eo, ..., e2k-1, eo, ... . e2k_1 form an orthonormal basis of R4k for all s. To prove this define ma- trices

P _{(<ei, ej>),}

Q

and use (5.16) to obtain

776 J. L. M. BARBOSA AND I. A. DELGADO

(5.17) Q'=-P+QtA+R,

R I-Q-tQ+AR+RtA.

By our choice of initial data in (5.15) we have P(s0) = R(so) = _{I and} Q(so) = _{0. On the other hand, since A + tA = I one observes that P(s) =} R(s) = Iand Q(s) = _{0 is one solution for (5.17) with that initial conditions}

and hence the only such solution.

Now we setX= io 1iei as in (2.5). This function will define a ruled 2k-submanifold of S4k-1 whose mean curvature vector H will satisfy the equation

2kH = _{X + X} = _{pi (e' + ei).}

It now follows from the equation AtA = a I that I H I is constant. Let's assume A defined on the whole real line. Then X extends to a map

X: R X S2k1 - S4k-

which will define an isometric ruled immersion into 54k-1 with mean cur-

vature of constant length a/4k2. Observe that X depend only on the map A up to the choice of initial conditions Eo in (5.15). Let F (k, a) represent the set of such isometric immersions and let Ok) be the set of differentia- ble curves A: R -+ _{0(2k)/U(k). Then we have proved the following proposi-}

tion.

PROPOsITION. (5.18) There is a mapfrom e(k) X 0(4k) onto (k,a).

This map is of course not 1-1. Let now X1 = Ej_k_o tpiei and

X2= =0 q

be two local expressions of X given as in (2.5). By changing the parameter s of X1 or x2 we may assume

Xl(sO, tl, . _{t2k-1) = X(so,} , . . ., 0).

Observe that (XI, Xtl, 9 .. ., Xt2k-1l at the point (so, tl, * ,t2k- 1), {go .. *,

f2k-1} and {eo, ..., e2k-1} span the same subspace of R2k. Hence

2k

fj =S _bTee.

Since {eo, **., e2k 1

}

we get

B = (bit) E 0(2k).

On the other hand by derivation we obtain

2k n

fi=e bje e + E bj ee. ₀

Since ee E span {fo, ... 1 } {eO, .* . . e2k e6, .. ., e2k} is an orthonor- mal basis of R4k and _(figfk> = _{0 it follows that bit = (fi,} ee> = 0. Now we have fj = E2ek _{bj1 ek,} where bjy is constant. And so fj? = 2k o bk ek. By computation we get

2k

(fi; fi> = e by (ee em > bim

Therefore, if A = ((e "; ej>) is associated to X1 andA = _((fig f>) is associ- ated to X2, then

A = BA tB.

Let OC(k) represent the quotient space of C(k) by the equivalence relation "A e C(k) is equivalent to A E C (k) if there is B E 0(2k) such that A(s) =

B A(s)tB for all s e R." Now we may represent F(k,a) as the product of 0 C(k) X 0(4k).

To finish the proof of (1.3) we just need to observe that if two elements of F(k,a) differ by a rigid motion of 54k-1 then we can always take local

representations for both X = Ei_ p (ppei and Y = E7o pifi such thatfi = Bei andf' = Bei. Then it follows that the function A(s) will be the same for both and they will differ just by initial conditions Eo on (5.15). Let F repre- sent the quotient space of F(k,a) by the equivalence relation 'X - Y if Y =

AX, A e 0(4k)," then

F = OC(k) .

And so the Theorem 1.3 follows.

778 J. L. M. BARBOSA AND J. A. DELGADO

values of m whose codimensions can not be reduced. Proposition (5.24) shows we can have this immersions with nonzero mean curvature of con- stant length. It also shows that n + 1 can be odd if the codimension of Mn+l is greater than n.

Let a: R - _{S2N+l be a curve of constant curvatures. The Frenet equa-} tions for a as a curve in R2Nf+2 are

a/' _El

(5.20)

Ej = kj- I Ej- 1 + kj Ej+ 1, 1 C j c 2N + 2 1~~j?~2N+2, where {Ej, 1 < j < 2N + 2} is the Frenet frame of a as a curve in R2N+2 and ko = k2N+2 -0. It is known that a has constant curvatures in S2N+ 1 if and only if it has constant curvatures in R2N+2. _{Thus kj are constant in} (5.20). Assume k2N+1 * 0.

LEMMA. _{(5.21) The vectors Ej (s), with jodd are tangent to} S2N+1 at a(s); the ones with j even form a constant angle with ca(s). The curvatures

kl, ...,k2N+I satisfy the equation

3 .2kN+ =Ik3k5.. .5k2N+1 + k2k5k7.. .k2N+1 +

(5.22)

+ k2k4k7 . .. k2N+ 1 + ...+ k2k4k6 * 42N

Furthermore this is the only integrability condition for the existence of aZ with constant curvatures k1,... ,k2N+1 (k2N+1 * 0) in R2N+2 that lies in some Euclidean sphere of radius one.

Proof. The proof of the first part of Lemma (5.21) is easy and it can be done by induction. To see that (5.22) is true one considers the function

( ) Y(s) = e(s) + k E2(s) + k2(S) _{E4(s) +} (5.23) l()k(s)k3(s) k

+ k2(s)k4(s) ... k2(s)

kI(s)k3(s ) . k2N+ 1 (s)

lies in a sphere of center Y. Since k2N+1 * 0 we can not reduce the codi- mension of a. But a lies also in S2N+ 1. Therefore Y = 0 and (5.22) follows. On the other hand let ae be a curve in R2N+2 whose curvatures are constant and satisfy (5.22). If k2N+l * 0 we can use (5.20), (5.22) and

(5.23) to show that

I

Let F (n,N), N > n > I and 6n < 2N+5, be the subset of R2(N-n) defined by

x1 > 1;XX2> 1; j > 0, 1 C j C 2(N-n); 0 < _xj < 1, 2 C j C n + 1.

PROPOSITION. (5.24) Given n, N with N > n 2 1 and 6n c 2N + 5, there is a 1-1 correspondence from F (n,N) into the set of ruled isometric immersions X: R X Sn -- 52+1 with mean curvature of constant length.

Proof. Let az: R 52N+1 c R2N+2 be a curve with constant curva- tures. The Frenet equations for ai as a curve in R2N+2 are given by (5.20). Since 6n -3 c 2N+ 2 we can define n + 1 vector fields along a by

e0o = a; ej=E6j-3 1 j n.

Finally, we set

n

X(S,9 tl , ... ., tn) E (PSi(t, I tn) ei(s),

where so =(so, (po ..p., I ,Pn) is a local parametrization of a neighborhood of (1, 0, ..., 0) E Sn c Rn+1. From Lemma (5.21) e>, 1 c j < n are tangent to 52N+1, Hence X is a ruled submanifold of 52N+1. In what follows we impose some restrictions on the choice of the curvatures kj of a in order to make X to be an isometry and to have constant mean curvature. Then we show that there is a bijection between F (n, N) and the set of kj that satisfy the system of equations defined by the given restrictions and the equation (5.22). From (5.27) it will follow that we always can construct aZ in S2N+t for such values of kj and so the proposition is proved.

Let's assume that the set of curvatures of a satisfy the following equa- tions

k6j_4 + k6j_3 = 1,

(5.25)

780 J. L. M. BARBOSA AND J. A. DELGADO

where 1 c j c n. By using (5.20) we get from a simple computation and (5.25) that X is an isometry and it has mean curvature with constant length

k2-1.

To see that there are solutions for the system formed by the equations (5.22) and (5.25) we observe that we can rewrite (5.22) as

(5.26) A k2(1-k2k2) + B = O,

where

1 if N = 2 A =

.k2 N > 3 .kN+1 ...

k2k if N= 2,

B

=

+... +k 2k2... k2Ng if N2>3

(5.27) Now it is easy to verify that (5.25) and (5.26) can be solved for k2, k6129 k k26j-4, 1 < < n, for any nonzero values of the other variables in the set F(n, N) defined by

k1 > 1,0 <k6j3 < _{1, kjk3> 1,1 j < n.}

Finally it is trivial to see that, up to the choice of the variables, F (n, N) and F(n, N) are the same.

UNIVERSIDADE FEDERAL DO CEARA

REFERENCES

[1l J. L. M., Barbosa, M. Dajczer, and L. P. Jorge; Minimal ruled submanifolds in spaces of constant curvature, to appear in the Indiana Univ. Math. J.