An estimate for the curvature of bounded submanifolds

An Estimate for the Curvature of Bounded Submanifolds Author(s): L. Jorge and D. Koutroufiotis

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Source: American Journal of Mathematics, Vol. 103, No. 4 (Aug., 1981), pp. 711-725 Published by: The Johns Hopkins University Press

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OF BOUNDED SUBMANIFOLDS

By L. JORGE and D. KOUTROUFIOTIS*

1. Statement of the result. All manifolds considered in this paper shall be connected, of class C"' (smooth), and dimension at least 2. Im- mersions will also be smooth, and of codimension at least 1.

If M and M are Riemannian manifolds and (p: M - _{M is an isometric} immersion with the property that sp(M) lies in a ball, we intend to estimate the supremum of the sectional curvature K of M in terms of the sectional curvature K of M and the radius of the ball. This is our result:

THEOREM A. Let Ml' be a complete Riemannian manifold whose scalar curvature is bounded below; let M" +q be a Riemannian manifold with q c n -1, and Bx a closed normal ball in Mt4+q, of radius X. Sup- pose Sp:M" - M"+q j is an isometric immersion with the property that ep(M"l) is contained in Bx. Take a positive number 6.

(i) If maxBx K = 62 and X < ir/26, then minBx K+ 62 [cotan(Xb6)12 < supm K,

(ii) If maxBx K = 0, then minBx K + 1/X2 < _{supM K,}

(iii) If maxB, K_ =62, then minBx K+ 62[cotanh(X6)j2 ' supM K.

Note that (ii) is the limit case of (iii), as 6 tends to zero.

These estimates are sharp in the sense that, taking for M,,+q each of the three standard space forms, and for M"l = yO(Mn) the boundary of the corresponding normal ball Bx, we obtain equality.

Theorem A implies and extends several older non-immersibility theorems [14, 16, 7, 11] as well as some recent ones [2, 5]; we shall discuss them in Part 4 of this paper. The restriction on the codimension of the im- mersion is, of course, due to the use of a variant of Otsuki's lemma who also obtained results in this direction [15]. The well-known Chern-Kuiper extension [4] of a result of Topkins' [17] is the origin of this long series of

Manuscript received May 6, 1980.

*Work done under support of C.N.Pq., Brazil, at the Federal University of Ceara.

American Journal of Mathematics, Vol. 103, No. 4, pp. 711-725 711 0002-9327/81/1034-0711 $1.50

712 L. JORGE AND D. KOUTROUFIOTIS

papers on isometric immersibility of a compact manifold; we are able to consider here complete but not necessarily compact manifolds by making use of a powerful theorem of Omori's [13], as in [2] and [5]. From the proof of our theorem we also deduce, for hypersurfaces under appropriate assumptions, the existence of points where the second fundamental form is definite, thus generalizing known results [12] of this type.

Finally, the inequalities obtained in the proof of Theorem A permit us to give, modulo Omori's theorem, a simple proof of a recent result by Jorge and Xavier [8] concerning bounded isometric immersions of bounded mean curvature. We shall consider this in Part 5.

2. Preliminaries. Here we assemble all the formulae and proposi- tions that we shall use in the proof of Theorem A.

Let M and M be Riemannian manifolds, (p:M - M an isometric im- mersion. Henceforth, we shall tacitly make the usual identification of X E TpM with dop(X). In particular, if g:M - R is smooth and we consider the compositionf = g o s?, then we have at p E M for every X E _TpM:

(gradf, X> = df(X) = dg(X) = <gradg, X >,

so that

gradg = gradf + (gradg)', (2.1)

where (grad g) l is perpendicular to TpM.

Take now a point x E M and consider in particular the function g:M - R _{given by}

g(x) [d(xo, x)]2, (2.2) 2

where d is the Riemannian distance function on M. This g is smooth at x if x lies inside a normal ball around xo, and we can connect xo with x by a unique minimizing geodesic 'y:'y(O) = _{xo, oy(s) = x,} I y'(0)I = 1. We have d(xo, x) = s and grad g at x is perpendicular to the hypersphere d(xo,) s; we obtain from this

Denote by V, respectively V, the Riemannian connection of M, respectively M. We wish to compute the Hessian form V 2f of the function

f

(Vx gradf, Y> Vx gradf - _{a(X, grad f), Y> < Vx gradf, Y>}

= X<gradf, Y> - _{<gradf, VxY> (< Vxgradg, Y>}

+ <(gradg)', VxY>

= V2g(X, Y) + <(gradg)', o(X, Y)> = V2g(X, Y)

+ < grad g, c (X, Y) >.

Thus,

V2fp(X, Y) = _{V2g,,O(p)(X, Y) + (gradg, (x(X, Y)>s,(p).} (2.4)

We wish to estimate the Hessian of the function (2.2) in terms of the curvature of the ambient manifold. The estimate below, known in various formulations, is obtained using well-known comparison methods. We give its proof for the sake of completeness of exposition.

(2.5) LEMMA. Let N be an m-dimensional Riemannian manifold with sectional curvature K, xo and x points of N so that x does not lie in the cut locus of xo. Let -y: [0, 1] - N be the minimizing geodesic segment connecting xo with x, parametrized by arc-length. Take a positive number 6. For any unit vector X E TxN, perpendicular to y' (1), the Hessian of the function (2.2) at x satisfies V 2g(X, X) -, u(l), where

h5.cotan(16) if max K = 62 and l < r/6

ydl) t I 1 if max K 0

16. cotanh(16) if max K = 62

-Y

Proof. Take a vector X E T,N as in the lemma, and a geodesic

,B:(-E, +E)-N so that O(O)-x, S'(O) = X. For e small, we can lift f to

714 L. JORGE AND D. KOUTROUFIOTIS

expj0(lv(s)) = 3(s), s E (-E, +E). Consider the smooth variation of the geodesic -y(t)

f:(-E, +E) X [0, l1 _{N, f(s, t) = expj0(tv(s)).}

The vector field along -y(t)

J(t)

as

is a Jacobi field which satisfies

J(0) = 0, J(I) = X, (J(t), -y'(t)>3 0

The third equality is a consequence of the assumption that X is perpen- dicular to -y'(1).

Consider now the function g along the geodesic f. We compute easily (g o 3)"(0) = V2g(X, X). On the other hand,

1

_af

/ a2 1 (go

f3)(s)

Ij[j

2 0 at

2 0 at

2

where E stands for the energy of the geodesic -ys t -f(s, t). Thus, (g 0)"(0) = (1/2)lE"(0). Now using the fact that the variation has fixed left endpoint, and that 3 is a geodesic, we have (1/2)E"(0) = I,(J, J), where II is the index form of -y = -yo; that is, for a vector field V along -y,

I,(V, V) = {<V', V'> - _{<R(y', V)'y', V>}dt,}

where V' is the covariant derivative of V along -y and R is the curvature tensor of N. Consult [10], p. 75, for the above second variation formula for the energy of -y. We use here the sign convention for R adopted in [10], p. 51.

It follows now from the above that V 2g(X, X) = II,(J, J).

gSi? if max K = 62

-Y

N= Rni if maxK = 0

-Y

Hn" if maxK =- 62

-Y

where S", is the sphere of curvature 62 and H"1 the hyperbolic space of cur- vature -62. Pick a point E E N and a geodesic j(t), t arc-length, with _y(0) = - ; set y(l) = xi. We have I < 7r/6 in case max) K = 62, SO y con- tains no points conjugate to x-0 on (0, 1]. Set e,,,(t) = -y'(t) and let {eI(t),

en,(t)} be an orthonormal basis field, parallel along -y. Set e,(t) = y'(t) and let {Ie(t), ..., e#(t)} be an orthonormal basis field, parallel along y. If J(t) = 7I _{gi(t)ei(t), define (4J)(t) =} eI gi(t)e(t). It is easy to see, using (J, -y'> = 0 =(J, ( y'> and the curvature assump- tions, that the index forms I and I of -y and y satisfy I,(J, J) > I,(4J, 4J).

Define now a vector field W along y by setting W(l) = (4J)(l) and let- ting W(t) be the parallel-translate of W(l) along y, from xl to xiT. Clearly,

W is a unit vector field, perpendicular to -. Consider the Jacobi field

sin(6t)

W(t) if maxK = 62andl < - sin(61) 'Y6

J(t)= W(t) if maxK=0

sinh(lt) W(t) if max K =- _62. sinh(61) '

JsatisfiesJ(0) = 0 = (bJ)(0), J(l) = (4J)(l), (J(t),y '(t)> 0 O. We con- clude from the Index Lemma (see for example [3], p. 24), since y contains no points conjugate to xi0 on (0, 11, that l,(J, J) < _{11(4J, (J). Therefore,}

V2g(X, X) > lI,(J, J). We compute now I,(J, J) and obtain the an- nounced estimate. This ends the proof of Lemma (2.5).

The main tool in the proof of our result will be the following theorem due to Omori [13].

716 L. JORGE AND D. KOUTROUFIOTIS

propertiesf(p) >f(p(), IgradfIp < -, V2f(X,X) < EtXt2foreveryXE

TpM.

Finally we mention a variant of what is generally known as Otsuki's lemma, which we shall use further on in the manner of [7]. Its proof is contained in the proof of the lemma on p. 28 of [9].

(2.6) Let a::Rn X Rn - _{Rq, q} < n - 1, be a bilinear symmetric mapping satisfying ox(X, X) ? Ofor X ? 0. There exist linearly indepen- dent vectors X, Y so that a(X, X) = a(X, Y) and ox(X, Y) = 0.

3. The proof. Denote by xo the center of the closed normal ball Bx. Let d be the distance in M and g the function (2.2); it is smooth on BX since the cut locus of xo lies outside Bx. Setf = g o _{(p; nowf is a smooth} function on M because (p(M) C BA, andf ' X2/2. We may assume that

infM K > -oo since, otherwise, the hypothesis that the scalar curvature of

M is bounded below would entail supM K = +oo, and therefore the estimates in Theorem A would be satisfied trivially. Thus, we may apply Omori's theorem to the functionf. Pick a pointp- E M so that (p(j-) = x- ? x0. For every positive integer m, there exists a point Pn, E M satisfying

1 _-)

,1

f(Pm) 2f(p) = - [d(xo,x)]2> 0, tgradfp,,, <-9

2 Pin m

v2f(X, X) <-1X12 for allX Tp",M. M (3.1)

m Ir

Set xn, = (P(P,,1). Let -yM,1 be the geodesic segment of length Sn,9

emanating from xo and ending at x,,1, which realizes the distance between

xo and xM :d(xo, x,,,) = sni. Put d(xo, xJ) = s; we have 0 < s < S_ < .

From (2.3) we know that

I

I

-Igrad g * I oe(X, X)I < _{(grad g, oe(X, X)>,}

so, using (2.4), we deduce from the third inequality in (3.1) that, at Pn, and for all X E TP,, M,

1X12.

v

Sn,1I U(X, X) I <-

From now on, for the sake of economy of notation, we omit the bar over V and rewrite the above inequality for all X ? 0 at pill in the form

1 [V2g(X,X) I 1 IaU(X,X)I(

s,,,

L

x12

We wish to estimate the Hessian of g at X E TpM. To this purpose, we decompose X as

X = XT + XN, (3.3)

where XT is normal to grad g and XN is collinear with grad g. Lemma (2.5) supplies an estimate for V2g(XT, XT); we show, using the second ine-

quality in (3.1), that the remaining terms in

V2g(X, X) _ _{V2g(XT, XT) V2g(XT, XN)} V 2g(XN, XN)

=

_~

_~

IX12 IXxI2 IXI 2

(3.4)

are negligible for n large. This is a consequence of the fact that, by con-

struction, the tangent space to so(M) at x,, tends to become perpendicular

to the radial geodesics as m tends to infinity. More precisely, we have at

Xnz

<X, grad g>

XN='n Si

Snm

and so, by (2.1) and the second inequality in (3.1),

I XN I < 1 . (3.5)

It follows from this and (3.3) that

s

- I _{< (3.6)}

718 L. JORGE AND D. KOUTROUFIOTIS

Now, at every point x E BX, the symmetric bilinear form V 2g on T,,M can

be estimated:

IV2g(X,Y)j = <LXX, Y>I ' IILx IXI X IYI,

where Lx is the linear transformation X - Vx grad g on TXM. By con- tinuity and compactness of Bx, we have sup{ I Lx I , x E Bx} = c < oo.

Thus, I V2g(X, Y) I < c IXI X I YI on Bx. Using this, (3.5) and I XT I < IXI, we obtain from (3.4)

V2g(X,X) V2g(XT,XT)_ 3c _{for all non-zero}

XETp ,M.

IX12X2 msni "

Now we have at x,,,, according to Lemma (2.5), (3.7)

V 2g(XT, XT) IXrTI (3.8)

> IA s,,,).8

Recall that 0 < s c s,,i. Thus, for m large enough, the left-hand side of (3.6) is positive. This enables us to combine (3.6), (3.7), (3.8) and obtain

V2g(X, X) ( _)(1 __ _2 3c

t ( )

IX12 JmSm Msm a Xm~op

(3.9)

We restrict the radius of Bx further in the first case of Lemma (2.5) to ensure that u(sm) is bounded below by a positive number:

(6s cotan(Gsm) bs-cotan(6X) > 0 if ₂₆

A(SM) = I

16sm cotanh(6sm) 2 bicotanh(6X) > 0.

From (3.9) we deduce now

for all non-zero vectors X E TpmM and m large enough. It follows from (3.2) that a(X, X) ? 0 for all non-zero X E TpmM, and Lemma (2.6) is ap- plicable: We pick vectors X, Y as in (2.6), write down the corresponding inequalities (3.2), and multiply them term-by-term. Applying (3.9), we obtain

1 [2(sI) (1 m-) - m

Q2<

Y)|

sm2 mm msm m |X12|

y12

(a(X, X), a(Y, Y)> - _{Ia(X, y)1 2}

IX12. I y12

<

(C!(X, X), c(Y, Y)) - (I(X, y) 1 2 IXI2. 1 yI2 - <X, y>2

Thus, by the Gauss equation, we have at .p(p.), for m large, a plane a = X A Y satisfying

0 < Srn2 [,sm)(1 - m ) _ncm

112

sm

~

(3.11)

It follows that

minK+ I[.]2 <supK. (3.12)

BX Sm2 M

We take now a convergent subsequence of the sequence {sm}, which we denote again by {sm }:

O<Ts? limsm=XcX.

m -00

From (3.12) we obtain

min K + [y(X)]2 C sup K.

720 L. JORGE AND D. KOUTROUFIOTIS

However, in the three cases under consideration,

(62[cotan(bX)12

X\ yX1 2 X 21

[(2[cotanh(bX)12

since the functions involved are monotonically decreasing, and the proof of Theorem A is complete.

4. Applications. Theorem A applied to the unit sphere Sn+q C

Rn +q+l yields immediately:

THEOREM 1. Let Mn be a complete Riemannian manifold with scalar curvature bounded below, and p:Mn - _{Sn+q, q ' n} - 1, an

isometric immersion with the property that #p(Mn) lies inside a geodesic ball of radius X < Ir/2. Then 1 + (cotanX)2 < _{supm K.}

This implies, under the above restriction on codimension, that a complete Mn with sectional curvature K c 1 and scalar curvature bounded below, which is isometrically immersed in Sn+q, must have accumulation points on every great hypersphere; and must in fact intersect every great hypersphere if it is compact. This was proved by Hasanis [5] using an estimate for the supremum of K which is somewhat weaker than ours.

An Hadamard manifold M is a simply-connected complete Rieman- nian manifold with non-positive sectional curvature. Recall that every point p of an Hadamard manifold is a pole, that is, expp: TpM - _{M is a} diffeomorphism.

THEOREM 2. Let Mn be a complete Riemannian manifold with

scalar curvature bounded below, Mn+q an Hadamard manifold with q c

n - _{1, and sp: Mn _ Mn} +q an isometric immersion satisfying K(U) ' K(U) at every point of Mn and for every plane a through it. Then #p(Mn) is un- bounded.

Proof. We argue indirectly. If #p(Mn) were bounded, it would be contained in a normal ball. But then, as we have shown in the proof of Theorem A, there would exist a point in Mn and a plane a through it so that inequalities (3.11) hold; this is a contradiction.

isometric immersion of Mn into an Hadamard manifold Mn+q so that K(a) ' K(or) everywhere; this result is due to O'Neill [14].

Note that results similar to Theorem 2 can be obtained by assuming merely that Mn+q possesses a pole.

The case Mn+q = Rn+q of Theorem 2 is due to Baikousis and Koufogiorgos [2], and Jacobowitz [7] if Ml' is assumed compact. Here is an interesting implication in this case: a complete M2 with Gaussian cur- vature satisfying -00 < a < K < 0, isometrically immersed in R3, is ex-

trinsically unbounded.

THEOREM 3. Let Mn+q, q ' n - 1, be an Hadamard manifold with sectional curvature satisfying a < K < b < 0; let Mn be a complete Riemannian manifold with scalar curvature bounded below and sectional curvature satisfying K c a - _{b. If} Sc : _ Mn +q is an isometric immer- sion, then #o(Mn) is unbounded.

Proof. Say sp(Mn) C B, and

a ' minK c maxK = -62 < _{b < 0,} 6 > 0.

BX BX

Then,

a - b < a - _{b[cotanh(GX)]2} c min K + 62[cotanh(6X)]2 c sup K.

BX M

Thus, there exists a point in Mn and a plane a through it so that a - _b <

K(o), which is a contradiction.

The case maxBx K = b = 0 is dealt with similarly, using inequality (ii) of Theorem A. This completes the proof.

In particular, if Mn is compact and all the hypotheses of Theorem 3 hold, there exists no isometric immersion of Mn into M' +q; this is due to J. D. Moore [11]. The special case K = const. < 0 is Stiel's theorem [16].

THEOREM 4. Let M be a Riemannian manifold and let M be a com- plete hypersurface of M, whose sectional curvature is bounded below. Suppose that M lies in a closed normal ball Bx of M. If either maxBx K = 62 with 6 > 0 and X < r/(26) or maxBx K c 0, then M has points where the second fundamental form is definite.

722 L. JORGE AND D. KOUTROUFIOTIS

hand side is positive at some point of M for all tangent vectors X, a fact we proved later on (inequality (3.10)).

Theorem 4 was proven by S. B. Myers ([12], p. 214) in case M is assumed compact. The case where M is the euclidean space is due to Baikousis and Koufogiorgos [2]. Clearly, inequality (3.2) contains infor- mation about the second fundamental form even for codimension larger than one; we shall not pursue this however.

5. Immersions of bounded mean curvature. The estimates derived in the proof of Theorem A permit us to give a succinct proof of the follow- ing result, due to Jorge and Xavier [8].

THEOREM H. Let M be a complete Riemannian manifold whose scalar curvature is bounded below, and <p:M - M an isometric immer- sion into the Riemannian manifold M. Suppose <p(M) is contained in a normal ball Bx, and the mean curvature vector H of <p satisfies

I

where Ho is a positive constant. Take a positive number 6.

(i) If maxJ3 K

=

(iii) If maxBx K =-62, then 6 < Ho and X 2 (1/6) arctanh(6/H0).

Note that no codimension assumption is necessary.

We need the lemma below, which enables us to apply Omori's theorem.

LEMMA. Let M and M be Riemannian manifolds with dim M < dim M; assume the scalar curvature of M is bounded below. If there exists an isometric immersion p: M - _{M satisfying}

I

Proof of the Lemma. Suppose the scalar curvature R of M satisfies inf R = a > _{-xo. Take a point p E M and an orthonormal basis {X1}1=I} of TpM, where n = dim M. We have the following identity, obtained from the Gauss equation by contraction:

R = E K(Xi A Xj) + n2IH12-S, i?j

S s n(n - 1) maxK + n2HO 2+ a = const., B

so S is bounded on M. Now, for an arbitrary plane X A Y at so(p),

IK(X A Y) l _{I K(X A Y) I +} I (c(X, X), a(Y, Y)>-I a(X, Y) 1 21

s max K + 2S < _const.

B

Proof of Theorent H. The notation is the same as in Part 3. By vir- tue of the previous lemma and the assumptions, we may again apply Omori's theorem to the functionf = g o sp, and construct the sequence of points pm which satisfy (3.1). Taking an orthonormal basis {Xi}1=, of TpmM, we have from (2.4) for the Laplacian of f at Pm:

n

_~~~~~~~n

Af = V2g(Xi, X) + (grad g, nH) <-.

From I HI < Ho_ we obtain by virtue of the Cauchy-Schwarz inequality:

-nSmHo < (grad g, nH>,

and from (3.9),

n[!t(sm) (1 - m:) - < E V2g(X1,X,).

Combining the last three inequalities, we have at pm:

IM(S ) I1- )- _ i, - < <SI H0

M M msm m

Now we let m go to infinity, pick a convergent subsequence of the se- quence {sn }, and obtain, as in the proof of Theorem A, a positive number X satisfying X s X and A(X) s XHO. This last inequality, together with the definition (2.5) of 14, imply immediately the assertion in case (ii). In case (i), we obtain for X < ir/a:

724 L. JORGE AND D. KOUTROUFIOTIS

If X < _Or/(26), so cotan(X6) > 0, hence tan(X6) 2 6/Ho and the assertion follows from the monotonicity of the tangent function. In case (iii) we have

6 < 6 cotanh(X6) c 6 cotanh(X6) c Ho,

so 6 < Ho and tanh(X6) 2 6/H_. Since 6/H0 < 1, we can take inverses, and the assertion follows from the monotonicity of the hyperbolic tangent function.

The case M = RN and dim M = 2 of Theorem H is due to Aminov [1], the case M = RN to Hasanis and Koutroufiotis [6], the case M = SN

to Hasanis [5]. Theorem H implies in particular that a complete manifold with scalar curvature bounded below, isometrically and minimally im- mersed in an Hadamard manifold M, must be unbounded in M.

A consequence of case (iii) may be noteworthy: a compact sub- manifold of the hyperbolic space of sectional curvature -1, must satisfy max IHI > 1.

UNIVERSIDADE FEDERAL DO CEARA UNIVERSITY OF IOANNINA

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