Internat. J. Math. & Math. Sci. VOL. 16 NO. 4 (1993) 725-732
725
UNIQUE
ISOMETRIC
IMMERSION
INTO
HYPERBOLIC SPACE
M. BELTAGY
Mathematics
Department
Facultyof ScienceTanta
UniversityEgypt
(Received
February6, 1992 andinrevisedformAugust
24,1992)
ABSTRACT.
A
result concerningequivalenceofisometric immersions withrelatedGauss
maps intohyperbolicspacehas been established.KEY
WORDS ANDPHRASES.
Isometric immersion, rigidity, second fundamental tensor,Lorentz
manifold,Minkowskispace.1991 AMS
SUBJECT
CLASSIFICATION CODES.
53C40, 53C42.1. INTRODUCTION.
Equivalencebetween isometric immersions ofa Riemannian manifold into another one has been aninteresting fruitful areaof research for alongtime
([1], [4], [6], [7], [12]).
The concept of equivalence between twoisometricimmersions may begivenasfollows.Let
fl:M--N
andf2:M--,N
be two differentisometric immersionsofaRiemannian manifold M into another Riemannian manifold N. The immersionJ’l
will be equivalent to12
(up to an isometry randN)
if thereexists anisometryr:N-Nsuch that rJ’l
)’2"
WemaywriteJ’l
I2
(rood. r).Ifthe above mentioned concept is satisfied for each pair ofisometric immersions ofM into N,wesaythat M isuniquely isometrically immersed inN.
In
classical differentialgeometryMis said toberigid.If for each pair ofisometric immersions
fI:M--,N
and’2:MN"
there exists a continuous curve rs, 8e[0,hiinthe groupGofisometriesofNsuchthat ro idandra1 2’
wesaythatM isuniquely continuouslyisometricallyimmersedinto N
[8].
In
[1]
thefollowingremarkable result concerningisometric immersions into(n+
1)-Euclideanspace En
+
has beenproved.THEOREM
1.1. Let M be a connected, orientable C-Riemannian n-manifold.Suppose
that the
Gauss
mapsof each pair j’ and of isometricimmersions ofM intoEn+
differby an isometry ofEn+
1.
Then M isuniquely isometricallyimmersed in En+
if anyof the followingconditionsis satisfied.
(a)
Mis compact,(b)
Thereexistsat leastonepointmeMwhere allsectionalcurvaturesarepositive,(c)
Thereexists atleast onepoint rneMsuchthat1"(or
)
isnot minimal atrn,andMhasAs far as we are concerned, the corresponding study in hyperbolic space Hn+1 has not yet been considered. Accordingly, we devote the present work to deal with the rigidity problemof
hypersurfaces inhyperbolicspace with constant sectional curvatnre asareceivingspace. The main result will be" stated
n
the nextsection.
From now on all manifolds and maps are sufficiently smooth for computationstomakesense.2. PRELIMINARIES
AND
RESULTS.We begin this section by giving a briefword about the model ofhyperbolic space we shall usethroughout thepaper.
In
Rn+2 with the natural basis O, el, e2,...,en+
1’ n>_1, we consider a nondegeneratesymmetricbilinearform
n+l
b(z,y)= -Zoy
o+
y
zkyk r, yERn+2.
k:l
The pairVn
+
2(Rn +
2,b
iscalled the Mmkowski space.Let
0(1,n+
1)denote theorthogonalgroup ofb,i.e.,where
0(1,n
+
1) {AEGl(n+2,R):b(Az,Av) b(r,y)} {A Gl(n+
2,R):A’SA S}0
ln+l
The fact that A’SA=S implies that det A= +l.
A
matrixA=(aij
in 0(1,n+l) belongs to the identity component ifandonly ifdet A andaoo
>1. Let G denote the identity component of O(1,n+
1).Now,
the hypersurface in Vn+2 defined by b(z,y)=-! is the disjoint union of two connected components:Hn+I
{z:Zo>
1} andKn+
{z(:Zo
< -1}.0(1,a
+
1) acts transitivelyon whileGacts transitivelyonHa+
[9].
Let zeHa
+
1,
i.e., b(z,y) andzo_>1. Thetangent space
TzH
a,+ isgiven,through the identification by parallel displacement in Va+
2,
by the subspace of all vectors aEVa+
2 such that b(z,a)=0. Therestrictionof toTzH
a+ is positivedefinite. Thus the formbrestricted to thetangent space at each point ofHa+
gives rise to acomplete Riemannian metric on//a+
which isobviously
invariant under G.Moreover,
(Ha+1,b)
is of constant sectional curvatures K 1. Hence Ha+
will be taken to be the model in which our immersions occur.For
UNIQUE ISOMETRIC IMMERSION INTO HYPERBOLIC SPACE 727
Thehypersurface + in
v
n+:2definedby b(z,y) isconnected Lorentz. We call.
n+the conjugatehypersurfaceofboth Hn+ and Kn+
1.
(S
Fig. 1).Figure
Let I:M--,Hn+l be an immersion of the orientable manifold M. The Gauss map
:
M
n+
ofthis immersion is definedtobethe parallel translationoftheunitnormal IV{m} of M, asahypersurfaceofHn+
1,
at /’(m) uptothe origin0ofVn+
2.
Hence,theGauss
image ofM willlieonthe conjugatehypersurface n+
ofHn+
I.
Now,
westateourresult.THEOREM 2.1. Let f and
]
be two C isometric immersionsof a connected, orientable Riemannian n-manifold M(n>2) into the (n+ 1)-hyperbolic space Hn+1C
Vn+2.
Suppose
that the Gauss maps andg
differ by amatrixaeG. Then with anyoneof the conditions(i)-(iii)
below,/’and] differbyanisometryrof(i)
Mhasatleast onepointmeMwhereallsectionalcurvaturesaregreaterthan 1.(ii)
Miscompact.(iii)
Thereexists atleastonepointmeMsuch thatI (or
)
is notminimal at m,and thereisnopoint in M with A 0
(or
0),
whereAisthesecondfundamental tensor.Assume that a where is an element of G, then we can show that the immersion
f a
.:
MHn+
has the Gaussmap of the form +/- a.
The signdependsonwhether a is orientation preservingor not.Hence
+ which will beused, just forsimplicity,insteadof
=o.
In
the light of the above discussions, let us take3
and notice that if-
we maychange the orientation to have again
3
.
In
this way, N(m)=(m)
for every meM up to aparallel translation in Vn
+
2.
Let D be the Levi-Civita connection of Yn+2 and be the induced connection on //n+1.
Applying Weingarten equation to I(M) and (M) as immersedhypersurfacesofItn
+
1,
wehave,(,X 17 f, (-AX),
(2.1a)
7
,X7,
]X),(2.1b)
where Aand ] arethe second fundamental tensors of
I
and ],respectively, and X isa tangentvector of M. Ifwe write
D,f,X
N in terms ofits tangent component to Itn+
and the normalone,wehve
Dr,
XN= Vf,X N+ Sinceb(N,f.X) 0, thenI,X b(N,l(:)) b(D
f XN,f(x})+bN,I,X) 0
=
b(Df,
X N,f(z)) 0g =0.
(2.2b)
To be consistent with the referenceswe dependon, weshall replace b(X,Y) by <X,Y> in all thefollowingcomputations.
From (2.2a)
and(2.2b),
wehaveFromequations
(2.1)-(2.3)
wehaveAs N
,
then equations(2.4)
giveDr,
x
N= XTf,X N
O7,
x
Nv],
x
N(2.3a)
D
f
x
N f,( AX)(2.4a)
D-/
,x
f-/*(
]x)(2.4b)
f,(AX)
y
,(I
X)(2.5)
uptoaparalleltranslationin
V"
+
2.
LEMMA
2.1. Aand] have thesamenull space at each point ofM.PROOF. It is sufficient to show that AX=O if and only if iX=0. If AX=O, then
f,(AX)
0 andso]
,(]
x)=0. Since]
is animmersion, then],
is injective and hence ]x
0. Theconverseis direct.LEMMA
2.2. A2]2.
PROOF.
For
arbitrary vectorsXandYtangentto M,wehave<
]2X,
y> < iX,4Y> <],(2Y)
>(2.6)
<f,(AX),f,(AY)> <AX, AY> <A2X,
y>.Hence
the result.LEMMA
2.3. (tr A)A (tr A)A(2.7)
PROOF. TheRicci tensorSof M,asbeingpreserved byisometries,maybewrittenas
[9]
S /(trA)A- A2
="
/(tr A )A i2where istheRicci tensorofHn
+
I.
Then(trA)A-A2 (trA)A ]
2.
UsingLemma 2.2,weobtainthe requited result.
LEMMA
2.4.(i)
(trA)2 (tr2)2
(ii)
IftrA#0at m,then2 + Aatm.(iii)
The choice of the sign in(ii)
is constant in each component of the open set V {m I(tr A)(m)#0}.PROOF. Part (i).
Thisbecomes direct fromLemma2.3 whentakingthe trace of both sides of equation(2.7).
Part (ii).
If (tr A) (m)#O, then by part(i)
we have that (tr2)
(m)#0. Using equation(2.8),
wehave [(tr A)/(tr )](m) +i. Applying equation(2.7),
weobtainthat ] +Aatm.UNIQUE ISOMETRIC IMMERSION INTO HYPERBOLIC SPACE 729 m. Hencethe subset
v
{ml(trA)(m) 0}is anopen subsetof M.Nowconsiderthe followingcases"
(a)
Let A at m andassume that ineach neighborhoodV of m thereexists apoint m whereA(rni)=-l(mi).
In
this way, we obtain a sequence of points{mi}
for whichA(mi)
-
(rni).
Taking thelimitofbothsidesasm mwehave lirnA(mi)
lirn A(rni).
mf-.,m rnf-.,m
By the continuity of both A and
,
wehave A(m)=-(m)
which contradicts the assumption andsoA nearm.(b)
IfA-
atm, wefollowasimilardiscussion. 3.A RIGIDITY RESULT.
In
this section we give a general rigidity result for hypersurfaces in Hn+ similar toTheorem 6.4p.45-46
[9].
For
notations and detailswerefer to[9].
Althoughthefollowingresult has beenalready given in[13],
we write its full proofas it is too important for the subsequentresults.
THEOREM 3.1. Let Mbeaconnected n-dimensional Riemannian manifold and letfand be isometric immersions of M into H
n+IcV
n+2 with fields of unit normals and,
,respectively. If thesecond fundamental forms h and of
!
and(with
respect to and),
respectively,coincideon M, then thereisanisometry rofHn
+
such thatr f.PROOF. We follow the same procedure of
[9]
and start with the local version of the theorem.Assume
that zo is a point ofm.
We
have two different frames(el,e2,-..,en,,v)
and’
2’’
n,
,
at Zo, wherey f(z) and (z),respectively,such that<ei,ej >
<’’i,’j>
<i,j<_n<,ei> <,Ti> =0, <Y,ei> <,’i> =0, <V,>
<,
> =0.As G acts transitively on I-In+
1,
then there exists an orthogonal matrix rEG which is anisometry of Itn+l such that r maps the frame
(el,e2,...,,y)
at zo upon the framef
,g2,",
,
atthesamepointzo.In
terms of local coordinates in acoordinate neighborhood U about to, simple calculationsusing
Gauss
and Weingarten equations give the following two systems of partial differential equationsi
gijy+
k
Ffj
ek+
hij
j
j
+
rtj
z
t+ hij
3/=
-taJ
zt
(7)
Themaindifferencebetween these twosystems and those of
[9]
isthat thesystemsherearenon-homogeneous.
In
spite of this difference, the existence and uniqueness principle of solutions is stillworking.Since the two systems (I) and
(T)
above have the same initial conditions at zo, then byuniquenesswehave
As ej
Oy[Ozj,
j0-9/0rj,
and since Y(Zo) (o), we have that y on U and the local version isnowproved.The global versionof the above theorem can be proved exactly as that ofimmersions into Euclidean spaceEn
+
[9].
Notice that forimmersions into Euclidean space En
+
1,
the required isometry isa resultant of translation in addition to an orthogonal matrix r6-0(n+1). From the above theorem, it becomes clear that in thecaseofimmersions intohyperbolicspace Hn+
theisometryneededto establish the proof is obtained purely from an orthogonal matrix of G acting on //n+
byrestriction.
4. PROOF OF THEOREM2.1.
In
the light of Theorem 3.1, ifwe consider V to be the component of Von which ]a,
then thereexists an isometry r of//n+
such that r7
!
onVl.
In
the following wegiven moresignificanceofr.Let us consider a point m6_V and take X
6-Tram
to be an eigenvector of A(=])
correspondingtoanon-zeroeigenvalue. Then equation
(2.5)
givesI,(x)
],(x)
(4.1)
Cgnsequently,
if(rant A)(m) n, then all theeigenvaluesofA are non-zerosandsoI,(Tm M)=
7
,(Tm M) up to a parallel translation in Vn+
2.
Now consider the following two different cases:Case
(i).
(m) I(m),7
,(TinM) I,(TmM) and (m) N(m).In
thiscase, the neededisometry ofHn+
turns outto betheidentitymap.Case
(ii).
I(m)7
(m),]
,(Tm M)//I,(TmM) and (m)//N(m).Clearly, (m) and l(m) should be antipodal points of
.
This situation will not occur unless ]’(M)CHn+
and7
(M)Cgn+
which contradictsthe assumption that both of theimmersions!
and
7
are in//n+
1.
Consequently,thiscaseshouldbedisregarded. Using Theorem 3.1,wehave thefollowing.LEMMA
4.1 Ifv
is a component ofv
where ] A and rank A n at one point m6-V then I(x)7
()onV1.Onthe other handif A ] and (rant A)(m) n, then l,(X) ,(X) V X6.TmM.
In
this case I,(Tm M) will also be parallel to ,(Tm M) and as mentioned above I(m) shouldcoincide with
]
(m). The isometry ofUn+
needed is different fromtle
identity.In
fact it issomesortofreflection.
LEMMA
4.2. Ifv
2is a componentofVwhere] -Aandrank A n at onepoint m6-V2 then the matrix reG which satisfies r(m) m,
r(N)
N and r(X) X V X6-Tm M has thepropertythatr
]
(z) I()onV2.In
thefollowing, weshallrestrict ourselves to thecaseofLemma4.1 mentioned aboveand giveahinttoshow howtodeal with thecaseofLemma
4.2.Letusdefinethe real-valuedfunction FonMasfollows F(a)-- <f(a),c >,
wherecisaconstant vector. Followingcomputations similar tothatin
([9],
p.342)
wehave thator
UNIQUE ISOMETRIC IMMERSION INTO HYPERBOLIC SPACE 731
where A-n and 0 is he mean curvature vector
=l
(r A)N of M as a hypersurface of//n
+
1.
Thenewoperator iselliptic and hasCocoefficients.Weare now inaposition
o
proveTheorem2.1.PROOF OF
PANT (i).
Le be a poin ofvhere
all sectional curvatures are greaterthan -1. Then rank A rank n and A is definite at m. Let m(V where V is the componentof
v
where,
A. Hence .(X)= f.(X) for eachtangent vector X(TmM. From the abovediscussions,the identitymap willbe therequiredisometry which takes (v1)
to f(V1).
Let W be the largest open set of M containing V such that z z. Following the same
procedures of
[1]
we conclude that forevery boundary point p(Ow, thereexistsa neighborhoodQ ofl0 such that for each q Qo
v,
A at qwhich means that Q intersectsthe componentv
only.Hence
wemay writew
t3QcVl
t3V’2" For pointsinv
1,wehaveW 0 asX
AonV1’ and for pointsin Q- V1, wehaveW 0. Consequently, W ron the wholeofWoQ.Consider thereal-valuedfunction9:woQ-,Rdefinedby
g(z) <f(z)-f(z), c>. Applyingthe operatortog, wehave
Ag= A <f(x)-f(x), c> A </(z), c> A <f(z), c> =n<r/,c> --n<,c> =0 on Wt
O.
Since 0 on W, then a 0 on Wo 0 bythe uniquecontinuation principleofsecondorder elliptic operators with 6,00
coefficients
([3],
[0], []).
Hence /’(-)7
(-) onwoo
contradicting the maximality ofw.
Consequently,w
will bethe whole ofM andso f(x)= for all points ofM andwehave that M andwehave that M isrigid which ends theproofof part(i).
Consideringm(V2where
X
-A, weapplyr,giveninLemma
4.2,to//n+
andcarryout theproof ta&ingasmis invariant underr.
PROOF OF
PART (ii).
This partof the theoremmay beconsidered as acorollaryofpart(i)
inthe followingsense.It has been provedin
([2],
[5])
that if M is compact and isometrically immersed in Hn+1,
thenthereexistsapoint ofm
allof whose sectional curvaturesarepositive. Applying part(i)
we obtainthedesiredresult.PROOF OF
PART (iii). In
this case, as trA 0at rnandrankA#0everywhereonM,thesameprocedurecanbefollowed butthe requiredisometry ofhrn
+
will be different fromthoseof parts(i)
and(ii).
ACKNOWLEDGEMENT. The authorisgratefultothe referee forhisvaluable suggestions.
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