Real Hypersurfaces in CP^2 AND CH^2 whose structure Jacobi operator is Lie D-parallel

0.09 MB·English

by  K. Panagiotidou

CP2 CH2
Real hypersurfaces in and whose structure
D
Jacobi operator is Lie -parallel
KonstantinaPanagiotidouandPhilipposJ.Xenos
MathematicsDivision-SchoolofTechnology,AristotleUniversityofThessaloniki,Thessaloniki54124,Greece
2
1 E-mail:[email protected],[email protected]
0
2

n ABSTRACT.In[3],[7]and[8]resultsconcerningtheparallelnessoftheLiederivativeofthestructureJacobi
a
operatorofarealhypersurfacewithrespecttoξandtoanyvectorfieldXwereobtainedinbothcomplex
J
projectivespaceandcomplexhyperbolicspace.Inthepresentpaper,westudytheparallelnessoftheLie
2
1 derivativeofthestructureJacobioperatorofarealhypersurfacewithrespecttovectorfieldXǫDinCP2and
CH2.Moreprecisely,weprovethatsuchrealhypersurfacesdonotexist.
]
G
Keywords:Realhypersurface,LieD-parallelness,StructureJacobioperator,Complexprojectivespace,
D
Complexhyperbolicspace.
.
h
t MathematicsSubjectClassification(2000):Primary53C40;Secondary53C15,53D15.
a
m
[

1
1 Introduction
v
0
4 A complex n-dimensional Kaehler manifold of constant holomorphic sectional curvature c is
5
calledacomplexspaceform,whichisdenotedbyM (c). Acompleteandsimplyconnectedcom-
2 n
. plex space form is complex analyticallyisometric to a complexprojectivespace CPn, a complex
1
0 EuclideanspaceCnoracomplexhyperbolicspaceCHnifc>0,c=0orc<0respectively.
2
LetM be a realhypersurfaceina complexspace formM (c), c 6= 0. Thenan almostcontact
1 n
: metric structure (ϕ,ξ,η,g) can be defined on M induced from the Kaehler metric and complex
v
i structure J on M (c). The structure vector field ξ is called principalif Aξ = αξ, where A is the
X n
shapeoperatorofMandα = η(Aξ)isasmoothfunction. ArealhypersurfaceissaidtobeaHopf
r
a
hypersurfaceifξisprincipal.
Theclassificationproblemofrealhypersurfacesincomplexspaceformsisofgreatimportancein
DifferentialGeometry.ThestudyofthiswasinitiatedbyTakagi(see[10]),whoclassifiedhomoge-
neousrealhypersurfacesinCPnandshowedthattheycouldbedividedintosixtypes,whicharesaid
tobeoftypeA1,A2,B,C,DandE. Berndt(see[1])classifiedhomogeneousrealhypersurfacesin
CHnwithconstantprincipalcurvatures.
TheJacobioperatorwithrespecttoXonMisdefinedbyR(·,X)X,whereRistheRiemmanian
curvatureofM.ForX =ξ theJacobioperatoriscalledstructureJacobioperatorandisdenotedby
l = R(·,ξ)ξ. It hasa fundamentalrole in almostcontactmanifolds. Manydifferentialgeometers
havestudiedrealhypersurfacesintermsofthestructureJacobioperator.
ThestudyofrealhypersurfaceswhosestructureJacobioperatorsatisfiesconditionsconcernedto
theparallelnessofitisaproblemofgreatimportance. In[6]thenonexistenceofrealhypersurfaces
ThefirstauthorisgrantedbytheFoundationAlexandrosS.Onasis.GrantNr:GZF044/2009-2010.
1
innonflatcomplexspaceformwithparallelstructureJacobioperator(∇l = 0)wasproved. In[9]
a weaker condition (D-parallelness, where D = ker(η)), that is ∇ l = 0 for any vector field X
X
orthogonalto ξ, was studied and it was proved the nonexistenceof such hypersurfacesin case of
CPn (n ≥ 3). Theξ-parallelnessofstructureJacobioperatorincombinationwithotherconditions
wasanotherproblemthatwasstudiedbymanyauthorssuchasKi,Perez,Santos,Suh([4]).
The Lie derivative of the structure Jacobi operator is another condition that has been studied
extensively.Moreprecisely,in[7]provedthenon-existenceofrealhypersurfacesinCPn,(n≥3),
whoseLiederivativeofthestructureJacobioperatorwithrespecttoanyvectorfieldXvanishes(i.e.
L l = 0). On the other hand, real hypersurfacesin CPn, (n ≥ 3), whose Lie derivative of the
X
structureJacobioperatorwithrespecttoξ vanishes(i.e. L l = 0,Lieξ-parallel)areclassified(see
ξ
[8]). IveyandRyanin[3]extendsomeoftheaboveresultsinCP2 andCH2. Moreprecisely,they
provedthatinCP2andCH2thereexistnorealhypersurfacessatisfyingconditionL l=0,forany
X
vectorfield X, butrealhypersurfacessatisfyingconditionL l = 0 existand theyclassified them.
ξ
Additional,theyprovedthatthereexistnorealhypersurfacesinCPn orCHn,(n ≥ 3),satisfying
conditionL l=0,foranyvectorfieldX.
X
Followingthenotionof[8],thestructureJacobioperatorissaidtobeLieD-parallel,whenthe
LiederivativeofitwithrespecttoanyvectorfieldXǫDvanishes. Sothefollowingquestionraises
naturally:
”Dothereexistrealhypersurfacesinnon-flatM2(c)withLieD-parallelstructureJacobiopera-
tor?”
In this paper, we study the abovequestionin CP2 and CH2. The conditionof Lie D-parallel
structureJacobioperator,i.e. L l =0withXǫD,implies:
X
∇ (lY)+l∇ X =∇ X +l∇ Y, (1.1)
X Y lY X
whereYǫTM.
Weprovethefollowing:
Main Theorem: There exist no real hypersurfaces in CP2 and CH2 equipped with Lie D-
parallelstructureJacobioperator.
2 Preliminaries
Throughoutthis paper all manifolds, vector fields e.t.c are assumed to be of class C∞ and all
manifolds are assumed to be connected. Let M be a connected real hypersurface immersed in a
nonflatcomplexspace form(M (c),G) with almostcomplexstructureJ ofconstantholomorphic
n
sectionalcurvaturec. LetNbealocallydefinedunitnormalvectorfieldonMandξ = −JN. For
avectorfieldXtangenttoMwecanwriteJX = ϕ(X)+η(X)N,whereϕX andη(X)N arethe
tangentialandthenormalcomponentofJX respectively. TheRiemannianconnection∇inM (c)
n
and∇inMarerelatedforanyvectorfieldsX,YonM:
∇ X =∇ X +g(AY,X)N
Y Y
∇ N =−AX
X
2
wheregistheRiemannianmetriconMinducedfromGofM (c)andAistheshapeoperatorofM
n
inM (c). Mhasanalmostcontactmetricstructure(ϕ,ξ,η)inducedfromJonM (c)whereϕisa
n n
(1,1)tensorfieldandηa1-formonMsuchthat([2])
g(ϕX,Y)=G(JX,Y), η(X)=g(X,ξ)=G(JX,N).
Thenwehave
2
ϕ X =−X +η(X)ξ, η◦ϕ=0, ϕξ =0, η(ξ)=1 (2.1)
g(ϕX,ϕY)=g(X,Y)−η(X)η(Y), g(X,ϕY)=−g(ϕX,Y) (2.2)
∇ ξ =ϕAX, (∇ ϕ)Y =η(Y)AX −g(AX,Y)ξ (2.3)
X X
Since the ambient space is of constant holomorphicsectional curvature c, the equations of Gauss
andCodazziforanyvectorfieldsX,Y,ZonMarerespectivelygivenby
c
R(X,Y)Z = [g(Y,Z)X−g(X,Z)Y +g(ϕY,Z)ϕX (2.4)
4
−g(ϕX,Z)ϕY −2g(ϕX,Y)ϕZ]+g(AY,Z)AX −g(AX,Z)AY
c
(∇ A)Y −(∇ A)X = [η(X)ϕY −η(Y)ϕX −2g(ϕX,Y)ξ] (2.5)
X Y
4
whereRdenotestheRiemanniancurvaturetensoronM.
Relation(2.4)impliesthatthestructureJacobioperatorlisgivenby:
c
lX = [X −η(X)ξ]+αAX −η(AX)Aξ (2.6)
4
ForeverypointP ǫ M,thetangentspaceT M canbedecomposedasfollowing:
P
T M =span{ξ}⊕ker(η),
P
whereker(η) = {X ǫ T M : η(X) = 0}. Duetotheabovedecomposition,thevectorfieldAξ
P
canbewritten:
Aξ =αξ+βU, (2.7)
whereβ =|ϕ∇ ξ|andU =−1ϕ∇ ξ ǫker(η),providedthatβ 6=0.
ξ β ξ
3 Some PreviousResults
LetMbeanon-HopfhypersurfaceinCP2orCH2(i.e.M2(c),c6=0).Thenthefollowingrelations
holdsoneverythree-dimensionalrealhypersurfaceinM2(c).
Lemma3.1 LetMbearealhypersurfaceinM2(c). ThenthefollowingrelationsholdonM:
AU =γU +δϕU +βξ, AϕU =δU +µϕU, (3.1)
3
∇ ξ =−δU +γϕU, ∇ ξ =−µU +δϕU, ∇ ξ =βϕU, (3.2)
U ϕU ξ
∇UU =κ1ϕU +δξ, ∇ϕUU =κ2ϕU +µξ, ∇ξU =κ3ϕU, (3.3)
∇UϕU =−κ1U −γξ, ∇ϕUϕU =−κ2U −δξ, ∇ξϕU =−κ3U −βξ, (3.4)
whereγ,δ,µ,κ1,κ2,κ3aresmoothfunctionsonM.
Proof:Let{U,ϕU,ξ}beanorthonormalbasisofM.Thenwehave:
AU =γU +δϕU +βξ AϕU =δU +µϕU,
whereγ,δ,µaresmoothfunctions,sinceg(AU,ξ)=g(U,Aξ)=βandg(AϕU,ξ)=g(ϕU,Aξ)=
0.
Thefirstrelationof(2.3),becauseof(2.6)and(3.1),forX = U,X = ϕU andX = ξ implies
(3.2),owingto(2.7).
Fromthewellknownrelation:Xg(Y,Z)=g(∇ Y,Z)+g(Y,∇ Z)forX,Y,Zǫ{ξ,U,ϕU}
X X
weobtain(3.3)and(3.4),whereκ1,κ2andκ3aresmoothfunctions. (cid:3)
BecauseofLemma3.1theCodazziequationimplies:
Uβ−ξγ = αδ−2δκ3 (3.5)
c
2 2
ξδ = αγ+βκ1+δ +µκ3+ −γµ−γκ3−β (3.6)
4
Uα−ξβ = −3βδ (3.7)
ξµ = αδ+βκ2−2δκ3 (3.8)
(ϕU)α = αβ+βκ3−3βµ (3.9)
c
2
(ϕU)β = αγ+βκ1+2δ + −2γµ+αµ (3.10)
2
Uδ−(ϕU)γ = µκ1−κ1γ−βγ−2δκ2−2βµ (3.11)
Uµ−(ϕU)δ = γκ2+βδ−κ2µ−2δκ1 (3.12)
WerecallthefollowingProposition([3]):
Proposition3.2 There doesnotexist real non-flathypersurfacein M2(c), whose structure Jacobi
operatorvanishes.
4 AuxiliaryRelations
IfMisarealnon-flathypersurfaceinCP2orCH2(i.e. M2(c),c6=0),weconsidertheopensubset
WofpointsP ǫM,suchthatthereexistsaneighborhoodofeveryP,whereβ =0andNtheopen
subsetofpointsQǫM, suchthatthereexistsaneighborhoodofeveryQ, whereβ 6= 0. Since, β
is a smooth functionon M, then W∪N is an open and dense subset of M. In W ξ is principal.
Furthermore,weconsiderV,ΩopensubsetsofN:
V={Q ǫ N:α=0 in a neighborhood of Q},
Ω={Q ǫ N:α6=0 in a neighborhood of Q},
4
whereV∪ΩisopenanddenseintheclosureofN.
Proposition4.1 Let M be a real hypersurface in M2(c), equipped with Lie D-parallel structure
Jacobioperator.Then,Visempty.
Proof: Let {U,ϕU,ξ} be an orthonormal basis on V. The following relations hold, because of
Lemma3.1
′ ′ ′ ′
AU =γ U +δ ϕU +βξ, AϕU =δ U +µϕU, Aξ =βU (4.1)
′ ′ ′ ′
∇ ξ =−δ U +γ ϕU, ∇ ξ =−µU +δ ϕU, ∇ ξ =βϕU, (4.2)
U ϕU ξ
′ ′ ′ ′ ′
∇ U =κ ϕU +δ ξ, ∇ U =κ ϕU +µξ, ∇ U =κ ϕU, (4.3)
U 1 ϕU 2 ξ 3
′ ′ ′ ′ ′
∇ ϕU =−κ U −γ ξ, ∇ ϕU =−κ U −δ ξ, ∇ ϕU =−κ U −βξ, (4.4)
U 1 ϕU 2 ξ 3
whereγ′,δ′,µ′,κ′,κ′,κ′ aresmoothfunctionsonV.
1 2 3
From(2.6)forX =U andX =ϕU,takingintoaccount(4.1),weobtain:
c c
2
lϕU = ϕU lU =( −β )U. (4.5)
4 4
Relation(1.1),becauseof(4.2),(4.3)(4.4)and(4.5)implies:
′
δ = 0,for X =ϕU and Y =ξ (4.6)
c
′ ′ 2
(µ −κ )( −β ) = 0,for X =ϕU and Y =ξ (4.7)
3
4
′ ′
κ = γ ,for X=U and Y =ξ. (4.8)
3
OnV,relations(3.5)-(3.12),takingintoaccount(4.6),become:
c
′ ′ ′ ′ ′ ′ ′ 2
βκ +µκ + = γ µ +γ κ +β (4.9)
1 3 3
4
′ ′
κ = 3µ (4.10)
3
c
′ ′ ′
(ϕU)β = βκ + −2γ µ (4.11)
1
2
′ ′ ′ ′ ′ ′ ′
(ϕU)γ = κ γ +βγ +2βµ −µκ . (4.12)
1 1
Dueto(4.7),weconsidertheopensubsetsV:
c
V1 ={P ǫ V:β2 6= in a neighborhood of P},
4
c
V′ ={P ǫ V:β2 = in a neighborhood of P},
1
4
whereV1∪V′1isopenanddenseintheclosureofV. SoinV1 weobtain:µ′ =κ′3.
OnV1,becauseofrelations(4.8)and(4.10),weobtainµ′ =κ′3 =γ′ =0.
Relation (1.1), for X = U and Y = ϕU, due to (4.3), (4.4) and (4.5), implies: κ′ = 0.
1
Substitutingin(4.9)µ′ = κ′3 = γ′ = κ′1 = 0,leadsto: β2 = 4c,whichisimpossibleonV1. SoV1
isemptyandβ2 = c holdsonV.
4
OnV,becauseof(4.8)and(4.10),wehaveγ′ =κ′ =3µ′. Substitutingthelasttworelationsin
3
(4.9),implies: βκ′ = 9µ′2. Differentiationofβ2 = c withrespecttoϕU andtakingintoaccount
1 4
5
(4.13), γ′ = 3µ′ and βκ′ = 9µ′2, yields: c = −6µ′2, which is a contradictionbecause β2 = c.
1 4
Hence,V=∅. (cid:3)
InwhatfollowsweworkinΩ.
Byusing(2.6),becauseof(3.1),weobtain:
c c
2
lU =( +αγ−β )U +αδϕU lϕU =αδU +(αµ+ )ϕU (4.13)
4 4
Relation(1.1)becauseof(3.2),(3.3)and(3.4)implies:
c
2
δ(ακ3+ −β ) = 0 for X =U and Y =ξ (4.14)
4
c
2
( +αµ)(κ3−γ)+αδ = 0 for X =U and Y =ξ (4.15)
4
c
2 2
( +αγ−β )(µ−κ3)−αδ = 0 for X =ϕU and Y =ξ (4.16)
4
c
δ(ακ3+ ) = 0 for X =ϕU and Y =ξ (4.17)
4
Dueto(4.17),weconsidertheopensubsetsΩ1andΩ′1ofΩ:
Ω1 ={Q ǫ Ω: δ 6=0 in a neighborhood of Q},
′
Ω ={Q ǫ Ω: δ =0 in a neighborhood of Q},
1
whereΩ1∪Ω′1isopenanddenseintheclosureofΩ.
InΩ1,from(4.14)and(4.17),wehave:β =0,whichisacontradiction,thereforeΩ1 =∅. Thus
wehave:δ =0inΩandrelationsfromLemma3.1,(4.13),(4.15)and(4.16)becomerespectively:
AU =γU +βξ, AϕU =µϕU, Aξ =αξ+βU (4.18)
∇ ξ =γϕU, ∇ ξ =−µU, ∇ ξ =βϕU (4.19)
U ϕU ξ
∇UU =κ1ϕU, ∇ϕUU =κ2ϕU +µξ, ∇ξU =κ3ϕU, (4.20)
∇UϕU =−κ1U −γξ, ∇ϕUϕU =−κ2U, ∇ξϕU =−κ3U −βξ, (4.21)
c c
2
lU =( +αγ−β )U lϕU =(αµ+ )ϕU (4.22)
4 4
c
2
( +αγ−β )(µ−κ3)=0 (4.23)
4
c
( +αµ)(κ3−γ)=0. (4.24)
4
Owingto(4.23),weconsidertheopensubsetsΩ2andΩ′2ofΩ:
Ω2 ={Q ǫ Ω: µ6=κ3 in a neighborhood of Q},
′
Ω2 ={Q ǫ Ω: µ=κ3 in a neighborhood of Q},
whereΩ2∪Ω′2isopenanddenseintheclosureofΩ.
SoinΩ2,wehave:
β2 c
γ = − and (4.21) implies: lU =0. (4.25)
α 4α
6
Dueto(4.24)weconsiderΩ21andΩ′21theopensubsetsofΩ2:
c
Ω21 ={Q ǫ Ω2 : µ=− in a neighborhood of Q},
4α
c
′
Ω21 ={Q ǫ Ω2 : µ6=− in a neighborhood of Q},
4α
where Ω21 ∪Ω′21 is open and dense in the closure of Ω2. So in Ω21, (4.22) implies: lϕU = 0
andbecauseofProposition3.2weobtainΩ21 = ∅,thusinΩ2, wehaveµ 6= −4cα andasaresult:
lϕU 6=0. Furthermore,becauseof(4.24)κ3 =γ.
Lemma4.2 LetMbearealhypersurfaceinM2(c),equippedwithLieD-parallelstructureJacobi
operator.ThenΩ2isempty.
Proof:OnΩ2,relation(1.1)forX =ϕU andY =U owingto(4.20)and(4.21)implies:κ2lϕU =
0. Soκ2 =0. Duetothelast,relations(3.8)and(3.12)imply:
Uµ=ξµ=0. (4.26)
Relation (1.1) for X = U and Y = ϕU taking into account (4.20), (4.21), (4.22) and (4.25)
implies:µ=−γandκ1 =0.Substitutionin(3.6)oftherelationswhichholdonΩ2implies:γ =0
andthisresultsin: β2 = c. DifferentiationofthelastwithrespecttoϕU,becauseof(3.10)leadsto
4
c=0,whichisacontradiction.ThiscompletestheproofofthepresentLemma. (cid:3)
Summarizing,inΩwehave:δ =0andµ=κ3. Dueto(4.24),weconsiderΩ3andΩ′3theopen
subsetsofΩ:
c
Ω3 ={Q ǫ Ω: µ6=− in a neighborhood of Q},
4α
c
′
Ω ={Q ǫ Ω: µ=− in a neighborhood of Q},
3
4α
whereΩ3∪Ω′3 isopenanddenseintheclosureofΩ. Sinceµ 6=−4cα,dueto(4.22)and(4.24)we
obtain:lϕU 6=0andγ =µ=κ3,inΩ3.
Lemma4.3 LetMbearealhypersurfaceinM2(c),equippedwithLieD-parallelstructureJacobi
operator.ThenΩ3isempty.
Proof:InΩ3relations(3.6),(3.9),(3.10)and(3.11)becomerespectively:
c
2 2
αγ+βκ1+ = β +γ (4.27)
4
(ϕU)α = αβ−2βγ (4.28)
c
2
(ϕU)β = 2αγ+βκ1+ −2γ (4.29)
2
(ϕU)µ=(ϕU)γ = 3βγ. (4.30)
Relation(1.1)forX = Y = ϕU,becauseof(4.22),(4.28)and(4.30)implies: γ(2α−γ)= 0.
Owingtothelastrelation,weconsiderΩ31andΩ′31theopensubsetsofΩ3:
Ω31 ={Q ǫ Ω3 : γ 6=0 in a neighborhood of Q},
′
Ω31 ={Q ǫ Ω3 : γ =0 in a neighborhood of Q},
7
whereΩ31∪Ω′31isopenanddenseintheclosureofΩ3.InΩ31,γ =2α.Differentiationofthelatter
withrespecttoϕU andtakingintoaccount(4.28)and(4.30)leadsto:αβ =0,whichisimpossible.
SoΩ31isempty.
ResumingonΩ3 wehave: γ = µ = κ3 = 0andrelation(3.8)implies: κ2 = 0. Relation(1.1)
forX =U andY = ϕU,becauseof(4.20),(4.21)and(4.22)yields: κ1 =0andsorelation(4.27)
implies:β2 = c. DifferentiationofthelastalongϕU andbecauseof(4.29)leadstoc=0,whichis
4
acontradictionandthiscompletestheproofofLemma4.3. (cid:3)
SoonΩthefollowingrelationshold:
c
δ =0, µ=κ3 =− ,
4α
andrelation(4.22),becauseofthelastoneimplies: lϕU =0.
Relation(1.1),forX = U andY = ϕU, becauseof(4.20)and(4.21)implies: κ1lU = 0. So
κ1 =0,duetoProposition3.2.
Duetotheaboverelations,onΩrelations(3.9),(3.10)and(3.11)becomerespectively:
cβ
(ϕU)α = αβ+ (4.31)
2α
cγ c
(ϕU)β = αγ+ + (4.32)
2α 4
cβ
(ϕU)γ = βγ− (4.33)
2α
Relation(1.1)forX =ϕU andY =U takingintoaccount(4.20),(4.21)and(4.22)yields:
c
2
(ϕU)( +αγ−β ) = 0 (4.34)
4
c
2
κ2( +αγ−β ) = 0 (4.35)
4
c
2
(µ+γ)( +αγ−β ) = 0. (4.36)
4
Dueto(4.35),weconsider:Ω4andΩ′4theopensubsetsofΩ:
c
2
Ω4 ={Q ǫ Ω: +αγ−β =0 in a neighborhood of Q},
4
c
′ 2
Ω ={Q ǫ Ω: +αγ−β 6=0 in a neighborhood of Q},
4
4
whereΩ4∪Ω′4isopenanddenseintheclosureofΩ. SoinΩ4wehave:γ = βα2 − 4cα andbecause
of(4.22)lU =0,whichisimpossibleduetoProposition3.2.Therefore,Ω4 =∅
SoinΩwehave:κ2 =0andrelation(4.36)implies: µ=−γ.
Lemma4.4 LetMbearealhypersurfaceinM2(c),equippedwithLieD-parallelstructureJacobi
operator.ThenΩisempty.
Proof:InΩrelations(3.8)and(3.12)yields:Uα=ξα=0.
Usingtheaboverelations,weobtain:
[U,ξ]α=Uξα−ξUα=0,
c
[U,ξ]α=(∇ ξ−∇ U)α= (ϕU)α.
U ξ
2α
8
Combiningthelasttworelationsandtakingintoaccount(4.31),wehave: c= −2α2andsoµ = α
2
andγ = −α. Relation(4.33),becauseof(4.31)andthelasttworelationsimply: α = 0,whichis
2
impossibleinΩ. ThiscompletestheproofofLemma4.4. (cid:3)
WeleadtothefollowingduetoLemmas4.1and4.4:
Proposition4.5 Every realhypersurfacein M2(c), equippedwith Lie D-parallelstructure Jacobi
operatorisaHopfhypersurface.
5 ProofofMainTheorem
Since M is a Hopf hypersurface, due to Theorem 2.1, ([5]) , we have that α is a constant. We
consideraunitvectorfieldZ ǫker(η), suchthatAZ = λZ,thenAϕZ = νϕZ. Then{ξ,Z,ϕZ}
isanorthonormalbasisandthefollowingrelationholdsonM,(Corollary2.3,[5]):
α c
λν = (λ+ν)+ (5.1)
2 4
Therelation(2.6)implies:
c c
lZ =( +αλ)Z lϕZ =( +αν)ϕZ (5.2)
4 4
Relation(1.1)forX = Z andY = ϕZ andforX = ϕZ andY = Z,becauseof(5.2)andtaking
theinnerproductofthemwithξimpliesrespectively:
c
(λ+ν)( +αν) = 0 (5.3)
4
c
(ν+λ)( +αλ) = 0 (5.4)
4
I.Supposethatα6=0.
Dueto(5.3)and(5.4),weconsidertheopensubsetofM:
M1 ={P ǫ M :λ6=−ν in a neighborhood of P}.
Because of (5.3) and (5.4) in M1 we have λ = ν = −4cα. In M1 (5.1) yields c = 0, which is a
contradiction.Therefore,M1 =∅.
Hence,inMwehave: λ = −ν. Substitutionofthelatterin(5.1)impliesc = −4λ2. Fromthe
lastrelationweconcludethat:c<0andλ=constant. Theonlyhypersurfacethatwehaveinthis
case is of type B in CH2. Substitutingthe eigenvaluesof this hypersurfacein λ = −ν leads to a
contradiction.
II.Supposeα=0.
Relation(5.3)impliesλ = −ν andsofromrelation(5.1)weobtain: c = −4λ2. Fromthelast
two relations, we conclude that the only case which occurs is that of a real hypersurfacein CH2
withthreedistinctconstanteigenvalues.SoitshouldbeoftypeBinCH2,butforsuchhypersurface
αcannotvanish.Soweleadtoacontradictionandthiscompletestheproofofourmaintheorem.
9
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