Real Hypersurfaces Equipped with $xi$-parallel Structure Jacobi Operator in CP^2 or CH^2

0.09 MB·English

by  K. Panagiotidou

REAL HYPERSURFACESEQUIPPEDWITH ξ-PARALLELSTRUCTURE
JACOBIOPERATOR INCP2 ORCH2
KonstantinaPanagiotidouandPhilipposJ.Xenos
MathematicsDivision-SchoolofTechnology,AristotleUniversityofThessaloniki,Greece
E-mail:[email protected],[email protected]
2
1
0
2 ABSTRACT. Theξ-parallelnessconditionofthestructureJacobioperatorofrealhypersurfaceshas

n beenstudiedincombinationwithadditionalconditions.Inthepresentpaperwestudythree
a dimensionalrealhypersurfacesinCP2orCH2equippedwithξ-parallelstructureJacobioperator.
J
3 WeprovethattheyareHopfhypersurfacesandifadditionalη(Aξ)6=0,wegivetheclassification
1 ofthem.

]
G Keywords:Realhypersurface,ξ-parallelstructureJacobioperator,Complexprojectivespace,
D Complexhyperbolicspace.
.
h
t MathematicsSubjectClassification(2000):Primary53B25;Secondary53C15,53D15.
a
m
[ 1 Introduction

1
v A complex n-dimensional Kaehler manifold of constant holomorphic sectional curvature
0
1 c is called a complex space form, which is denoted by M (c). A complete and simply
n
9
connected complex space form is complex analytically isometric to a complex projective
2
. space CPn, a complex Euclidean space Cn or a complex hyperbolic space CHn if c >
1
0
0,c = 0orc< 0respectively.
2
1 Thestudy ofreal hypersurfaces in anonflat complex space form isaclassical problem
:
v in Differential Geometry. Let M be a real hypersurface in M (c). Then M has an almost
n
i
X contact metricstructure (ϕ,ξ,η,g). Thestructure vector fieldξ iscalled principal ifAξ =
r
a αξ holds onM,whereAistheshape operator ofM inMn(c)andαisasmoothfunction.
Arealhypersurface iscalledHopfhypersurface ifξ isprincipal.
Takagi in [14] classified homogeneous real hypersurfaces in CPn and Berndt in [1]
classified Hopf hypersurfaces with constant principal curvatures in CHn. Let M be a real
hypersurface inM (c), c 6= 0. Thenwestatethefollowing theorems duetoOkumura[11]
n
forCPn andMontielandRomero[9]forCHn respectively.
Theorem1.1 Let M be a real hypersurface of M (c) , n ≥ 2, c 6= 0. If it satisfies Aϕ−
n
ϕA = 0,thenMislocallycongruent tooneofthefollowinghypersurfaces:
• IncaseCPn
(A )ageodesic hypersphere ofradiusr,where0 < r < π,
1 2
1
(A )atubeofradiusroveratotallygeodesicCPk,(1 ≤ k ≤ n−2),where0< r <
2
π.
2
• IncaseCHn
(A )ahorosphere inCHn,i.eaMontieltube,
0
(A )ageodesic hypersphere oratubeoverahyperplane CHn−1,
1
(A )atubeoveratotallygeodesic CHk (1 ≤ k ≤ n−2).
2
Since 2006 many authors have studied real hypersurfaces whose structure Jacobi op-
erator is parallel (∇l = 0). Ortega, Perez and Santos [12] proved the nonexistence of
real hypersurfaces in non-flat complex space form with parallel structure Jacobi operator
∇l = 0. Perez, Santos andSuh[13]continuing theworkof[12]considered aweakercon-
dition(D-parallelness), thatis∇ l = 0foranyvectorfieldX orthogonaltoξ. Theyproved
X
thenon-existence ofsuchrealhypersurfaces inCPm,m ≥ 3.
KimandKiin[7]classifiedrealhypersurfacesif∇ l = 0andSϕ =ϕS. KiandLiu[5]
ξ
provedthatrealhypersurfaces satisfying∇ l = 0andlS = SlareHopfhypersurfacespro-
ξ
videdthatthescalarcurvature isnon-negative. Ki,et.al. in[6]classifiedrealhypersurfaces
satisfying∇ l = 0and∇ S = 0. Kimet.al. in[8]studiedtherealhypersurfacessatisfying
ξ ξ
g(∇ ξ,∇ ξ) = µ2 =const, 6µ2 + c 6= 0 and classified those whose l is ξ−parallel. Cho
ξ ξ 4
andKi[3]classifiedrealhypersurfaces satisfying Al = lAand∇ l = 0.
ξ
RecentlyIveyandRyan,in[4]studiedrealhypersurfaces inM (c).
2
Motivated by all the above conclusions we study real hypersurfacs in CP2 or CH2
equipped with ξ-parallel structure Jacobi operator, i.e. ∇ l = 0. More precisely, the fol-
ξ
lowingrelationholds:
(∇ l)X = 0. (1.1)
ξ
Weprovethefollowingtheorem
Main Theorem: Let M be a connected real hypersurface in CP2 or CH2 with ξ-
parallel structure Jacobi operator. ThenM isaHopfhypersurface. Further, ifη(Aξ) 6= 0,
then:
• inthecaseofCP2,M islocallycongruent to
ageodesic sphere, where0 < r < π andr 6= π,
2 4
• inthecaseofCH2,M islocallycongruent
toahorosphere,
ortoageodesic sphere
ortoatubeoverthehyperplane CH1.
2
2 Preliminaries
∞
Throughout this paper all manifolds, vector fields e.t.c. are assumed to be of class C
and all manifolds are assumed to be connected. Furthermore, the real hypersurfaces are
supposed tobeorientedandwithoutboundary.
Let M be a real hypersurface immersed in a nonflat complex space form (M (c),G)
n
withalmostcomplexstructure Jofconstant holomorphic sectional curvature c. LetN bea
unit normal vector field on M and ξ = −JN. For a vector field X tangent to M we can
write JX = ϕ(X) + η(X)N, where ϕX and η(X)N are the tangential and the normal
component of JX respectively. The Riemannian connection ∇ in M (c) and ∇ in M are
n
relatedforanyvectorfieldsX,Y onM:
∇ X = ∇ X +g(AY,X)N
Y Y
∇ N = −AX
X
where g is the Riemannian metric on M induced from G of M (c) and A is the shape
n
operatorofM inM (c). M hasanalmostcontactmetricstructure(ϕ,ξ,η)inducedfromJ
n
onM (c)whereϕisa(1,1)tensorfieldandη a1-formonM suchthat(see[2]
n
g(ϕX,Y)= G(JX,Y), η(X) = g(X,ξ) = G(JX,N).
Thenwehave
ϕ2X = −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 holomorphic sectional curvature c, the equations of
GaussandCodazziforanyvectorfieldsX,Y,Z onM arerespectively givenby
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
whereRdenotestheRiemanniancurvature tensoronM.
3
ForeverypointPǫM,thetangent spaceT M canbedecomposed asfollowing:
P
T M = span{ξ}⊕kerη
P
where ker(η) = {X ǫ T M : η(X) = 0}. Due to the above decomposition,the vector
P
fieldAξ isdecomposed asfollows:
Aξ = αξ +βU (2.6)
whereβ = |ϕ∇ ξ|andU = −1ϕ∇ ξ ǫ ker(η),provided thatβ 6= 0.
ξ β ξ
3 Auxiliaryrelations
Let M be a real hypersurfaces in CP2 or CH2, i.e. M (c), c 6= 0. We consider the open
2
subsetNofM suchthat:
N = {P ǫ M : β 6= 0, inaneighborhood ofP}.
Furthermore, weconsider V,ΩopensubsetsofNsuchthat:
V= {P ǫ N : α = 0, inaneighborhood ofP},
Ω = {P ǫ N : α6= 0, inaneighborhood ofP},
whereV∪ΩisopenanddenseintheclosureofN.
Lemma3.1 Let M be a real hypersurface in M (c), equipped with ξ-parallel structure
2
Jacobioperator. ThenVisempty.
Proof: Let{U,ϕU,ξ}bealocalorthonormal basisonV. Therelation (2.6)takes theform
Aξ = βU. Thefirstrelation of(2.3)forX = ξ,takingintoaccount thelatterimplies
∇ ξ = βϕU.
ξ
Relation(1.1)forX = ξ,because oftheaboverelation yields:
∇ (lξ)= l∇ ξ ⇒ βϕU = 0,
ξ ξ
whichleadstoacontradiction andthiscompletes theproofofLemma3.1. (cid:3)
InwhatfollowsweworkonΩ,whereα 6= 0andβ 6= 0.
Lemma3.2 Let M be a real hypersurface in M (c), equipped with ξ-parallel structure
2
Jacobioperator. Thenthefollowingrelations holdinΩ:
β2 c κ c
AU = ( − + )U +βξ, AϕU = − ϕU (3.1)
α 4α α 4α
4
β2 c κ c
∇ ξ = βϕU, ∇ ξ = ( − + )ϕU, ∇ ξ = U (3.2)
ξ U ϕU
α 4α α 4α
c
∇ U = κ ϕU, ∇ U = κ ϕU, ∇ U = κ ϕU − ξ (3.3)
ξ 1 U 2 ϕU 3
4α
β2 c κ
∇ ϕU = −κ U−βξ, ∇ ϕU = −κ U−( − + )ξ, ∇ ϕU = −κ U (3.4)
ξ 1 U 2 ϕU 3
α 4α α
κκ = 0, (ξκ) = 0, (3.5)
1
whereκ,κ ,κ ,κ aresmoothfunctions onM.
1 2 3
Proof: Let{U,ϕU,ξ}bealocalorthonormal basisofΩ.
The first relation of (2.3) for X = ξ implies: ∇ ξ = βϕU and so relation (1.1) for
ξ
X = ξ,takingintoaccountthelatter, gives:
lϕU = 0. (3.6)
Relation (2.4)forX = ϕU andY = Z = ξ gives: lϕU = cϕU +αAϕU, whichbecause
4
of(3.6)impliesthesecond of(3.1). Relation(2.4)forX = U andY = Z = ξ,wehave:
c
lU = U +αAU −βAξ (3.7)
4
Thescalarproducts of(3.7)withϕU andU,becauseof(2.6)andthesecondof(3.1)imply
thefirstof(3.1),whereκ= g(lU,U).
Thefirstrelation of(2.3), forX = U andX = ϕU. taking intoconsideration relations
(3.1),givestherestofrelation(3.2).
From the well known relation: Xg(Y,Z) = g(∇ Y,Z)+g(Y,∇ Z) for X,Y,Z ǫ
X X
{ξ,U,ϕU}weobtain(3.3)and(3.4),whereκ ,κ ,κ aresmoothfunctions inΩ.
1 2 3
Ontheotherhand
ξκ = ξg(lU,U)
⇒ ξκ= g(∇ (lU),U)+g(lU,∇ U)
ξ ξ
⇒ ξκ= g((∇ l)U +l(∇ U),U)+g(lU,∇ U)
ξ ξ ξ
⇒ ξκ= g(l(∇ U),U)+g(lU,∇ U)
ξ ξ
Theaboverelationbecause of(3.3),(3.6)and(3.7)yields:
ξκ= g(κ lϕU,U)+g(lU,κ ϕU) ⇒ ξκ = 0
1 1
5
Ontheotherhand:
ξg(lϕU,U) = 0
⇒ g(∇ (lϕU),U)+g(lϕU,∇ U) = 0
ξ ξ
⇒ g((∇ l)ϕU +l(∇ ϕU),U)+g(lϕU,∇ U)= 0
ξ ξ ξ
Fromtheaboveequation because of(1.1),(2.6),(3.4),(3.6)andκ = g(lU,U)weobtain:
g(l(−κ U −βξ),U) = 0 ⇒ κκ = 0
1 1
(cid:3)
Relation(2.5)forX ǫ{U,ϕU}andY = ξ,because ofLemma3.2yields:
β2 c κ
Uβ = ξ( − + ) (3.8)
α 4α α
Uα = ξβ (3.9)
β2κ c κ c β2
1
= κ+βκ + ( − + ) (3.10)
2
α 4α α 4α α
κ β2 c β2 c κ
(ϕU)β = 1 +β2+ ( − + ) (3.11)
α 4α α 4α α
4α2κ β
3
ξα = (3.12)
c
3c
(ϕU)α = β(κ +α+ ) (3.13)
1
4α
Furthermore,relation(2.5),forX = U andY = ϕU,duetoLemma3.2and(3.10),implies:
cβκ
1
(ϕU)κ = − +κβ +κκ −cβ (3.14)
2
4α
4κ α
Uα = 3 (β2+κ) (3.15)
c
Usingtherelations (3.9)-(3.15) andLemma3.2weobtain:
c c
[U,ξ]( )= (∇ ξ−∇ U)
U ξ
4α 4α
c cβ β2 c κ 3c
⇒ [U,ξ]( )= − ( − + −κ )(κ +α+ ) (3.16)
4α 4α2 α 4α α 1 1 4α
c c c
[U,ξ]( ) = (U(ξ ))−(ξ(U ))
4α 4α 4α
c β2 κ
⇒ [U,ξ]( ) = −βκ2−β(Uκ )+( + )(ξκ ) (3.17)
4α 3 3 α α 3
6
Similarly:
c β2 κ κ c β2
[U,ϕU]( ) = κ κ ( + )+βκ ( − + )
2 3 3
4α α α α 2α α
cβκ 3c
3
+ (κ +α+ ) (3.18)
4α2 1 4α
c 2κ β3κ κ β3 5cκ β cβκ κ
[U,ϕU]( ) = 3 1 + 3 + 3 (β2+κ)− 1 3
4α α2 α 4α3 4α2
cβκ 5c2βκ cβ βκκ κ
3 3 3 3
− − − (Uκ )− + ((ϕU)κ)
4α 16α3 4α2 1 α α
β2 κ
+( + )((ϕU)κ ) (3.19)
3
α α
c c β2 κ
[ϕU,ξ]( )= −κ (κ + )( + )−β2κ (3.20)
3 1 3
4α 4α α α
c 2κ κ β2 7cβ2κ c2κ cκκ
[ϕU,ξ]( ) =− 1 3 −κ β2− 3 + 3 + 3
4α α 3 4α2 16α2 2α2
cβ
−β(ϕU)κ +κκ + (ξκ ). (3.21)
3 3 4α2 1
Duetothefirstrelationof(3.5),weconsider Ω theopensubsetofΩsuchthat:
1
Ω = {P ǫ Ω : κ 6= 0, in a neighborhood of P}.
1 1
SoinΩ ,wehave: κ = 0.
1
InΩ relation (3.14),sinceκ = 0,yields:
1
κ =−4α (3.22)
1
andfromrelation (3.10),takingintoaccount(3.22),weget:
cβ c2
κ = −4β − + (3.23)
2 4α2 16α2β
From(3.20)and(3.21),using(3.12),(3.22)and(3.23)weobtain:
3cβ2κ c2κ
3 3
β(ϕU)κ = − + (3.24)
3 2α2 16α2
From(3.18), (3.19),using(3.15),(3.22), (3.23)and(3.24),weobtain:
κ (4α2 −c) = 0. (3.25)
3
7
′
Becauseof(3.25),letΩ betheopensubsetofΩ suchthat:
1 1
′
Ω = {P ǫ Ω : κ 6= 0, in a neighborhood of P}.
1 1 3
SoinΩ′ weobtain: c = 4α2. Differentiation ofthelatterwithrespecttoξ,impliesξα= 0
1
′
whichbecause of(3.12) leads toκ = 0, whichisimpossible. SoΩ isempty and κ = 0
3 1 3
inΩ .
1
Lemma3.3 Let M be a real hypersurface in M (c), equipped with ξ-parallel structure
2
Jacobioperator. ThenΩ isempty.
1
Proof: WeresumethatinΩ wehave:
1
κ = κ = 0 (3.26)
3
andrelations (3.22),(3.23)and(3.24)hold.
Relations(3.8),(3.9),(3.12)and(3.15),becauseof(3.5)and(3.26),yield:
Uα = Uβ = ξα= ξβ = 0 (3.27)
InΩ ,combining(3.16)and(3.17)andtakingintoaccount (3.22)and(3.26),weobtain:
1
c β2 c
( −α)( − +4α) = 0 (3.28)
4α α 4α
Owingto(3.28),letΩ betheopensubsetofΩ ,suchthat:
11 1
Ω = {P ǫ Ω :c 6= 4α2, inaneighborhood ofP}.
11 1
From(3.28) in Ω , wehave: 4α = −β2 + c . Differentiation ofthe latter along ϕU, be-
11 α 4α
causeof(3.11),(3.13),(3.22),(3.26)andthelastrelationyieldsc = 0,whichisimpossible.
Hence,Ω isempty.
11
So in Ω the relation c = 4α2 holds. Due to the last relation and (3.22), the relation
1
(3.11)becomes:
(ϕU)β = −(α2+2β2). (3.29)
From (3.27) we have [U,ξ]β = U(ξβ)−ξ(Uβ) ⇒ [U,ξ]β = 0. On the other hand, from
(3.2) and (3.3) we obtain [U,ξ]β = (∇ ξ −∇ U)β ⇒ [U,ξ]β = 1(3α2 + β2)(ϕU)β.
U ξ α
Thelasttworelations imply(ϕU)β = 0. Therefore, from (3.29)weobtain α2 +2β2 = 0,
whichisacontradiction. Hence,Ω isempty. (cid:3)
1
SinceΩ isempty,inΩwehaveκ = 0. Sofromrelations(3.20)and(3.21)weobtain:
1 1
κ
β(ϕU)κ = 3 [c2−24cβ2 +12cκ+16α2κ].
3 16α2
8
Furthermore, the combination of relations (3.18) and (3.19), using (3.10) and (3.14), im-
plies:
cβκ
(β2+κ)(ϕU)κ = 3[16α2 −24(β2 +κ)+9c].
3 16α2
Fromthelasttworelations weobtain:
κ [c2κ+12cκ2 +12cβ2κ+16α2β2κ−16cα2β2+16α2κ2−8c2β2] = 0
3
Duetotheaboverelation, weconsider Ω theopensubsetofΩ,suchthat:
2
Ω = {P ǫ Ω : κ 6= 0, inaneighborhood ofP},
2 3
soinΩ thefollowingrelation holds:
2
c2κ+12cκ2 +12cβ2κ+16α2β2κ−16cα2β2+16α2κ2−8c2β2 = 0. (3.30)
Differentiating (3.30)withrespecttoξ andusing(3.5),(3.9),(3.12)and(3.15)weobtain:
8α2β2κ−8cα2β2+8α2κ2+3cβ2κ−2c2β2+3cκ2−4cα2κ
−2c2κ = 0 (3.31)
From(3.30)and(3.31)weobtain:
5cκ+6κ2 +6β2κ−4cβ2 +8α2κ = 0. (3.32)
Differentiating (3.32) with respect to ξ and using (3.5), (3.9), (3.12) and (3.15) we have:
4κα2 = (2c−3κ)(β2 +κ). The last relation with (3.32) imply: κ = 0. Substituting the
latter in (3.30) gives c = −2α2. Differentiation ofthe last relation with respect toϕU and
takingintoaccount(3.13),c = −2α2 andκ = 0resultsinα= 0,whichisimpossible.
1
SoΩ isemptyandinΩweget: κ = 0.
2 3
Lemma3.4 Let M be a real hypersurface in M (c), equipped with ξ-parallel structure
2
Jacobioperator. ThenΩisempty.
Proof: WeresumethatinΩthefollowingrelation holds:
κ = κ = 0. (3.33)
1 3
Relations(3.8),(3.9),(3.12),(3.15),because of(3.5)and(3.33)yield:
Uα = Uβ = ξα= ξβ = 0 (3.34)
InΩthecombination of(3.16),(3.17)andtakingintoaccount(3.33),implies:
c
(4α2 +3c)(β2 +κ− )= 0 (3.35)
4
9
Dueto(3.35),weconsider Ω theopensubsetofΩsuchthat:
3
c
Ω = {P ǫ Ω :β2+κ6= , inaneighborhood ofP}.
3
4
SoinΩ thefollowingrelationholds:
3
4α2
c = − . (3.36)
3
Differentiation of(3.36)withrespecttoϕU implies:
(ϕU)α = 0. (3.37)
Because of (3.34) we have [U,ξ]β = U(ξβ)−ξ(Uβ) ⇒ [U,ξ]β = 0. On the other hand
due to (3.2), (3.3) and (3.33) weget [U,ξ]β = (∇ ξ −∇ U)β ⇒ [U,ξ]β = (β2 − c +
U ξ α 4α
κ)(ϕU)β. Combinationofthelastrelations imply:
α
(ϕU)β = 0 (3.38)
From(3.11),owingto(3.33),(3.36)and(3.38)yields2β2 = κ+ α2. Differentiation ofthe
3
lastrelationwithrespecttoϕU andtakingintoaccount(3.37)and(3.38)imply(ϕU)κ = 0.
Sofrom(3.14),becauseofthelatterand(3.33),weobtainκ(β+κ )= cβ. Thecombination
2
ofthelatterwith(3.10)andtakingintoaccount (3.33),(3.36)and2β2 = κ+ a2 imply:
3
α2 =18β2 κ = 5β κ= −4β2 (3.39)
2
Therelations ofLemma3.2inΩ ,becauseof(3.36)and(3.39)become:
3
α α
AU = U +βξ, AϕU = ϕU (3.40)
6 3
α α
∇ ξ = βϕU, ∇ ξ = ϕU, ∇ ξ = − U, (3.41)
ξ U ϕU
6 3
α
∇ U = 0, ∇ U = 5βϕU, ∇ U = ξ, (3.42)
ξ U ϕU
3
α
∇ ϕU = −βξ, ∇ ϕU = −5βU − ξ, ∇ ϕU = 0. (3.43)
ξ U ϕU
6
Therelation(2.4),becauseof(3.36),(3.39)and(3.40)implies: R(U,ϕU)U = 23β2ϕU.
OntheotherhandR(X,Y)Z = ∇ ∇ Z−∇ ∇ Z−∇ Z,becauseof(3.34),(3.36),
X Y Y X [X,Y]
(3.39) and(3.41)-(3.43) yields: R(U,ϕU)U = 26β2ϕU. The combination the last two re-
lationsimpliesβ = 0,whichisimpossible inΩ .
3
SoΩ isemptyandinΩthefollowingrelation holds
3
c
β2+κ= . (3.44)
4
InΩ(3.10)becomes:
κ+βκ = 0. (3.45)
2
10