DOI:10.1142/S0129167X14500591
Real hypersurfaces in the complex quadric with Reeb parallel shape operator
Young Jin Suh Department of Mathematics College of Natural Sciences Kyungpook National University
Daegu 702-701, South Korea yjsuh@knu.ac.kr
Received 10 February 2014 Accepted 19 May 2014 Published 12 June 2014
First, we introduce the notion of shape operator of Codazzi type for real hypersurfaces in the complex quadricQm = SOm+2/SOmSO2. Next, we give a complete proof of non-existence of real hypersurfaces inQm=SOm+2/SOmSO2with shape operator of Codazzi type. Motivated by this result we have given a complete classification of real hypersurfaces inQm=SOm+2/SOmSO2with Reeb parallel shape operator.
Keywords: Reeb parallel hypersurface; K¨ahler structure; complex conjugation; complex quadric.
Mathematics Subject Classification 2010: 53C40, 53C55
1. Introduction
Now let us consider the other Hermitian symmetric space with rank 2. We can give some examples of Riemannian symmetric space SUm+2/S(U2Um) and SU2,m/S(U2Um), which are both Hermitian symmetric spaces and quaternionic K¨ahler symmetric spaces. We denote byJthe K¨ahler structure and byJthe quater- nionic K¨ahler structure onSU2,m/S(U2Um). The rank ofSU2,m/S(U2Um) is 2 and there are exactly two types of singular tangent vectorsX ofSU2,m/S(U2Um) which are characterized by the geometric propertiesJ X ∈JXandJ X⊥JX, respectively.
As another kind of Hermitian symmetric space with rank 2 of compact type, we can give a complex quadricQm=SOm+2/SOmSO2, which is a complex hyper- surface in complex projective spaceCPm. So this complex quadric admits both a complex conjugation structureAand a Kaehler structure J, which anti-commutes with each other, that is,AJ =−J A. Then form≥2 the triple (Qm, J, g) is a Her- mitian symmetric space of compact type with rank 2 and its maximal sectional curvature is equal to 4.
When the shape operatorSofM inQmsatisfying (∇XS)Y = (∇YS)X for any X, Y onM inQm, we say that the shape operator is of Codazzi type. In this paper, we want to give a non-existence property of real hypersurface of Codazzi type in the complex quadricQmwith parallel shape operator as follows:
Theorem 1.1. There do not exist any real hypersurfaces in complex quadricQm, m≥3,with shape operator of Codazzi type.
Since the parallel shape operator S naturally satisfy the Codazzi type shape operator, we can also assert the following.
Theorem 1.2. There do not exist any real hypersurfaces in complex quadricQm, m≥3,with parallel shape operator.
When the shape operatorS ofM in Qm is parallel, that is,∇ξS= 0 along the direction the structure vector fieldξ, we say that the shape operator isReeb parallel.
Moreover, we say that the Reeb principal curvature is constant if the function α defined by α = g(Sξ, ξ) is constant. Motivated by these results, we have given a complete classification of real hypersurfaces in complex quadric Qm with Reeb parallel shape operator as follows:
Theorem 1.3. Let M be a real hypersurface in complex quadric Qm, m≥3 with Reeb parallel shape operator and non-vanishing constant Reeb curvature. Thenm= 2k, andM is locally congruent to a tube over a totally geodesic complex projective spaceCPk in Q2k.
2. The Complex Quadric
For more details in this section we refer to [4,7]. The complex quadricQm is the complex hypersurface inCPm+1which is defined by the equationz12+· · ·+zm+22 = 0, wherez1, . . . , zm+2are homogeneous coordinates onCPm+1. We equipQmwith the Riemannian metric which is induced from the Fubini Study metric onCPm+1 with constant holomorphic sectional curvature 4. The K¨ahler structure onCPm+1 induces canonically a K¨ahler structure (J, g) on the complex quadric. For each z ∈Qm we identifyTzCPm+1 with the orthogonal complement Cm+2Cz of Cz in Cm+2 (see Kobayashi and Nomizu [5]). The tangent spaceTzQm can then be identified canonically with the orthogonal complementCm+2(Cz⊕Cρ) ofCz⊕Cρ in Cm+2, whereρ∈νzQmis a normal vector ofQminCPm+1at the pointz.
The complex projective spaceCPm+1is a Hermitian symmetric space of the spe- cial unitary group SUm+2, namely CPm+1 =SUm+2/S(Um+1U1). We denote by o= [0, . . . ,0,1]∈CPm+1 the fixed point of the action of the stabilizerS(Um+1U1).
The special orthogonal group SOm+2 ⊂ SUm+2 acts on CPm+1 with cohomo- geneity one. The orbit containing o is a totally geodesic real projective space RPm+1⊂CPm+1. The second singular orbit of this action is the complex quadric Qm = SOm+2/SOmSO2. This homogeneous space model leads to the geometric
interpretation of the complex quadricQm as the Grassmann manifoldG+2(Rm+2) of oriented 2-planes inRm+2. It also gives a model ofQmas a Hermitian symmetric space of rank 2. The complex quadricQ1is isometric to a sphereS2 with constant curvature, and Q2 is isometric to the Riemannian product of two 2-spheres with constant curvature. For this reason, we will assumem≥3 from now on.
For a unit normal vector ρ of Qm at a point z ∈ Qm we denote byA = Aρ the shape operator ofQm in CPm+1 with respect toρ. The shape operator is an involution on the tangent spaceTzQm and
TzQm=V(Aρ)⊕J V(Aρ),
whereV(Aρ) is the +1-eigenspace andJ V(Aρ) is the (−1)-eigenspace ofAρ. Geo- metrically this means that the shape operatorAρ defines a real structure on the complex vector spaceTzQm, or equivalently, is a complex conjugation on TzQm. Since the real codimension ofQm inCPm+1 is 2, this induces anS1-subbundle A of the endomorphism bundle End(T Qm) consisting of complex conjugations.
There is a geometric interpretation of these conjugations. The complex quadric Qmcan be viewed as the complexification of them-dimensional sphereSm. Through each pointz∈Qmthere exists a one-parameter family of real forms of Qmwhich are isometric to the sphereSm. These real forms are congruent to each other under action of the center SO2 of the isotropy subgroup ofSOm+2 at z. The isometric reflection ofQmin such a real formSmis an isometry, and the differential atz of such a reflection is a conjugation onTzQm. In this way, the familyAof conjugations onTzQm corresponds to the family of real formsSmofQm containingz, and the subspacesV(A)⊂TzQmcorrespond to the tangent spacesTzSm of the real forms SmofQm.
The Gauss equation for Qm⊂CPm+1 implies that the Riemannian curvature tensor R of Qm can be described in terms of the complex structure J and the complex conjugationsA∈A:
R(X, Y)Z =g(Y, Z)X−g(X, Z)Y +g(JY, Z)JX −g(JX, Z)JY −2g(JX, Y)JZ +g(AY, Z)AX−g(AX, Z)AY +g(JAY, Z)JAX −g(JAX, Z)JAY. Note that J and each complex conjugationA anti-commute, that is, AJ =−J A for eachA∈A.
Recall that a non-zero tangent vector W ∈ TzQm is called singular if it is tangent to more than one maximal flat in Qm. There are two types of singular tangent vectors for the complex quadricQm:
(1) If there exists a conjugationA∈Asuch that W ∈V(A), thenW is singular.
Such a singular tangent vector is calledA-principal.
(2) If there exist a conjugationA∈Aand orthonormal vectorsX, Y ∈V(A) such that W/W = (X+J Y)/√
2, then W is singular. Such a singular tangent vector is called A-isotropic.
For every unit tangent vector W ∈ TzQm there exist a conjugation A ∈ A and orthonormal vectorsX, Y ∈V(A) such that
W = cos(t)X+ sin(t)J Y
for some t ∈ [0, π/4]. The singular tangent vectors correspond to the values t = 0 and t = π/4. If 0 < t < π/4 then the unique maximal flat containing W is RX ⊕RJ Y. Later, we will need the eigenvalues and eigenspaces of the Jacobi operatorRW =R(·, W)W for a singular unit tangent vectorW.
(1) If W is an A-principal singular unit tangent vector with respect to A ∈ A, then the eigenvalues ofRW are 0 and 2 and the corresponding eigenspaces are RW⊕J(V(A)RW) and (V(A)RW)⊕RJ W, respectively.
(2) IfW is anA-isotropic singular unit tangent vector with respect toA∈Aand X, Y ∈V(A), then the eigenvalues ofRW are 0, 1 and 4 and the corresponding eigenspaces areRW⊕C(J X+Y),TzQm(CX⊕CY) andRJ W, respectively.
3. The Totally Geodesic CPk⊂ Q2k
We now assume thatmis even, saym= 2k. The map
CPk →Q2k ⊂CP2k+1, [z1, . . . , zk+1]→[z1, . . . , zk+1, iz1, . . . , izk+1] provides an embedding ofCPk intoQ2k as a totally geodesic complex submanifold.
We define a complex structurej onC2k+2 by
j(z1, . . . , zk+1, zk+2, . . . , z2k+2) = (−zk+2, . . . ,−z2k+2, z1, . . . , zk+1).
Note thatij=ji. We can then identifyC2k+2 withCk+1⊕jCk+1 and get TzCPk ={X+ijX|X ∈Ck+1Cz}.
Now consider the standard embedding of Uk+1 into SO2k+2 which is determined by the Lie algebra embedding
uk+1→so2k+2, C+iD→
C −D
D C
,
whereC, D∈Mk+1,k+1(R). The action ofUk+1onQ2kis of cohomogeneity one and CPkis the orbit of this action containing the pointz= [1,0, . . . ,0, i,0, . . . ,0]∈Q2k, where theisits in the (k+ 2)nd component. We now fix a unit normal vectorρof Q2k at z and denote the corresponding complex conjugation Aρ ∈A byA. Then we can write alternatively
TzCPk={X+ijX|X ∈V(A)}.
Note that the complex structureionC2m+2 corresponds to the complex structure onTzQ2m via the obvious identifications.
We are now going to calculate the principal curvatures and principal curvature spaces of the tube with radiusraroundCPk inQ2k. For this we use the standard
Jacobi field method as described in [1, Sec. 8.2]. Let N = (X +J Y)/√ 2 be a unit normal vector of CPk in Q2k, where X, Y ∈ V(A) are orthonormal. The normal Jacobi operatorRN leaves the tangent spaceTzCPk and the normal space νzCPkinvariant. When restricted toTzCPk, the eigenvalues ofRN are 0 and 1 with corresponding eigenspacesC(J X+Y) andTzCPkC(J X+Y). The corresponding principal curvatures on the tube of radiusrare 0 and tan(r), and the corresponding principal curvature spaces are the parallel translates ofC(J X+Y) andTzCPk C(J X+Y) along the geodesicγ inQ2k withγ(0) =z and ˙γ(0) =N fromγ(0) to γ(r). We denote the latter parallel translate by W1. When restricted to νzCPk RN, the eigenvalues of RN are 1 and 4 with corresponding eigenspacesνzCPk CN and RJ N. We denote the first parallel translate by W2. The corresponding principal curvatures on the tube of radiusr are−cot(r) and −2 cot(2r), and the corresponding principal curvature spaces are the parallel translates ofνzCPkCN and RJ N along γ from γ(0) to γ(r). This shows in particular that the tube is a Hopf hypersurface.
For 0 < r < π/2 this process leads to real hypersurfaces in Q2k, whereas for r = π/2 we get another totally geodesic CPk ⊂ Q2k. The two totally geodesic complex projective spaces are precisely the two singular orbits of theUk+1-action onQ2k, and the tubes of radius 0< r < π/2 are the principal orbits of this action.
Using the homogeneity of the tubes we can now conclude that the tubeM of radius 0< r < π/2 has four distinct constant principal curvatures and the property that the shape operator leaves invariant the maximal complex subbundle C of T M. Moreover, all principal curvature spaces in C areJ-invariant. Summing up all the properties mentioned above, we have the following.
Proposition 3.1 ([3]). LetM be the tube of radius0< r < π/2around the totally geodesicCPk in Q2k. Then the following statements hold:
(1) M is a Hopf hypersurface.
(2) Every unit normal vector N of M is A-isotropic and therefore can be written in the form N = (X+J Y)/√
2 with some orthonormal vectorsX, Y ∈V(A) andA∈A.
(3) The principal curvatures and corresponding principal curvature spaces ofM are principal curvature eigenspace multiplicity
0 C(J X+Y) 2
tan(r) W1 2k−2
−cot(r) W2 2k−2
−2 cot(2r) RJ N 1
(4) Each of the two focal sets of M is a totally geodesic CPk ⊂Q2k. (5) The Reeb flow of M is an isometric flow.
(6) The shape operator S and the structure tensor fieldφsatisfy Sφ=φS.
(7) M is a homogeneous hypersurface of Q2k. More precisely, it is an orbit of the Uk+1-action onQ2k isomorphic to Uk+1/Uk−1U1,anS2k−1-bundle overCPk. For the complex projective space CPm a full classification was obtained by Okumura in [6]. He proved that the Reeb flow on a real hypersurface in CPm = SUm+1/S(UmU1) is isometric if and only if M is an open part of a tube around a totally geodesic CPk ⊂ CPm for some k ∈ {0, . . . , m−1}. For the complex 2- plane GrassmannianG2(Cm+2) =SUm+2/S(UmU2) the classification was obtained by Berndt and Suh in [2]. The Reeb flow on a real hypersurface in G2(Cm+2) is isometric if and only if M is an open part of a tube around a totally geodesic G2(Cm+1)⊂G2(Cm+2). In this paper, we investigate this problem for the complex quadric Qm = SOm+2/SOmSO2. In view of the previous two results a natural expectation might be that the classification involves at least the totally geodesic Qm−1 ⊂Qm. But, surprisingly, in [3] Berndt and Suh have proved the following result:
Theorem 3.1. Let M be a real hypersurface of the complex quadric Qm, m ≥3.
The Reeb flow onM is isometric if and only if m is even, say m= 2k, and M is an open part of a tube around a totally geodesicCPk⊂Q2k.
4. Some General Equations
LetM be a real hypersurface ofQmand denote by (φ, ξ, η, g) the induced almost contact metric structure. Note that ξ = −J N, whereN is a (local) unit normal vector field of M. The tangent bundle T M of M splits orthogonally into T M = C ⊕Rξ, whereC= ker(η) is the maximal complex subbundle ofT M. The structure tensor field φrestricted toC coincides with the complex structureJ restricted to C, andφξ = 0.
At each pointz∈M we define
Qz={X∈TzM|AX∈TzM for allA∈Az}, the maximalA-invariant subspace ofTzM.
Lemma 4.1. For eachz∈M we have (i) IfNz isA-principal, thenQz=Cz.
(ii) IF Nz is not A-principal, there exist a conjugation A ∈ A and orthonormal vectorsX, Y ∈V(A)such thatNz= cos(t)X+ sin(t)J Y for some t∈(0, π/4].
Then, we haveQz=CzC(J X+Y).
Proof. First assume thatNzisA-principal. Then there exists a conjugationA∈A such thatNz∈V(A), that is,ANz=Nz. Then we haveAξz=−AJ Nz=J ANz= J Nz=−ξz. It follows thatA restricted toCNz is the orthogonal reflection in the lineRNz. Since all conjugations inAdiffer just by a rotation on such planes we see that CNz is invariant under A. This implies that Cz = TzQmCNz is invariant under A, and henceQz=Cz.
Now assume thatNz is notA-principal. Then there exist a conjugationA∈A and orthonormal vectors X, Y ∈ V(A) such that Nz = cos(t)X + sin(t)J Y for somet∈(0, π/4]. The conjugationArestricted toCX⊕CY is just the orthogonal reflection inRX⊕RY. Again, since all conjugations in Adiffer just by a rotation on such invariant spaces we see thatCX⊕CY is invariant underA. This implies thatQz=TzQm(CX⊕CY) =CzC(J X+Y) is invariant underA, and hence Qz=CzC(J X+Y).
We see from the previous lemma that the rank of the distributionQis in general not constant onM. However, ifNzis notA-principal, thenN is notA-principal in an open neighborhood ofz ∈M, andQ defines a regular distribution in an open neighborhood ofz.
We assume now thatM is a Hopf hypersurface. Then, we have Sξ=αξ
with the smooth functionα=g(Sξ, ξ) onM. When we consider a transformJ X of the Kaehler structureJ onQm for any vector fieldX onM inQm, we may put
J X=φX+η(X)N
for a unit normalN toM. Then we consider the Codazzi equation
g((∇XS)Y −(∇YS)X, Z) =η(X)g(φY, Z)−η(Y)g(φX, Z)−2η(Z)g(φX, Y) +g(X, AN)g(AY, Z)−g(Y, AN)g(AX, Z) +g(X, Aξ)g(J AY, Z)−g(Y, Aξ)g(J AX, Z).
PuttingZ=ξ, we get
g((∇XS)Y −(∇YS)X, ξ) =−2g(φX, Y)
+g(X, AN)g(Y, Aξ)−g(Y, AN)g(X, Aξ)
−g(X, Aξ)g(J Y, Aξ) +g(Y, Aξ)g(J X, Aξ).
On the other hand, we have g((∇XS)Y −(∇YS)X, ξ)
=g((∇XS)ξ, Y)−g((∇YS)ξ, X)
= (Xα)η(Y)−(Y α)η(X) +αg((Sφ+φS)X, Y)−2g(SφSX, Y).
Comparing the previous two equations and puttingX=ξyields Y α= (ξα)η(Y)−2g(ξ, AN)g(Y, Aξ) + 2g(Y, AN)g(ξ, Aξ).
Reinserting this into the previous equation yields g((∇XS)Y −(∇YS)X, ξ)
=−2g(ξ, AN)g(X, Aξ)η(Y) + 2g(X, AN)g(ξ, Aξ)η(Y)
+ 2g(ξ, AN)g(Y, Aξ)η(X)−2g(Y, AN)g(ξ, Aξ)η(X) +αg((φS+Sφ)X, Y)−2g(SφSX, Y).
Altogether this implies
0 = 2g(SφSX, Y)−αg((φS+Sφ)X, Y)−2g(φX, Y) +g(X, AN)g(Y, Aξ)−g(Y, AN)g(X, Aξ)
−g(X, Aξ)g(J Y, Aξ) +g(Y, Aξ)g(J X, Aξ)
+ 2g(ξ, AN)g(X, Aξ)η(Y)−2g(X, AN)g(ξ, Aξ)η(Y)
−2g(ξ, AN)g(Y, Aξ)η(X) + 2g(Y, AN)g(ξ, Aξ)η(X).
At each pointz∈M we can chooseA∈Az such that N = cos(t)Z1+ sin(t)J Z2
for some orthonormal vectorsZ1, Z2∈V(A) and 0≤t≤π4 (see [7, Proposition 3]).
Note thattis a function onM. First of all, sinceξ=−J N, we have N = cos(t)Z1+ sin(t)J Z2,
AN = cos(t)Z1−sin(t)J Z2, ξ= sin(t)Z2−cos(t)J Z1, Aξ= sin(t)Z2+ cos(t)J Z1. This impliesg(ξ, AN) = 0 and hence
0 = 2g(SφSX, Y)−αg((φS+Sφ)X, Y)−2g(φX, Y) +g(X, AN)g(Y, Aξ)−g(Y, AN)g(X, Aξ)
−g(X, Aξ)g(J Y, Aξ) +g(Y, Aξ)g(J X, Aξ)
−2g(X, AN)g(ξ, Aξ)η(Y) + 2g(Y, AN)g(ξ, Aξ)η(X).
We have J Aξ = −AJ ξ = −AN, and inserting this into the previous equation implies the following.
Lemma 4.2. LetM be a Hopf hypersurface inQmwith(local)unit normal vector field N. For each point in z ∈ M we choose A ∈ Az such that Nz = cos(t)Z1+ sin(t)J Z2 holds for some orthonormal vectorsZ1, Z2∈V(A)and0≤t≤ π4. Then
0 = 2g(SφSX, Y)−αg((φS+Sφ)X, Y)−2g(φX, Y) + 2g(X, AN)g(Y, Aξ)−2g(Y, AN)g(X, Aξ) + 2g(ξ, Aξ){g(Y, AN)η(X)−g(X, AN)η(Y)} holds for all vector fields X, Y onM.
We will now apply this result to get more information on Hopf hypersurfaces for which the normal vector field isA-principal everywhere.
Lemma 4.3. Let M be a Hopf hypersurface in Qm such that the normal vector field N is A-principal everywhere. Then α is constant. Moreover, if X ∈ C is a principal curvature vector ofM with principal curvatureλ,then2λ=αandφX is a principal curvature vector ofM with principal curvature αλ+22λ−α.
Proof. Let A ∈ A such that AN = N. Then we also have Aξ = −ξ. In this situation, we get
Y α= (ξα)η(Y).
Since gradMα= (ξα)ξ, we can compute the Hessian hessMαby
(hessMα)(X, Y) =g(∇XgradMα, Y) =X(ξα)η(Y) + (ξα)g(φSX, Y).
As hessMαis a symmetric bilinear form, the previous equation implies (ξα)g((Sφ+φS)X, Y) = 0
for all vector fieldsX, Y onM which are tangential toC.
Now let us assume thatSφ+φS= 0. SinceAN =N andAξ=−ξ, Lemma4.2 implies
S2φX+φX = 0.
For every principal curvature vectorX∈ C such thatSX =λX this implies SφX =−φSX =−λφX.
Moreover, by applyingS to both sides, it follows that S2φX=−λSφX=λ2φX.
Substituting these two equations into the above formula gives (λ2+ 1)φX = 0 for any vector fieldX onM, which becomes a contradiction, and thereforeSφ+φS= 0 cannot occur.
It follows thatξα= 0 and thus gradMα= 0. SinceM is connected we conclude that the principal curvature α is constant. The remaining part of the statement follows easily from the equation
(2λ−α)SφX = (αλ+ 2)φX in Lemma4.2.
Lemma 4.4. Let M be a Hopf hypersurface inQm, m≥3,such that the normal vector fieldN isA-isotropic everywhere. Thenαis constant.
Proof. Assume thatN isA-isotropic everywhere. At each point ofM we can then find A ∈ A such that N = (X0 +J Y0)/√
2 for some orthonormal vectors X0,
Y0∈V(A). ThenN, ξ, AN, Aξare pairwise orthonormal andC Q=RAN⊕RAξ.
As in Lemma4.2, we get
(ξα)g((Sφ+φS)X, Y) = 0 for all vector fieldsX, Y onM which are tangential toC.
Now let us assume thatSφ+φS= 0. For every principal curvature vectorX∈ C such that SX = λX this implies SφX =−φSX =−λφX. We assume X = 1 and putY =φX. Using Lemma 4.2, we get
1≤λ2+ 1 =g(X, AN)2+g(X, Aξ)2=XCQ2≤1,
where XCQ denotes the orthogonal projection of X onto C Q. This implies XCQ2= 1 for all principal curvature vectorsX ∈ CwithX= 1. This is only possible ifC =C Q, or equivalently, ifQ = 0. Sincem ≥3 this is not possible.
Hence we must haveSφ+φS = 0 everywhere and thereforeξα= 0, which implies gradMα= 0. Since M is connected this implies thatαis constant.
5. Proof of Theorems 1.1 and 1.2
Now let us denote by S the shape operator of a real hypersurfaceM in complex quadric Qm. If a real hypersurfaceM in Qm has with shape operator of Codazzi type, that is, (∇XS)Y = (∇YS)X for anyX andY onM, then by the equation of Codazzi we have
0 =η(X)g(φY, Z)−η(Y)g(φX, Z)−2η(Z)g(φX, Y) +g(X, AN)g(AY, Z)−g(Y, AN)g(AX, Z)
+g(X, Aξ)g(J AY, Z)−g(Y, Aξ)g(J AX, Z). (5.1) From this, let us putX =ξ, we know that
0 =g(φY, Z) +g(ξ, AN)g(AY, Z)−g(Y, AN)g(Aξ, Z)
+g(ξ, Aξ)g(J AY, Z)−g(Y, Aξ)g(J Aξ, Z). (5.2) Here, let us putZ=ξ, we have
0 =g(ξ, AN)g(AY, ξ)−g(Y, AN)g(Aξ, ξ) +g(ξ, Aξ)g(J AY, ξ)−g(Y, Aξ)g(J Aξ, ξ)
= 2{g(ξ, AN)g(AY, ξ)−g(Aξ, ξ)g(Y, AN)}. Sinceg(Aξ, N) = 0, it follows that
g(Aξ, ξ)g(AN, Y) =g(AJ N, J N)g(AN, Y) = 0.
This gives that cos 2t= 0 or g(AN, Y) = 0 for any tangent vector field Y on M. Then it follows that
AN =N or t= π 4. From this we assert the following lemma.
Lemma 5.1. Let M be a real hypersurface in complex quadric Qm, m ≥3, with shape operator of Codazzi type. Then the unit normal vector field N is either A- principal orA-isotropic.
Then let us consider the first case as follows:
Case (1)N:A-principal, that is,AN =N Equation (5.2) gives the following
0 =g(φY, Z) +g(ξ, Aξ)g(J AY, Z)−g(Aξ, Y)g(J Aξ, Z)
=g(φY, Z)−g(J AY, Z), (5.3)
where in the second equality we have used that g(ξ, Aξ) =g(JN,AJN) =
−g(JN,JAN) =−g(JN,JN) =−1 andg(J Aξ, Z) =−g(JAJN, Z) =−g(AN, Z) = g(N, Z) = 0 for any vector fields Y and Z on M in Qm. Thus, we knowg(φY, Z)
= g(JAY, Z). Then the left side is skew-symmetric, but by the anti-commuting property ofAJ =−JA, the right side becomes
g(JAY, Z) =−g(AY,JZ) =g(Y,JAZ),
that is,JAbecomes symmetric. This gives us a contradiction. So we conclude that there do not exist any real hypersurface inQmwith parallel shape operator for the A-principal unit normal vector field.
As a second we consider the next case as follows:
Case (2)N:A-isotropic.
In this case the unit normal vector fieldN can be putN= √1
2(Z1+J Z2) forZ1, Z2∈V(A), whereV(A) denotes a (+1)-eigenspace of the complex conjugationA.
Then it follows that AN = 1
√2(Z1−JZ2), AJN =− 1
√2(JZ1+Z2) and JN = 1
√2(JZ1−Z2).
Then it gives that
g(ξ, Aξ) =g(JN,AJN) = 0, g(ξ,AN) = 0 and g(AN, N) = 0. (5.4) From this, we know that AN is a tangent vector. Then by putting X = AN into (5.1), we have
0 =−η(Y)g(φAN, Z)−2η(Z)g(φAN, Y) +g(AY, Z)
−g(Y, Aξ)g(JA2N, Z)
=−η(Y)g(φAN, Z)−2η(Z)g(Aξ, Y) +g(AY, Z) +η(Z)g(Y, Aξ)
=−η(Y)g(Aξ, Z)−η(Z)g(Aξ, Y) +g(AY, Z), (5.5) where in the third equality we have used
g(φAN, Z) =g(JAN, Z) =−g(AJN, Z) =g(Aξ, Z).
Then, Eq. (5.4) means that
g(AY, Z) = 0
for anyY, Z∈H, whereHdenotes the complex subbundle ofT M orthogonal to the Reeb vector fieldξ. From this, together with (5.4) the complex conjugation on the complex quadricQmcan be expressed in such a way that
A=
0 0 ∗ · · · ∗ 0 0 ∗ · · · ∗
∗ ∗ 0 · · · 0 ... ... 0 · · · 0
∗ ∗ 0 · · · 0
. (5.6)
But we know that the complex conjugation is involutive, that is, A2 =I. So the expression (6.6) gives us a contradiction. Accordingly, forA-isotropic normal vector field N there do not exist any hypersurfaces in complex quadric Qm with shape operator of Codazzi type.
Summing up these two cases, we conclude that there do not exist any real hypersurfaces in complex quadric Qm with shape operator of Codazzi type. This completes the proof of our Theorem1.1. Naturally, parallel shape operator satisfies the shape operator of Codazzi type. Accordingly, as a corollary of Theorem1.1, we get Theorem1.2.
6. Proof of Theorem 1.3
Before going to prove Theorem1.3, first let us show that the shape operator of the tube of radiusrover a complex projective spaceCPkinQ2kis Reeb parallel or not.
In order to do this, let us mention that the shape operatorSof the tube commutes with the structure tensor φ, that is, Sφ = φS as in Proposition 3.1. Then by a paper due to Berndt and Suh (see [3, p. 1350050-14]), the expression of covariant derivative for the shape operator ofM in complex quadricQmbecomes
(∇XS)Y ={dα(X)η(Y) +g((αSφ−S2φ)X, Y) +δη(Y)ρ(X) +δg(BX, φY) +η(BX)ρ(Y)}ξ
+{η(Y)ρ(X) +g(BX, φY)}Bξ+g(BX, Y)φBξ
−ρ(Y)BX −η(Y)φX−η(BY)φBX, where we have put
AY =BY +ρ(Y)N, ρ(Y) =g(AY, N)
for a complex conjugationA∈A. PuttingX =ξand usingαconstant andρ(ξ) = 0 for theA-isotropic unit normal vector fieldN ofM in Q2k, we have
(∇ξS)Y ={δg(Bξ, φY) +η(Bξ)ρ(Y)}ξ
+{η(Y)ρ(ξ) +g(Bξ, φY)}Bξ+g(Bξ, Y)φBξ
−ρ(Y)Bξ−η(Y)φξ−η(BY)φBξ
={g(Bξ, φY)−ρ(Y)}Bξ
={−g(φBξ, Y) +g(Y, φBξ}Bξ
= 0,
where in the third equality we have used
ρ(Y) =g(AY, N) =g(Y, AN)
=g(Y, AJ ξ)
=−g(Y, J Aξ) =−g(Y, J Bξ)
=−g(Y, φBξ).
So we conclude that a real hypersurfaceM inQ2k with commuting shape operator, that is,Sφ=φS, has a parallel shape operator along the Reeb direction∇ξS= 0.
Now let us prove our Theorem1.3in the introduction. Let us assume∇ξS= 0.
Then by puttingX=ξin the equation of Codazzi, we have
−g((∇YS)ξ, Z) =g(φY, Z) +g(ξ, AN)g(AY, Z)−g(Y, AN)g(Aξ, Z) +g(ξ, Aξ)g(J AY, Z)−g(Y, Aξ)g(J Aξ, Z).
SinceM is Hopf, we know that
(∇YS)ξ=∇Y(Sξ)−S(∇Yξ)
=∇Y(αξ)−S∇Yξ
= (Y α)ξ+αφSY −SφSY.
From this, together with the above equation, it follows that 0 =η(Z)Y α+αg(φSY, Z)−g(SφSY, Z)
+g(φY, Z) +g(ξ, AN)g(AY, Z)−g(Y, AN)g(Aξ, Z)
+g(ξ, Aξ)g(J AY, Z)−g(Y, Aξ)g(J Aξ, Z). (6.1) From this, puttingZ=ξand usingM is Hopf andg(Aξ, N) = 0 in Sec. 5, we have
0 =Y α+g(ξ, AN)g(AY, ξ)−g(Y, AN)g(Aξ, ξ) +g(ξ, Aξ)g(J AY, ξ)−g(Y, Aξ)g(J Aξ, ξ)
=Y α−2g(Y, AN)g(ξ, Aξ), (6.2)
where we have used that g(Aξ, N) = 0 in Sec. 4. So in Eq. (6.2) the function α should be constant if the unit normal vector fieldN isA-principal orA-isotropic,
because AN =N or g(ξ, ξ) = 0 for a complex conjugation A∈A. Now conversely let us prove the following.
Lemma 6.1. LetM be a real hypersurface in complex quadricQm. Then the Reeb curvature function α =g(Aξ, ξ) is constant if and only if the unit normal vector fieldN is either A-principal orA-isotropic.
When the unit normal vector fieldN ofM inQm isA-principal and the shape operator is Reeb parallel, by usingAN =N we know
g(ξ, Aξ) =g(J N, AJ N) =−g(J N, J N) =−1.
So, Eq. (6.1) becomes
αg(φSY, Z)−g(SφSY, Z) +g(φY, Z)−g(J AY, Z) = 0.
This formula can be written as follows:
0 =αg(φSZ, Y)−g(SφSZ, Y) +g(φZ, Y)−g(J AZ, Y)
=−αg(SφY, Z) +g(SφSY, Z)−g(φY, Z)−g(J AY, Z).
Then taking sum and substracting from above two equations gives the following respectively
αg((φS−Sφ)Y, Z) = 2g(J AY, Z) (6.3) and
αg((φS+Sφ)Y, Z)−2g(SφSY, Z) + 2g(φY, Z) = 0 (6.4) WhenN is A-principal, let us choose a complex conjugation A∈A such that AN =N. Then naturallyρ(X) =g(AX, N) = 0 andφBξ=φAξ=φJ AN=−φξ.
So, it follows that
2SφS−α(φS+Sφ) = 2φ. (6.5)
Then includingN is notA-principal, we want to introduce the following.
Lemma 6.2 ([3]). Let M be a Hopf hypersurface in Qm, m≥3. Then the tensor field
2SφS−α(φS+Sφ) leaves Q andC Qinvariant and we have
2SφS−α(φS+Sφ) = 2φonQ and
2SφS−α(φS+Sφ) = 2δ2φonC Q. Now, first we want to prove the following proposition.
Proposition 6.1. There do not exist any hypersurfaces in complex quadric Qm withA-principal normal vector field and Reeb parallel Ricci tensor.
Proof. Before going to give our proof, let us mention the following formulas J AY =J(BY +ρ(Y)N)
=φBY +η(BY)N+ρ(Y)J N
=φBY −ρ(Y)ξ+η(BY)N,
g(J AY, Z) =g(φBY −ρ(Y)ξ, Z) =g(φBY, Z)−ρ(Y)η(Z) and
g(AY, Z) =g(BY +ρ(Y)N, Z) =g(BY, Z).
So the Codazzi equation becomes
(∇XS)Y −(∇YS)X =η(X)φY −η(Y)φX−2g(φX, Y)ξ +g(X, AN)BY −g(Y, AN)BX +g(X, Aξ){φBY −ρ(Y)ξ}
−g(Y, Aξ){φBX−ρ(X)ξ}.
From this, puttingX =ξ, and using the Reeb parallel, we have the following for anyA-principal unit normalN
SφSY −(Y α)ξ−αφSY =−(∇YS)ξ
=φY +g(ξ, Aξ){φBY −ρ(Y)ξ} −g(Y, Aξ)φBξ, (6.6) where we have putAξ=Bξand AX=BX+ρ(X)N. From this, taking an inner product withξ, we have
Y α=ρ(Y) =g(AY, N) =g(Y, AN) =g(Y, N) = 0.
From this, together with (6.5), (6.6) andφBξ= 0 for N isA-principal, we have α
2g((Sφ−φS)Y, Z) =−g(φBY, Z).
Then by (6.3), the left side of the above equation becomes α2g((Sφ−φS)Y, Z) = g(J AY, Z) =g(φBY, Z). So it follows thatφBY = 0 for any tangent vector fieldY onM inQm. This gives that φ2BY =−BY +η(BY)ξ= 0, which implies
AY =BY =η(BY)ξ=g(Bξ, Y)ξ=g(Aξ, Y)ξ.
This means rankA= 1, which gives a contradiction, becauseAis a symmetric real structure satisfyingA2=I.
Then by virtue of Lemma6.1and Proposition6.1, we are now considering only the case thatN isA-isotropic forM inQm. Naturally, we can assert the following.
Proposition 6.2. LetM be a Hopf real hypersurface in complex quadricQm, m≥3 with non-vanishing constant Reeb curvature. If the unit normal N is A-isotropic and the shape operator is Reeb parallel, thenM is locally congruent to a tube over a totally geodesic CPk inQ2k.
Proof. By Lemma 6.1, we know that the Reeb curvatureα is constant, because N isA-isotropic. Moreover, the unit normalN can be put asN = √1
2(Z1+J Z2), whereZ1, Z2∈V(A). Accordingly, it follows that
g(ξ, Aξ) =g(J N, AJ N) = 0, g(ξ, AN) = 0, g(AN, N) = 0, because AN =√1
2(Z1−J Z2),AJ N =−√12(J Z1+Z2) andJ N =√1
2(J Z1−Z2).
From this, using∇ξS = 0 in the equation of Codazzi, we have αg(φSX, Z)−g(SφSX, Z) =−g(φX, Z) +g(X, AN)g(Aξ, Z)
−g(ξ, AN)g(AX, Z) +g(X, Aξ)g(J Aξ, Z)
−g(ξ, Aξ)g(J AX, Z). (6.7) On the distributionQwe know thatAX∈TxM,x∈M for anyA∈A. So it follows that g(X, AN) =g(AX, N) = 0 and
g(J Aξ, Z) =−g(J AJ N, Z) =−g(AN, Z) =−g(N, AZ) = 0.
On the other hand, by Lemma4.2Sec. 4 due to Berndt and Suh [3], we know that SφS= α
2(Sφ+φS) +φ. (6.8)
From this, together with (6.6), it follows that
−α
2g((Sφ−φS)X, Z) = 0.
So from the assumption we have that the shape operator S commutes with the structure tensor φ, that is,Sφ = φS on the distribution Q. Then together with (6.8), on the distributionQwe get the following
SφS−αSφ=φ.
When we consider a principal curvature vectorX∈ Q such that SX =λX, then the principal curvature λ becomes a solution ofx2−αx−1 = 0. Moreover, this equation has two distinct roots, we may putλ= cotr,µ=−tanrandα= 2 cotr.
Now let us continue our discussion on the distribution C − Q. Then by Lemma6.2in Sec. 6, we know that
2SφS−α(Sφ+φS) = 0 (6.9)
because δ = 0 for an A-isotropic normal vector field N. Now let us differentiate g(ξ, AN) = 0. Then it follows that
g( ¯∇Xξ, AN) +g(ξ,( ¯∇XA)N+A∇¯XN) = 0, which gives that
0 =g(φSX, AN)−g(ξ, ASX)
=g(φSX, AN) +g(J N, ASX)
=g(φSX, AN) +g(N, AφSX) +η(SX)g(N, Aξ)
=−2g(SφAN, X) (6.10)
for any vector fieldX on the distribution C − Q. So SφAN = 0 is equivalent to SAξ= 0. From this, together with (6.9), we have
αSφAξ= 0.
So we get SφAξ = 0 from the assumption. This means that Sφ = φS on the distribution C − Q = Span[Aξ, φAξ]. Consequently, we conclude that the shape operatorS commutes with the structure tensor φ for a Hopf hypersurface M in Qm. This means that the Reeb flow of M is isometric. Then by Theorem 3.1 we give a complete proof of our proposition.
Acknowledgment
This work was supported by grants Proj. Nos. NRF-2011-220-1-C00002 and NRF- 2012-R1A2A2A-01043023 from National Research Foundation of Korea.
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