Bull Braz Math Soc, New Series 39(3), 355-370
© 2008, Sociedade Brasileira de Matemática
Parallel and semi-parallel hypersurfaces of S
n× R
Joeri Van der Veken* and Luc Vrancken
Abstract. We give a complete classification of totally umbilical, parallel and semi- parallel hypersurfaces of the Riemannian product spaceSn×R.
Keywords: hypersurface, totally umbilical, parallel, semi-parallel.
Mathematical subject classification: 53B25.
1 Introduction
Totally umbilical, parallel and semi-parallel submanifolds are natural general- izations of totally geodesic submanifolds. They are important to study because these families of submanifolds provide nice examples and because they give information about the ambient space. In [3] one can find initial work on the geo- metry of hypersurfaces of the Riemannian product spacesSn×RandHn×R.
In the present paper we give a full classification of totally umbilical, parallel and semi-parallel hypersurfaces ofSn ×R. By comparing our classifications of totally umbilical hypersurfaces (Theorem 4) and of parallel hypersurfaces (Theorem 6), we remark that, unlike for submanifolds of real space forms, to- tal umbilicity does not imply parallelism. In fact, we find a correspondence between totally umbilical hypersurfaces of Sn ×R and solutions of the one- dimensional Sine-Gordon equation from physics. Moreover, our classifications of totally umbilical hypersurfaces (Theorem 4) and of semi-parallel hypersur- faces (Theorem 5) include examples of so-called rotation hypersurfaces, which were introduced in [6].
Received 20 July 2007.
*The author is a postdoctoral researcher supported by the Research Foundation – Flanders (F.W.O.).
2 Preliminaries
LetF: Mn→ Men+1be an isometric immersion of Riemannian manifolds with Levi Civita connections ∇ and e∇ respectively. Denote by N a unit normal vector field on the hypersurface and letX,Y, ZandWbe arbitrary vector fields tangent toMn. We define the shape operatorSbySX = −e∇XN, and the second fundamental formh byh(X,Y)= hSX,Yi = hX,SYi. The formula of Gauss states that
e∇XY = ∇XY +h(X,Y)N. (1) Moreover, the equations of Gauss and Codazzi are given respectively by (cfr. [2])
hRe(X,Y)Z,Wi = hR(X,Y)Z,Wi +h(X,Z)h(Y,W)
−h(Y,Z)h(X,W), (2) heR(X,Y)Z,Ni =(∇h)(X,Y,Z)−(∇h)(Y,X,Z), (3) whereRandReare the Riemann-Christoffel curvature tensors of MnandMen+1 respectively. We use the following sign convention: R(X,Y)Z = ∇X∇YZ −
∇Y∇XZ − ∇[X,Y]Z. The covariant derivative ofhis defined by
(∇h)(X,Y,Z)=X[h(Y,Z)] −h(∇XY,Z)−h(Y,∇XZ). (4) We say that Mn is totally geodesic in Men+1 if h = 0, that Mn is totally umbilical in Men+1 if h is a scalar multiple of the metric at every point, that Mn is parallel in Men+1 if ∇h = 0 and that Mn is semi-parallel in Men+1 if R∙h=0, where
(R∙h)(X,Y,Z,W)= −h(R(X,Y)Z,W)−h(Z,R(X,Y)W). (5) Parallel hypersurfaces of real space forms were classified by H.B. Lawson in [7], whereas the classification of semi-parallel hypersurfaces of real space forms was obtained by J. Deprez for Euclidean space in [4] and by F. Dillen for spaces of non-zero constant sectional curvature in [5].
Denote by En+2 the Euclidean space of dimension n + 2. We define the Riemannian product manifoldSn×Ras the following subset ofEn+2, equipped with the induced metric:
Sn×R=
(x1, . . . ,xn+2)∈En+2|x12+x22+. . .+xn+12 =1 .
ThenSn×R is the Riemannian product of the unit sphereSn(1) and the real line. Remark thatξ =(x1, . . . ,xn+1,0)is a unit normal vector field onSn×R
inEn+2. IfX andY are vector fields onSn×R, we denote by XSn andYSn the projections ofX andY onto the tangent space toSn(1). From (1), we find that the Levi Civita connectione∇ofSn×Ris given bye∇XY =DXY+ hXSn,YSniξ, where D is the covariant derivative in En+2. This expression yields that the curvature tensor eRofSn×Ris determined by
hRe(X,Y)Z,Wi = hYSn,ZSnihXSn,WSni − hXSn,ZSnihYSn,WSni. Now let F: Mn → Sn×Rbe a hypersurface with unit normal N. Let T denote the projection of the coordinate vector field∂xn+2 onto the tangent space to Mn and denote by θ a function on Mn such that cosθ = hN, ∂xn+2i. This means that∂xn+2 = T +cosθN.The equations of Gauss and Codazzi, (2) and (3), reduce to
hR(X,Y)Z,Wi = hSX,WihSY,Zi − hSX,ZihSY,Wi + hX,WihY,Zi − hX,ZihY,Wi
+ hY,TihW,TihX,Zi + hX,TihZ,TihY,Wi
− hX,TihW,TihY,Zi − hY,TihZ,TihX,Wi, (6)
∇XSY − ∇YSX−S[X,Y] =cosθ (hY,TiX− hX,TiY), (7) whereX,Y, Z andW are vector fields tangent to Mn. Moreover, by using the fact that∂xn+2 is parallel inSn×R, one obtains
∇XT =cosθ SX, X[cosθ] = −hSX,Ti. (8) These equations appear in the following existence and uniqueness theorem for immersions of hypersurfaces intoSn×R:
Theorem 1 ([3]). Let Mn be a simply connected Riemannian manifold with Levi Civita connection∇and curvature tensor R. Let S be a field of symmetric operators Sp: TpMn →TpMn, and let T andθ be a vector field and a smooth function on Mnsuch thatkTk2=sin2θ. Assume that equations(6),(7)and(8) are satisfied. Then there exists an isometric immersion F : Mn →Sn×Rwith unit normal N, such that the shape operator with respect to this normal is given by S and such that∂xn+2 =T +cosθ N. Moreover, the immersion is unique up to global isometries of Sn×Rpreserving the orientations of bothSnandR.
We will now recall the definition of a special class of hypersurfaces ofSn×R, proposed in [6], namelyrotation hypersurfaces. Consider a three-dimensional
subspace P3ofEn+2containing thexn+2-axis. Then(Sn×R)∩P3is a cylin- derS1×R. Let P2 be a two-dimensional subspace of P3, also through the xn+2-axis. Denote by J the group of isometries of En+2 which leave Sn×R globally invariant and which leaveP2pointwise fixed. Finally, letαbe a curve in S1×R which does not intersect P2. Then the rotation hypersurface Mn ofSn×Rwith profile curveα and axis P2 is defined as the J-orbit ofα. It is clear from the definition that the velocity vector of α is proportional to T, unlessαlies in a plane orthogonal to∂xn+2, in which caseT =0. In the follow- ing, we will always assume that P3 is spanned by∂x1,∂xn+1 and∂xn+2 and that P2 is spanned by∂x1 and∂xn+2. In [6] it was proved that there exists a local orthonormal frame{e1, . . . ,en} on Mn, with T = kTke1, such that the shape operatorStakes the form
S =
λ
μ . ..
μ
.
Moreover, ifαis not a vertical line inS1×R, it can be locally parametrized as α(s)= coss,0, . . . ,0,sins,a(s)
, and we have λ= − a00(s)
1+a0(s)23/2, μ= − a0(s)cots
1+a0(s)21/2. (9) Ifαis a vertical lineα(s)= cosc,0, . . . ,0,sinc,s
for some real constantc, we have
λ=0, μ= −cotc. (10)
Finally, we mention the following characterization:
Theorem 2 ([6]). Take n ≥ 3and let F : Mn → Sn×Rbe a hypersurface with shape operator
S=
λ
μ . ..
μ
,
with λ 6= μ. Suppose that ST = λT and assume that there is a functional relationλ(μ). Then Mnis an open part of a rotation hypersurface.
3 Totally umbilical hypersurfaces
In this section, we classify totally umbilical hypersurfaces ofSn×R. Let us first note that there are only few totally geodesic hypersurfaces ofSn×R.
Theorem 3. Let Mnbe a totally geodesic hypersurface of Sn×R. Then Mn is an open part of a hypersurfaceSn(1)× {t0}for t0 ∈ R, or of a hypersurface Sn−1×R.
Proof. LetMnbe totally geodesic inSn×R. It follows from the equation of Codazzi (7) that there are two cases to consider, namelyT =0 and cosθ =0.
In the first case, Mn is everywhere orthogonal to ∂xn+2. This gives the first family of hypersurfaces mentioned in the theorem.
In the second case, Mn is everywhere tangent to ∂xn+2 and we have a hy- persurface of type Mˉn−1×R, where Mˉn−1 is a hypersurface ofSn(1). It is easy to see thatMˉn−1×Ris totally geodesic inSn×Rif and only if Mˉn−1 is totally geodesic inSn(1). Hence, we obtain the second family of hypersurfaces
of the theorem. ¤
In the following proposition we remark a correspondence between totally umbilical hypersurfaces and solutions of the one-dimensional Sine-Gordon equation from physics.
Proposition 1. Let Mn be a totally umbilical hypersurface of Sn ×Rwith angle functionθ and let p be a point of Mnwheresinθ 6= 0. Then there exist local coordinates(u, v1, . . . , vn−1)on an open neighbourhood U of p in Mnsuch thatθ only depends on u and such thatφ := 2θ satisfies the one-dimensional Sine-Gordon equation
φ00+sinφ =0. (11)
Conversely, starting with an open subset U ⊆Rn with coordinates(u, v1, . . . , vn−1)and a solutionφ(u)of(11), which is nowhere zero on U, we can putθ = φ2 and we can define a functionλand a Riemannian metric on U such that there exists an isometric immersion F: U →Sn ×Rwith shape operator S = λid and angle functionθ.
In the proof of this proposition, we will use the following result.
Proposition 2 ([8]). Let M =N1×f N2be a warped product of semi-Rieman- nian manifolds with Levi Civita connection∇ and curvature tensor R. Denote
by RN1 and RN2 the lifts of the curvature tensors of N1 and N2 respectively.
If X, Y and Z are lifts of vector fields on N1and U, V and W are lifts of vector fields on N2, then
(i) R(X,Y)Z is the lift of RN1(X,Y)Z,
(ii) R(X,U)Y = Hf(X,Yf )U, where Hf is the Hessian of f , (iii) R(X,Y)U =R(V,W)X =0,
(iv) R(U,X)V = hU,fVi∇X(grad f),
(v) R(U,V)W = RN2(U,V)W −hgradff,grad2 fi(hV,WiU− hU,WiV). Proof of Proposition 1. Assume that Mn is totally umbilical inSn×Rwith shape operatorS =λid. Since sinθ 6=0 at p, there exists an open neighbour- hoodU of p in Mn on which sinθ is nowhere zero. Let X be a vector field tangent toU. Then the equation of Codazzi (7) and the second equation of (8)
yield (
X[λ] = −cosθhX,Ti,
X[cosθ] = −λhX,Ti. (12) Moreover, the first equation of (8) yields that the orthogonal complement of span{T}is integrable. Indeed, ifX,Y ⊥T, then we obtain
h[X,Y],Ti = h∇XY − ∇YX,Ti = − hY,∇XTi + hX,∇YTi =0. This means that we can choose coordinates (u, v1, . . . , vn−1) on U such that
∂u = sinTθ and∂u ⊥∂vi. The system (12) yields that∂viλ=∂viθ =0, such that λandθ are functions ofuonly, and that
(λ0 = −cosθsinθ,
θ0 = λ. (13)
Remark that the functionθ satisfies the equationθ00 = −cosθsinθ.After the substitutionφ =2θ, we obtain (11).
Conversely, let us start with an open partUofRn, with coordinates(u, v1, . . . , vn−1)and with a non-vanishing solutionφ(u)of the Sine-Gordon equation (11).
OnU, we define a Riemannian metric g=du2+
Xn−1 i,j=1
gi jdvidvj,
the functionθ (u)= φ(2u), the vector fieldT =sinθ ∂uand the field of operators S =θ0id. These data satisfy the equation of Codazzi (7) and the second equation of (8). From Theorem 1, we know that there exists an isometric immersion of (U,g)into Sn×Rwith shape operator S, structure vector field T and angle functionθ if and only if the equation of Gauss (6) and the first equation of (8) are satisfied. These equations are equivalent to
hR(∂u,X)∂u,Yi = − cos2θ +(θ0)2
hX,Yi, (14) hR(X,Y)∂u,Zi =0, (15) hR(X,Y)Z,Wi = 1+(θ0)2
(hX,Wi hY,Zi − hX,Zi hY,Wi), (16)
∇∂u∂u =0, (17)
sinθ∇X∂u=(cosθ )θ0X, (18)
whereX,Y,Z andW are vector fields onU orthogonal toT,Ris the Riemann- Christoffel curvature tensor of(U,g) and ∇ is the Levi Civita connection of (U,g). From (18) we find
∂ugi j = ∂u
∂vi, ∂vj
=
∇∂u∂vi, ∂vj
+
∂vi,∇∂u∂vj
= 2(cotθ )θ0
∂vi, ∂vj
=2(cotθ )θ0gi j.
Hence,gi j =sin2θci j(v1, . . . , vn−1)and the metricgtakes the form of a warped product metric
g=du2+sin2θ (u) Xn−1 i,j=1
ci j v1, . . . , vn−1
dvidvj =du2+sin2θgc. (19) Equations (17) and (18) are now satisfied and the question is whether we can choose the functions ci j such that the curvature tensor R of (U,g) satisfies equations (14), (15) and (16).
From Proposition 2, we obtain forX ⊥T R(∂u,X)∂u = Hsinθ(∂u, ∂u)
sinθ X = −(θ0)2+cotθ (θ00)
X = − (θ0)2+cos2θ X. Hence, equation (14) is satisfied for any choice ofci j. On the other hand, we have forX,Y,Z ⊥T
R(X,Y)Z = Rc(X,Y)Z− kgrad(sinθ )k2
sin2θ hY,ZiX− hX,ZiY
= Rc(X,Y)Z−(cot2θ )(θ0)2 hY,ZiX− hX,ZiY ,
where Rc is the curvature tensor associated to the metric gc. Hence, equation (15) is satisfied and equation (16) is equivalent to
Rc(X,Y)Z =
1+ (θ0)2 sin2θ
hY,ZiX− hX,ZiY
or, equivalently,
Rc(X,Y)Z = (θ0)2+sin2θ
gc(Y,Z)X −gc(X,Z)Y .
Remark that(θ0)2+sin2θ is constant, sinceφ = 2θ satisfies the equation (11). Hence, the last relation is satisfied if and only if the metricgchas constant
curvaturec=(θ0)2+sin2θ. ¤
In [9], the first named author classified totally umbilical surfaces in S2×R by means of an explicit parametrization. It turns out that the totally umbilical surfaces inS2×R, which are not totally geodesic, are rotation (hyper)surfaces.
The following theorem is a straightforward generalization of that result.
Theorem 4. Let Mnbe a totally umbilical hypersurface ofSn×R, with angle functionθ and let p be a point of Mn where sinθ 6= 0. Then there exist co- ordinates(u, v1, . . . , vn−1)on an open neighbourhood U of p in Mn such that θ only depends on u, the shape operator is S=θ0id, and
(θ0)2+sin2θ =c, (20)
where c is a strictly positive real constant. Moreover, Mnis locally isometric to a rotation hypersurface with profile curve
α(u)= √1c
sinθ (u),0, . . . ,0, θ0(u),√ cZ
sinθdu
. (21)
Conversely, all rotation hypersurfaces with profile curve(21), whereθ and c satisfy(20), are totally umbilical inSn×R.
Proof. Let Mn be totally umbilical in Sn×R with shape operator S=λid and angle functionθ. Since sinθ 6=0 at p, there exists an open neighbourhood U of p in Mn on which sinθ is nowhere zero. In the proof of Proposition 1, we obtained that there exist local coordinates(u, v1, . . . , vn−1)onU such that λand θ only depend on u and satisfy (13). This implies that λ2+sin2θ = (θ0)2+sin2θ = cis constant. Remark that c > 0 since sinθ is nowhere zero onU.
As the functionθ satisfies (20), the proof of Proposition 1 shows that in the local coordinates (u, v1, . . . , vn−1), the induced metric on U is g = du2 + sin2θgc(v1, . . . , vn−1), where gc is a Riemannian metric of constant sectional curvature c. It follows from Theorem 1 that there exists, up to isometries of Sn×R, a unique immersion F: U → Sn ×R such that F is isometric; the projection of∂xn+2 on F(U)isF∗(sinθ ∂u); the angle between the unit normal N and∂xn+2 isθ and the shape operator isS =θ0id. A straightforward compu- tation yields that the immersion
F(u, v1, . . . , vn−1)=√1c
ϕ1sinθ (u), . . . , ϕnsinθ (u), θ0(u),√ cZ
sinθ (u)du , where ϕ1(v1, . . . , vn−1), . . . , ϕn(v1, . . . , vn−1)
is a parametrization of the unit sphereSn−1(1)⊂En, satisfies these four conditions. ¤ Remark. Changing the coordinate u tou, withˉ ∂uˉ = sinθ ∂u, equation (20) becomes
(θ0)2+sin4θ =csin2θ.
After puttingθ =arctan(f), we obtain f0
1+ f2 2
+ f
p1+ f2
!4
=c f p1+ f2
!2
,
or, equivalently,
f0 fp
c+(c−1)f2 = ±1. Direct integration yields
f = 4ce±√cˉu+d 4c(1−c)+e2(±√cu+dˉ ).
A similar approach to the original equation(θ0)2+sin2θ = cwith respect to theu-coordinate, would lead to an elliptic integral.
4 Semi-parallel hypersurfaces
In this section, we give a classification of semi-parallel hypersurfaces ofSn×R.
First, we characterize them in terms of their shape operator.
Lemma 1. Let Mn be a semi-parallel hypersurface ofSn ×Rand define T andθ as above. Then there exists a local orthonormal frame field{e1, . . . ,en} on Mn, with respect to which the shape operator S takes one of the following forms:
(i) S =
λ
. ..
λ
,
(ii) S =
λ
μ . ..
μ
, withλμ = −cos2θ and moreover, if n ≥ 3,
T = kTke1,
(iii) S =
0
λ . ..
λ μ
. ..
μ
, withλμ= −1and e1=T =∂xn+2.
Proof. Let Mn be a hypersurface ofSn×R and denote by {e1, . . . ,en} an orthonormal tangent frame satisfyingSei =λiei. Assume thatT =Pn
i=1Tiei. By using the equation of Gauss (6), we obtain
R(ei,ej)ej = (λiλj +1−(Tj)2)ei +TiTjej −TiT, R(ei,ej)ek = TiTkej −TjTkei,
wherei, j andkare assumed to be mutually different. Ifn<3, the second for- mula does not make sense, but the first one is still correct. From these equations, we can computeR∙h:
(R∙h)(ei,ej,ei,ei) = 0,
(R∙h)(ei,ej,ei,ej) = λj −λi
λiλj +1−(Ti)2−(Tj)2 , (R∙h)(ei,ej,ek,ei) = λi −λk
TjTk, (R∙h)(ei,ej,ek,el) = 0.
Again, we assumei, j,kandlto be mutually different, such that the last equa- tions are meaningless for low dimensions.
Now assume thatMnis semi-parallel, then we obtain λi −λj
λiλj+1−(Ti)2−(Tj)2
= λi −λk
TjTk =0
for all mutually different indicesi, jandk. If all the eigenvalues ofSare equal, we are in case (i) of the lemma. From now on, suppose that Shas at least two different eigenvalues and fix indicesi and j such thatλi 6= λj. Remark that T 6= 0, because forT = 0, we have that Mn ⊆ Sn× {t0}is totally geodesic.
We consider two cases, namely T ∈/ span{ei,ej} and T ∈ span{ei,ej}. In the first case, we haven ≥ 3 and there exists an index k different fromi and j such that Tk 6= 0. From (λj −λi)TkTi = 0, we find that Ti = 0, and from(λi −λj)TkTj = 0, we find thatTj = 0. Hence T is perpendicular to span{ei,ej}and the equation(λi −λj)(λiλj +1−(Ti)2−(Tj)2)=0 yields λiλj = −1. In the second case, it follows from(λi −λj)(λiλj +1−(Ti)2− (Tj)2)=0 thatλiλj+1− kTk2=0 and henceλiλj = −cos2θ. We conclude the following: ifλi andλj are different eigenvalues ofS, then eitherλiλj = −1 andT ⊥span{ei,ej}orλiλj = −cos2θ andT ∈span{ei,ej}.
Now assume thatS has exactly two distinct eigenvalues, say Sei = λei for i ∈ {1, . . . ,k}and Sei = μei for i ∈ {k +1, . . . ,n}. If λμ = −1, we have that T is perpendicular to span{ei,ej} for every i ∈ {1, . . . ,k} and for every j ∈ {k +1, . . . ,n}. But this implies that T = 0, a contradiction. Hence we haveλμ= −cos2θ, which yields thatT ∈span{ei,ej}for everyi ∈ {1, . . . ,k}
and for every j ∈ {k+1, . . . ,n}. This is only possible ifk =1 (orn−k =1, but then we switch the role of λ andμ). Moreover, if n ≥ 3, we have that T = kTke1. This is case (ii) of the lemma.
Now assume thatShas at least three different eigenvalues, sayλ,μandν. If λμ = λν = −1, we have μ = ν, which is a contradiction. If λμ = −1 and λν= −cos2θ, we find
μν= cos2θ λ2 ≥0,
which is only possible ifν = cosθ = 0. Finally, if λμ = λν = −cos2θ, we obtain that eitherμ = ν, which is a contradiction, orλ =cosθ = 0. We conclude that one of the eigenvalues is zero and that cosθ = 0. Assume that λ1=0. Sinceλ1λi =0 for everyi ∈ {2, . . . ,n}, we haveT ∈ span{e1,ei}for everyi ∈ {2, . . . ,n}. Hence, T =∂xn+2 lies in the direction ofe1. Now letλi andλj be mutually different eigenvalues withi,j 6=1. SinceT ⊥span{ei,ej} we haveλiλj = −1. It follows that there can be only two different nonzero eigenvalues and that their product is−1. This is case (iii) of the lemma. ¤
Lemma 1 enables us to give a full description of semi-parallel hypersurfaces ofSn×R:
Theorem 5. Let Mn be a semi-parallel hypersurface ofSn ×R. Then there are four possibilities:
(i) n=2and M2is flat, (ii) Mnis totally umbilical,
(iii) Mnis an open part of a rotation hypersurface for which the profile curve is either a vertical line, or can be parametrized as
α(s)=
coss,0, . . . ,0,sins,± Z s
s0
pCcos2σ−1dσ
,
(iv) Mn⊆ ˉMn−1×R, whereMˉn−1is a semi-parallel hypersurface ofSn(1). As mentioned above, the classification of semi-parallel hypersurfaces of Sn(1)is given in [5].
Proof of Theorem 5. LetMnbe a semi-parallel hypersurface ofSn×Rwith shape operatorS. According to Lemma 1, there are three possible forms ofSto consider.
In the first case,Mn is totally umbilical by definition. This gives case (ii) of the theorem.
Now assume that we are in the second case of Lemma 1. Ifn =2, thenM2is a general flat surface inS2×Rand we are in case (i) of the theorem. Ifn ≥3, the form ofS is similar to the one given in Theorem 2. In the present case, the relationλμ = −cos2θ is not a functional relation in the strict sense, because θ can be a non-constant function. But from (8) we see that θ does not vary in directions orthogonal to T. By looking at the proof in [6], we see that this is actually enough to obtain that Mn is a rotation hypersurface. Moreover, the equalityλμ= −cos2θdetermines the profile curve of the rotation hypersurface.
First remark that formulae (10) yield that this equality is satisfied in the case that αis a vertical line. In this case we are in case (iv) of the theorem, where Mˉn−1 is a hypersphere ofSn(1). Ifαis not a vertical line, formulae (9) give
λμ= a0(s)a00(s)cots (1+a0(s)2)2 .
On the other hand, we have
−cos2θ = sin2θ−1=
∂xn+2, T kTk
2
−1=
∂xn+2, α0 kα0k
2
−1
= a0(s)2
1+a0(s)2 −1= − 1 1+a0(s)2.
Thus the equationλμ = −cos2θ is equivalent to(a0(s)2)0 +2 tans a0(s)2 =
−2 tans,for which the general solution is given bya0(s)2=Ccos2s−1, with C∈R.This covers case (iii) of the theorem.
In the last case of Lemma 1,∂xn+2 is everywhere tangent toMnand hence we are dealing with an open part of a product hypersurfaceMˉn−1×R, whereMˉn−1 is a hypersurface of Sn(1). Since Sn(1)is a totally geodesic hypersurface of Sn×R, we have that the shape operatorSˉ ofMˉn−1inSn(1)satisfiesSXˉ = SX forX tangent toMˉn−1, such thatSˉ takes the form
Sˉ =
λ
. ..
λ μ
. ..
μ
,
with λμ = −1. It was proven in [5] that this condition is equivalent to the condition thatMˉn−1is semi-parallel inSn(1). ¤ 5 Parallel hypersurfaces
We will now give a full classification of parallel hypersurfaces ofSn×R. One can easily verify that a parallel hypersurface has to be semi-parallel and hence in view of Theorem 5, there are only four cases to consider.
Theorem 6. Let Mnbe a parallel hypersurface ofSn×R. Then there are two possibilities:
(i) Mnis an open part of a totally geodesic hypersurfaceSn× {t0},
(ii) Mnis an open part of a Riemannian product Mˉn−1×R, whereMˉn−1is a parallel hypersurface ofSn(1).
Proof. Since Mn has to be semi-parallel, we just have to find the parallel hypersurfaces in the four families of Theorem 5.
Case (i). Parallel surfaces inS2×Rare classified in [1]. The flat ones are open parts of Riemannian products of a circle inS2(1)andR. This is a special case of the second case in the theorem.
Case (ii). IfMnis totally umbilical inSn×R, with shape operatorS=λid, a straightforward computation shows thatMnis parallel if and only ifλis constant.
But then the first equation of (13) yields that either cosθ =0 or sinθ =0. This means hat Mnis an open part of either Mˉn−1×Ror ofSn× {t0}. In the latter case, Mn is totally geodesic and we are in the first case of the theorem. In the former case, the second equation of (13) givesλ=0 and hence we obtain that Mˉn−1has to be a totally geodesic hypersurface ofSn(1). This is again a special case of the second case of the theorem.
Case (iii). Assume thatMnis a rotation hypersurface withλμ= −cos2θ. We may assume thatn ≥ 3, because forn = 2, we are dealing with a flat surface and this case was treated above. TakeXandYperpendicular toT. The equation of Codazzi forX andY gives X[μ]Y −Y[μ]X =0, yielding thatμis constant in directions perpendicular toT. Now letα(s)be the profile curve of Mn and choose a vector fieldX(s)alongα(s)which satisfies the following conditions:
X(s) ⊥ α0(s)(or equivalentlyX(s) ⊥T), X(s)is parallel alongα(s)inEn+2 and kX(s)k = 1. Such a vector field clearly exists, since it is sufficient to choose X orthogonal to the subspace P3, appearing in the construction of the rotation hypersurface, and tangent toMn. The formula of Gauss (1) yields that DTX = ∇TX, such thatX is also parallel alongα(s)inMn. If we now assume Mnto be parallel, we obtain
0=(∇h)(T,X,X)=T[h(X,X)] −2h(∇TX,X)=T[μ]. This implies thatμis constant.
Now take X perpendicular to T, then the equation of Codazzi for X andT gives
∇XST − ∇TSX−S[X,T] =cosθkTk2X
⇒ ∇X(λT)− ∇T(μX)−S(∇XT − ∇TX)=cosθsin2θX
⇒ X[λ]T +λ∇XT −μ∇TX−S(cosθ SX)+S(∇TX)=cosθsin2θX
⇒ X[λ]T +λμcosθ X −μ∇TX−μ2cosθX +μ∇TX =cosθsin2θ X
⇒ X[λ]T −cosθ (μ2+1)X =0
⇒ cosθ =0.
This means that that Mn is an open part of a Riemannian product Mˉn−1×R.
The second equation of (8) yields ST = 0 and henceλ = 0. Remark that the conditionλμ = −cos2θ = 0 is automatically satisfied. The shape operators ofMˉn−1 ⊂ Sn(1)as a submanifold ofEn+1are S1 =μid and S2 =id. If we change the basis of the normal plane by an appropriate rotation, they become Sˉ1=p
μ2+1 id andSˉ2=0. HenceMˉn−1is an open part of a sphere of radius 1/p
μ2+1 and we are in a special case of of the second case of the theorem.
Case (iv). Finally, we assume that Mn is an open part of Mˉn−1×R, where Mˉn−1is a semi-parallel hypersurface ofSn(1). It is easy to see thatMˉn−1×R is parallel inSn×Rif and only if Mˉn−1is parallel inSn(1). This is the second
case of the theorem. ¤
References
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[2] B.-Y. Chen. Geometry of submanifolds. M. Dekker, New York (1973).
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[6] F. Dillen, J. Fastenakels and J. Van der Veken.Rotation hypersurfaces inSn×R andHn×R, to appear in Note Mat.
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Joeri Van der Veken
Katholieke Universiteit Leuven Departement Wiskunde Celestijnenlaan 200 B B-3001 Leuven BELGIUM
E-mail: [email protected] Luc Vrancken
Université de Valenciennes LAMATH
ISTV2
Campus du Mont Houy 59313 Valenciennes Cedex 9 FRANCE
E-mail: [email protected]
