Ann Glob Anal Geom (2012) 42:333–347DOI 10.1007/s10455-012-9315-3

Relative nullity distributions, an affine immersionfrom an almost product manifoldand a para-pluriharmonic isometric immersion

Sanae Kurosu

Received: 12 July 2011 / Accepted: 1 March 2012 / Published online: 17 March 2012© Springer Science+Business Media B.V. 2012

Abstract We give a characterization for an affine immersion from an almost product man-ifold the almost product structure of which is adjoint or skew-adjoint with respect to its affinefundamental form by the relative nullity distribution.

Keywords Cylinder theorem · Relative nullity distribution · Almost product manifold ·Para-complex manifold · Para-pluriharmonic map

Mathematics Subject Classification 53A15 · 53C40

1 Introduction

A para-complex manifold can be considered as a special class of an almost product manifoldand a para-complex geometry is a topic with many analogies with the complex geometryand also with differences. A para-Kähler manifold is a para-complex manifold with a para-hermitian metric which is originally introduced by Rozenfeld [10] and Libermann [9]. Sucha manifold plays some important roles in supersymmetric field theories as well as in stringtheory, see [3] for example. On the other hand, an affine immersion from a para-complexmanifold is studied in [4,8] for example.

In this paper, we study the relative nullity distribution of an affine immersion from analmost product manifold with an affine connection and characterize such an affine immer-sion to an affine space. This paper is organized as follows. In Sect. 2, we prepare notions andnotations for an affine immersion and a splitting tensor. We study an affine immersion froman almost product manifold with an affine connection and its relative nullity distribution inSect. 3. We also give a characterization for such an affine immersion to an affine space interms of its relative nullity distribution. In Sect. 4, we study an isometric immersion from

S. Kurosu (B)Department of Mathematics, Tokyo University of Science, 1-3 Kagurazaka, Shinjuku-ku,Tokyo 162-8601, Japane-mail: kurosu@rs.kagu.tus.ac.jp

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an almost product manifold with a metric, which includes an almost product Riemannianmanifold. We show a cylinder theorem for such a hypersurface as a corollary of a result inthe previous section. In Sect. 5, we introduce a notion of a para-Kähler manifold as a specialcase of an almost product manifold with a metric. We prove a cylinder theorem for a para-pluriharmonic isometric hypersurface from a para-Kähler manifold to a pseudo-Euclideanspace, which can be considered as a para-Kähler version of a result given by Abe [1] andFuruhata [5].

2 Preliminaries

Throughout this paper, all manifolds are assumed to be connected and all objects and mor-phisms are assumed to be smooth. Let M be a manifold, T M its tangent bundle and T ∗Mits cotangent bundle.

We use letters E, ˜E to denote vector bundles over M . The fibre of a vector bundle Eat x ∈ M is denoted by Ex , the space of all cross sections of E by Γ (E) and the set ofall connections on E by C(E). We denote by Ap(E) = Γ (∧pT ∗M ⊗ E) the space of allE-valued p-forms on M . Let Hom(˜E, E) be the vector bundle of which fibre Hom(˜E, E)x

at x ∈ M is the vector space HomR(˜Ex , Ex ) of linear maps from ˜Ex to Ex .Next we prepare a notation of an affine immersion and introduce the relative nullity space.

Let M and ˜M be manifolds, f : M → ˜M an immersion. For a subbundle N of T ˜M alongf , if

T f (x)˜M = f∗x Tx M ⊕ Nx

for each x ∈ M , then we call such an immersion an immersion with transversal bundle N .Let πT M : T ˜M → f∗T M and πN : T ˜M → N be projection homomorphisms.

We denote by C0(T M) the set of all torsion free affine connections on M . Let (M,∇) and( ˜M, ˜∇) be manifolds with ∇ ∈ C0(T M) and ˜∇ ∈ C0(T ˜M).

For an immersion f : M → ˜M with transversal bundle N , if it holds that

πT M ˜∇X f∗Y = f∗∇X Y

for any X, Y ∈ �(T M), we say such a morphism ( f, N ) : (M,∇) → ( ˜M, ˜∇) an affineimmersion with transversal bundle N and denote it by f : (M,∇) → ( ˜M, ˜∇) for simplicity.When for an immersion f : (M,∇) → ( ˜M, ˜∇), there is a subbundle N of T ˜M along fsuch that f : (M,∇) → ( ˜M, ˜∇) is an affine immersion with transversal bundle N , we callf : (M,∇) → ( ˜M, ˜∇) an affine immersion. For an affine immersion f : (M,∇) → ( ˜M, ˜∇)

with transversal bundle N , define the affine fundamental form α ∈ A1(Hom(T M, N )), theshape tensor A ∈ A1(Hom(N , T M)) and the transversal connection ∇N ∈ C(N ) by

α(X, Y ) := πN ˜∇X f∗Y,

f∗ Aξ X := −πT M ˜∇X ξ,

∇NX ξ := πN ˜∇X ξ

for X, Y ∈ �(T M) and ξ ∈ Γ (N ). Since both ˜∇ and ∇ are torsion free, α is symmetric, thatis, α(X, Y ) = α(Y, X) holds for any X, Y ∈ Γ (T M). Then we can write the Gauss and theWeingarten formulas as

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˜∇X f∗Y = f∗∇X Y + α(X, Y ),

˜∇X ξ = − f∗ Aξ X + ∇NX ξ

for X, Y ∈ Γ (T M) and ξ ∈ Γ (N ).The equations of Gauss, Codazzi and Ricci are given by

πT M ˜RX,Y Z = RX,Y Z − Aα(Y,Z) X + Aα(X,Z)Y,

πN ˜RX,Y Z = (∇Xα)(Y, Z) − (∇Y α)(X, Z),

πT M ˜RX,Y ξ = −(∇X A)ξ Y + (∇Y A)ξ X,

πN ˜RX,Y ξ = RNX,Y ξ − α(X, Aξ Y ) + α(Y, Aξ X),

where ˜R, R, RN are the curvatures of ˜∇,∇,∇N , respectively, (∇X A) and (∇X α) are definedby

(∇Xα)(Y, Z) = ∇NX (α(Y, Z)) − α(∇X Y , Z) − α(Y,∇X Z),

(∇X A)ξ Y = ∇X (Aξ Y ) − Aξ∇X Y − A∇NX ξ Y

for X, Y, Z ∈ Γ (T M) and ξ ∈ Γ (N ).

Definition 1 For an affine immersion f : (M,∇) → ( ˜M, ˜∇) with transversal bundle N , thesubspace of Tx M defined by

Δ(x) := {X ∈ Tx M | α(X, Y ) = 0, for all Y ∈ Tx M}is called the relative nullity space of f at x ∈ M , the distribution x → Δ(x) is called therelative nullity distribution and its dimension dim Δ(x) is called the index of relative nullity.

Note that the relative nullity distribution does not depend on the choice of transversalbundle. Put ν0 := min{Δ(x)|x ∈ M}. Then {x ∈ M | dim Δ(x) = ν0} is an open set.

From now on, we always assume that dim Δ(x) is constant on M and let E be a distribu-tion such that Tx M = Δ(x) ⊕ E(x) for each x ∈ M . According to the decomposition, wedenote by prE : T M → E the projection homomorphism.

Definition 2 For an affine immersion f : (M,∇) → ( ˜M, ˜∇) with transversal bundle N ,define the splitting tensor C E by

C ES X := −prE∇X S

for any S ∈ Γ (Δ) and X ∈ Γ (E).

In the case where f is an isometric immersion from a Riemannian manifold and E is theorthogonal complement of Δ in T M, C E is studied in [2,5] for example and is often calledthe conullity operator in these papers.

Definition 3 Let M be a manifold and ∇ ∈ C0(T M). A distribution P is totally geodesic if∇X Y ∈ Γ (P) holds for any X, Y ∈ Γ (P). A totally geodesic distribution P is complete withrespect to ∇ if the induced connection on the maximal integral manifold of P is complete.

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We note that if πN ˜R = 0, then Δ is a totally geodesic distribution from the Codazziequation. In fact, we have

α(∇T S, X) = ∇NT (α(S, X)) − (∇T α)(S, X) − α(S,∇T X)

= −(∇T α)(S, X)

= −πN ˜RT,X S − (∇Xα)(T, S)

= −πN ˜RT,X S

for any S, T ∈ Γ (Δ) and X ∈ �(T M).We also have

α(Y, C ET X) = −α(Y,∇X T ) = −∇N

X (α(Y, T )) + (∇Xα)(Y, T ) + α(T,∇X Y )

= (∇Xα)(Y, T )

= πN ˜RX,Y T + (∇Y α)(T, X)

= πN ˜RX,Y T + α(X, C ET Y ) (2.1)

for any T ∈ Γ (Δ) and X, Y ∈ Γ (E).

3 An affine immersion from an almost product manifold and its characterization

In this section, we introduce an affine immersion from an almost product manifold and derivesome fundamental properties for such an affine immersion.

Let (M, J ) be an almost product manifold, that is, a real n-dimensional manifold M withan almost product structure J . Note that J satisfies J = ±idT M and J 2 = idT M .

Let (M,∇) be an almost product manifold with ∇ ∈ C0(T M) such that ∇ J = 0, ( ˜M, ˜∇)

a real manifold (dim ˜M = n + p) with ˜∇ ∈ C0(T ˜M) and f : (M,∇) → ( ˜M, ˜∇) an affineimmersion with transversal bundle N such that

α(J X , Y ) = δα(X, JY ), δ = ±1 (3.1)

for any X, Y ∈ �(T M). Note that δ is constant on M .For an almost product manifold (M, J ), put

P = 1

2(idT M + J ), Q = 1

2(idT M − J ).

Then we have

P + Q = idT M , P2 = P, Q2 = Q, P Q = Q P = 0, J = P − Q.

Thus, P and Q define two complementary distributions ˜P and ˜Q globally. Any eigenvec-tor corresponding to the eigenvalue +1 is in ˜P and any eigenvector corresponding to theeigenvalue −1 is in ˜Q. Put p := dim ˜P and q := dim ˜Q.

The Eq. (3.1) yields

α|˜Px טQx

= α|˜Qx טPx

= 0, if δ = 1, (3.2)

α|˜Px טPx

= α|˜Qx טQx

= 0, if δ = −1 (3.3)

for each x ∈ M .

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Lemma 1 The property that an affine immersion satisfies the condition (3.1) does not dependon the choice of transversal bundle.

Lemma 2 For an affine immersion f : (M,∇) → ( ˜M, ˜∇) with transversal bundle N whichsatisfies (3.1), the relative nullity space is J -invariant at each point.

Proof Fix a point x ∈ M . For any U ∈ Δ(x) and Y ∈ Tx M , we have

α(JU , Y ) = δα(U, JY ) = 0

from (3.1). Since Y is taken arbitrary, we see that JU ∈ Δ(x). ��From now on, we always assume that E is a J -invariant distribution such that Tx M =

Δ(x) ⊕ E(x) for each x ∈ M .

Lemma 3 Let f : (M,∇) → ( ˜M, ˜∇) be an affine immersion with transversal bundle Nwhich satisfies (3.1). For the splitting tensor, we have

C EJ S X = JC E

S X, (3.4)

α(Y, C ET J X) = δα(Y, C E

J T X) + πN ˜RJ X,Y T + δπN ˜RX,Y J T (3.5)

for any T, S ∈ Δ(x) and X, Y ∈ E(x), x ∈ M.

Proof From the definition of the splitting tensor, we get

α(Y, C EJ S X) = −α(Y,∇X J S) = −α(Y, J∇X S) = −δα(JY ,∇X S)

= δα(JY , C ES X) = δ2α(Y, JC E

S X) = α(Y, JC ES X)

and

α(Y, C EJ S X − JC E

S X) = 0 (3.6)

for any S ∈ Γ (Δ) and X, Y ∈ Γ (E). By (3.6), we obtain

C EJ S X − JC E

S X ∈ Δ(x) ∩ E(x)

for any S ∈ Δ(x) and X ∈ E(x), x ∈ M , that is, (3.4) holds.It follows from (2.1) and (3.4) that

α(Y, C ET J X) = πN ˜RJ X,Y T + α(J X , C E

T Y )

= πN ˜RJ X,Y T + δα(X, JC ET Y )

= πN ˜RJ X,Y T + δα(X, C EJ T Y )

= δα(Y, C EJ T X) + πN ˜RJ X,Y T + δπN ˜RX,Y J T

for any T ∈ Δ(x), X, Y ∈ E(x), x ∈ M . ��Corollary 1 Under the same assumption as in the previous lemma, if πN ˜R = 0, then

C ET J X = δC E

J T X = δ JC ET X (3.7)

holds for any T ∈ Δ(x), X ∈ E(x), x ∈ M.

For an affine immersion from an almost product manifold, we denote by pΔ(x) (resp.qΔ(x)) the dimension of an eigenspace corresponding to the eigenvalue +1 (resp. −1) ofJ |Δ(x). Note that pΔ(x) ≤ p, qΔ(x) ≤ q and pΔ(x) + qΔ(x) = dim Δ(x) hold for eachx ∈ M .

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Lemma 4 For an affine immersion f : (M,∇) → ( ˜M, ˜∇) with transversal bundle N whichsatisfies (3.1), we assume that Δ is complete with respect to ∇ and πN ˜R = 0.

(1) If dim Δ(x) = n − 1 at x ∈ M, then C ET = 0 for any T ∈ Δ(x).

(2) Assume that δ = −1. If ˜Px ⊂ Δ(x) or ˜Qx ⊂ Δ(x) holds at x ∈ M, then C ET = 0 for

any T ∈ Δ(x).(3) If δ = 1, dim Δ(x) = n − 2 and pΔ(x) = p − 1 hold at x ∈ M, then C E

T = 0 for anyT ∈ Δ(x).

Proof We first prove (1). Since dim E(x) = 1 at x ∈ M , there exists Y ∈ Tx M such thatE(x) = Span{Y }. In this case, we have C E

T Y = aY for some a ∈ R and a = 0 from Lemma2.4 in [6].

Next we show (2). By the assumptions, J |E(x) = εidE(x), ε ∈ {±1} holds at x ∈ M .Then (3.7) implies

εC ET = C E

T J = δ JC ET = −JC E

T = −εC ET

and C ET = 0 for any T ∈ Δ(x).

To prove (3), we take Y, Z ∈ Tx M such that JY = Y, J Z = −Z and E(x) = Span{Y, Z}.For T ∈ Δ(x), put

C ET Y = aY + bZ , C E

T Z = cY + d Z ,

where a, b, c, d ∈ R. Then we have

C ET JY = C E

T Y = (aY + bZ). (3.8)

It follows from (3.7) and (3.8) that

aY + bZ = C ET JY = δ JC E

T Y = J (aY + bZ) = aY + b(−1)Z

and b = 0. Since b = 0, C ET Y = aY holds and a = 0 by Lemma 2.4 in [6]. Analogously,

we obtain c = d = 0 when we calculate the both hand sides of C ET J Z = δ JC E

T Z . Hence,C E

T = 0 for any T ∈ Δ(x), x ∈ M . ��Hereafter in this paper, we denote by (Rn+p, D) an (n + p)-dimensional affine space with

the standard affine connection D.To define a cylindrical immersion, we prepare the following:For manifolds M, Mi and ∇ ∈ C0(T M),∇ i ∈ C0(T Mi ), i = 1, 2, we say that (M,∇)

and (M1,∇1) × (M2,∇2) are affine isomorphic if there exists a local diffeomorphism Φ :M1 × M2 → M such that

Φ∗(∇1X1

Y1 + ∇2X2

Y2) = ∇Φ∗(X1+X2)Φ∗(Y1 + Y2)

for each xi ∈ Mi , any Xi ∈ Txi Mi and Yi ∈ Γ (T Mi ), i = 1, 2, where we use the samenotation to denote Xi ∈ Txi Mi and the lift of Xi ∈ Txi Mi to T(x1,x2)(M1 × M2) for eachxi ∈ Mi .

Definition 4 An affine immersion f : (M,∇) → (Rn+p, D) is r-cylindrical if there existan (n − r)-dimensional manifold M ′ and ∇′ ∈ C0(T M ′) such that (M,∇) is affine isomor-phic to (M ′,∇′) × (Rr , D) with respect to Φ : M → M ′ × R

r and an affine immersionf ′ : (M ′,∇′) → (Rn−r+p, D) such that

f ∼= ( f ′ × idRr ) ◦ Φ.

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Ann Glob Anal Geom (2012) 42:333–347 339

From Lemma 4, we have the following characterization for an affine immersion from analmost product manifold to an affine space.

Proposition 1 For an affine immersion f : (M,∇) → (Rn+p, D) with transversal bundleN which satisfies (3.1), if M is complete with respect to ∇ and the index of relative nullity isn − 1 everywhere, then f is (n − 1)-cylindrical.

Since ∇ J = 0, we have

J∇X Z = ∇X J Z = ∇X Z (resp. − ∇X Z)

for any X ∈ Tx M, Z ∈ Γ (˜P) (resp. Z ∈ Γ (˜Q)), x ∈ M . Hence, ˜P and ˜Q are paralleldistributions with respect to ∇. Using a cylinder theorem in [6] (Proposition 3.4), we get

Proposition 2 For an affine immersion f : (M,∇) → (Rn+p, D) with transversal bundleN which satisfies (3.1), if ˜P ⊂ Δ (resp. ˜Q ⊂ Δ) holds and ˜P (resp. ˜Q) is complete withrespect to ∇, then f is p (resp. q)-cylindrical.

If δ = −1, we have the following from Lemma 4, (2).

Proposition 3 For an affine immersion f : (M,∇) → (Rn+p, D) with transversal bundleN which satisfies (3.1), if δ = −1, M is complete with respect to ∇, dim Δ(x) = r is constantand ˜Px ⊂ Δ(x) or ˜Qx ⊂ Δ(x) holds for each x ∈ M, then f is r-cylindrical.

Definition 5 A foliation P on M and an immersion f : M → Rn+p , we say that f is ruled

with respect to P if the image of any leaf of P under f is contained in (rankP)-dimensionalaffine subspaces of R

n+p . We call f completely ruled with respect to P if in addition theleaves are complete (rankP)-dimensional affine subspaces of R

n+p .

From (3.3), we get

Proposition 4 For an affine immersion f : (M,∇) → (Rn+p, D) with transversal bundleN which satisfies (3.1), if δ = −1, then f is ruled with respect to both ˜P and ˜Q. Moreover,if M is complete with respect to ∇, then f is completely ruled.

Next we will characterize an affine immersion to an affine space which satisfies (3.1) andthe index of relative nullity of which is n − 2.

Proposition 5 For an affine immersion f : (M,∇) → (Rn+p, D) with transversal bundleN which satisfies (3.1), assume that M is complete with respect to ∇ and the index of relativenullity is n − 2 everywhere.

(1) If δ = 1 and pΔ(x) = p − 1 hold for each x ∈ M, then f is (n − 2)-cylindrical.(2) Assume that δ = −1. If ˜Px ⊂ Δ(x) or ˜Qx ⊂ Δ(x) holds for each x ∈ M, then f is

(n − 2)-cylindrical.(3) If δ = −1 and pΔ(x) = p − 1 hold and Δ(x) is not parallel with respect to ∇ for

each x ∈ M, then there exists an (n − 1)-dimensional distribution ˜Δ such that f iscompletely ruled with respect to ˜Δ.

Proof The statements (1) and (2) follow from Lemma 4. To prove (3), we take Y, Z ∈ Tx Msuch that E(x) = Span{Y, Z} and JY = Y, J Z = −Z . For T ∈ Δ(x), put

C ET Y = aY + bZ , C E

T Z = cY + d Z ,

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where a, b, c, d ∈ R. From (3.7), we get

2aY = 0, −2d Z = 0,

that is, a = d = 0.Since Δ(x) is not parallel with respect to ∇, b = 0 or c = 0 holds. First we assume that

c = 0 holds. If bc ≥ 0, then the eigenvalues of C ET are ±√

bc and are equal to zero since±√

bc ∈ R. Hence b = 0 holds.When bc ≤ 0, since it holds that

C EJ T

(

Y +√

−b

cZ

)

= −C ET J

(

Y +√

−b

cZ

)

= −C ET

(

Y +√

−b

c(−Z)

)

= −bZ +√

−b

ccY

=√

−b

cc

(

Y +√

−b

cZ

)

,

we have b = 0. Thus, we get ImC ET = KerC E

T = Span{Y } and dim(ImC ET ∩ KerC E

T ) = 1.If we assume that b = 0, then we obtain c = 0, ImC E

T = KerC ET = Span{Z} and

dim(ImC ET ∩ KerC E

T ) = 1 by a similar way as above.Next we claim that for any X ∈ KerC E

T , it holds that α(X, X) = 0. Since ImC ET = KerC E

T ,there exists ˜X ∈ E(x) such that X = C E

T˜X . By (2.1), we get

α(X, X) = α(C ET

˜X , C ET

˜X) = α(˜X , C ET C E

T˜X) = 0

since C ET is nilpotent for any T ∈ Δ(x), x ∈ M .

We claim now that for any S ∈ Δ(x), it holds that KerC ET ⊂ KerC E

S . Because both C ET

and C ES are nilpotent, we get

0 = C E(T +S)C

E(T +S) = C E

T C ES + C E

S C ET (3.9)

and for X ∈ KerC ET , it holds that

C ET C E

S X = −C ES C E

T X = 0, (3.10)

that is, C ES X ∈ KerC E

T .Fix a point x ∈ M and define L : E(x) → E(x) by

LW := C ET C E

S W

for W ∈ E(x). Let X ∈ KerC ET and V ∈ E(x) such that E(x) = Span{X, V }. By (2.1),

(3.9) and (3.10), we get

L X = 0,

α(V, LV ) = −α(LV , V ) = 0,

α(X, LV ) = α(C ES C E

T X , V ) = 0,

that is, C ET C E

S = 0. Thus we see that KerC ET ⊂ KerC E

S .

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Ann Glob Anal Geom (2012) 42:333–347 341

Therefore, we get the one-dimensional distribution

x →⋂

S∈Δ(x),C ES =0

(ImC ES ∩ KerC E

S ).

Hence, from Proposition 2.7 in [7], f is completely ruled with respect to the (n − 1)-dimen-sional distribution ˜Δ given by

x → ˜Δ(x) := Δ(x) ⊕⋂

S∈Δ(x),C ES =0

(ImC ES ∩ KerC E

S ).

��Note that the distribution ˜Δ defined in Proposition 5, (3) does not depend on the choice

of E such that T M = Δ ⊕ E .

4 A cylinder theorem for a certain isometric hypersurface from an almost productmanifold with a metric

In this section, we study an almost product manifold with a pseudo-Riemannian metric.Let (M, J ) be an almost product manifold, g a pseudo-Riemannian metric such that

g(J X, JY ) = εg(X, Y ), ε = ±1 (4.1)

for any X, Y ∈ Tx M, x ∈ M . Note that ε is constant on M .An almost product manifold with a Riemannian metric which satisfies ε = 1 is called an

almost product Riemannian manifold, see [12].The Eq. (4.1) yields

g|˜Px טQx

= 0, g|˜Qx טPx

= 0, if ε = 1,

g|˜Px טPx

= 0, g|˜Qx טQx

= 0, if ε = −1

for each x ∈ M .In this section, we consider an isometric immersion f : M → ˜M from an almost product

manifold with a metric which satisfies (4.1) to a pseudo-Riemannian manifold ˜M the secondfundamental form of which satisfies (3.1).

Let g be a pseudo-Riemannian metric on ˜M . Since f is an isometric immersion, we have

g(Aξ J X , Y ) = g(α(J X , Y ), ξ) = δg(α(X, JY ), ξ)

= δg(Aξ X, JY ) = εδg(J Aξ X, Y )

for any X, Y ∈ Tx M, ξ ∈ T ⊥x M, x ∈ M , where we denote by T ⊥

x M the normal space of anisometric immersion f at x ∈ M . Since g is non-degenerate, we get

Aξ J X = εδ J Aξ X (4.2)

for any X ∈ Tx M and ξ ∈ T ⊥x M, x ∈ M . From this equation, we get

Lemma 5 If εδ = 1, then Im Aξ |˜Px⊂ ˜Px and Im Aξ |˜Qx

⊂ ˜Qx hold for any ξ ∈ T ⊥x M, x ∈

M. If εδ = −1, then Im Aξ |˜Px⊂ ˜Qx and Im Aξ |˜Qx

⊂ ˜Px hold for any ξ ∈ T ⊥x M, x ∈ M.

By the Gauss equation, we get

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342 Ann Glob Anal Geom (2012) 42:333–347

Lemma 6 If ˜M is flat, then it holds that

RJ X,JY = εRX,Y (4.3)

for any X, Y ∈ Tx M, x ∈ M.

From now on, we introduce the following notation for an isometric hypersurface f : M →R

n+1k from an almost product manifold with a metric to a pseudo-Euclidean space of index

k. Let ξ be a normal vector field such that |g(ξ, ξ)| = 1, S a (1, 1)-tensor field defined by

SX := Aξ X

and h a symmetric bilinear form defined by

h(X, Y )ξ := α(X, Y )

for any X, Y ∈ Tx M and ξ ∈ T ⊥x M as above, x ∈ M .

For an isometric hypersurface we get,

Lemma 7 For an isometric hypersurface f : M → Rn+1k from a pseudo-Riemannian man-

ifold, if dim ImSx ≤ 1 at x ∈ M, then M is flat at x.

Proof If dim ImSx = 0 at x ∈ M , then x is a geodesic point and ∇ is flat at x ∈ M .Next, we assume that dim ImSx = 1 at x ∈ M . Then, there exist V ∈ Tx M and η :

Tx M → R such that

SX = η(X)V

for any X ∈ Tx M . By the Gauss equation, we have

RX,Y Z = SXh(Y, Z) − SY h(X, Z)

= η(X)g(SY, Z)V − η(Y )g(SX, Z)V

= η(X)η(Y )g(V, Z)V − η(Y )η(X)g(V, Z)V

= 0

for any X, Y, Z ∈ Tx M, x ∈ M . ��

We remark that this lemma holds for any isometric hypersurface from a pseudo-Riemann-ian manifold to a flat pseudo-Riemannian manifold in general.

Next we will consider the following four cases 1, 2, 3 and 4 for an isometric hypersurfacef : M → R

n+1k from an almost product manifold with a metric and give a characterization

for such a hypersurface.

Case 1. ε = 1, δ = 1.From (3.2) and (4.3), we get

0 = RX,Y Z = h(Y, Z)SX − h(X, Z)SY = −h(X, Z)SY

for any X, Z ∈ ˜Px , Y ∈ ˜Qx , x ∈ M . Hence, ˜Px ⊂ Δ(x) or ˜Qx ⊂ Δ(x) holds at x ∈ M .

Proposition 6 For an isometric hypersurface f : M → Rn+1k , we assume that ε = δ = 1.

Then ˜Px ⊂ Δ(x) or ˜Qx ⊂ Δ(x) holds for each x ∈ M.

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Ann Glob Anal Geom (2012) 42:333–347 343

Case 2. ε = 1, δ = −1.From (3.3) and (4.3), we get

0 = RX,Y Z = h(Y, Z)SX − h(X, Z)SY = −h(Y, Z)SX

for any X, Z ∈ ˜Px , Y ∈ ˜Qx , x ∈ M . From this and (3.3), either ˜Px ⊂ Δ(x) or ˜Qx ⊂ Δ(x)

holds at x ∈ M .First, we assume that ˜Px ⊂ Δ(x) at x ∈ M . If dim ImS|

˜Qx≥ 2, then for any X ∈ ˜Qx

such that SX = 0, there exists Y ∈ ˜Qx such that SX, SY are linearly independent. On theother hand, we have

0 = RX,Y Z = h(Y, Z)SX − h(X, Z)SY

for any X, Y, Z ∈ ˜Qx , x ∈ M , which is a contradiction. Hence, we obtain dim ImS|˜Qx

< 2

and dim ImSx ≤ 1 at x ∈ M . When we assume that ˜Py ⊂ Δ(y) at y ∈ M , we havedim ImSy ≤ 1 at y ∈ M by a similar way as above. Thus, we can prove

Proposition 7 For an isometric hypersurface f : M → Rn+1k , we assume that ε = 1 and

δ = −1. Then dim Δ(x) ≥ n − 1 holds for each x ∈ M. Moreover, if M is complete andthere are no geodesic points, then f is (n − 1)-cylindrical.

Proof The first part follows since dim ImSx ≤ 1 implies that dim Δ(x) ≥ n − 1 for eachx ∈ M . Moreover, if M is complete and there are no geodesic points, then dim Δ(x) = n −1holds for each x ∈ M and the result follows from Proposition 1. ��Case 3. ε = −1, δ = 1.

From (3.2) and (4.3), we have

0 = RX,Y Z = h(Y, Z)SX − h(X, Z)SY

for any X, Y, Z ∈ ˜Px , x ∈ M . If dim ImS|˜Px

≥ 2 at x ∈ M , then for X ∈ ˜Px such thatSX = 0, there exists Y ∈ ˜Px such that SX, SY and are linearly independent at x ∈ M . Inthis case, we have h(Y, Z) = h(X, Z) = 0 for any Z ∈ ˜Px . If SX = 0 for X ∈ ˜Px , weget h(X, Z) = g(SX, Z) = 0. By (3.2), h(U, Z) = 0 holds for any U ∈ ˜Qx . Hence, wecan conclude that Z ∈ Δ(x). Since Z is taken arbitrary, we have ˜Px ⊂ Δ(x), which is acontradiction. Hence dim ImS|

˜Px≤ 1 at x ∈ M .

We also have dim ImS|˜Qx

≤ 1 at x ∈ M by the same argument as above. Therefore wehave dim ImSx ≤ 2 for each x ∈ M . Hence we can prove

Proposition 8 For an isometric hypersurface f : M → Rn+1k , we assume that ε = −1 and

δ = 1. Then dim Δ(x) ≥ n − 2 for each x ∈ M. Moreover, if M is complete and not flat foreach point, then f is (n − 2)-cylindrical.

Proof Since M is not flat at x , then dim ImSx = 2, that is, dim Δ(x) = n − 2 for eachx ∈ M . From Proposition 5, (1), we see that f is (n − 2)-cylindrical. ��Case 4. ε = δ = −1.

By a similar way as Case 3, we get dim ImS|˜Px

≤ 1 and dim ImS|˜Qx

≤ 1 for each x ∈ Mand we obtain the following:

Proposition 9 For an isometric hypersurface f : M → Rn+1k , we assume that ε = −1

and δ = −1. Then dim Δ(x) ≥ n − 2 for each x ∈ M. Moreover, if M is complete andnot flat and Δ(x) is not parallel with respect to ∇ for each x ∈ M, then there exists an(n − 1)-dimensional distribution ˜Δ such that f is completely ruled with respect to ˜Δ.

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344 Ann Glob Anal Geom (2012) 42:333–347

Proof Since M is not flat at x , then dim ImSx = 2, that is, dim Δ(x) = n − 2 for eachx ∈ M . Therefore the statement follows from Proposition 5 (3). ��

5 A cylinder theorem for a para-pluriharmonic isometric hypersurfacefrom a para-Kähler manifold

We prepare the following definitions:

Definition 6 An almost para-complex structure J on a manifold M is an almost productstructure such that the two eigenspaces T ±M := Ker(idT M ∓ J ) of J have the same dimen-sion. An almost para-complex structure is called integrable if the distributions T ±M areintegrable. An integrable almost para-complex structure is called a para-complex structureand a manifold M endowed with a para-complex structure is called a para-complex manifold.

We note that an almost para-complex manifold is an almost product manifold such thatp = q.

Next, we introduce a notion of a para-Kähler manifold and prove a cylinder theorem fora para-pluriharmonic isometric hypersurface from a complete para-Kähler manifold.

Definition 7 For a para-complex manifold (M, J ) and a pseudo-Riemannian metric g, wecall (M, J, g) a para-Kähler manifold if ∇ J = 0 and g(J X, JY ) = −g(X, Y ) hold for anyX, Y ∈ Tx M, x ∈ M , where ∇ is the Levi-Civita connection of g.

We remark that the pseudo-Riemannian metric on a para-Kähler manifold is neutral. Fora para-Kähler manifold (M, J, g), we have

RJ X,JY = −RX,Y (5.1)

for any X, Y ∈ Tx M, x ∈ M .We note that a para-pluriharmonic isometric immersion is an isometric immersion from

a para-Kähler manifold such that

α(J X, Y ) = α(X, JY )

for any X, Y ∈ Tx M, x ∈ M , that is, δ = 1 in the Eq. (3.1). For a para-pluriharmonic map,see [11].

An isometric immersion from a para-Kähler manifold to a pseudo-Riemannian manifoldis para-pluriharmonic if and only if its shape tensor A satisfies

Aξ J X = −J Aξ X

for any X ∈ Tx M, ξ ∈ T ⊥x M, x ∈ M .

From now on, we always assume that M is a real 2m-dimensional para-Kähler manifoldand ˜M is a pseudo-Riemannian manifold.

Lemma 8 For an isometric immersion f : M → ˜M, the following are equivalent:

(1) f is para-pluriharmonic,(2) α(X, Y ) = 0 for any X ∈ ˜Px and Y ∈ ˜Qx , x ∈ M,(3) Im Aξ |˜Px

⊂ ˜Qx and Im Aξ |˜Qx⊂ ˜Px for any ξ ∈ T ⊥

x M, x ∈ M.

As a corollary of Proposition 8, we have

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Ann Glob Anal Geom (2012) 42:333–347 345

Theorem 1 For an isometric para-pluriharmonic hypersurface f : M → R2m+1k , if M is

complete and not flat for each point, then f is (2m − 2)-cylindrical.

Lemma 9 For any para-pluriharmonic isometric immersion f : M → ˜M, the mean curva-ture is identically zero.

Proof Fix a point x ∈ M and let e1, . . . , e2m be a basis of Tx M such that ei+m =Jei , g(ei , e j ) = δi j , g(ei , Je j ) = 0, g(Jei , Je j ) = −δi j for i, j = 1, . . . , m. Let H bethe mean curvature vector field. By a straightforward calculation, we get

2m H =2m∑

l=1

g(el , el)α(el , el)

=m

∑

k=1

g(ek, ek)α(ek, ek) +m

∑

j=1

g(Je j , Je j )α(Je j , Je j )

=m

∑

k=1

g(ek, ek)α(ek, ek) +m

∑

j=1

(−g(e j , J 2e j ))α(e j , J 2e j )

=m

∑

k=1

g(ek, ek)α(ek, ek) −m

∑

j=1

g(e j , e j )α(e j , e j ) = 0.

��Next, we will consider a condition for an isometric hypersurface from a para-Kähler manifoldto a pseudo-Euclidean space to be para-pluriharmonic.

We note that the curvature tensor satisfies

RX,Y = 0

for any X, Y ∈ ˜Px or X, Y ∈ ˜Qx , x ∈ M since (5.1) holds.Hereafter in this paper, we always consider an isometric hypersurface and recall a notation

for an isometric hypersurface introduced in the previous section. For an isometric hypersur-face f : M → R

2m+1k from a para-Kähler manifold to a pseudo-Euclidean space of index

k, let ξ be a normal vector field such that |g(ξ, ξ)| = 1, S a (1, 1)-tensor field defined bySX := Aξ X and h a symmetric bilinear form defined by h(X, Y )ξ := α(X, Y ) for anyX, Y ∈ Tx M and ξ ∈ T ⊥

x M as above, x ∈ M .First we assume dim ImS|

˜Px≥ 2 at x ∈ M . Then for X ∈ ˜Px such that SX = 0, there

exists Y ∈ ˜Px such that SX, SY are linearly independent. In this case, we get

h(Y, Z) = h(X, Z) = 0

for any Z ∈ ˜Qx and f is para-pluriharmonic from Lemma 8 since if SX = 0 for X ∈ ˜Px ,we always have h(X, Z) = 0 for any Z ∈ Tx M . By a similar way, f is para-pluriharmonicif dim ImS|

˜Qx≥ 2 at x ∈ M .

If dim ImS|˜Px

= dim ImS|˜Qx

= 0, x is a geodesic point.

If dim ImS|˜Px

= 1 and dim ImS|˜Qx

= 0, we have ˜Px ⊂ Δ(x) and h(X, Y ) = 0 for any

X ∈ ˜Px and Y ∈ ˜Qx and f is para-pluriharmonic from Lemma 8.If dim ImS|

˜Px= 0 and dim ImS|

˜Qx= 1, we have ˜Qx ⊂ Δ(x) and h(X, Y ) = 0 for any

X ∈ ˜Px and Y ∈ ˜Qx and f is para-pluriharmonic from Lemma 8.

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346 Ann Glob Anal Geom (2012) 42:333–347

If dim ImS|˜Px

= dim ImS|˜Qx

= 1, put F(x) := ImS|˜Px

∪ ImS|˜Qx

. Then we obtain

F(x) ∩ ˜Px = {0}, F(x) ∩ ˜Qx = {0}.In fact, if F(x) ⊂ ˜Px , then it holds that

h(X, Y ) = g(SX, Y ) = 0

for any X ∈ Tx M and Y ∈ ˜Px , that is, Y ∈ Δ(x). Since Y ∈ ˜Px is taken arbitrary, we get˜Px ⊂ Δ(x), which is a contradiction since we assume that ImS|

˜Px= {0}. Hence, we have

F(x) ∩ ˜Qx = {0}. By a similar way, we get F(x) ∩ ˜Px = {0}.From now on, we will consider the following two subcases:

Subcase 1 ImS|˜Px

= ImS|˜Qx

.

In this case M is flat from Lemma 7 since dim ImSx = 1 holds. Since ImS|˜Px

∩˜Px = {0}, fcannot be para-pluriharmonic in this case.Subcase 2 ImS|

˜Px= ImS|

˜Qx.

Let e1, e2 ∈ Tx M such that e1 ∈ ˜Px , e2 ∈ ˜Qx , Se1 and Se2 are linearly independent.If g(e1, e2) = 0, put u1 := (e1 + e2)/

√|2g(e1, e2)| and u2 := Ju1 = (e1 −e2)/

√|2g(e1, e2)|. Then, u1, u2 are mutually orthogonal.From the definition of the mean curvature vector H , we have

2m H = g(Su1, u1)g(u1, u1) + g(Su2, u2)g(u2, u2)

= g(Su1, u1) − g(Su2, u2)

= h(u1, u1) − h(Ju1, Ju1).

Then, H ≡ 0 is equivalent to that f is para-pluriharmonic in this case.If g(e1, e2) = 0, for v2 ∈ ˜Qx ∩ Δ(x) such that g(e1, v2) > 0, put

e1 := e1, e2 := e2 + v2.

Then, we have

g(e1, e2) = g(e1, e2) + g(e1, v2) = g(e1, v2) = 0.

So, we can conclude that H ≡ 0 is equivalent to that f is para-pluriharmonic in this casefrom the previous argument.

Therefore, we have

Proposition 10 For an isometric immersion f : M → R2m+1k , if M is not flat for each point,

then the mean curvature is identically zero if and only if f is para-pluriharmonic.

From the above and Theorem 1, we have the following cylinder theorem.

Theorem 2 For a hypersurface f : M → R2m+1k (m > 2), if M is complete and not flat for

each point and the mean curvature is identically zero, then f is (2m − 2)-cylindrical.

We note that this theorem can be considered as a ‘para-complex version’ of a cylindertheorem for a complete minimal real Kähler hypersurface to a pseudo-Euclidean space givenby [1,5].

Acknowledgements The author would like to express her sincere gratitude to Professor Naoto Abe for hisvaluable advice and encouragement. She also wishes to express her hearty thanks to the referee for his kindsuggestions to complete the revised version.

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