on a certain class of hypersurfaces in semi-euclidean spaces

SOOCHOW JOURNAL OF MATHEMATICS

Volume 25, No. 2, pp. 221-234, April 1999

ON A CERTAIN CLASS OF HYPERSURFACES IN SEMI-EUCLIDEAN SPACES BY KADRI ARSLAN1 , RYSZARD DESZCZ 2 AND RIDVAN EZENTAS.3 Dedicated to Professor Dr. Leopold Verstraelen on his 50th birthday

1. Introduction A semi-Riemanninan manifold (M g), n = dim M metric 16] if R
R=0

3, is said to be semisym(1)

holds on M . It is clear that every semisymmetric manifold realizes the following condition

R
S = 0:

(2)

For precise de nitions of the symbols used, we refer to Section 2. A semiRiemannian manifold (M g), n 3, realizing (2) is called Ricci-semisymmetric. Evidently, there exist non-semisymmetric Ricci-semisymmetric manifolds. However, under some additional assumptions, (1) and (2) are equivalent for certain manifolds (e.g. see 1]). As a proper generalization of locally symmetric spaces (rR = 0) semisymmetric manifolds were studied by many authors (see e.g. 16]). For hypersurfaces of Euclidean spaces, (1) was rst studied in 14]. Since then, (1), (2) and other similar curvature conditions, we say curvature conditions of semisymmetry type, Received Feburary 25, 1998

This paper is prepared during the second named author's visit to the Uluda
g University, Bursa, Turkey in December 1997. The second author is supported by the Scientic and Technical Research Council of Turkey (TU BITAK) for NATO-CP Advanced Fellowships Programme. 221

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KADRI ARSLAN, RYSZARD DESZCZ AND RIDVAN EZENTAS.

were studied by many authors, see e.g. 7], 8], 15] and 17]. An example of a curvature condition of semisymmetry type is the following

S
R = 0:

(3)

A natural extension of such curvature conditions form curvature conditions of pseudosymmetry type (10], 19]). Conformally at manifolds ful lling (3) were investigated in 2]. There exist also nonconformally at semi-Riemannian manifolds ful lling (3) (see Theorem 4.2 of 2]). Hypersurfaces in Euclidean spaces En+1 , n 3, realizing (3) were studied in 18]. The main results of 18] is the following:

Theorem 1.1. (18], Theorem) Let M be a hypersurface in a Euclidean

space En+1 , n 3. Then (3) holds on M if and only if M is a hypercylinder.

This is related with the following result. Theorem 1.2. (2], Theorem 5.1) Every conformally at hypersurface M in a semi-Euclidean space Ens +1 , n 4 fullling (3) is cylindrical. Evidently, the converse of the statement is also true. Let H be the second fundamental tensor of a hypersurface M in a Euclidean space En+1 , n 3. We can easily check that M is cylindrical at a point x 2 M if and only if the relation H 3 = tr(H )H 2 (4) holds point at x. Thus Theorem 1.1 is equivalent to the following statement.

Theorem 1.3. Let M be a hypersurface in a Euclidean space En+1 , n 3.

Then (3) holds on M if and only if the second fundamental tensor H of M satised (4) at every point of M .

In this paper we extend the above result to the case when the ambient space is a semi-Euclidean space. Namely, we will prove the following statement (see Theorem 3.1): Let M be a hypersurface in a semi-Euclidean space Ens +1 , n 4. Then (3) holds on M if and only if the second fundamental tensor H of M satises (4) at every point of M . We also prove (see Theorem 3.2) that: Every hypersurface M in E5s satisfying (4) is semisymmetric.

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223

In Section 2 we state that there exist non-conformally at hypersurfaces in Ens +1 , n 4, which realize (4). The paper is organized as follows in Section 2 we x the notations and we give preliminary results. Next, in Section 3 we prove our main results, i.e. Theorems 3.1 and 3.2.

2. Preliminary Results Let (M g) be a connected n-dimensional, n 3, semi-Riemannian manifold of class C 1 . We de ne on M the endomorphisms R(X Y ), and X ^ Y by

R(X Y )Z = rX rY ]Z ; rX Y ]Z (X ^ Y )Z = g(Y Z )X ; g(X Z )Y respectively, where r is the Levi-Civita connection of (M g) and X Y Z 2 (M ), (M ) being the Lie algebra of vector elds on M . Furthermore, we de ne the Riemann-Christoel curvature tensor R and the (0,4)-tensor G of the manifold (M g) by

R(X1 X2 X3 X4 ) = g(R(X1 X2 )X3 X4 ) G(X1 X2 X3 X4 ) = g((X1 ^X2 )X3 X4 ) respectively. We denote by S and  the Ricci tensor and the scalar curvature of (M g), respectively. The Ricci operator S of (M g) is de ned by g(S X Y ) = S (X Y ). For (0 k)-tensor eld T on M , k 1, we de ne (0 k)-tensor S
T by (S
T )(X1 : : : Xk ) = ;T (S X1 X2 : : : Xk ) ;


; T (X1 : : : Xk;1 S Xk ): (5) Furthermore, for a (0 k)-tensor eld T on M , k 1, we de ne (0 k + 2)-tensors R
T and Q(g T ) by (R
T )(X1 : : : Xk  X Y ) = (R(X Y )
T )(X1 : : : Xk ) = ;T (R(X Y )X1 : : : X2 : : : Xk ) ;


; T (X1 : : : Xk;1 R(X Y )Xk ) Q(g T )(X1 : : : Xk  X Y ) = ((X ^ Y )
T )(X1 : : : Xk ) = ;T ((X ^ Y )X1 : : : X2 : : : Xk ) ;


; T (X1 : : : Xk;1 (X ^ Y )Xk )

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respectively. For (0,2)-tensors A and B we de ne their Kulkarni-Nomizu product A^~ B (3]) by (A^~ B )(X1 X2 X3 X4 ) = A(X1 X4 )B (X2 X3 ) + A(X2 X3 )B (X1 X4 ) ;A(X1 X3 )B (X2 X4 ) ; A(X2 X4 )B (X1 X3 ): For a symmetric (0,2)-tensor A we de ne the endomorphism X ^A Y of (M ) by (X ^A Y )Z = A(Y Z )X ; A(X Z )Y: We note that X ^g Y = X ^ Y . Furthermore, for a (0 k)-tensor eld T , k and the tensor eld A we de ne the tensor Q(A T ) by

1,

Q(A T )(X1 : : : Xk  X Y ) = ((X ^A Y )
T )(X1 : : : Xk ) = ;T ((X ^A Y )X1 X2 : : : Xk ) ;


; T (X1 : : : Xk;1 (X ^A Y )Xk ): Let M be a hypersurface immersed isometrically in a semi-Riemannian manifold (N g~), dim N = n + 1 4 and let g be the induced metric tensor on M from the metric g~. The hypersurface M is said to be quasi-umbilical at a point x 2 M (12]) if the second fundamental tensor H of M satis es at this point the following equality H = g + w  w   2 R where w is a covector at x. If such condition is ful lled at every point of M , then M is called a quasi-umbilical hypersurface. In particular, the hypersurface M is said to be cylindrical (resp. umbilical) at a point x 2 M if H is of rank one, (resp. if H is proportional to the metric tensor g) at this point. That is H = w,  2 R, (rsep. H = g,  2 R) holds at x, where w is a covector at x. If such condition is ful lled at every point of M then M is called a cylindrical, respectively umbilical hypersurface. Let now M be a hypersurface immersed isometrically in a semi-Euclidean space Ens +1 , of index s, n 3. Let g be the induced metric tensor on M from the metric of the ambient space Ens +1 . We denote by gij , Rhijk , Sij and Sjk = gks Sjs the local components of the metric g, the curvature tensor R, the Ricci tensor S and the Ricci operator S of (M g), respectively. Further, we denote by Hij the local components of the second fundamental tensor of M in Ens +1 and let

ON A CERTAIN CLASS OF HYPERSURFACES IN SEMI-EUCLIDEAN SPACES

225

Hij2 = Hipgpq Hjq and Hij3 = Hip2 gpq Hjq be the local components of the tensors H 2 and H 3 , respectively and let tr(H ) = gij Hij . The Gauss equation of M in Ens +1 we can write in the form

Rhijk = "(Hhk Hij ; Hhj Hik ) " = 1:

(6)

From this, by contraction with ghk , we obtain

Sij = "(tr(H )Hij ; Hij2 ):

(7)

Using (6) and (7) we can prove that the following relations are satis ed on M (12], Proposition 3.1) (a) R
R = Q(S R)

(b) R
R = ; 2" Q(H 2 H ^~ H ):

(8)

Further, let (S
R)hijk denote the local components of the tensor S
R. Using (5) we can present (3) in the following form (S
R)hijk = ;ShpRpijk ; SipRhpjk ; SjpRhipk ; Skp Rhijp = 0:

Lemma 2.1. Let M be a hypersurface in a semi-Euclidean space

n 3.

(9) Ens +1 ,

(i) If at a point x 2 M rank (H ) = 1 then (4) holds at x. (ii) If at a point x 2 M (4) is fullled then (3) holds at x.

Proof. (i) The relation rank(H ) = 1, we can present in the following form Hij = wiwj  2 R (10) where wj are the local components of a covector w at x. From (10), by contraction with gij , we get tr(H ) = ws ws ws = gsj wj : (11) Transvecting (10) with wi and using (11) we obtain

wsHsj = wswsbj = tr(H )wj : Further, transvecting (10) with Hki and using (12) we nd

Hjk2 = tr(H )wj wk = tr(H )Hjk

(12)

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KADRI ARSLAN, RYSZARD DESZCZ AND RIDVAN EZENTAS.

which, by transvection with Hlj , leads to (4). (ii) Transvecting (6) with Slh we obtain

SlpRpijk = "(Slp HpkHij ; SlpHpj Hik ):

(13)

Next, transvecting (7) with Hki and using (4) we get

SlpHpk = "(tr(H )Hlk2 ; Hlk3 ) = 0 which reduces (13) to Slp Rpijk = 0. Making use of (9), this implies (3). This completes the proof of the Lemma. Combining Lemma 2.1 with Theorem 1.2 (i.e. with Theorem 5.1 of 2]) we obtain the following: Corollary 2.1. The conditions (3) and (4) are equivalent on every conformally at hypersurface in a semi-Euclidean space Ens +1 , n 4. We nish this section with the following

Remark 2.1. Einstein hypersurfaces M , dim M 4, in semi-Riemannian

spaces of constant curvature such that their second fundamental tensor H satisfy H 2 = 0 were investigated in 13]. We note that H 2 = 0 implies (4). Moreover, (8) reduces to R
R = 0. Let M , n = dim M = 2k 4, be an Einstein hypersurface in a semi-Riemannian space of constant curvature N 2k+1 (c). In 13] it was shown that if the second fundamental tensor H of M has maximal rank, which means that rank (H ) = k at every point of M in that case, and if H 2 = 0 holds on M then the ambient space reduces to a semi-Euclidean space E2sk+1 (13], Corollary 2.3). Moreover, we note that if M has maximal rank (k 2) then M is a non-conformally at hypersurface. In fact, if we suppose that M is an Einstein conformally at hypersurface then R = 12
G holds on M . Applying this in the Gauss equation (6) we get

G Hhk Hij ; Hhj Hik = " 12 hijk

(14)

whence tr(H )Hij = "
4 gij , and tr(H )Hij2 = "
4 Hij i.e.  = 0 or H = 0 hold on M . Evidently, if  = 0 then (14) implies rank (H )  1. So, if  = 0 or H = 0 hold on M then rank (H )  1, which is a contradiction.

ON A CERTAIN CLASS OF HYPERSURFACES IN SEMI-EUCLIDEAN SPACES

227

3. Non-Cylindrical Hypersurfaces In this section we will consider non-cylindrical hypersurfaces M in Ens +1 , n 3. Transvecting (7) with Hki = gis Hsk we obtain

Dij = HipSpj = "(tr(H )Hij2 ; Hij3 ):

(15)

Applying (6) and (15) in (9) we get (H ^~ D)hijk = Hij Dhk + Hhk Dij ; Hik Dhj ; Hhj Dik = 0:

(16)

By making use of Lemma 3.1 (2]) we have the follwing identity on M Q(D H ^~ D) = ; 21 Q(H D^~ D): Since H ^~ D = 0, the last relation reduces to Q(H D^~ D) = 0. Further, we assume that at a point x 2 M rank(H ) > 1. Now Proposition 4.1 of 4] states that

D^~ D = H ^~ H

(17)

holds at x, where  and  are certain real numbers and  6= 0.

Proposition 3.1. Let M be a hypersurface in Ens +1 , n 3, fullling (3). If at a point x 2 M we have rank(H ) > 1 and , dened by (17), vanishes then (4)

holds at x.

i.e.

Proof. Since  = 0, (17) yields rank(D)  1. We suppose that rank(D) = 1, Hij3 ; tr(H )Hij2 = lilj

2 R ; f 0g (18) where li are the local components of a non-zero covector l at x. Now (16) turns into

lhlk Hij + li lj Hhk ; lhlj Hik ; li lk Hhj = 0: (19) We prove now that from (19) it follows that the tensor H has at the point x the decomposition of the form

H = v  v + w  w   2 R

(20)

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KADRI ARSLAN, RYSZARD DESZCZ AND RIDVAN EZENTAS.

where v and w are certain covectors at x. Let V j be the local components of the vector V at x such that = V s ls 6= 0. Transvecting (19) with V h V k we obtain

2Hij + V  V sHrslilj ; ki lj ; li kj = 0 kj = V sHsj whence

r s Hij = ~li lj + 1 (li kj + ki lj ) ~ = ; V V 2Hrs :

(21)

I. Let ~ = 0. We put vi = 12 (li + ki ) and wi = 21 (li ; ki ), respectively. Now (21) takes the form Hij = 2 (vivj ; wiwj ): (22) II. Let ~ 6= 0. Then we have
Hij = ~ li lj + 1 ~ ki lj + 1 ~ likj
= ~ li lj + 1 ~ ki lj + 1 ~ li kj + 21 ~2 ki kj ; 21 ~2 ki kj

= ~ li + 1 ~ ki lj + 1 ~ kj ; 21 ~2 ki kj : (23) So, from (22) and (23) it follows that the tensor H has at the point x the decomposition of the form (20). Now, applying to (20) Lemma 1.1 of 11], we obtain H 3 ; tr(H )H 2 = H 2 R: Comparing this with (18) we get Hij = li lj . Because rank(H ) > 1, the last relation implies = = 0 or = 0 and l = 0, a contradiction. Thus we have D = 0, i.e. (4) holds at x. This completes the proof of the Proposition.

Proposition 3.2. Let M be a hypersurface in Ens +1 , n 3, fullling (3). If at a point x 2 M the condition: rank(H ) > 1 is satised then

1 tr(D)D = ;  H 2 + 1 tr(H )  H 2  2  holds at x where  and  are dened by (24).

(24)

Proof. From (17) we have

Dij Dhk ; Dhj Dik =  (Hij Hhk ; Hhj Hik ):

(25)

ON A CERTAIN CLASS OF HYPERSURFACES IN SEMI-EUCLIDEAN SPACES

229

Transvecting this with Hli we obtain

Blj Dhk ; Blk Dhj =  (Hlj2 Hhk ; Hlk2 Hhj )

(26)

where Blj = Hli Dij . From (26), by symmetrization in l k, we obtain

2H ) Blj Dhk + Bhj Dlk ; Blk Dhj ; Bhk Dlj =  (Hlj2 Hhk + Hhj2 Hlk ; Hlk2 Hhj ; Hhk lj

whence

Q(D B ) =  Q(H H 2 ):

From (16), by contraction with ghk , we get B = 12 tr(D)H + 12 tr(H )D which implies Q(D B ) = Q(D 12 tr(D)H ) = ; 12 tr(H )Q(H D): Appying this to (27) we obtain ; 21 tr(D)Q(H D) =  Q(H H 2 ) whence Q(  H 2 + 21 tr(D) D H ) = 0:

(27) (28)

But this, in view of Lemma 3.4 of 9], gives  H 2 + 1 tr(D)D = H 2 R: (29)  2 Now (26) takes the form Blj Dhk ; Blk Dhj = (Hlj Hhk ; Hlk Hhj ) ; 21 tr(D)(Dlj Hhk ; Dlk Hhj ): (30) Applying in this (28) we get 1 tr(D)(H D + H D ; H D ; H D ) lj hk hk lj lk hj hj lk 2 + 12 tr(H )(Dlk Dhk ; Dlk Dhj ) = (Hlj Hhk ; Hlk Hhj )

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KADRI ARSLAN, RYSZARD DESZCZ AND RIDVAN EZENTAS.

which, by (16), reduces to 1 tr(H )(D D ; D D ) = (H H ; H H ): lj hk lk hj lj hk lk hj 2 This, by making use of (25), turns into ( 12 tr(H )  ; )(Hlj Hhk ; Hlk Hhj ) = 0 whence = 21 tr(H )  . Now (29) reduces to (24), which completes the proof.

Proposition 3.3. Let M be a hypersurface in Ens +1 , n 3, fullling (3). If at a point x 2 M we have rank(H ) > 1 then , dened by (17), must vanishes

at x.

Proof. I. We assume, in addition, that tr(D) = 0 is ful lled at x. Now (24)

reduces to

(H 2 ; 21 tr(H )H ) = 0:

(31)

D = " 14 tr(H )H:

(32)

Applying this in (15) we obtain

Since tr(D) = 0, (32) implies tr(H ) = 0. If  6= 0 then tr(H ) = 0 and (31) reduces to H 2 = 0 whence D = 0 and, by (25), rank(H )  1, a contradiction. II. We assume now that tr(D) 6= 0 holds at x. Applying (24) to (16) we obtain

;  (Hij Hhk2 + Hhk Hij2 ; Hik Hhj2 ; Hhj Hik2 ) + tr(H )  (Hij Hhk ; Hik Hhj ) = 0:

We suppose that  6= 0. Now the last equality turns into

2 + H H 2 ; H H 2 ; H H 2 = tr (H )(H H ; H H ): Hij Hhk ij hk hk ij ik hj hj ik ik hj

(33)

Contracting this with gij we obtain

H 3 = tr(H )H 2 + 12 (tr(H 2) ; (tr(H ))2 )H:

(34)

D = ; 12 "(tr(H 2 ) ; (tr(H ))2 H:

(35)

Further, (16) and (34) yields

ON A CERTAIN CLASS OF HYPERSURFACES IN SEMI-EUCLIDEAN SPACES

231

Substituting this in (16) we obtain (tr(H 2 ) ; (tr(H ))2 )(Hij Hhk ; Hik Hhj ) = 0 whence tr(H 2 ) ; (tr(H ))2 = 0. Thus (35) reduces to D = 0 and, by (25), rank(H )  1, a contradiction. Hence the above discussion shows that  = 0. This completes the proof of the Proposition. From Propositions 3.1 and 3.3 if follows immediately

Proposition 3.4. Let M be a hypersurface in Ens +1 , n 3, fullling (3). If at a point x 2 M the condition rank(H ) > 1 is satised then (4) holds at x. Finally, using Proposition 3.4 and Corollary 2.1 we can prove the follwing: Theorem 3.1. Let M be a hypersurface in a semi-Euclidean space Ens +1 , n 4. Then (3) holds on M if and only if the second fundamental tensor H of M satises (4) at every point of M . in 5] (Lemma 2.3) it was shown that the identity

R
S = Q(H tr(H )H 2 ; H 3 )

(36)

holds on every hypersurface M in a semi-Euclidean space Ens +1 , n this and Theorem 1.4 we obtain the following:

Corollary 3.1. Every hypersurface M in Ens +1 , n

3. Using

3, fullling (3) is a

Ricci-Semisymmetric manifold. The converse of the statement is not true. E.g. the hypersurface M = p S En;p in En+1, p 2, n ; p 2, is a semisymmetric (and hence Riccisemisymmetric) hypersurface does not satisfy (4) (14], 17]). From (36), in view of Lemma 3.4 of 9], it follows that a hypersurface M in n +1 Es , n 3, is a Ricci-semisymmetric manifold if and only if at every point of M , at which the tensor H is non-zero, the following relation is satis ed

H 3 ; tr(H )H 2 = H 2 R: Thus we have

(37)

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KADRI ARSLAN, RYSZARD DESZCZ AND RIDVAN EZENTAS.

Corollary 3.2. A Ricci-semisymmetric hypersurface M in Ens +1 , n

4, satises (3) if and only if at every point of M , at which H is non-zero, the number , dened by (37), vanishes. Let x be a point of a hypersurface M in Ens +1 , n 3, such that its second fundamental tensor H has at x the following form

H = v  v + w  w v w 2 Tx (M )   2 R ; f0g:

(38)

Lemma 1.1 and Theorem 3.1 of 11] state that the following relations are ful lled at x:

Q(H 2 H ^~ H ) = 0 H 3 ; tr(H )H 2 = H =  ((g(V W ))2 ; g(V V )g(W W )) H 2 = 2 g(V V )v  v +  2 g(W W )w  w + g(V W )(v  w + w  v) R
R=0 where the vectors V W 2 Tx (M ) are related to the covectors v w by v(X ) = g(V X ) and w(X ) = g(W X ), respectively, and X 2 Tx(M ). Let M be the hypersurface in E5s satisfying at every point x 2 M the following two conditions: rankH = 2 and H 2 = 0. Applying these assumptions to the above relations we state that = 0, i.e. (4), and g(V W ) = g(V V ) = g(W W ) = 0 hold at x. Finally we prove the following: Theorem 3.2. Every hypersurface M in E5s satisfying (4) is semisymmetric.

Proof. It is clear, by making use of (8)(b), that our assertion is true at all umbilical points of M . Let now x 2 M be a non-umbilic point. As it was shown

in 6], the relations 1 tr(H )H ^~ H + G ; H ^~ H 2 ; g^~ (H 3 ; tr(H )H 2 ; H ) = 0 2
 = 13 tr(H )( 2"  ; tr(H 2)) + tr(H 3)  = 31 (tr(H 2 ) ; (tr(H ))2 ) ; 6"  (39)

hold at x. Further, from (39) we obtain (see 6], Lemma 3.1)

H 4 = tr(H )H 3 + H 2 + H + g  2 R

(40)

ON A CERTAIN CLASS OF HYPERSURFACES IN SEMI-EUCLIDEAN SPACES

and  and  must satis ed the equations
 = 31 tr(H 3) ; tr(H )(tr(H 2)) ;  respectively. Now (40), by (4), reduces to

233

 = 21 ((tr(H ))2 ; tr(H 2 ))

(41)

H 2 + H + g = 0:

(42)

Note that if  = 0 then (42) implies  =  = 0. In that case (39) turns into 1 tr(H )H ^~ H = H ^~ H 2 (43) 2 whence Q(H H ^~ H 2 ) = 0, which, in virtue of Lemma 3.1 of 2], yields Q(H 2 H ^~ H ) = 0. But this reduces (8) to R
R = 0. If  6= 0 then (42) gives H 3 + H 2 + H = 0, whence, by (4), we get ( + tr(H ))H 2 = ;H:

(44)

If  + tr(H ) 6= 0 then (44) yields H 2 = ; +tr (H ) H , which reduces (8) to R
R = 0. If  + tr(H ) = 0 then (44) yields  = 0. Now (42) turns into H 2 = ;  H , which reduces (8) again to R
R = 0. Our theorem is thus proved.

Acknowledgment The second author would like to express his thanks to NATO and TU BITAK for their supports.

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R

Department of Mathematics, Uludag University, Gorukle Kampusu, 16059 Bursa, TURKEY. 2 Department of Mathematics, Agricultural University of Wroclaw, ul. Grunwaldzka 53, PL-50357 Wroclaw, POLAND. 13