MATEMATIQKI VESNIK

59 (2007), 189–196

UDK 514.763

originalni nauqni rad research paper

SOME CURVATURE CONDITIONS OF THE TYPE 2 × 4 ON THE SUBMANIFOLDS SATISFYING CHEN’S EQUALITY

Ana Hinić

Abstract. Submanifolds of the Euclidean spaces satisfying equality in the basic Chen’s inequality have, as is known, many interesting properties. In this paper, we discuss on such submanifolds the curvature conditions of the form E_{2} · F_{4} = 0, where E_{2} is the Ricci or the Einstein curvature operator, F_{4} is any of the standard curvature operators R, Z, P, K, C, and E_{2} acts on F_{4} as a derivation.

1. Introduction

This paper is a continuation of the paper [21] by M. Petrović-Torgašev and the present author. Hence, in this paper all definitions, denotations and the properties from [21] are accepted. Only some necessary definitions and statements from the mentioned previous paper are briefly repeated.

Let M^{n} be an n-dimensional submanifold of a Euclidean space E^{m} of dimension m = n + p (n ≥ 2, p ≥ 1), which we suppose to be ideal in the sense of Chen (see [5]), thus which satisfies the so called the basic Chen’s equality.

Let R, S, G and τ be respectively the Riemann-Christoffel curvature tensor, the Ricci tensor, the Einstein tensor and the scalar curvature of M^{n}. The corresponding curvature operators in the tangent space T (M^{n}) are denoted by R(X, Y) etc.

Besides, we shall consider the next, standard curvature operators on M^{n}: the concircular curvature operator Z(X, Y) (n ≥ 2), the projective curvature operator P (X, Y) (n ≥ 2), the Weyl conformal curvature operator C(X, Y) (n ≥ 3), the conharmonic curvature operator K(X, Y) (n ≥ 3), each of them acting in the tangent space T_{x}(M^{n}), where x is a fixed or variable point of M^{n}.

In the next simple proposition we collect the conditions for a manifold M^{n} satisfying Chen’s equality to be flat with respect to different curvature operators.

Proposition. ([15]) If n ≥ 2, then M^{n} is flat if and only if n = 2 and σ = ab, or n ≥ 3 and σ = a = b = 0.

AMS Subject Classification: Primary 53B25, Secondary 53C40.

Keywords and phrases: Submanifolds, curvature conditions, basic equality.

This paper is supported by the Ministry of science of Republic of Serbia, Grant No. 144030.

If n ≥ 3, then M^{n} is of constant curvature if and only if it is flat.

If n ≥ 3, then M^{n} is conformally flat if and only if n = 3, or n ≥ 4 and σ = ab.

If n ≥ 3, then M^{n} if conharmonically flat if and only if it is flat, or n = 3 and σ = ab + μ^{2}.

In the sequel, for a manifold M^{n}, we shall consider the curvature conditions of the form E · F = 0, where E = E_{2} is one of the operators S, G, F = F_{4} ∈ M = {R, Z, P, C, K}, and E_{2} · F_{4} is defined as

(E_{2} · F_{4}(X, Y))U = E_{2}(F_{4}(X, Y)U) − F_{4}(E_{2}X, Y)U
− F_{4}(X, E_{2}Y)U − F_{4}(X, Y)(E_{2}U),

where X, Y, U ∈ T_{x}(M^{n}).

Recall that semisymmetric manifolds are defined by relation R·R = 0, meaning that

(R(X, Y) · R)(U, V)W = R(X, Y)(R(U, V)W) − R(R(X, Y)U, V)W − − R(U, R(X, Y)V)W − R(U, V)(R(X, Y)W) = 0,

for all tangent vector fields X, Y, U, V, W on M^{n}.

The semisymmetric manifolds and similar curvature conditions have been investigated in many papers (see for instance, [1–3], [7–10], [14–15], [18–21], [23–34] etc).

2. Main results

Throughout this section, we shall suppose that a submanifold M^{n} in an Euclidean space E^{m} (m = n + p, p ≥ 1, n ≥ 2), satisfies the basic Chen’s equality, and we investigate in such submanifolds several curvature conditions of the form E_{2} · F_{4} = 0, where E_{2} equals S or G, and F_{4} is one of the operators R, Z, P, K, C. In general case the operators S ·F_{4} and G·F_{4} do not coincide, so we have to discuss the corresponding curvature conditions separately.

In the simplest case n = 2, we get that R = \frac{τ}{2} B, Z = P = 0, S = \frac{τ}{2} I, G = 0, so that G · R = G · Z = G · P = S · Z = S · P = 0. But, as is easily seen, S · R = 0 if and only if R_{1} = 0. So, the case n = 2 is mostly trivial and we shall usually suppose that n ≥ 3.

If F is one of the operators R, Z, K, C, and (E_{2} · F)_{ijk} = (E_{2} · F (e_{i}, e_{j}))e_{k} (i, j, k = 1, …, n), then, by linearity, it is easily seen that E_{2} · F = 0 if and only if (E_{2} · F)_{ijk} = 0 for any choice of indices i, j, k = 1, …, n. Moreover, if n ≥ 3, then by a straightforward calculation, one can see that E_{2} · F = 0 if and only if the next system of equations holds: (E_{2} · F)_{121} = (E_{2} · F)_{122} = (E_{2} · F)_{131} = (E_{2} · F)_{133} = (E_{2} · F)_{232} = (E_{2} · F)_{233} = (E_{2} · F)_{343} = 0. So we obtain a system of equations which completely describes the curvature condition E_{2} · F = 0:

α F_{1} = β F_{1} = α F_{2} = β F_{3} = γ F_{2} = γ F_{3} = γ F_{0} = 0.  (1)

Some curvature conditions of the type 2 × 4 191

Note that the last equation exists only for n ≥ 4.

Theorem 1. If n ≥ 2, then S · R = 0 if and only if M^{n} is a totally geodesic plane, or n = 2 and R_{1} = 0.

Proof. If M^{n} is a totally geodesic plane, or n = 2 and R_{1} = 0, then obviously S · R = 0.

Next, suppose that n ≥ 2 and S · R = 0.

If n = 2 then α = −R_{1}, and equation (S · R)_{121} = 0 gives R_{1} = 0.

If n ≥ 4, then equation (S · R)_{343} = 0 gives μ = 0, and consequently α = β = −R_{1}. Then by equation (S · R)_{121} = 0 we obtain R_{1} = 0, σ = ab = −a^{2}, and immediately σ = a = b = 0, thus M^{n} is a totally geodesic plane.

Finally, assume that n = 3. Then α = a μ − R_{1}, β = b μ − R_{1}, γ = μ^{2}, and equations (S · R)_{133} = 0 and (S · R)_{233} = 0 give a μ = b μ = 0, and consequently α = −R_{1}, β = −R_{1}. But then equation (S · R)_{121} = 0 also gives R_{1} = 0. If μ ≠ 0, then a = b = 0, a contradiction. If μ = 0, then a = −b, and σ = ab again gives σ = a = b = 0, thus M^{3} is a totally geodesic plane.

In the proof of Theorem 2, we shall need the next lemma which is proved in [21].

Lemma 1. If n ≥ 3 and Z_{1} = 0, then M^{n} is a totally geodesic plane.

Theorem 2. If n ≥ 3, then S · Z = 0 if and only if M^{n} is a totally geodesic plane.

Proof. If M^{n} (n ≥ 3) is a totally geodesic plane, then obviously S · Z = 0. Next, assume that n ≥ 3, S · Z = 0 and M^{n} is not totally geodesic. Then by

Lemma 1, Z_{1} ≠ 0, and by equations (S · Z)_{121} = 0 and (S · Z)_{122} = 0 we obtain α = β = 0, so that a = ±b, σ = ab + (n − 2) aμ. If now μ = 0, then σ = −a^{2}, and immediately M^{n} is a totally geodesic plane, a contradiction. If a = b, then σ = (2n − 3) a^{2}, and by equation (S · Z)_{133} = 0 we obtain γ Z_{2} = 0. But since then

γ = 4(n − 2) a^{2}, Z_{2} = \frac{2(n^{2} − 7n + 8)}{n(n − 1)} a^{2},

we find that a = 0, thus σ = a = b = 0, and M^{n} is totally geodesic, again a contradiction.

Theorem 3. If n ≥ 3, then the following conditions are equivalent: (1^{∘}) S · C = 0; (2^{∘}) G · C = 0; (3^{∘}) M^{n} is conformally flat.

Proof. The implications (3^{∘}) ⟹ (1^{∘}) and (3^{∘}) ⟹ (2^{∘}) are obvious.

If n = 3, then M^{3} is conformally flat, so in the sequel we can assume that n ≥ 4.

(1^{∘}) ⟹ (3^{∘}). Assume that S · C = 0 and M^{n} is not conformally flat, thus C_{1} ≠ 0 and C_{2} ≠ 0. Then by equations (S · C)_{121} = 0, (S · C)_{122} = 0 and (S · C)_{232} = 0, we get α = β = γ = 0, so M^{n} is Ricci flat. But then it must be totally geodesic, which is a contradiction.

(2^{∘}) ⟹ (3^{∘}). Assume that n ≥ 4, G · C = 0 and M^{n} is not conformally flat, thus R_{1} ≠ 0. Then equations (G · C)_{121} = 0, (G · C)_{122} = 0 and (G · C)_{343} = 0 immediately give α_{0} = β_{0} = γ_{0} = 0. Hence, M^{n} is Einstein, which yields that M^{n} is a totally geodesic plane, a contradiction.

In the proof of Theorem 4, we shall need another lemma which is also proved in [21].

Lemma 2. If n ≥ 3 and K_{1} = K_{2} = 0, then M^{n} is conharmonically flat.

Theorem 4. If n ≥ 3, then the following conditions are equivalent: (1^{∘}) S · K = 0; (2^{∘}) G · K = 0; (3^{∘}) M^{n} is conharmonically flat.

Proof. The condition (3^{∘}) obviously implies (1^{∘}) and (2^{∘}).

(1^{∘}) ⟹ (3^{∘}). Assume that n ≥ 3 and S · K = 0.

If n ≥ 4, then by equation (S·K)_{343} = 0 we get μ = 0, hence K_{2} = −R_{1}/(n−2), α = β = −R_{1}, while equation (S · K)_{131} = 0 gives R_{1} = 0, so that M^{n} is a totally geodesic plane.

Let n = 3. Then

K_{1} = K_{2} = −R_{1} + μ^{2}, α = a μ − R_{1}, β = b μ − R_{1}, γ = μ^{2}.

If K_{1} ≠ 0, then by the first, second and the fifth equation we get α = β = γ, and hence μ = 0, R_{1} = 0. But this means that M^{3} is totally geodesic. If K_{1} = 0, then by Lemma 2, M^{3} is conharmonically flat.

(2^{∘}) ⟹ (3^{∘}). Suppose that G · K = 0. If K_{2} ≠ 0, then by equations (G · K)_{131} = 0, (G · K)_{232} = 0, (G · K)_{133} = 0, we get α_{0} = β_{0} = γ_{0} = 0, so that M^{n} is Einstein. But then it must be totally geodesic, which contradicts to K_{2} ≠ 0. Hence, K_{2} = 0, i.e. R_{1} = (n − 2) μ^{2}. If in addition, K_{1} = 0, then M^{n} is conharmonically flat by Lemma 2. Assuming that K_{1} ≠ 0, by equations (G · K)_{121} = 0, (G · K)_{122} = 0, we find α_{0} = β_{0} = 0, thus

(n − 2) a μ − R_{1} = (n − 2) b μ − R_{1} = \frac{τ}{n}.  (2)

From the first equation in (2) we get a = ±b. If μ = 0, then R_{1} = 0, and M^{n} is totally geodesic, which contradicts to K_{1} ≠ 0. Hence μ ≠ 0, which implies a = b. Then R_{1} = −2(n−2) a^{2}, and equation (2) easily get μ = 0, which is a contradiction again.

This completes the proof.

Theorem 5. If n ≥ 3, then following conditions are equivalent: (1^{∘}) G·R = 0; (2^{∘}) G · Z = 0; (3^{∘}) M^{n} is a totally geodesic plane.

Proof. The condition (3^{∘}) obviously implies (1^{∘}) and (2^{∘}).

(1^{∘}) ⟹ (3^{∘}). Assume that n ≥ 3 and G · R = 0.

If μ = 0, then R_{1} = σ + a^{2}, τ = −2R_{1}, α_{0} = −\frac{n − 2}{n}R_{1}, so that equation (G · R)_{121} = 0 immediately gives σ = a = 0. Then b = 0 too, so M^{n} is a totally geodesic plane. Next, assume that μ ≠ 0, and M^{n} is not totally geodesic. Note that γ_{0} can be written as

γ_{0} = \frac{2σ + (n − 2) a^{2} + (n − 2) b^{2} + 2(n − 3) ab}{n}.

Since σ ≥ 0, and the expression (n − 2) a^{2} + (n − 2) b^{2} + 2(n − 3) ab ≥ 0 (n ≥ 3), and equals zero if and only if a = b = 0, we have that γ_{0} > 0 since M^{n} is not totally geodesic. Now, using the equations (G · R)_{133} = 0, (G · R)_{233} = 0, we find a μ = b μ = 0, hence by adding (a + b) μ = μ^{2} = 0, thus μ = 0, which is a contradiction.

(2^{∘}) ⟹ (3^{∘}). Now assume that n ≥ 3, G · Z = 0, and M^{n} is not totally geodesic. Then γ_{0} > 0, and using the equations (G · Z)_{133} = 0, (G · Z)_{233} = 0, we find that Z_{2} = Z_{3} = 0. Hence, a = ±b. If μ = 0, then τ = −2R_{1} = −2(σ + a^{2}),

α_{0} = −\frac{n − 2}{n} R_{1}, Z_{1} = \frac{(n − 2)(n + 1)}{n(n − 1)} R_{1},

and by equation (G · Z)_{121} = 0 we get R_{1} = 0, thus Z_{1} = 0. But this by Lemma 1 contradicts to assumption that M^{n} is not totally geodesic. If a = b ≠ 0, then

α_{0} = −(n − 2)(n − 3) a^{2}, σ = (n^{2} − 5n + 5) a^{2},
R_{1} = (n − 1)(n − 4) a^{2}, τ = 2n(n − 1) a^{2},
Z_{1} = (n − 2)(n − 3) a^{2}.

But then equation (G · Z)_{121} = 0 gives α_{0}Z_{1} = 0, thus (n − 3) a = 0. Since a ≠ 0, we then get n = 3, Z_{1} = 0, and Lemma 1 gives that M^{3} is totally geodesic, again a contradiction.

This completes the proof.

Further, we shall consider conditions of the form E_{2} · P = 0, where E_{2} equals S or G. If n = 2 then P = 0 so that S · P = G · P = 0.

If n ≥ 3, then by a straightforward calculation, it can be seen that E_{2} · P = 0, thus (E_{2} · P)_{ijk} = (E_{2} · P (e_{i}, e_{j}))e_{k} = 0 for any choice of indices i, j, k = 1, …, n if and only if the following conditions hold: (E_{2} · P)_{121} = (E_{2} · P)_{122} = (E_{2} · P)_{131} = (E_{2} · P)_{133} = (E_{2} · P)_{232} = (E_{2} · P)_{233} = (E_{2} · P)_{343} = 0. Hence, E_{2} · P = 0 if and only if the following system of equations holds:

\{ \begin{matrix}α P_{1} = 0, β P̃_{1} = 0, α P_{2} = 0, γ P̃_{2} = 0, \\ β P_{3} = 0, γ P̃_{3} = 0, γ P_{0} = 0 (n ≥ 4). \end{matrix}  (3)

Discussing this system in the particular cases E_{2} = S and E_{2} = G, we get the next two theorems.

Theorem 6. If n ≥ 3, then S · P = 0 if and only if one of the following cases occurs: (1^{∘}) M^{n} is totally geodesic; (2^{∘}) n = 3, a = b ≠ 0 and σ = 3a^{2}.

Proof. It is easy to prove that S · P = 0 if for instance n = 3, a = b ≠ 0 and σ = 3 a^{2}.

Conversely, assume that n ≥ 3 and S · P = 0. If n ≥ 4, then equation γ P_{0} = 0 gives that μ = 0, so that equation α P_{1} = 0

implies R_{1} = 0, and M^{n} is a totally geodesic plane.

Next, suppose that n = 3. Then the first and the second equation give that R_{1} = ±a μ = ±b μ, and consequently a = ±b. If μ = 0, then R_{1} = 0, and M^{n} is totally geodesic. If a = b, then the first equation implies (σ + a^{2})(σ − 3a^{2}) = 0. If σ = a = 0, then M^{3} is totally geodesic. Otherwise, we have σ = 3a^{2}, so we find the case (2^{∘}).

Theorem 7. If n ≥ 3, then G · P = 0 if and only if M^{n} is a totally geodesic plane.

Proof. Assume that n ≥ 3, G · P = 0 and M^{n} is not totally geodesic. If n ≥ 4, then γ_{0} > 0, and the seventh equation gives μ = 0. Then τ = −2R_{1},

α_{0} = −\frac{n − 2}{n} R_{1}, while the first equation gives R_{1} = 0, which contradicts to M^{n} is not totally geodesic.

Finally, assume that n = 3. From the fourth and the sixth equations, we obviously get a = ±b. If μ = 0, then the first equation gives R_{1} = 0, contradicting to M^{3} is not totally geodesic. If a = b, then α_{0} = −(σ + a^{2})/3, R_{1} + a μ = σ + a^{2}, and the first equation again gives σ = a = 0, thus again the same contradiction.

This ends the proof.

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196 A. Hinić (received 17.04.2007, in revised form 25.07.2007) Dep. of Math., Faculty of Sci., Radoja Domanovića 12, 34000 Kragujevac, Serbia.

E-mail: anahinic@kg.ac.yu