MATEMATIQKI VESNIK

59 (2007), 143–150

UDK 514.763

originalni nauqni rad research paper

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

Miroslava Petrović-Torgašev and 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 the curvature conditions of the form E · S = 0 on such submanifolds, where E is any of the standard 4-covariant curvature operators, S is the Ricci curvature operator, and E acts on S as a derivation.

1. Introduction

1. Let M^{n} be an n-dimensional submanifold of a Euclidean space E^{m} of dimension m = n + p (p ≥ 1, n ≥ 2). Let g be the Riemannian metric induced on M^{n} from the standard metric on E^{m}, ∇ the corresponding Levi Civita connection on M^{n}, and R, S and τ respectively the Riemann-Christoffel curvature tensor, the Ricci tensor and the scalar curvature of M^{n}. We use the sign convention given by R(X, Y) = [∇_{X}, ∇_{Y}] − ∇_{[X,Y]} and the normalization of the scalar curvature given by τ = ∑_{i,j=1}^{n} K(e_{i} ∧e_{j}), where K denotes the sectional curvature and e_{i} ∧e_{j} is the plane section of T M^{n} spanned by the vectors e_{i} and e_{j} for i ≠ j of an orthonormal tangent frame field e_{1}, …, e_{n} on M^{n}.

Consider the real function inf K on M^{n} defined for every x ∈ M by

(inf K)(x) := inf{ K(π) : π is a 2-plane in T_{x}(M^{n}) }.

Since the set of 2-planes at a certain point is compact, this infimum is actually a minimum. B. Y. Chen proved in [5] the following basic inequality between the intrinsic scalar invariants inf K and τ of M^{n}, and the extrinsic scalar invariant |H|, being the length of the mean curvature vector field H of M^{n} in E^{m}.

Theorem A. ([5]). Let M^{n} (n ≥ 2), be any submanifold of E^{m} (m = n + p, p ≥ 1). Then

inf K ≥ \frac{1}{2} \{τ − \frac{n^{2}(n − 2)}{n − 1} |H|^{2}\} .  (1)

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 the Republic of Serbia, grant No.

144030.

Equality holds in (1) at a point x if and only if with respect to suitable local orthonormal frames e_{1}, …, e_{n} ∈ T_{x}M^{n} and e_{n+1}, …, e_{n+p} ∈ T_{x}^{⊥}M^{n}, the Weingarten maps A_{t} with respect to the normal sections ξ_{t} = e_{n+t} (t = 1, …, p) are given by

A_{1} = ( \begin{matrix}a & 0 & 0 & … & 0 \\ 0 & b & 0 & … & 0 \\ 0 & 0 & μ & … & 0 \\ ⋮ & ⋮ & ⋮ & & ⋮ \\ 0 & 0 & 0 & … & μ \end{matrix} ) , A_{t} = ( \begin{matrix}c_{t} & d_{t} & 0 & … & 0 \\ d_{t} & −c_{t} & 0 & … & 0 \\ 0 & 0 & 0 & … & 0 \\ ⋮ & ⋮ & ⋮ & & ⋮ \\ 0 & 0 & 0 & … & 0 \end{matrix} ) (t > 1),

where μ = a + b. For any such frame, inf K(x) is attained by the plane e_{1} ∧ e_{2}.

The purpose of the present paper is to study submanifolds M^{n} of E^{m} for which the basic inequality (1) at all points is actually an equality. Such submanifolds are called ideal submanifolds.

If S is the Ricci operator of a manifold M^{n} in a Euclidean space, the Ricci curvatures Ric_{i} = S_{ii} (i = 1, …, n) of M^{n} are given by

Ric_{i} = Ric(e_{i}) = ∑_{j=1}^{n} K_{ij},

where K_{ij} = K(e_{i} ∧ e_{j}) (i, j = 1, …, n) are the corresponding sectional curvatures. If, in particular, M^{n} is satisfying the basic equality in (1), then we have

K_{12} = ab − σK_{1j} = a μ, K_{2j} = b μ, K_{ij} = μ^{2},

for i, j > 2, where σ = ∑_{t=2}^{p}(c_{t}^{2} + d_{t}^{2}). In this case the Ricci curvatures Ric_{i} of M^{n} (i = 1, …, n) are given by

Ric_{1} = (n − 2) a μ + K_{12}, Ric_{2} = (n − 2) b μ + K_{12},
Ric_{3} = ⋯ = Ric_{n} = (n − 2) μ^{2},

and we also have S(e_{i}, e_{j}) = 0 if i ≠ j. The scalar curvature τ = 2ab − 2σ + (n − 1)(n − 2) μ^{2}. In the sequel, we shall also denote Ric_{1} = α, Ric_{2} = β, Ric_{3} = γ.

2. Next, we recall several curvature operators which we shall use in the sequel. The concircular curvature operator Z(X, Y) is defined for n ≥ 2 by

Z(X, Y) = R(X, Y) − \frac{τ}{n(n − 1)} B(X, Y),

where τ = τ(x) is the scalar curvature of M^{n}, and the operator B(X, Y)U = (X ∧ Y)U = g(U, Y)X − g(U, X)Y (X, Y, U ∈ T_{x}(M^{n})). Note that, in components, B_{ijkl} = g_{il}g_{jk} − g_{ik}g_{jl}.

The projective curvature operator P (X, Y) is defined for n ≥ 2 by

P (X, Y)U = R(X, Y)U − \frac{1}{n − 1}B(X, Y)(SU).

Some curvature conditions of the type 4 × 2 145

The Weyl’s conformal curvature operator C is defined for n ≥ 3 by

C(X, Y) = R(X, Y) − \frac{1}{n − 2}\{SX ∧ Y + X ∧ SY \} + \frac{τ}{(n − 1)(n − 2)}B(X, Y).

As is well known, every submanifold of dimension n = 3 is conformally flat, that is C = 0 identically holds on M^{3}.

The conharmonic curvature operator K(X, Y) is defined for n ≥ 3 by

K(X, Y) = R(X, Y) − \frac{1}{n − 2}\{SX ∧ Y + X ∧ SY \}.

The Einstein curvature operator is defined for n ≥ 2 by G = S − \frac{τ}{n} I, where I is the identity operator on T_{x}(M^{n}).

As is well-known, every manifold M^{2} is of constant sectional curvature (that is, its curvature tensors Z and P vanish), and a manifold M^{n} (n ≥ 3) is conharmonically flat if and only if it is conformally flat, and its scalar curvature τ identically vanishes.

3. Now, assume that M^{n} is a submanifold of the Euclidean space E^{m} (m = n+p, p ≥ 1, n ≥ 2) satisfying Chen’s basic equality. If we denote R_{ijk} = R(e_{i}, e_{j})e_{k} (i, j, k = 1, …, n), then a straightforward calculation of the curvature operator R(X, Y)U gives that, in the orthonormal basis {e_{1}, …, e_{n}} of the tangent space T_{x}(M^{n}), we have

\{ \begin{matrix}R_{121} = R_{1} e_{2}, R_{122} = −R_{1} e_{1}, \\ R_{1p1} = R_{2} e_{p} (p ≥ 3), R_{1pp} = −R_{2} e_{1} (p ≥ 3), \\ R_{2p2} = R_{3} e_{p} (p ≥ 3), R_{2pp} = −R_{3} e_{2} (p ≥ 3), \\ R_{pqp} = R_{0} e_{q} (p, q ≥ 3, p ≠ q), \end{matrix}  (2)

where R_{1} = −K_{12} = σ − ab = ∑_{t=2}^{p}(c_{t}^{2} + d_{t}^{2}) − ab, R_{2} = −K_{1j} = −a μ, R_{3} = −K_{2j} = −b μ and R_{0} = −K_{ij} = −μ^{2} (i, j > 2). Besides, R_{ijk} = 0 if all i, j, k are mutually distinct.

Moreover, if E = E_{4} is any of the curvature operators R, Z, B, C, K, we get the similar equations for the values E_{ijk} = E(e_{i}, e_{j})e_{k} (i, j, k = 1, …, n), with the corresponding functions E_{i} (i = 0, 1, 2, 3). We only give in short the exact values for the functions E_{i} (i = 0, 1, 2, 3) in each of these cases.

(B): B_{1} = B_{2} = B_{3} = B_{0} = −1 at any point x ∈ M^{n}.

(Z): \{ \begin{matrix}Z_{1} = R_{1} + \frac{τ}{n(n−1)}, Z_{2} = −a μ + \frac{τ}{n(n − 1)}, \\ Z_{3} = −b μ + \frac{τ}{n(n − 1)}, Z_{0} = −μ^{2} + \frac{τ}{n(n − 1)}. \end{matrix}

(K): K_{1} = \frac{n − 4}{n − 2} R_{1} + μ^{2}, K_{2} = K_{3} = −\frac{R_{1}}{n − 2} + μ^{2}, K_{0} = μ^{2}.

(C): \{ \begin{matrix}C_{1} = \frac{n − 3}{n − 1} R_{1}, C_{2} = C_{3} = −\frac{n − 3}{(n − 1)(n − 2)} R_{1}, \\ C_{0} = \frac{2 R_{1}}{(n − 1)(n − 2)}. \end{matrix}

We note that each of the functions R_{0}, B_{0}, Z_{0}, K_{0}, C_{0} exists only if n ≥ 4.

We also note that the corresponding equations for the projective curvature operator P differ of the previous because the corresponding curvature tensor is not antisymmetric in the last two indices. In fact, denoting P_{ijk} = P (e_{i}, e_{j})e_{k} (i, j, k = 1, …, n), we find the following equations:

\{ \begin{matrix}P_{121} = P_{1} e_{2}, P_{122} = P̃_{1} e_{1}, P_{1p1} = P_{2} e_{p} (p ≥ 3), \\ P_{1pp} = P̃_{2} e_{1} (p ≥ 3), P_{2p2} = P_{3} e_{p} (p ≥ 3), \\ P_{2pp} = P̃_{3} e_{2} (p ≥ 3), P_{pqp} = P_{0} e_{q} (p, q ≥ 3, p ≠ q), \end{matrix}  (3)

where

\{ \begin{matrix}P_{1} = \frac{n−2}{n−1} (R_{1} + a μ), P̃_{1} = −\frac{n−2}{n−1} (R_{1} + b μ), \\ P_{2} = −\frac{R_{1}+a μ}{n−1}, P̃_{2} = \frac{μ}{n−1}[a − (n − 2)b], \\ P_{3} = −\frac{R_{1}+b μ}{n−1}, P̃_{3} = \frac{μ}{n−1} [b − (n − 2)a], \\ P_{0} = −\frac{μ^{2}}{n−1}. \end{matrix}

Besides, P_{ijk} = 0 if all i, j, k are mutually distinct, and P_{0} exists only if n ≥ 4.

Finally, denoting any of the tensors S, G by E, we find that the Ricci operator S and the Einstein tensor G act as follows:

Ee_{1} = α e_{1}, Ee_{2} = β e_{2}, Ee_{p} = γ e_{p} (p ≥ 3),

where, if E = S: α = (n − 2) a μ − R_{1}, β = (n − 2) b μ − R_{1}, γ = (n − 2) μ^{2}, and, if E = G: α_{0} = (n−2) a μ−R_{1}−τ/n, β_{0} = (n−2) b μ−R_{1}−τ/n, γ_{0} = (n−2) μ^{2}−τ/n.

In [15] it was proved that a submanifold M^{n} (n ≥ 3) satisfying Chen’s equality is flat if and only if it is totally geodesic. The last condition means that a = b = σ = 0, thus that a = b = c_{t} = d_{t} = 0 (t = 2, …, p). There, it was also proven that M^{n} is Einstein if and only if n = 2, or it is flat.

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 1. 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.

If n ≥ 2, then M^{n} is of constant curvature if and only if n = 2, or 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}.

2. Main results

Throughout this section we suppose that M^{n} is a submanifold in the Euclidean space E^{m} (m = n + p, p ≥ 1, n ≥ 2) satisfying the basic Chen’s equality, and we investigate on such a submanifold several curvature conditions of the form E·F_{2} = 0, where E is any of the curvature operators R, Z, P, K, C, F_{2} is any of the operators S, G, and the operation E · F_{2} is defined as

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

for all tangent vectors X, Y, U ∈ T_{x}(M^{n}).

Similar curvature conditions of the form E_{4}·F_{2} = 0, and of the form E_{4}·F_{4} = 0 have been investigated in many papers (see e.g. [1–3],[7–10], [14–15], [18–20], [22– 33], etc.).

Since obviously E · G = E · S, we shall consider only the case F_{2} = S, thus we shall discuss exactly the operations R · S, Z · S, P · S, C · S and K · S.

In the most simple case n = 2, we get that S = \frac{τ}{2} I (moreover, Z = P = 0), so that R · S = Z · S = P · S = 0. Hence, this case is trivial and we can suppose that n ≥ 3.

If n ≥ 3 and E is any of the curvature operators R, Z, K, C, then obviously E·S = 0 if and only if (E·S)_{ijk} = (E(e_{i}, e_{j})·S)e_{k} = 0 for all indices i, j, k = 1, …, n. It is also not difficult to see that the above condition is satisfied if and only if the following three equations hold:

(E · S)_{121} = (E · S)_{131} = (E · S)_{232} = 0.

Hence, E · S = 0 if and only if the next system of equations is satisfied:

(α − β) E_{1} = 0, (α − γ) E_{2} = 0, (β − γ) E_{3} = 0.  (4)

By an easy discussion of the corresponding system in any of the cases E = R, Z, K, C we get the following theorems.

Theorem 1. If n ≥ 3, then R · S = 0 if and only if one of the following three cases occurs: (1^{0}) μ = 0; (2^{0}) σ = a = 0, b ≠ 0; (3^{0}) σ = b = 0, a ≠ 0.

Proof. By direct calculations, it is easy to check that R · S = 0 in any of the cases (1^{0}), (2^{0}), (3^{0}).

Next, assume that R · S = 0 and μ ≠ 0. Supposing that a, b ≠ 0, by equations (R · S)_{131} = 0 and (R · S)_{232} = 0, we get

R_{1} = −(n − 2) a μ = −(n − 2) b μ,

and consequently a = b, σ = −(2n − 5) a^{2}. Therefore σ = a = 0, a contradiction. Hence, ab = 0. If a = 0, then the third equation gives b σ = μ σ = 0, thus σ = 0 (because μ ≠ 0), so we have the case (2^{0}). If b = 0, we similarly have the case (3^{0}).

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

Proof. Note that equation Z_{1} = 0 in the developed form reads:

(n + 1)σ = −(n − 1) a^{2} − (n − 1)b^{2} − (n − 3) ab.

Since n ≥ 3, the previous equality is possible only if σ = a = b = 0, which means that M^{n} is totally geodesic n-plane in E^{m}.

Theorem 2. If n ≥ 3, then Z · S = 0 if and only if one of the following cases occurs: (1^{0}) M^{n} is a totally geodesic plane; (2^{0}) n ≥ 4, a = b and σ = (n^{2} − 5n + 5) a^{2}.

Proof. If n ≥ 4, σ = (n^{2} − 5n + 5) a^{2}, a = b, then Z_{2} = Z_{3} = 0, α = β, and immediately Z · S = 0.

Conversely, assume that n ≥ 3, Z · S = 0, and M^{n} is not totally geodesic plane. Then by Lemma 1, Z_{1} ≠ 0. By equation (Z · S)_{121} = 0, we get a = ±b. Supposing that μ = 0, we have τ = −2 R_{1} = −2(σ + a^{2}), Z_{1} = \frac{(n+1)(n−2)}{n(n−1)} R_{1}, and by (Z · S)_{131} = 0 we get a contradiction R_{1} = σ + a^{2} = 0, M^{n} is a totally geodesic plane. If a = b, then

\begin{matrix}Z_{2} = −\frac{2}{n(n − 1)} \{σ − (n^{2} − 5n + 5) a^{2}\}, \\ R_{1} + (n − 2) b μ = σ + (2n − 5) a^{2} > 0, \end{matrix}

and by equation (Z · S)_{131} = 0, we obtain Z_{2} = 0, i.e. σ = (n^{2} − 5n + 5) a^{2}. Since M^{n} is not totally geodesic, the case n = 3 is excluded, so that n ≥ 4, and we have the case (2^{0}).

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

Proof. If K_{1} = K_{2} = 0, then easily R_{1} = (n − 2) μ^{2} and (n − 3) μ^{2} = 0. If μ = 0, then R_{1} = 0, and we obtain that M^{n} is totally geodesic plane. If n = 3, then R_{1} = μ^{2}, and we again get that M^{n} is conharmonically flat.

Theorem 3. If n ≥ 3, then K · S = 0 if and only if one of the following cases occurs: (1^{0}) M^{n} is conharmonically flat; (2^{0}) a = b ≠ 0, σ = (4n − 7) a^{2}.

Proof. If a = b and σ = (4n − 7) a^{2}, then K_{2} = 0 and α = β, so that the condition K · S = 0 is obviously satisfied.

Conversely, assume that n ≥ 3 and K · S = 0. If K_{2} ≠ 0, then by equations (K · S)_{131} = 0 and (K · S)_{232} = 0 we obtain

R_{1} = −(n − 2) a μ = −(n − 2) b μ,

and hence a = ±b. If a = b, then by R_{1} = −2(n − 2) a^{2}, we easily get that M^{n} is totally geodesic, contradicting to K_{2} ≠ 0. If a = −b, μ = 0, then σ = −a^{2}, and M^{n} is totally geodesic, again a contradiction. Hence K_{2} = 0, i.e. R_{1} = (n − 2) μ^{2}. If, in addition, we assume that M^{n} is not conharmonically flat, then by Lemma 2, K_{1} ≠ 0. By equation (K · S)_{121} = 0, we then have a = ±b. If a = b, then

K_{2} = \frac{(4n − 7) a^{2} − σ}{n − 2} = 0,

and hence σ = (4n − 7) a^{2}, so we have the case (2^{0}). If a = −b, μ = 0, then by K_{2} = 0, we get R_{1} = 0, σ = −a^{2}, σ = a = b = 0, contradicting to K_{1} ≠ 0.

Theorem 4. If n ≥ 3, then C · S = 0 if and only if M^{n} is conformally flat.

Proof. Suppose that C · S = 0 and M^{n} is not conformally flat. Then n ≥ 4 and R_{1} ≠ 0. By equations (C · S)_{121} = 0, (C · S)_{131} = 0, we then get α = β = γ, i.e. a = ±b and R_{1} = −(n − 2) b μ. If μ = 0, then we get a contradiction R_{1} = 0. If a = b, then we get σ = −(2n − 5) a^{2}, and hence σ = a = 0, thus again a contradiction R_{1} = 0.

Next, assume that n ≥ 3 and consider the condition P · S = 0. It is also not difficult to see that (P · S)_{ijk} = 0 for any choice of indices i, j, k = 1, …, n if and only if the next equations are satisfied: (P · S)_{121} = (P · S)_{122} = (P · S)_{131} = (P · S)_{133} = (P · S)_{232} = (P · S)_{233} = 0. Hence, the complete system of equations for the operation P · S reads:

\{ \begin{matrix}(α − β) P_{1} = (α − β) P̃_{1} = (α − γ) P_{2} = 0, \\ (α − γ) P̃_{2} = (β − γ) P_{3} = (β − γ) = P̃_{3} = 0. \end{matrix}  (5)

Therefore, it is not difficult to get the following result.

Theorem 5. If n ≥ 3, then P · S = 0 if and only if M^{n} is a totally geodesic plane.

Proof. Assume that n ≥ 3 and P · S = 0. Then, by equations (P · S)_{121} = 0, (P · S)_{122} = 0, we find that a = ±b. If μ = 0, then by equation (P · S)_{131} = 0 we have R_{1} = 0, and M^{n} is a totally geodesic plane. If a = b, then by the same equation we have (σ + a^{2}){σ + (2n − 5) a^{2}} = 0, thus σ = a = b = 0, so M^{n} is again totally geodesic.

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(received 17.04.2007, in revised form 25.07.2007)

Dep. of Math., Faculty of Sci., Radoja Domanovića 12, 34000 Kragujevac, Serbia.

E-mails: mirapt@kg.ac.yu, anahinic@kg.ac.yu