Conformal Kaehler submanifolds

This paper presents two results in the realm of conformal Kaehler submanifolds. These are conformal immersions of Kaehler manifolds into the standard ﬂat Euclidean space. The proofs are obtained by making a rather strong use of several facts and techniques developed in [2] for the study of isometric immersions of Kaehler manifolds into the standard hyperbolic space. Let f : M 2 n → R 2 n + p denote a conformal Kaehler submanifold . Thus ( M 2 n , J ) is a Kaehler manifold of complex dimension n ≥ 2 and f a conformal immersion into Euclidean space that lies in codimension p . Thus, there is a positive function λ ∈ C ∞ ( M ) such that the Kaehler metric and the one induced by f relate by h , i f = λ 2 h , i M 2 n . Conformal Kaehler submanifolds laying in the low codimensions p = 1 and p = 2 have already been considered in [1]. In this paper, we are interested in higher codimensions although not too large in comparison to the dimension of the manifold. Our ﬁrst result, provides a necessary condition for the existence of a conformal immersion in codimension at most n − 3 in terms of the sectional curvature of the Kaehler manifold.

Theorem 1.Let f : M 2n → R 2n+p , p ≤ n − 3, be a conformal Kaehler submanifold.Then at any x ∈ M 2n there is a complex vector subspace V 2m ⊂ T x M with m ≥ n − p such that the sectional curvature of M 2n satisfies K M (S, JS)(x) ≤ 0 for any S ∈ V 2m .
Notice that the conclusion of Theorem 1 remains valid if the Euclidean ambient space R 2n+p is replaced by any locally conformally flat manifold of the same dimension.
Our second result characterizes a submanifold, in terms of the degree of positiveness of the sectional curvature, as being locally the Example 2 presented below.
The light-cone V m+1 ⊂ L m+2 of the standard flat Lorentzian space is any one of the two connected components of the set of all light-like vectors, namely, {v ∈ L m+2 : v, v = 0, v = 0} endowed with the (degenerate) induced metric.The Euclidean space R m can be realized as an umbilic hypersurface of V m+1 as follows: Take vectors v, w ∈ V m+1 such that v, w = 1 and a linear isometry C : R m → {v, w} ⊥ .Let ψ : R m → V m+1 ⊂ L m+2 be defined by Then ψ is an isometric embedding of R m as an umbilical hypersurface in the light cone which is the intersection of V m+1 with an affine hyperplane.Namely, we have that The normal bundle of ψ is N ψ R m = span {ψ, w} and its second fundamental form is Proposition 9.9 in [3] gives an elementary correspondence between the conformal immersions in Euclidean space and the isometric immersions into the light cone which goes as follows: Associated to a given conformal immersion f : M m → R m+p with conformal factor λ ∈ C ∞ (M) there is the associated isometric immersion defined by Conversely, any isometric immersion F : M m → V m+p+1 \ Rw ⊂ L m+p+2 gives rise to an associated conformal immersion f : M m → R m+p given by ψ • f = π • F with conformal factor 1/ F, w .Here π : V m+p+1 \ Rw → R m+p is the projection π(x) = x/ x, w .
Example 2. Let the Kaehler manifold M 2n be the Riemannian product of one hyperbolic plane and a set of two-dimensional round spheres such that be a connected conformal Kaehler submanifold.Assume that at a point x 0 ∈ M 2n there is a complex tangent vector subspace V 2m ⊂ T x 0 M with m ≥ p + 1 such that the sectional curvature of M 2n satisfies K M (S, JS)(x) > 0 for any 0 = S ∈ V 2m .Then p = n − 2 and f (M) is an open subset of the submanifold given by Example 2.

The proofs
Let V 2n and L p , p ≥ 2, be real vector spaces such that there is J ∈ Aut(V ) which satisfies J 2 = −I and L p is endowed with a Lorentzian inner product , .Then let W p,p = L p ⊕ L p be endowed with the inner product of signature (p, p) defined by Let α : V 2n × V 2n → L p be a symmetric bilinear form and β : V 2n × V 2n → W p,p the associated bilinear form given by We denote the vector subspace of W p,p generated by β by and say that β is surjective if S(β) = W p,p .The (right) kernel β is defined by is the linear transformation defined by B X Y = β(X, Y ).The set RE(β) of regular elements of β is easily seen to be an open dense subset of V 2n , for instance see Proposition 4.4 in [3].
It is said that If β is flat and X ∈ RE(β) we have from Proposition 4.6 in [3] that Proposition 4. Let β : V 2n × V 2n → W p,p , p ≤ n, be flat and surjective.Then Proof: This is condition (9) in Proposition 11 of [2].
Let V 2n be endowed with a positive definite inner product ( , ) with respect to which J ∈ Aut(V ) is an isometry.Assume that there is a light-like vector w ∈ L p such that α(X, Y ), w = −(X, Y ) for any X, Y ∈ V 2n . (4) Let U s 0 ⊂ L p be the s-dimensional vector subspace given by where π 1 : W p,p → L p denotes the projection onto the first component of W p,p .From Proposition 9 in [2] we know that In addition, if S(β) is a degenerate vector subspace then 1 ≤ s ≤ p − 1 and there is a light-like vector v ∈ U s 0 such that Proposition 5. Let the bilinear form β : V 2n × V 2n → W p,p be flat and the vector subspace S(β) degenerate.Then L = span {v, w} ⊂ L p is a Lorentzian plane.Moreover, choosing v such that v, w = −1 and setting Proof: We obtain from (6) that From (4) and the fact that J is an isometry with respect to ( , ), we also have that In particular, we have β(X, X), (w, 0) < 0 for any 0 = X ∈ V 2n , which jointly with (9) implies that v and w are linearly independent and thus span a Lorentzian plane.Since w is light-like and v satisfies v, w = −1, we have a positive definite induced inner product, we obtain from (15), ( 17) and (18) that α(S, S), α(JS, JS) − α(S, JS) where ψ is given by (1).Differentiating F, F = 0 once gives F ∈ Γ(N F M) and twice yields that the second fundamental form α Let the bilinear forms γ, β : Proposition 7. Let N F M(x) ⊕ N F M(x) be endowed with the inner product defined by (ξ, ξ), (η, η) = ξ, η − ξ, η .
Then the bilinear form β is flat and Proof: The proof is straightforward using that β(X, JY ) = −β(JX, Y ), that the curvature tensor satisfies R(X, Y )JZ = JR(X, Y )Z for any X, Y, Z ∈ T x M and the Gauss equation for f ; for details see the proof of Proposition 16 in [2] Proof of Theorem 1: It suffices to show that the vector subspace S(β) is degenerate since then the proof follows from the Gauss equation jointly with Proposition 6 and Proposition 7. If S(β) is not degenerate, and since we have the result given by Proposition 7, then Proposition 4 yields dim N (β) ≥ 2n − 2p − 4 > 0. But this is a contradiction since from (19) we have that N (β) = 0.
Proposition 8. Let the bilinear form β : V 2n × V 2n → W p,p , s ≤ n, be flat.Assume that the vector subspace S(β) is nondegenerate and that (12) holds.For p ≥ 4 assume further that there is no non-trivial J-invariant vector subspace Then s = n and there is an orthogonal basis {X i , JX i } 1≤i≤n of V 2n such that: (ii) The vectors {β(X j , X j ), β(X j , JX j )} 1≤j≤n form an orthonormal basis of S(β).
Proof of Theorem 3: Theorem 1 gives that p = n − 2. In an open neighborhood U of x 0 in M 2n there is a complex vector subbundle V ⊂ T M such that V (x 0 ) = V 2m and K M (S, JS) > 0 for any 0 = S ∈ V .At any point of U the vector subspace S(β) is nondegenerate.In fact, if otherwise then by Proposition 6 there is a point y ∈ U and a complex vector subspace P 2ℓ ⊂ T y M with ℓ ≥ 2 such that the sectional curvature satisfies K M (S, JS) ≤ 0 for any 0 = S ∈ P 2ℓ , in contradiction with our assumption.By Proposition 8, there is at any y ∈ U an orthogonal basis {X j , JX j } 1≤j≤n of T y M such that both parts hold.By part (ii) the vectors (ξ j , 0) = β(X j , X j ) ∈ N F M(y), 1 ≤ j ≤ n, are orthonormal.Then the argument used for the proof of Lemma 18 in [2] gives that F | U has flat normal bundle, that rank A ξ j =2 for 1 ≤ j ≤ n and that the normal vector fields ξ 1 , . . ., ξ n are smooth on connected components of an open dense subset of U.Moreover, we obtain from the Codazzi equation and the use of the de Rham theorem that M 2n is locally a Riemannian product of surfaces Having that the codimension is n = p + 2 and that α F (Y i , Y j ) = 0 if Y i ∈ (E i ) and Y j ∈ (E j ), i = j, then by Theorem 8.7 in [3] there are isometric immersions g 1 : M 2 1 → L 3 and g j : M 2 j → R 3 , 2 ≤ j ≤ n, such that F (x 1 , . . ., x n ) = (g 1 (x 1 ), g 2 (x 2 ), . . ., g n (x n )).
Since F (M) ⊂ V 3n−1 ⊂ L 3n then F, F = 0. Hence g j * X j , g j = g * X j , g j = 0 and thus g j = r j with −r 2 1 + n j=2 r 2 j = 0.This gives that where 1/c i = r 2 i and, by continuity, this also holds for F (M).