CLASSIFICATION OF COMPLETE REAL KAHLER¨ EUCLIDEAN SUBMANIFOLDS IN CODIMENSION THREE
DISSERTATION
Presented in Partial Fulfillment of the Requirements for the Degree
Doctor of Philosophy
in the Graduate School of the Ohio State University
By
Wing San Hui, M.Phil.
Graduate Program in Mathematics
The Ohio State University
2011
Dissertation Committee:
Professor Fangyang Zheng, Advisor
Professor Andrzej Derdzinski
Professor Bo Guan c Copyright by
Wing San Hui
2011 ABSTRACT
We show that if the second fundamental form α of a real K¨ahlerEuclidean sub- manifold f : M 2n → R2n+p of codimension p ≥ 3 splits orthogonally as α = α0 ⊕ γ, with image of γ spans a rank 2 subbundle of the normal bundle and satisfies some symmetry with respect to the complex structure, then the submanifold can be ex- tended to a real K¨ahlerEuclidean submanifold f˜ : M˜ 2n+2 → R2n+p with 2 higher real dimensions. Using this result, we can describe a class of codimension 3 real
K¨ahlerEuclidean submanifold that can be extended to a real K¨ahlerhypersurface.
In addition, in codimension 3, we describe some of the the non-minimal situations by showing that f is a cylinder over a real K¨ahlercurve, surface or threefold.
ii ACKNOWLEDGMENTS
I am greatly indebted to my advisor Professor Fangyang Zheng for his patience, his continual guidance and constant encouragement and help throughout the period of my graduate studies. I am grateful for the numerous enlightening discussions and suggestions as well as the intellectual support he has provided.
I would like to thank Professor Andrzej Derdzinski and Professor Bo Guan, for kindly agreeing to serve on my committee and for their invaluable time and profes- sional insights.
I would like to thank Professor Herbert Clemens, Professor Jeffery McNeal and their students for the group study and invaluable discussions. Special thank to Pro- fessor Clemens for his constant encouragement and help in many ways.
I am thankful for the Department of Mathematics for their support expressed through several teaching and research fellowships that I enjoyed.
Last but not the least, many thanks to Na and my parents, for the support and help through these years.
iii VITA
2001 ...... B.Sc. in Mathematics, The Chinese University of Hong Kong, Hong Kong, China
2003 ...... M.Phil. in Mathematics, The Chinese University of Hong Kong, Hong Kong, China
2008 ...... MAS. in Statistics, The Ohio State University, Columbus, Ohio
2003-Present ...... Graduate Teaching Associate, Department of Mathematics, The Ohio State University, Columbus, Ohio
PUBLICATIONS
Florit, L.; Hui, W.; Zheng, F.: On real K¨ahlerEuclidean submanifolds with nonneg- ative Ricci or holomorphic curvature. J. Eur. Math. Soc. (JEMS) 7, 1-11 (2005)
iv FIELDS OF STUDY
Major Field: Mathematics
Specialization: Differential Geometry
v TABLE OF CONTENTS
Abstract ...... ii
Acknowledgments ...... iii
Vita ...... iv
CHAPTER PAGE
1 Introduction ...... 1
2 Preliminaries ...... 4
3 Main Results ...... 8
3.1 Symmetric complex bilinear map, S, with dimC img S ≤ 2 . . . . . 8 3.2 Symmetric bilinear map on real vector space with linear complex structure...... 11 3.3 Real K¨ahlerEuclidean submanifolds of codimension 3 ...... 22
Bibliography ...... 34
vi CHAPTER 1
INTRODUCTION
It is well-known that any complete Riemannian manifold M can be isometrically em- bedded in an Euclidean space [N56]. The interplay between the extrinsic and intrinsic geometry as well as the underlying topology of the manifold are extensively studied in the literature. Here we are particularly interested in the special case when M is a complete K¨ahler manifold, namely, a complete Riemannian manifold M 2n which is also a complex manifold of complex dimension n, such that both the Riemannian metric and the Levi-Civita connection compatible with the complex structure. Let
2n 2n+p f : M → R be the isometric embedding. Ideally, one would hope that the target space is a complex Euclidean space and f is a holomorphic map. However, from Calabi’s thesis [C53], it was known that such a case is very rare. In fact, Calabi gave a complete description of what kind of K¨ahlermanifold can be holomorphically and isometrically immersed into a complex Euclidean space (or more generally a com- plex space form). Most K¨ahlermanifolds do not admit such a map. So we have to study the hybrid case of M 2n being K¨ahlerwhile f is just a smooth (isometric) map into a real Euclidean space. Following terminologies in this area, we will call such a manifold a real K¨ahlerEuclidean submanifold, or simply a real K¨ahlersubmanifold.
In the joint work with L. Florit and F. Zheng [FHZ05], we observed that the sec- ond fundamental form of a real K¨ahlerEuclidean submanifold obeys some additional symmetry conditions, and when the codimension p is relatively small, or when the
1 curvature tensor of M 2n is restrictive, the manifold tends to be rather special. Fol- lowing that work, Florit and Zheng continued the study and obtained classification results for complete real K¨ahlersubmanifolds in codimension 1 and 2 [FZ05], [FZ07],
[FZ08]. It turns out that in codimension 1, any complete real K¨ahlersubmanifold
2n 2n+1 2 3 f : M → R must be a cylinder. That is, f = g × I, where g : N → R is 2n−2 ∼ n−1 a complete surface in Euclidean 3-space and I is the identity map of R = C . For codimension 2, assuming n > 1, they showed that, either M is minimal, or it is
4 6 a cylinder over a N → R . Also, for N, under very mild additional assumptions, it is actually the product of two surfaces in Euclidean 3-spaces.
So in short, complete real K¨ahlersubmanifolds in codimension 1 or 2 are rather special: they are essentially cylinders over products of surfaces in Euclidean three spaces, unless they are minimal. Note that by a result of Dajczer and Gromoll [DG85], any minimal isometric immersion from a K¨ahlermanifold M into an Euclidean space
2n+p 2n+p R must be the real part of a holomorphic map from M into C . When p = 2 and n ≥ 3, a celebrated theorem of Dajczer and Gromoll [DG95] states that either 2n+2 ∼ n+1 M is holomorphic under some (isometric) identification R = C , or M is the total space of a holomorphic vector bundle over a Riemann surface, and f embeds
2n+2 each fiber onto a linear subvariety in R . In the latter case, they gave a precise description in terms of a Weierstrass type representation of the Gauss map.
The goal of this thesis is to give a systematic study of complete real K¨ahler submanifolds of codimension 3. We will try to give a complete classification of such manifolds. Notice that when the codimension is increased to 3, aside from the usual arising algebraic complexity, a major difficulty emerge from the new phenomenon of composition. For instance, since there are plenty of isometric embeddings from an
2n+2 2n+3 open subset of R into R , its composition with any codimension 2 real K¨ahler submanifold would automatically give a real K¨ahlersubmanifold of codimension 3.
2 Another type of composition that may occur in the codimension 3 case would be as follows. Let M 2n ⊆ N 2(n+1) be a holomorphic isometric embedding of K¨ahler manifolds of complex dimensions n and n + 1, respectively. If N can be a real
K¨ahlersubmanifold of codimension 1, then M would be a real K¨ahlersubmanifold of codimension 3. In order to classify the codimension 3 cases, we need to detect such compositions.
3 CHAPTER 2
PRELIMINARIES
2n 2n+p Let f : M → R be a real K¨ahlerEuclidean submanifold and let N be the normal bundle. For x ∈ M, let α be the second fundamental form of f at x. Extend
α bilinearly over C and still denote it by α,
α :(TxM) ⊗ C × (TxM) ⊗ C → N ⊗ C.
∼ Using the complex structure, J, of M, we get the decomposition (TxM)⊗C = V ⊕V , where V (the (1, 0)-part) is the eigenspace of J with eigenvalue i and V (the (0, 1)- part) is the eigenspace of J with eigenvalue −i. Write
H = α|V ×V and S = α|V ×V , then S : V × V → N ⊗ C is a symmetric complex bilinear map, while H : V × V →
N ⊗ C is a Hermitian bilinear map, that is,
H(Y, X) = H(X, Y ), ∀X,Y ∈ V. for the (1, 1) and (2, 0) parts of α, respectively. We will use the folowing notations, for any X, Y, ei, ej, e˜i, e˜j ∈ V ,
SXY = S(X,Y ),Sij = Seiej ,S˜i˜j = Se˜ie˜j
H = H(X, Y ),H = H ,H = H , XY ij eiej ˜i˜j e˜ie˜j and similarly fore ˆi, etc. 4 Definition 2.1. We define the kernel of H, ker H, and kernel of S, ker S, as follows:
ker H = {X ∈ V : HXY = 0, ∀ Y ∈ V }
ker S = {X ∈ V : SXY = 0, ∀ Y ∈ V }
Definition 2.2. We define the image of H, img H, and image of S, img S, as follows:
img H = span{X,Y ∈ V : HXY }
img S = span{X,Y ∈ V : SXY }
We also extend the inner product h , i on N bilinearly over C to N ⊗ C, and still denote it by h , i. Then the Gauss equation:
RABCD = hα(A, D), α(B,C)i − hα(A, C), α(B,D)i
2n also holds for any A, B, C, D ∈ (TxM) ⊗ C. Since M is K¨ahler, we have RXY ∗∗ = 0 if both X and Y in V . As in [FHZ05], we have the following:
2n 2n+p Proposition 2.3. Let f : M → R be a real K¨ahlerEuclidean submanifold. Fix a point x ∈ M 2n, and consider V,H,S as above. Then for any X,Y,Z,W ∈ V , it holds that
hHXW ,HY Z i = hHY W ,HXZ i
hHXW ,SYZ i = hHY W ,SXZ i
hSXW ,SYZ i = hSYW ,SXZ i
RXYZW = hHXW ,HZY i − SXZ , SYW
Suppose now M is complete. Let ∆x = {X ∈ TxM : α(X,TxM) = 0} be the relative nullity of f at x, and ν(x) = dimR ∆x be the index of relative nullity. Let ν0 be the minimum value of ν(x) on M and U = {x ∈ M : ν(x) = ν0} be an open subset
5 of M. It is well-known that ∆ is smooth and integrable in U with totally geodesic
2n 2n+p and complete leaves in both M and R .
Define the index of pluriharmonic nullity νJ = νJ (x) of f at x ∈ M by
νJ = dimC ker H.
Then we define Dx = ∆x ∩ J∆x, where J is the almost complex structure of M, so Dx has a linear complex structure. Let W 0 = ker H ∩ ker S, so that W 0 is a complex vector subspace of V . Then we
0 0 0 0 0 have Dx ⊗ C = W ⊕ W and write ν (x) = dimC W . Let ν0 be the minimum value 0 0 0 of ν (x) on U, and so U0 = {x ∈ U : ν (x) = ν0} is an open subset of U. Since J is parallel, D is also smooth and integrable in U0 with totally geodesic and complete leaves and D is called the complex relative nullity.
0 Let r = n − ν0 and fix any x ∈ U0. Considering the complexified tangent space ∼ TxM ⊗C, using J, we get the decomposition Tx⊗C = V ⊕V , that is V (the (1, 0) part) is the eigenspace of J with eigenvalue i and V (the (0, 1) part) is the eigenspace of J ∼ r with eigenvalue −i. Let W = C be the complex linear subspace of V perpendicular ⊥ 0 to Dx, that is W ⊕ W = Dx ⊗ C and we have V = W ⊕ W . ⊥ ⊥ Let C : D × D → D be the twisting tensor defined by CT X = −(∇X T )D⊥ where ∇ is the Levi-Civita connection of M and ()D⊥ is the orthogonal projection
⊥ onto D . Fix T ∈ D, for a basis B = {e1, . . . , er} of W , we write the complexified operator CT as r X CT (ei) = (Aijej + Bijej). j=1 With respect to the basis B, we have A and B are r × r complex matrices, while
H and S are Hermitian and complex symmetric matrices, respectively, with values in the complexification of the normal space of f at x. Then we have the following two main results from [FZ08] (Theorem 7 and Corollary 8):
6 Theorem 2.4. For any complete real K¨ahlerEuclidean submanifold f : M 2n →
2n+p R , the complex relative nullity D in U0 is a holomorphic foliation, that is B = 0.
Corollary 2.5. For any complete real K¨ahlerEuclidean submanifold f : M 2n →
2n+p R , for any T ∈ D, A is nilpotent, AS is symmetric, and AH = 0.
7 CHAPTER 3
MAIN RESULTS
3.1 Symmetric complex bilinear map, S, with dimC img S ≤ 2
In this section we will study the symmetric complex bilinear map with image di- mension 2. And we will use these results in the later sections. Let V be a complex vector space of complex dimension n and NC be another complex vector space. Let
S : V × V → NC be a symmetric complex bilinear map, that is,
S(X,Y ) = S(Y,X), ∀X,Y ∈ V, and let s be the complex dimension of img S. For s = 1, the diagonalization of symmetric bilinear form on complex vector space gives us,
Lemma 3.1. For s = 1, we have a basis {e1, . . . , en} that diagonalizes S. i.e, there exists k, 1 ≤ k ≤ n, such that S11 = ··· = Skk 6= 0, Sii = 0 for k + 1 ≤ i ≤ n and
Sij = 0 for all i 6= j.
Next, we will study the case when s = 2. For the special case n = 2, we have the following:
Lemma 3.2. For n = 2 and dim img S = 2, there exists a basis {e1, e2} for V such that we have one of the following two cases:
(1) {S11,S22} forms a basis for img S and S12 = 0. 8 (2) {S11,S12} forms a basis for img S and S22 = 0.
Proof. We pick any basis {e1, e2} for V . The only case that is not on our list is all
S11, S12 and S22 are nonzero. First, since dim img S = 2, if {S11,S22} are linearly independent, then S12 = kS11 + lS22 for some k, l ∈ C, and WLOG, we can assume k 6= 0. Then we consider the quadratic equation kx2 + x + l = 0, and let α and β be the roots of it, so we have
1 l α + β = − and αβ = . k k
Next we consider the vectorse ˜1 = e1 + αe2 ande ˜2 = e1 + βe2, then we have
S1˜2˜ = S11 + (α + β)S12 + αβS22
= [1 + k(α + β)]S11 + [αβ + l(α + β)]S22
= 0
Ife ˜1 6=e ˜2, then they will form a basis for V , and this happens iff α 6= β, which is equivalent to 1 − 4kl 6= 0. Compute S1˜1˜ and S2˜2˜:
2 S1˜1˜ = (1 + 2kα)S11 + (α + 2lα)S22 6= 0,
2 S2˜2˜ = (1 + 2kβ)S11 + (β + 2lβ)S22 6= 0,
1 this reduces to our case (1). For 1 − 4kl = 0, we have α = β = − = −2l,e ˜ =e ˜ 2k 1 2 and S2˜2˜ = 0. Then we will consider the basis {e1, e˜2}, and we get
S11 6= 0,
S2˜2˜ = 0, 1 S = S + αS = − S − 2lS 6= 0, 12˜ 11 12 2 11 22 and this reduces to our case (2). Finally, if {S11,S22} are linearly dependent, then
9 we can rescale e2 such that S22 = S11 and {S11,S12} must be linearly independent.
Then we consider the new basis {e˜1, e˜2} withe ˜1 = e1 + e2 ande ˜2 = e1 − e2, we have
S1˜1˜ = S11 + 2S12 + S22 = 2S11 + 2S12 6= 0,
S1˜2˜ = S11 − S22 = 0,
S2˜2˜ = S11 − 2S12 + S22 = 2S11 − 2S12 6= 0, and this reduces to our case (1).
For the general case n ≥ 2, we will have the following:
Corollary 3.3. For any n ≥ 2, if dim img S = 2, we have a pair of vectors {X,Y } such that span{SXX ,SXY ,SYY } = img S. Moreover, we can choose {X,Y } such that one of the following occurs:
1. {SXX ,SYY } forms a basis for img S and SXY = 0,
2. {SXX ,SXY } forms a basis for img S and SYY = 0.
Proof. By Lemma 3.2, we just need to find a pair {X,Y } such that span{SXX ,SXY ,SYY } = img S. Takes any basis {e1, . . . , en} for V , since dim img S = 2, by renumbering the ei’s, we can assume one of the following five cases happens:
1. span{S11,S12} = img S
2. span{S11,S22} = img S
3. span{S11,S23} = img S
4. span{S12,S13} = img S
5. span{S12,S34} = img S
10 In the first two cases, we are done. For the third case, we can assume S12, S13,
S22 and S33 are all parallel to S11, otherwise we are done. For the same reason, we can assume S22 and S33 are all parallel to S23, this forces S22 = S33 = 0. Then the space span{S22,S23,S33} is one dimensional, by the Lemma 3.1, we can diagonalize it and find Y = ae2 + be3 such that SYY = S23. Then, by taking X = e1, we get our X and Y .
In the forth case, we can assume S11 and S22 are parallel to S12, and S11 and S33 are parallel to S13. So this reduces to S11 = 0. Now, if one of S22 or S33 nonzero, we can reduce to the third case. So we can now assume S11 = S22 = S33 = 0.
Then consider X = e1 + ae2 and Y = e3 with a ∈ C to be chosen later. Then we have SXX = 2aS12 and SXY = S13 + aS23. Now, we just choose a 6= 0 such that
S13 6= −aS23 and get our the X and Y .
In the fifth case, we can assume S11 and S22 are parallel to S12, and S33 and S44 are parallel to S34. Then the space span{S11,S12,S22} is one dimensional, by Lemma
3.1, we can diagonalize it and find X = ae1 + be2 such that SXX = S12. Similarly, we can find Y = ce3 + de4 such that SYY = S34. And these are the X and Y we are looking for.
3.2 Symmetric bilinear map on real vector space with linear
complex structure.
In this section, we shall study the symmetric bilinear map on a real vector space with linear complex structure. Let T be a real vector space of dimension n with linear
2 complex structure J : T → T , J = − idT . Let N be a real vector space of dimension p with an inner product h, i. Let α : T × T → N be a symmetric bilinear map. We
11 extend α bilinearly over C, and still denote it by α,
α :(T ⊗ C) × (T ⊗ C) → N ⊗ C.
As in chapter 2, we get V, V,H,S and we will use similar notations for H and S here.
Again, we also extend the inner product h , i on N bilinearly over C to N ⊗ C, and still denote it by h , i.
In the view of Proposition 2.3, we will mainly work with those α that satisfy the following symmetry:
Definition 3.4. The symmetric bilinear map α (S and H) is called RKES-symmetric, if the corresponding H and S satisfies the following symmetries, for any X,Y,Z,W ∈ V
hHXW ,HY Z i = hHY W ,HXZ i (3.1)
hHXW ,SYZ i = hHY W ,SXZ i (3.2)
hSXW ,SYZ i = hSYW ,SXZ i (3.3)
We have the following key lemma, Lemma 7 in [FHZ05]:
0 Lemma 3.5. Suppose dimC img H = p ≤ n and H satisfies the equation (3.1), then
1. there exists a basis {e1, . . . , en} of V such that Hij = 0 if either i 6= j or
0 0 i = j > p , and for 1 ≤ i ≤ p , Hii 6= 0. Moreover, since Hii = Hii, it is real
and {wi = Hii/|Hii| : 1 ≤ i ≤ p} forms an orthonormal basis of a real vector subspace img H ∩ N of N.
2. if, in addition, S and H also satisfies the equation (3.2) and img H = N ⊗ C, 0 then, for the same basis above, we have Sij = 0 if either i 6= j or i = j > p ,
0 and for 1 ≤ i ≤ p , Sii collinear to wi.
We have the following easy observation:
12 0 Corollary 3.6. Suppose dimC img H = p ≤ n and H satisfies equation (3.1), then we always have the equality
dimC ker H = n − dimC img H.
Also, from the symmetry (3.2), we have the following:
Lemma 3.7. Suppose S and H satisfy equation (3.2), then we have
S(ker H,V ) ⊥ img H.
In particular, we have dimC img S|ker H×ker H ≤ p − dimC img H.
Proof. For any X ∈ ker H and Y,Z,W ∈ V , we have
hSXY ,HZW i = hSZY ,HXW i = hSZY , 0i = 0.
Next, we consider dimR N = p = 3 and H 6= 0, we also assume α is RKES- symmetric. Our goal is to find a good basis for V such that S and H can be described as simply as possible. Clearly, we have dimC img H ≤ 3. If dimC img H = 3, then by
Lemma 3.5, there exists a basis that diagonalizes H and S. For dimC img H = 2, by
Lemma 3.7, we have dimC img S|ker H×ker H ≤ 1, and we will have the following two cases:
Lemma 3.8. Suppose S and H satisfy the RKES-symmetry. If dimC img H = 2 and dimC img S|ker H×ker H = 0, then there exist a basis {e1, . . . , en} of V and an orthonor- mal basis, {w1, w2, w3}, of the real vector space N such that, for some a1, a3, b2, b3, c3 ∈
2 C with a3b3 = c3, we have H11 = w1, H22 = w2, S11 = a1w1 +a3w3, S22 = b2w2 +b3w3,
S12 = c3w3 and, for other i, j, Hij = 0 and Sij = 0.
13 Proof. The existence of a basis {e1, . . . , en} with {e3, . . . , en} a basis for ker H and
H11 = w1, H22 = w2 and H12 = 0 = H21 follows from Lemma 3.5. Take w3 ∈ N such that {w1, w2, w3} form an orthonormal basis of N. Then
hS11, w2i = hS11,H22i = hS21,H12i = 0
hS22, w1i = hS22,H11i = hS12,H21i = 0
Thus, we have S11 = a1w1 + a3w3 and S22 = b2w2 + b3w3 for some a1, a3, b2, b3 ∈ C.
For all i, j ≥ 3, we have Sij = 0 and, from Lemma 3.7, S1i,S2i ∈ span{w3}. Fix any i ≥ 3, then there exists k ∈ C such that S1i = kw3. On the other hand, we have
2 k = hS1i,S1ii = hS11,Siii = 0,
so we have k = 0. Hence S1i = 0 for any i ≥ 3. Similarly, we have S2i = 0 for any i ≥ 3. Finally, since
hS12,H11i = hS11,H21i = 0
hS12,H22i = hS22,H12i = 0, we have S12 = c3w3 for some c3 ∈ C and
2 c3 = hS12,S12i = hS11,S22i = a3b3.
Lemma 3.9. Suppose S and H satisfy the RKES-symmetry. If dimC img H = 2 and dimC img S|ker H×ker H = 1, then there exist a basis {e1, . . . , en} of V and an orthonormal basis, {w1, w2, w3}, of a real vector space N such that, for some a1, b2 ∈
C, we have H11 = w1, H22 = w2, S11 = a1w1, S22 = b2w2, S33 = w3 and, for other i, j, Hij = 0 and Sij = 0.
14 Proof. Again we have a basis {e1, . . . , en} with {e3, . . . , en} a basis for ker H such that
H11 = w1, H22 = w2, H12 = 0 = H21. Take w3 ∈ N such that {w1, w2, w3} form an orthonormal basis of N. By our assumption, we have img S|ker H×ker H = span{w3}, so by Lemma 3.1, we can find another basis for ker H that diagonalizes S|ker H×ker H .
For simplicity, we still call this basis {e3, . . . , en}. By rescaling and renumbering, we can assume S33 = w3. Now, by Lemma 3.7, for any i = 1, . . . , n and j = 4, . . . , n, we have Sij ⊥ span{w1, w2}. But we also have
hSij, w3i = hSij,S33i = hS3j,Si3i = 0,
so it gives Sij = 0. Next, fix any i = 1, 2, we have Si3 ⊥ span{w1, w2}, so Si3 ∈ span{w3} and we write Si3 = kiw3 = kiS33 for some ki ∈ C. Lete ˜1 = e1 − k1e3 and e˜2 = e2 − k2e3, we have H1˜1˜ = w1, H2˜2˜ = w2 and H1˜2˜ = 0 = H2˜1˜. We also have
S13˜ = S13 − k1S33 = 0
S23˜ = S23 − k2S33 = 0
Similar to the argument in Lemma 3.8, we have S1˜1˜ = a1w1 +a3w3, S2˜2˜ = b2w2 +b3w3 ˜ ˜ and S1˜2˜ = c3w3 for some a1, a3, b2, b3 ∈ C. But, for i, j = 1, 2, we also have
hSij, w3i = hSij,S33i = hS3j,Si3i = 0,
thus we get a3 = b3 = c3 = 0. Hence {e˜1, e˜2, e3, . . . , en} is the basis we want.
Next, for dimC img H = 1, by Lemma 3.7, we have dimC img S|ker H×ker H ≤ 2.
Lemma 3.10. Suppose S and H satisfy the RKES-symmetry. If dimC img H = 1 and dimC img S|ker H×ker H = 0, then there exist a basis {e1, . . . , en} of V and an orthonormal basis, {w1, w2, w3}, of a real vector space N such that we have one of the following case:
15 2 (1) for some a1, a2, a3 ∈ C with b 6= −1, we have H11 = w1, S11 = a1w1 + a2w2 +
a3w + 3 and, for other i, j, Hij = 0 and Sij = 0.
(2) for some a1, a2, a3 ∈ C with b 6= i, we have H11 = w1, S11 = a1w1 +a2w2 +a3w3,
S12 = w2 + iw3 and, for other i, j, Hij = 0 and Sij = 0.
Proof. Since dimC img H = 1, we have a basis {e1, . . . , en} with {e2, . . . , en} a basis for ker H such that H11 = w1. Take w2, w3 ∈ N such that {w1, w2, w3} form an orthonormal basis of N. By our assumption S|ker H×ker H = 0, we have Sij = 0 for all i, j ≥ 2. By Lemma 3.7, we have S1j ∈ span{w2, w3} for all j ≥ 2, we write
S1j = ajw2 + bjw3 with aj, bj ∈ C. If S1j = 0 for all j ≥ 2, then we get case (1).
Suppose the collection {S1j : j ≥ 2} are not all zero, by renumbering, we can assume S12 6= 0. Then we have
2 2 a2 + b2 = hS12,S12i = hS11,S22i = 0,
so we get either b2 = ia2 or b2 = −ia2. Replacing w3 by −w3 if needed, we can assume
−1 b2 = ia2 and S12 = a2(w2 + iw3) with a2 6= 0. Rescaling e2 by the factor a2 and still denote it e2 for simplicity, we have S12 = w2 + iw3. Now, for any j ≥ 3, we have
S1j = ajw2 + bjw3 and
aj + ibj = hS12,S1ji = hS11,S2ji = 0,
so we have bj = iaj and S1j = aj(w2 + iw3) = ajS12. Lete ˜j = ej − aje2 for j ≥ 3, we have S1˜j = S1j − ajS12 = 0, and this gives us case (2).
Lemma 3.11. Suppose S and H satisfy the RKES-symmetry. If dimC img H = 1 and dimC img S|ker H×ker H = 2, then there exist a basis {e1, . . . , en} of V and an orthonormal basis, {w1, w2, w3}, of a real vector space N such that we have one of the following case:
16 (1) for some a1 ∈ C, we have H11 = w1, S11 = a1w1, S22 = ··· = Skk = S1k+1 =
w2 + iw3 for some k with 2 ≤ k ≤ n − 1 and, for other i, j, Hij = 0 and Sij = 0.
(2) for some a1, b ∈ C, we have H11 = w1, S11 = a1w1 + b(w2 + iw3), S22 = ··· =
Skk = w2 + iw3 for some k with 2 ≤ k ≤ n and, for other i, j, Hij = 0 and
Sij = 0.
2 (3) for some a1, a2, a3, b, c ∈ C with a2 + a3b = c , we have H11 = w1, S11 =
a1w1 + a2w2 + a3w3, S12 = cw3, S22 = w2 + bw3 and, for other i, j, Hij = 0 and
Sij = 0.
Proof. Again we have a basis {e1, . . . , en} with {e2, . . . , en} a basis for ker H such that H11 = w1. Take w2, w3 ∈ N such that {w1, w2, w3} form an orthonormal basis of N. By our assumption and Lemma 3.1, S|ker H×ker H can be diagonalized, that is we can choose {e2, . . . , en} such that S22 = ··· = Skk 6= 0 for some k with 2 ≤ k ≤ n and for other i, j ≥ 2, Sij = 0.
Suppose k 6= n − 1 and the collection {S1j : j ≥ k + 1} are not all zero, by renumbering, we can assume S1k+1 6= 0 and, by Lemma 3.7, S1k+1 ∈ span{w2, w3}, then
hS1k+1,S1k+1i = hS11,Sk+1k+1i = 0 implies S1k+1 ∈ span{w2 + iw3} or S1k+1 ∈ span{w2 − iw3}. Replacing w3 by −w3 if needed and by rescaling ek+1, we can assume S1k+1 = w2 + iw3. Then for any j ≥ 2 and j 6= k + 1, we also have S1j ∈ span{w2, w3} and
hS1j,S1k+1i = hS11,Sk+1ji = 0, so there exists aj ∈ C such that S1j = ajS1k+1. Lete ˜j = ej − ajek+1, then we have
S1˜j = S1j − ajS1k+1 = 0
17 and, since Sk+1j = 0 and Sk+1k+1 = 0, we have S˜i˜j = Sij, for all i, j ≥ 2. So, for simplicity, we still denote this new choice of basis for ker H by {e2, . . . , en}. Then for any 2 ≤ j ≤ k, we have Sjj ∈ span{w2, w3} and
hSjj,S1k+1i = hS1j,Sjk+1i = 0,
so Sjj ∈ span{w2 + iw3} and, since Sjj 6= 0, by rescaling ej we get Sjj = w2 + iw3.
Next, we consider S11, we have
hS11,S22i = hS12,S12i = 0
b so S = a w + b(w + iw ) for some a , b . Lete ˜ = e − e , then, for any j ≥ 2 11 1 1 2 3 1 C 1 1 2 k+1 we have
b2 S = S − bS + e = a w 1˜1˜ 11 1k+1 4 k+1 1 1 b S = S − S = S , 1˜j 1j 2 k+1j 1j and we get case (1).
Next, suppose that the collection {S1j : j ≥ k + 1} are zero and 3 ≤ k ≤ n. So, we have S33 = S22 6= 0. But we also have S22 ∈ span{w2, w3} and
hS33,S22i = hS23,S23i = 0,
this implies, replace w3 by −w3 if needed, S22 = ··· = Skk = c(w2 + iw3) for some
0 6= c ∈ C. After rescaling, we have S22 = ··· = Skk = w2 + iw3. Then for any
2 ≤ j ≤ k, we can find 2 ≤ l ≤ k with l 6= j such that Sll = w2 + iw3 and we have
hS1j,Slli = hS1l,Sjli = 0,
k X thus S1j = aj(w2 + iw3) for some aj ∈ C. Lete ˜1 = e1 − alel, we have l=2
k X S1˜j = S1j − alSlj = S1j − ajSjj = 0. l=2 18 Also, we get
hS1˜1˜,S22i = hS12˜ ,S12˜ i = 0, this implies S1˜1˜ = a1w1 + b(w2 + iw3), and we get case (2).
Finally, the only case left is the collection {S1j : j ≥ 3} are zero and k = 2. In this case, by switching w2 and w3 if needed and rescaling e2, we get S22 = w2 + bw3 for some b ∈ C. Again we have S12 ∈ span{w2, w3}, so S12 = c2w2 + c3w3. Let e˜1 = e1 − c2e2, we have
S12˜ = S12 − c2S22 = (c3 − c2b)w3,
and we write c = c3 − c2b. Finally, we write S1˜1˜ = a1w1 + a2w2 + a3w3 and we have
hS1˜1˜,S22i hS12˜ ,S12˜ i ,
2 this implies a2 + a3b = c and we get case (3).
Lemma 3.12. Suppose S and H satisfy the RKES-symmetry. If dimC img H = 1 and dimC img S|ker H×ker H = 2, then there exist a basis {e1, . . . , en} of V and an orthonormal basis, {w1, w2, w3}, of a real vector space N such that we have one of the following cases:
2 (1) for some a1, b ∈ C with b 6= −1, we have H11 = w1, S11 = a1w1, S22 = w2+bw3,
S33 = −bw2 + w3 and, for other i, j, Hij = 0 and Sij = 0.
(2) for some a1, b ∈ C with b 6= i, we have H11 = w1, S11 = a1w1, S22 = w2 + bw3,
S23 = w2 + iw3 and, for other i, j, Hij = 0 and Sij = 0.
Proof. Again we have a basis {e1, . . . , en} with {e2, . . . , en} a basis for ker H such that H11 = w1. Take w2, w3 ∈ N such that {w1, w2, w3} form an orthonormal ba- sis of N. By our assumption and Lemma 3.3, we can choose X,Y ∈ ker H with span{SXX ,SXY ,SYY } = span{w2, w3}. By choosing {X + Y,Y } instead of {X,Y } 19 if needed, we can always assume span{SXX ,SXY } = span{w2, w3}. Then we define
SX : ker H → N by SX (Z) = S(X,Z), for any Z ∈ ker H. By our choice of X and
Lemma 3.7, we have img SX = span{w2, w3}. Hence ker SX is of codimension 2 in ker H, and we get the decomposition ker H = span{X,Y } ⊕ ker SX . Choose e2 = X, e3 = Y and {e4, . . . , en} to be a basis for ker SX . Then, for any any 1 ≤ i ≤ n and 4 ≤ j ≤ n, we have
hSij,H11i = Si1,H1j = 0
hSij,S22i = hSi2,S2ji = 0
hSij,S23i = hSi3,S2ji = 0,
since span{H11,S22,S23} = N, we must have Sij = 0.
Now, we may reselect our basis for span{e2, e3}, such that one of the cases in
Lemma 3.3 occurs, and still call it {e2, e3} for simplicity. If we are in case (1) of
Lemma 3.3, we have span{S22,S33} = span{w2, w3} and S23 = 0. By rescaling e2 and e3 and switching indexes if needed, we can assume S22 = w2 +bw3 and S33 = cw2 +w3. On the other hand, we have
c + b = hS22,S33i = hS23,S23i = 0,
so c = −b, and since S22 = w2 + bw3 and S33 = −bw2 + w3 are linearly independent, we have b2 6= −1. We also have
S − bS bS + S w = 22 33 , w = 22 33 . 2 1 + b2 3 1 + b2
Then, we consider S12 and S13, by Lemma 3.7, we have S12,S13 ∈ span{w2, w3} and we also have
hS12,S33i = hS13,S23i = 0
hS13,S22i = hS12,S23i = 0,
20 so, we get
hS ,S i b hS ,S i hS , w i = 12 22 , hS , w i = 12 22 , 12 2 1 + b2 12 3 1 + b2 −b hS ,S i hS ,S i hS , w i = 13 33 , hS , w i = 13 33 , 13 2 1 + b2 13 3 1 + b2 hence we have S12 = a2S22 and S13 = a3S33 for some a2, a3 ∈ C. Lete ˜1 = e1 − a2e2 − a3e3, then we have
S12˜ = S12 − a2S22 − a3S23 = 0
S13˜ = S13 − a2S23 − a3S33 = 0.
Also, we have
hS1˜1˜,S22i = hS12˜ ,S12˜ i = 0,
hS1˜1˜,S33i = hS13˜ ,S13˜ i = 0, this implies
hS1˜1˜, w2i = hS1˜1˜, w3i = 0, and hence we get S1˜1˜ = a1w1 for some a1 ∈ C. And this is our case (1).
Next, suppose we are in case (2) of Lemma 3.3, we have span{S22,S23} = span{w2, w3} and S33 = 0. By rescaling e2 and e3 and switching indexes if needed, we can assume
S22 = w2 + bw3. Then we write S23 = cw2 + dw3. Note that
2 2 c + d = hS23,S23i = hS22,S33i = 0,
so d = ±ic and, in particular, c 6= 0. By rescaling e3 and replacing w3 by −w3, b by
−b if needed, we have S22 = w2 + bw3 and S23 = w2 + iw3. Since S22 and S23 are linearly independent, we must have b 6= i, and
−iS + bS S − S w = 22 23 , w = 22 23 . 2 b − i 3 b − i 21 Then, consider S13, by Lemma 3.7, we have S13 ∈ span{w2, w3} and
hS13,S13i = hS11,S33i = 0, this implies S13 = a2(w2 + iw3) = a2S23 for some a2 ∈ C. Lete ˜1 = e1 − a2e2, we have
S13˜ = S13 − a2S23 = 0.
Next, consider S12˜ , again we have S12˜ ∈ span{w2, w3} and
hS12˜ ,S23i = hS13˜ ,S22i = 0,
this implies S12˜ = a3(w2 + iw3) = a3S23 for some a3 ∈ C. Lete ˆ1 =e ˜1 − a3e3, then we have
S12ˆ = S12˜ − a3S32 = 0
S13ˆ = S13˜ − a3S33 = 0
Also, we have
hS1ˆ1ˆ,S22i = hS12ˆ ,S12ˆ i = 0,
hS1ˆ1ˆ,S23i = hS13ˆ ,S13ˆ i = 0, this implies
hS1ˆ1ˆ, w2i = hS1ˆ1ˆ, w3i = 0, and hence we get S1ˆ1ˆ = a1w1 for some a1 ∈ C. This is our case (2).
3.3 Real K¨ahler Euclidean submanifolds of codimension 3
In this section, we will use the results in section 3.2 to give a classification of Real
K¨ahlerEuclidean submanifolds of codimension 3.
We have the following: 22 2n 2n+p Theorem 3.13. Let f : M → R , n ≥ p ≥ 3, be a real K¨ahlerEuclidean submanifold. Assume the second fundamental form splits orthogonally as α = α0 ⊕ γ with γ : TM × TM → L, where img γ = L ⊂ N is a vector subbundle of the normal bundle with rank 2, and for any X,Y ∈ TM, γ satisfies
γ(JX,Y ) = γ(X,JY ),
γ(JX,Y ) ⊥ γ(X,Y ),
kγ(JX,Y )k = kγ(X,Y )k ,
Suppose there exist a smooth rank 2 vector subbundle Γ ⊂ TM ⊕L with Γ∩TM = {0} so that ˜ ∇Z µ ∈ TM ⊕ L for all µ ∈ Γ and Z ∈ TM. Then f extends uniquely to a real K¨ahlersubmanifold
2n+2 2n+p F : M˜ → R .
Proof. The proof is similar to the proof of Theorem 1 in [DG97]. By assumption, we have
γ(JX,Y ) = γ(X,JY ), ∀X,Y ∈ TM, so, we can extend the complex structure J in TM to a complex structure, still denoted by J, in TM ⊕ L by
J(Z + γ(X,Y )) = JZ + γ(JX,Y ),
2n+2 for any X,Y,Z ∈ TM ⊗C. Now consider the total space of vector bundle π :Γ → 2n 2n+2 2n+3 M , we define the map F :Γ → R by
F (ξ) = f(π(ξ)) + ξ.
Then F is an immersion in a neighborhood of the 0-section of Γ2n+2q. Now, it remains to prove that this is K¨ahler . First, we claim that, over M, the complex structure J 23 0 ˜ in TM ⊕ L is parallel with respect to the induced connection, ∇ = (∇)TM⊕L, where
(v)TM⊕L denotes the orthogonal projection onto TM ⊕ L. Choose X,Y ∈ TM such that
kγ(X,Y )k = 1, then γ(X,Y ) and −γ(JX,Y ) form an orthonormal basis for L, call them ξ1 and ξ2 respectively, and we have
Jξ1 = −ξ2 and Jξ2 = ξ1.
For any X,Z ∈ TM,
0 ∇X JZ = ∇X JZ + γ(X,JZ)
= J∇X Z + Jγ(X,Z)
0 = J(∇X Z)
For the L direction, fix any X ∈ TM, we have the following,
0 ⊥ ∇X ξi = −Aξi X + (∇X ξi)L
So, to prove ξi are parallel, we need
⊥ ⊥ AJξi X = JAξi X and (∇X Jξi)L = J(∇X ξi)L.
⊥ We first compute (∇X ξi)L. Note that,
1 ∇⊥ ξ , ξ = X hξ , ξ i = 0 X i i 2 i i and
⊥ ⊥ ⊥ ∇X ξ1, ξ2 = X hξ1, ξ2i − ∇X ξ2, ξ1 = − ∇X ξ2, ξ1
24 So, we have the following:
⊥ ⊥ ⊥ (∇X ξ1)L = ∇X ξ1, ξ1 ξ1 + ∇X ξ1, ξ2 ξ2
⊥ = ∇X ξ1, ξ2 ξ2
⊥ ⊥ (∇X ξ2)L = ∇X ξ2, ξ1 ξ1
⊥ = − ∇X ξ1, ξ2 ξ1
These imply
⊥ ⊥ (∇X Jξ1)L = (∇X − ξ2)L
⊥ = ∇X ξ1, ξ2 ξ1
⊥ = ∇X ξ1, ξ2 Jξ2
⊥ = J( ∇X ξ1, ξ2 ξ2)
⊥ = J(∇X ξ1)L
Similarly, we will have
⊥ ⊥ (∇X Jξ2)L = J(∇X ξ2)L.
We consider ξ1, for any W ∈ TM, write γ(X,W ) = a1ξ1 + a2ξ2, hence γX, JW =
−a1ξ2 + a2ξ1, then
hAJξ1 X − JAξ1 X,W i = hAJξ1 X,W i − hJAξ1 X,W i
= hA−ξ2 X,W i − hAξ1 X, −JW i
= − hξ2, α(X,W )i + hξ1, α(X,JW )i
= − hξ2, γ(X,W )i + hξ1, γ(X,JW )i
= −a2 + a2
= 0
25 this proves AJξ1 X = JAξ1 X. Similarly, we can get AJξ2 X = JAξ2 X. This proves our claim. Next, note that, for any δ ∈ Γ,
TF (δ)F (Γ) = Tπ(δ)M ⊕ L(π(δ)), which depends only on π(δ) ∈ M. Thus, we have T Γ = π∗(TM ⊕ L) via parallel
2n+p transport in R , and the Levi-Civita connection for T Γ is the pullback of the 2n+p connection of TM ⊕ L. Also the parallel transport in R also extends the complex structure J on TM ⊕ L to the complex structure π∗(J) on π∗(TM ⊕ L) which is clearly parallel. This proves that F gives us a real K¨ahlersubmanifold.
2n 2n+3 Lemma 3.14. Let f : M → R , be a real K¨ahlerEuclidean submanifold, n ≥ 3.
Assume νJ = n − 1, we define pointwise
˜ Ω(x) = span{∇X η : X ∈ TxM, η ∈ img H}.
Let {e1, . . . , en} be the smooth frame that gives the diagonalization in Lemma 3.5, and H11 = w1 be the unit normal vector field that spans img H. If |hS11,H11i| 6= 1,
⊥ then dim img Aw1 = 2 and Ω forms a rank 2 subbundle of img Aw1 ⊕ (img H) .
˜ Proof. We claim that, for any X ∈ ker Aw1 , we have ∇X w1 = 0. Let X ∈ TM ⊗ C, write n X X = ajej + bjej. j=1 ⊥ We first compute ∇X w1. Observe that
1 ∇⊥ w , w = X hw , w i = 0, X 1 1 2 1 1 and, for any Y,Z ∈ V ,
∇Z Y ∈ V, ∇X Y ∈ V, ∇Z Y ∈ V, ∇X Y ∈ V.
26 Then for any j 6= i, using the Codazzi equation, we have
∇⊥ w = ∇⊥ α(e , e ) ej 1 ej 1 1
⊥ = α(∇ej e1, e1) + α(e1, ∇ej e1) + ∇e1 α(ej, e1) − α(∇e1 ej, e1) − α(ej, ∇e1 e1)
= 0
∇⊥ w = ∇⊥ w = 0 ej 1 ej 1
Therefore,
⊥ ⊥ ⊥ ∇X w1 = a1∇e1 w1 + b1∇e1 w1
Now, we have 0, for j 6= 1 hAw1 X, eji = hw1, α(X, ej)i = a1s + b1, for j = 1 0, for j 6= 1 hAw1 X, eji = hw1, α(X, ej)i = , a1 + b1s, for j = 1
where s = hSii,Hiii. Then X ∈ ker Aw1 will implies
a1s + b1 = 0
a1 + b1s = 0
⊥ If |s| 6= 1, then ai = bi = 0 and ∇X w1 = 0, for any X ∈ ker Aw1 . This proves our claim. In fact, from the above, we also see that dim ker Aw1 = 2n − 2, hence ⊥ ⊥ ⊥ img Aw1 is of rank 2. Note that ∇X w1, w1 = 0, ∇X w1 ⊂ (img H) . So, we have
⊥ ⊥ Ω ⊂ img Aw1 ⊕ (img H) is a rank 2 vector subbundle of img Aw1 ⊕ (img H) .
Using this lemma, we have the following result:
2n 2n+3 Theorem 3.15. Let f : M → R , n ≥ 3, be a real K¨ahlerEuclidean subman- ifold that is in the case (1) or (2) of Lemma 3.11. Then, there exist an open dense 27 subset U ⊂ M such that f extends uniquely to a real K¨ahlerhypersurface along each connected component of U.
Proof. First, in the case (1) or (2), we claim that the smooth frame {e1, . . . , en} satisfies s = |hS11,H11i|= 6 1. Suppose not, then rescale e1 by a factor of t with
2 −1 t = s , we can assume s = 1. Write e1 = v1 − iJv1, then we have
S11 = α(v1, v1) − α(Jv1, Jv1) − 2iα(v1, Jv1)
H11 = α(v1, v1) + α(Jv1, Jv1) since s = 1, we must have
α(v1, v1) = H11,
α(Jv1, Jv1) = 0,
⊥ α(v1, Jv1) ∈ (img H) which also means
S11 − H11 = 2iα(v1, Jv1)
Now, replacing e1 bye ˜1 = e1 + ek+1 in case (1) ore ˜1 = e1 + e2 in case (2) if needed, the new frame {e˜1, e2, . . . , en} still satisfies
S1˜1˜,H1˜1˜ = 1 and
S1˜1˜ − H1˜1˜ 6= 0.
Writee ˜1 =v ˜1 − iJv˜1, and we get
2iα(v ˜1,Jv˜1) 6= 0
On the other hand, we have
S1˜1˜ − H1˜1˜ ∈ span{w2 + iw3}, 28 which leads to a contradiction. So, we must have s = |hS11,H11i|= 6 1. Now, by Lemma 3.14, we have
˜ Ω = span{∇X η : X ∈ T M, η ∈ img H}
⊥ ⊥ is a rank 2 subbundle of img Aw1 ⊕(img H) . Denote (img H) by L. By assumption on H, this is a rank 2 subbundle of the normal bundle. And we can define γ :
TM × TM → L as follows, for any X,Y ∈ TM,
γ(X,Y ) = (α(X,Y ))L.
So the second fundamental form splits orthogonally as:
α(X,Y ) = hAw1 X,Y i w1 ⊕ γ(X,Y ).
We claim that γ satisfies the condition in Theorem 3.13. For any X,Y ∈ TM, let
Z = X − iJX ∈ V and W = Y − iJY ∈ V . By our assumption, we have
(HZW )L = 0 and h(SZW )L, (SZW )Li = 0.
On the other hand, we have
(HZW )L = γ(X − iJX, Y + iJY ) = γ(X,Y ) + γ(JX,JY ) + i(γ(X,JY ) − γ(JX,Y )), this implies
γ(X,Y ) = −γ(JX,JY ) and γ(JX,Y ) = γ(X,JY )
Also, we have
(SZW )L = γ(X − iJX, Y − iJY )
= γ(X,Y ) − γ(JX,JY ) − i(γ(X,JY ) + γ(JX,Y ))
= 2(γ(X,Y ) − iγ(JX,Y ))
29 and
1 h(S ) , (S ) i = hγ(X,Y ) − iγ(JX,Y ), γ(X,Y ) − iγ(JX,Y )i 4 ZW L ZW L = kγ(X,Y )k2 − kγ(JX,Y )k2 − 2i hγ(X,Y ), γ(JX,Y )i .
This implies
kγ(X,Y )k = kγ(JX,Y )k and γ(X,Y ) ⊥ γ(JX,Y ),
and proves our claim. Now, let Γ be its orthogonal complement in img Aw1 ⊕L, which can be viewed as a rank 2 subbundle of TM ⊕ L and Γ ∩ TM = {0}. In the view of
Theorem 3.13, we just need to show, for all µ ∈ Γ and Z ∈ TM,
˜ ∇Z µ ∈ TM ⊕ L,
By viewing TM ⊕ L as a subbundle of TM ⊕ N, this is equivalent to the statement
˜ ∇Z µ ⊥ w1.
Next, we know that µ ∈ Γ ⊂ TM ⊕ L, so
hµ, w1i = 0.
This implies D ˜ E D ˜ E ∇Z µ, w1 + µ, ∇Z w1 = Z hµ, w1i = 0.
But from the construction of Γ, we have
˜ µ ⊥ ∇Z w1, hence D ˜ E D ˜ E ∇Z µ, w1 = − µ, ∇Z w1 = 0.
Finally, apply Theorem 3.13, we get the K¨ahlerextension of f. 30 For the following cases, we get cylinders:
2n 2n+3 Theorem 3.16. Let f : M → R , n ≥ 3, be a real K¨ahlerEuclidean submanifold with νJ = n − 3. Then, each connected component Ui of U0 is isometric to a product
6 n−3 0 0 6 9 Ui = N × C and f|Ui = f × id splits, where f : N → R is a real K¨ahler n−3 ∼ 2(n−3) Euclidean submanifold and id : C = R is the identity map.
2n 2n+3 Theorem 3.17. Let f : M → R , n ≥ 3, be a real K¨ahlerEuclidean submanifold with νJ = n − 2 and suppose S|ker H×ker H = 0. Then, each connected component Ui
4 n−2 0 of U0 is isometric to a product Ui = N × C and f|Ui = f × id splits, where 0 4 7 n−2 ∼ 2(n−2) f : N → R is a real K¨ahlerEuclidean submanifold and id : C = R is the identity map.
2n 2n+3 Theorem 3.18. Let f : M → R , n ≥ 3, be a real K¨ahlerEuclidean submanifold with νJ = n − 2 and suppose S|ker H×ker H 6= 0. Then, each connected component Ui
6 n−3 0 of U0 is isometric to a product Ui = N × C and f|Ui = f × id splits, where 0 6 9 n−3 ∼ 2(n−3) f : N → R is a real K¨ahlerEuclidean submanifold and id : C = R is the identity map.
The above cases have been observed in [FZ08]. And here are the new cases:
2n 2n+3 Theorem 3.19. Let f : M → R , n ≥ 3, be a real K¨ahlerEuclidean submanifold that is in the case (1) of Lemma 3.10. Then, each connected component Ui of U0 is
2 n−1 0 0 2 5 isometric to a product Ui = N × C and f|Ui = f × id splits, where f : N → R n−1 ∼ 2(n−1) is a real K¨ahlerEuclidean submanifold and id : C = R is the identity map.
Proof. In this case, under the basis given by Lemma 3.10, A is just a 1 by 1 matrix and AH = 0 implies A = 0.
2n 2n+3 Theorem 3.20. Let f : M → R , n ≥ 3, be a real K¨ahlerEuclidean submanifold that is in the case (3) of Lemma 3.11. Then, each connected component Ui of U0 is 31 4 n−2 0 0 4 7 isometric to a product Ui = N × C and f|Ui = f × id splits, where f : N → R n−2 ∼ 2(n−2) is a real K¨ahlerEuclidean submanifold and id : C = R is the identity map.
Proof. In this case, under the basis given by Lemma 3.11, A is a 2 by 2 matrix and
AH = 0 implies 0 α A = . 0 β Also, A is nilpotent implies β = 0. Finally, we have 0 α a 0 0 α 2 AS2 = = . 0 0 0 1 0 0 Note that it is symmetric, α = 0 and hence A = 0.
2n 2n+3 Theorem 3.21. Let f : M → R , n ≥ 3, be a real K¨ahlerEuclidean submanifold that is in the case (1) of Lemma 3.12. Then, each connected component Ui of U0 is
6 n−3 0 0 6 9 isometric to a product Ui = N × C and f|Ui = f × id splits, where f : N → R n−3 ∼ 2(n−3) is a real K¨ahlerEuclidean submanifold and id : C = R is the identity map.
Proof. In this case, under the basis given by Lemma 3.12, A is a 3 by 3 matrix and
AH = 0 implies 0 α β A = 0 γ δ . 0 θ φ Then 0 α β 0 0 0 0 α −bβ AS2 = 0 γ δ 0 1 0 = 0 γ −bδ 0 θ φ 0 0 −b 0 θ −bφ 0 α β 0 0 0 0 bα β AS3 = 0 γ δ 0 b 0 = 0 bγ δ 0 θ φ 0 0 1 0 bθ φ
32 Note that they are symmetric, α = β = 0 and
−bδ = θ, δ = bθ this implies −b2δ = δ, but we have b2 6= −1, so δ = 0 and θ = −bδ = 0. Hence we have 0 0 0 A = 0 γ 0 , 0 0 φ since A is nilpotent, γ = φ = 0 and hence A = 0.
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