On Real Hypersurfaces of Type A in a Complex Space Form (III)

Hyang Sook Kim and Yong-Soo Pyo

Abstract We denote by M_{n}(c) a complex space form with the metric of constant holomorphic sectional curvature c and M a real hypersurface in M_{n}(c). Then M has an almost contact metric structure (ϕ, ξ, η, g) induced from the Kähler metric and the almost complex structure of M_{n}(c). We will give characterizations of homogeneous real hypersurfaces of type A under the condition that ∇_{ξ}A = f(Aϕ−ϕA)−df(ξ)I, 2f ≠ −g(Aξ, ξ) for a smooth function f without zero points, where I denotes the identity transformation and A mean the shape operator of M.

Mathematics Subject Classification: 53C40 Key words: complex space form, homogeneous real hypersurface, shape operator

1 Introduction

A complex n-dimensional Kähler manifold of constant holomorphic sectional curvature c is called a complex space form, which is denoted by M_{n}(c). A complete and simply connected complex space form consists of a complex projective space P_{n}C, a complex Euclidean space C^{n} or a complex hyperbolic space H_{n}C, according as c > 0, c = 0 or c < 0.

Now, let M be a real hypersurface of an n-dimensional complex space form M_{n}(c). Then M has an almost contact metric structure (ϕ, ξ, η, g) induced from the Kähler metric and the almost complex structure of M_{n}(c). Okumura [7] and Montiel and Romero [6] proved the following Theorem A. Let M be a real hypersurface of P_{n}C, n ≥ 2. If it satisfies

(1.1)  Aϕ − ϕA = 0,

then M is locally congruent to a tube of radius r over one of the following Kähler submanifolds:

(A_{1}) a hyperplane P_{n−1}C, where 0 < r < π/2,

(A_{2}) a totally geodesic P_{k}C (1 ≤ k ≤ n − 2), where 0 < r < π/2, where A is the shape operator in the direction of the unit normal C on M. Theorem B. Let M be a real hypersurface of H_{n}C, n ≥ 2. If it satisfies (1.1), then M is locally congruent to one of the following hypersurfaces:

Balkan Journal of Geometry and Its Applications, Vol.3, No.2, 1998, pp. 101-110 \begin{matrix}c \\ ◯ \end{matrix}Balkan Society of Geometers, Geometry Balkan Press

(A_{0}) a horosphere in H_{n}C,

(A_{1}) a tube of a totally geodesic hyperplane H_{n−1}C,

(A_{2}) a tube of a totally geodesic H_{k}C (1 ≤ k ≥ n − 2).

Such real hypersurfaces in Theorems A and B are said to be of type A. The following theorem is proved by Maeda and Udagawa [4] under the condition that the structure vector ξ is principal, and recently by Kimura and Maeda [3] and Ki, Kim and Lee [1] without the above assumption. Theorem C. Let M be a real hypersurface of M_{n}(c), c ≠ 0, n ≥ 2. If it satisfies

∇_{ξ}A = 0, g(Aξ, ξ) ≠ 0,

then M is of type A, where ∇ is the Riemannian connection on M.

In his previous paper [9], the second named auther proved the folllwing Theorem D. Let M be a real hypersurface of M_{n}(c), c ≠ 0, n ≥ 2. If it satisfies

∇_{ξ}A = a(Aϕ − ϕA), 2a ≠ −g(Aξ, ξ)

for some non-zero constant a, then M is of type A.

The purpose of this article is to generalize slightly Theorem D and to prove the following results. Theorem 1. Let M be a real hypersurface of M_{n}(c), c ≠ 0, n ≥ 2. If it satisfies

(1.2)  ∇_{ξ}A = f(Aϕ − ϕA) − df(ξ)I, 2f ≠ −g(Aξ, ξ)

for a smooth function f without zero points and the identity transformation I, then M is of type A. Theorem 2. Let M be a real hypersurface of M_{n}(c), c ≠ 0, n ≥ 2. If it satisfies

(1.3)  L_{ξ}(H + fg) = 0, 2f ≠ −g(Aξ, ξ)

for a smooth function f without zero points, then M is of type A, where L_{ξ} is the Lie derivative with respect to ξ and H is second fundamental form of M in M_{n}(c), namely H(X, Y) = g(AX, Y) for any vecror fields X and Y.

2 Preliminaries

First of all, we recall fundamental properties about real hypersurfaces of a complex space form. Let M be a real hypersurface of a complex n-dimensional complex space form (M_{n}(c), g) of constant holomorphic sectional curvature c, and let C be a unit normal vector field on a neighborhood in M. We denote by J the almost complex structure of M_{n}(c). For a local vector field X on the neighborhood in M, the images of X and C under the linear transformation J can be represented as

JX = ϕX + η(X)C, JC = −ξ,

where ϕ defines a skew-symmetric transformation on the tangent bundle T M of M, while η and ξ denote a 1-form and a vector field on the neighborhood in M, respectively. Then it is seen that g(ξ, X) = η(X), where g denotes the Riemannian metric tensor on M induced from the metric tensor on M_{n}(c). The set of tensors (ϕ, ξ, η, g) is called an almost contact metric structure on M. They satisfy the following properties :

ϕ^{2} = −I + η ⊗ ξ, ϕξ = 0, η(ξ) = 1,

where I denotes the identity transformation. Furthermore the covariant derivatives of the structure tensors are given by

(2.1)  ∇_{X}ξ = ϕAX, ∇_{X}ϕ(Y) = η(Y)AX − g(AX, Y)ξ

for any vector fields X and Y on M, where ∇ is the Riemannian connection on M and A denotes the shape operator of M in the direction of C.

Since the ambient space is of constant holomorphic sectional curvature c, the equations of Gauss and Codazzi are respectively given as follows:

(2.2)  \begin{aligned}R(X, Y)Z & = & \frac{c}{4}\{g(Y, Z)X − g(X, Z)Y \\ & + & g(ϕY, Z)ϕX − g(ϕX, Z)ϕY − 2g(ϕX, Y)ϕZ\} \\ & + & g(AY, Z)AX − g(AX, Z)AY, \end{aligned}

(2.3)  ∇_{X}A(Y) − ∇_{Y} A(X) = \frac{c}{4}\{η(X)ϕY − η(Y)ϕX − 2g(ϕX, Y)ξ\},

where R denotes the Riemannian curvature tensor of M and ∇_{X}A denotes the covariant derivative of the shape operator A with respect to X.

Next, we suppose that the structure vector field ξ is principal with corresponding principal curvature α, namely Aξ = αξ. Then it is seen in [2] and [5] that α is constant on M and it satisfies

(2.4)  2AϕA = \frac{c}{2}ϕ + α(Aϕ + ϕA).

3 Proof of Theorems

Let M be a real hypersurface of M_{n}(c), c ≠ 0, n ≥ 2. First of all, we shall give a sufficient condition for the structure vector field ξ to be principal. We suppose that ξ is principal, i.e., Aξ = αξ, where α is constant. Then, by (2.1) and (2.4), we get

∇_{X}A(ξ) = −\frac{c}{4}ϕX − \frac{1}{2}α(Aϕ − ϕA)X,

from which together with (2.3) it follows that we have

(3.1)  ∇_{ξ}A = −\frac{1}{2}α(Aϕ − ϕA).

Taking account of this property and the already known some facts, in order to prove our theorems, we shall assert the following Proposition 3.1. Let M be a real hypersurface of M_{n}(c), c ≠ 0, n ≥ 2. If it satisfies

(3.2)  ∇_{ξ}A = f(Aϕ − ϕA) − df(ξ)I

for a smooth function f without zero points, then ξ is principal, and hence df(ξ) = 0. By the assumption (3.2) and (2.3), it turns out to be

(3.3)  ∇_{Y} A(ξ) = f(Aϕ − ϕA)Y − df(ξ)Y − \frac{c}{4}ϕY.

Differentiating this equation with respect to X covariantly and using (2.1), we get

(3.4)  \begin{aligned}∇_{X}∇_{Y} A(ξ) & = & f\{∇_{X}A(ϕY) + g(Y, ξ)A^{2}X − g(AX, Y)Aξ \\ & − & g(AY, ξ)AX + g(AX, AY)ξ − ϕ∇_{X}A(Y)\} \\ & − & \frac{c}{4}\{g(Y, ξ)AX − g(AX, Y)ξ\} − ∇_{Y} A(ϕAX) \\ & + & df(X)(Aϕ − ϕA)Y \end{aligned}

for any vector fields X and Y. Since the Ricci formula for the shape operator A is given by

∇_{X}∇_{Y} A(Z) − ∇_{Y} ∇_{X}A(Z) = R(X, Y)(AZ) − A(R(X, Y)Z),

from (2.2), (2.3) and (3.4), it follows that (3.5)

∇_{X}A(ϕAY) − ∇_{Y} A(ϕAX) + f{∇_{X}A(ϕY) − ∇_{Y} A(ϕX)}
= −{fg(Y, ξ) + g(AY, ξ)}A^{2}X + {fg(X, ξ) + g(AX, ξ)}A^{2}Y
+ {fg(AY, ξ) + g(A^{2}Y, ξ)}AX
− {fg(AX, ξ) + g(A^{2}X, ξ)}AY
+ \frac{c}{4}\{fg(Y, ξ) + g(AY, ξ)\}X − \frac{c}{4}\{fg(X, ξ) + g(AX, ξ)\}Y
+ \frac{c}{4}\{g(AϕY, ξ)ϕX − g(AϕX, ξ)ϕY \} − \frac{c}{2}g(ϕX, Y)ϕAξ
+ df(Y)(Aϕ − ϕA)X − df(X)(Aϕ − ϕA)Y

for any vector fields X and Y.

Now, in order to prove Proposition 3.1, we shall express (3.5) with the simpler form. The inner product of (3.5) and ξ, combining with (2.3) and (3.2), implies

(3.6)  \begin{aligned}fg((AϕAϕ & − & ϕAϕA)X, Y) \\ & + & f^{2}\{g(X, ξ)g(AY, ξ) − g(Y, ξ)g(AX, ξ)\} \\ & − & df(ξ)\{g((Aϕ + ϕA)X, Y) + 2fg(ϕX, Y)\} \\ & + & f\{g(X, ξ)g(A^{2}Y, ξ) − g(Y, ξ)g(A^{2}X, ξ)\} \\ & + & 2\{g(AX, ξ)g(A^{2}Y, ξ) − g(AY, ξ)g(A^{2}X, ξ)\} \\ & − & df(X)g(AϕY, ξ) + df(Y)g(AϕX, ξ) = 0 \end{aligned}

for any vector fields X and Y. Since Y is any vector field, we get

{f(AϕAϕ − ϕAϕA) − df(ξ)(Aϕ + ϕA)}X − 2fdf(ξ)ϕX
+ {fg(X, ξ) + 2g(AX, ξ)}A^{2}ξ + {f^{2}g(X, ξ)
− 2g(A^{2}X, ξ)}Aξ − f{fg(AX, ξ) + g(A^{2}X, ξ)}ξ
+ df(X)ϕAξ + g(AϕX, ξ)∇f = 0

for any vector field X, where we denote by ∇f the gradient of the function f. On the other hand, taking account of (2.1) and the skew-symmetry of the transformation ϕ, we have

(3.7)  g((AϕAϕ − ϕAϕA)X, ϕX) = g(X, ξ)g(AϕAX, ξ).

Putting Y = ϕX in (3.6) and applying the above property, we get

(3.8) \begin{aligned}fg(X, ξ)\{g(AϕAX, ξ) & + & fg(AϕX, ξ) + g(A^{2}ϕX, ξ)\} \\ & + & 2\{g(AX, ξ)g(A^{2}ϕX, ξ) − g(AϕX, ξ)g(A^{2}X, ξ)\} \\ & − & df(ξ)\{g((Aϕ + ϕA)X, ϕX) + 2fg(ϕX, ϕX)\} \\ & − & df(X)g(Aϕ^{2}X, ξ) + df(ϕX)g(AϕX, ξ) = 0. \end{aligned}

Let T_{0} be a distribution defined by the subspace T_{0}(x) = {u ∈ T_{x}M : g(u, ξ(x)) = 0} of the tangent space T_{x}M of M at any point x, which is called a holomorphic distribution.

Now, suppose that the structure vector field ξ is not principal. Then we can put Aξ = αξ + βU, where U is a unit vector field in the holomorphic distribution T_{0}, and α and β are smooth functions on M. So we may consider the case that the function β does not vanish identically on M. Let M_{0} be the non-empty open subset of M consisting of points x at which β(x) ≠ 0. And we put AU = βξ + γU + δV, where U and V are orthonormal vector fields in T_{0}, and γ and δ are smooth functions on M_{0}. And let L(ξ, U) be a distribution spanned by ξ and U.

For any vector field X belonging to the holomorphic distribution T_{0}, (3.8) is simplified as

2{g(AX, ξ)g(A^{2}ϕX, ξ) − g(AϕX, ξ)g(A^{2}X, ξ)}
− df(ξ){g((Aϕ + ϕA)X, ϕX) + 2fg(ϕX, ϕX)}
+ β{df(X)g(X, U) + df(ϕX)g(ϕX, U)} = 0.

Furthermore, we can see that this equation holds for any vector field X. By the polarization of the above equation, we have

2{g(AX, ξ)g(A^{2}ϕY, ξ) − g(AϕX, ξ)g(A^{2}Y, ξ)
+g(AY, ξ)g(A^{2}ϕX, ξ) − g(AϕY, ξ)g(A^{2}X, ξ)}
−df(ξ){g((Aϕ + ϕA)X, ϕY) + g((Aϕ + ϕA)Y, ϕX)
+4fg(ϕX, ϕY)} + β{df(X)g(Y, U) + df(ϕX)g(ϕY, U)
+df(Y)g(X, U) + df(ϕY)g(ϕX, U)} = 0

for any vector fields X and Y. Hence we have

(3.9)  \begin{matrix}df(ξ)\{ϕ(Aϕ + ϕA)X + (Aϕ + ϕA)ϕX + 4fϕ^{2}X\} \\ −2\{g(AX, ξ)ϕA^{2}ξ + g(AϕX, ξ)A^{2}ξ − g(A^{2}ϕX, ξ)Aξ \\ −g(A^{2}X, ξ)ϕAξ\} + β\{df(X)U − df(ϕX)ϕU \\ +g(X, U)∇f + g(ϕX, U)df(ϕ)\} = 0. \end{matrix}

First, in order to prove Proposition 3.1, we shall show the following Lemma 3.2. The distribution L(ξ, U) is A-invariant on M_{0}, namely we have

(3.10)  AU = βξ + γU

on M_{0}. Proof. On the open subset M_{0}, by the forms Aξ = αξ+βU and AU = βξ+γU +δV, it turns out to be

A^{2}ξ = (α^{2} + β^{2})ξ + β(α + γ)U + βδV.

Thus we can rewrite (3.9) as

(3.11)  \begin{matrix}df(ξ)\{ϕ(Aϕ + ϕA)X + (Aϕ + ϕA)ϕX + 4fϕ^{2}X\} \\ +2\{αg(A^{2}ϕX, ξ) − (α^{2} + β^{2})g(AϕX, ξ)\}ξ \\ +2β\{g(A^{2}ϕX, ξ) − (α + γ)g(AϕX, ξ)\}U − 2βδg(AϕX, ξ)V \\ +2β\{g(A^{2}X, ξ) − (α + γ)g(AX, ξ)\}ϕU − 2βδg(AX, ξ)ϕV \\ +β\{df(X)U − df(ϕX)ϕU + g(X, U)∇f + g(ϕX, U)df(ϕ)\} = 0 \end{matrix}

for any vector field X. The inner product of (3.11) and ξ implies that

αg(ϕX, A^{2}ξ) − (α^{2} + β^{2})g(ϕX, Aξ) = 0

for any vector field X. This gives us

αA^{2}ξ − (α^{2} + β^{2})Aξ = 0

on M_{0} and hence we have

β{(αγ − β^{2})U + αδV } = 0.

Consequently, we have

(3.12)  β^{2} = αγ,  δ = 0

on M_{0}. So it completes the proof. □

Furthermore, by (3.12), we also get

(3.13)  A^{2}ξ = (α + γ)Aξ

on M_{0}.

Next, in order to prove Proposition 3.1, we shall prove the following Lemma 3.3. If it satisfies (3.2), then we have

(3.14)  AϕU = −λϕU,  λ = f + α + γ

on M_{0}. Proof. By the polarization of (3.8) and (3.13), we have

fg(X, ξ){g(AϕAY, ξ) + fg(AϕY, ξ) + g(A^{2}ϕY, ξ)}
+fg(Y, ξ){g(AϕAX, ξ) + fg(AϕX, ξ) + g(A^{2}ϕX, ξ)}
−df(ξ){g((Aϕ + ϕA)X, ϕY) + 4fg(ϕX, ϕY) + g((Aϕ + ϕA)Y, ϕX)}
−df(X)g(Aϕ^{2}Y, ξ) + df(ϕX)g(AϕY, ξ)
−df(Y)g(Aϕ^{2}X, ξ) + df(ϕY)g(AϕX, ξ) = 0

for any vector fields X and Y. Putting Y = ξ, we have

f{g(AϕAX, ξ) + fg(AϕX, ξ) + g(A^{2}ϕX, ξ)} = 0,

because AϕAξ is orthogonal to ξ. Since f has not zero points, we have

AϕAξ + fϕAξ + ϕA^{2}ξ = 0.

This equation, by (3.13), completes the proof. □

We remark here that the property f ≠ 0 is essential to derive the equation (3.14). Lastly, in order to prove Proposition 3.1, we have the following

Lemma 3.4. Assume that A^{2}ξ + hAξ = 0, where h is a smooth function on M_{0}. Then it satisfies

(3.15)  fλ^{2} + (4fγ − 2hγ + \frac{c}{4})λ − f^{2}γ − \frac{c}{4}(2h + 2α + γ) − βdh(ϕU) = 0

on M_{0}. Proof. Differentiating our assumption A^{2}ξ + hAξ = 0 with respect to X and taking account of (2.1), (2.3) and (3.3), we get

∇_{X}A(Aξ) + fA(Aϕ − ϕA)X + fh(Aϕ − ϕA)X + A^{2}ϕAX

+hAϕAX − df(ξ)(AX + hX) − \frac{c}{4}AϕX − \frac{c}{4}hϕX + dh(X)Aξ = 0

for any vector field X. The inner product of this equation with any vector field Y implies

g(∇_{X}A(Y), Aξ) + fg(A(Aϕ − ϕA)X, Y) + fhg((Aϕ − ϕA)X, Y)

+g(A^{2}ϕAX, Y) + hg(AϕAX, Y) − df(ξ)g(AX + hX, Y)

−\frac{c}{4}g(AϕX, Y) − \frac{c}{4}hg(ϕX, Y) + dh(X)g(Aξ, Y) = 0.

Exchanging X and Y in the above equation and substituting the second one from the first one, we have g(∇_{X}A(Y) − ∇_{Y} A(X), Aξ) + fg((A^{2}ϕ − 2AϕA + ϕA^{2})X, Y)

+g((A^{2}ϕA + AϕA^{2})X, Y) + 2hg(AϕAX, Y)

−\frac{c}{4}g((Aϕ + ϕA)X, Y) − \frac{c}{2}hg(ϕX, Y)

+dh(X)g(Aξ, Y) − dh(Y)g(Aξ, X) = 0

for any vector fields X and Y. Putting X = U and Y = ϕU in this equation and taking account of (2.3), (3.10), (3.12) and (3.14), we can easily see that the equation (3.15) holds. □

Now, we are in position to prove Proposition 3.1, namely, to prove the fact that under the condition (3.2), the structure vector ξ is principal. We suppose that the open set M_{0} is not empty. Then, differentiating the form Aξ = αξ + βU with respect to ξ covariantly on M_{0}, we have by (2.1)

∇_{ξ}A(ξ) = dα(ξ)ξ + αβϕU + dβ(ξ)U − βAϕU + β∇_{ξ}U.

This, combining with the assumption (3.2) and (3.14), implies

d(f + α)(ξ)ξ + dβ(ξ)U + β(2f + 2α + γ)ϕU + β∇_{ξ}U = 0.

From the inner product of ξ and U respectively, we get

(3.16)  ∇_{ξ}U = −(2f + 2α + γ)ϕU, d(f + α)(ξ) = 0, dβ(ξ) = 0

on M_{0}, where we have used that g(∇_{ξ}U, ξ) = 0 and g(∇_{ξ}U, U) = 0. By making use of (3.2) and (3.10), γ = g(AU, U) gives us to dγ(ξ) = −df(ξ). Therefore, from (3.14) and (3.16), we get dλ(ξ) = −df(ξ). Differentiating (3.14) with respect to ξ covariantly, and taking account of (2.1) and the above property, we get

∇_{ξ}A(ϕU) − g(AU, ξ)Aξ − λg(AU, ξ)ξ + (Aϕ + λϕ)∇_{ξ}U − df(ξ)ϕU = 0.

By (3.2), (3.10), (3.12), (3.14) and the first equation of (3.16), the above equation gives the following

(3.17)  (f + α + γ)(f + 2α + 2γ) = 0,  df(ξ) = 0

on M_{0}. Since f ≠ 0, we have α + γ ≠ 0 by the above equation.

Now, we consider the first case f + α + γ = 0. By (3.14) and (3.16), we get

(3.18)  AϕU = 0,  ∇_{ξ}U = γϕU.

Differentiating Aξ = αξ + βU with respect to any vector field X covariantly, and taking account of (2.1), (3.3) and the second equation (3.17), we get

\begin{matrix}f(Aϕ − ϕA)X − \frac{c}{4}ϕX + AϕAX − dα(X)ξ \\ −αϕAX − dβ(X)U − β∇_{X}U = 0. \end{matrix}

By taking the inner product of this equation with ξ and U respectively, we get

(3.19)  dα(X) = fβg(ϕX, U),  dβ(X) = (fγ − \frac{c}{4})g(ϕX, U),

where we have used (3.10) and the first equation of (3.18). Owing to β^{2} = αγ, it is easily seen that

2βdβ(X) = γdα(X) + αdγ(X),

from which together with (3.19), it turns out to be

β(fα + fγ − \frac{c}{2})g(ϕX, U) + αdf(X) = 0

for any vector field X, where we have used f + α + γ = 0. This implies β(f^{2} + \frac{c}{2}) + αdf(ϕU) = 0. Hence, by the first equation of (3.12) and (3.15), we get β = 0 on M_{0}, where we have used that λ = 0 and h = f. It leads to a contradiction.

Next, we consider the second case, that is, we suppose that f +2α+2γ = 0. Putting X = ξ and Y = U in (3.5) and from the inner product of ξ and U respectively, we obtain

βg(ϕ∇_{U}U, U) = (f + γ)(f + α + γ) + γ(f + α) + \frac{c}{4}

and

\begin{matrix}β(f + α + 2γ)g(ϕ∇_{U}U, U) = f(f + 2γ)(f + α + γ) \\ +γ^{2}(f + α) − \frac{c}{4}(f + α), \end{matrix}

where we have used (3.2), (3.10), (3.13), (3.14), (3.16) and df(ξ) = dγ(ξ) = 0. Combining of the above two equations, we get

(f + α + γ)(fα + 2fγ + 2αγ + 2γ^{2} + \frac{c}{2}) = 0.

By the supposition f + 2α + 2γ = 0, we have f^{2} = c. Therefore, we obtain α = 0, where we have used (3.15), f + 2α + 2γ = 0 and h = λ = \frac{f}{2} . Hence β = 0 on M_{0} by the first equation of (3.12). Therefore it also leads to a contradition.

Consequently, from these two cases it follows that the subset M_{0} is empty and hence the structure vector field ξ is principal. Thus, combining (3.1) with (3.2), we get df(ξ) = 0. It completes the proof of Proposition 3.1. □ Remark. Recently, Park[8] also give an another sufficient condition for the structure vector field ξ is principal. Proof of Theorem 1. By Proposition 3.1, the structure vector ξ is principal and df(ξ) = 0. Combining (3.1) with the assumption (1.2) of Theorem 1, we have

(2f + α)(Aϕ − ϕA) = 0,

which implies that Aϕ − ϕA = 0. Thus, owing to Theorems A and B the real hypersurface M is of type A.□ Proof of Theorem 2. Since L_{ξ}(H + fg)(X, Y) = g(∇_{ξ}A(X), Y) − fg((Aϕ − ϕA)X, Y) + df(ξ)g(X, Y) for any vector fields X and Y, by the assumption (1.3) of Theorem 2, we have

∇_{ξ}A = f(Aϕ − ϕA) − df(ξ)I.

Hence Theorem 2 is proved by Theorem 1. □

Acknowledgements. This work was supported by 1998 Inje University Research Grant.

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Department of Mathematics Inje University Kimhae 621-749, Korea email: mathkim@ijnc.inje.ac.kr

Division of Mathematical Sciences Pukyong National University Pusan 608-737, Korea email: yspyo@dolphin.pknu.ac.kr