Gen. Math. Notes, Vol. 5, No. 2, August 2011, pp.67-74
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
Bertrand Mate of Biharmonic Reeb Curves
in 3-Dimensional Kenmotsu Manifold
Talat Körpinar1 and Essin Turhan2
1
Fırat University, Department of Mathematics, 23119, Elazığ, Turkey
E-mail: talatkorpinar@gmail.com
2
Fırat University, Department of Mathematics, 23119, Elazığ, Turkey
E-mail: essin.turhan@gmail.com
(Received: 14-5-11/ Accepted: 27-6-11)
Abstract
In this article, we study biharmonic Reeb curves in 3-dimensional Kenmotsu manifold. Moreover, we apply biharmonic Reeb curves in special 3dimensional Kenmotsu manifold K. Finally, we characterize Bertrand mate
of the biharmonic Reeb curves in terms of their curvature and torsion in special
3-dimensional Kenmotsu manifold K.
Keywords: Kenmotsu manifold, biharmonic curve, Bertrand curve, Reeb
vector field.
1
Introduction
In the theory of space curves in differential geometry, the associated curves, the
curves for which at the corresponding points of them one of the Frenet vectors
of a curve coincides with the one of the Frenet vectors of the other curve
have an important role for the characterizations of space curves. The wellknown examples of such curves are Bertrand curves. These special curves are
very interesting and characterized as a kind of corresponding relation between
two curves such that the curves have the common principal normal i.e., the
Bertrand curve is a curve which shares the normal line with another curve.
These curves have an important role in the theory of curves.
Let (N, h) and (M, g) be Riemannian manifolds. A smooth map φ : N −→
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Talat Körpinar et al.
M is said to be biharmonic if it is a critical point of the bienergy functional:
Z
1
E2 (φ) =
|T (φ)|2 dvh ,
2
N
where the section T (φ) := tr∇φ dφ is the tension field of φ.
The Euler–Lagrange equation of the bienergy is given by T2 (φ) = 0. Here
the section T2 (φ) is defined by
T2 (φ) = −∆φ T (φ) + trR (T (φ), dφ) dφ,
(1.1)
and called the bitension field of φ. Obviously, every harmonic map is biharmonic. Non-harmonic biharmonic maps are called proper biharmonic maps.
In this article, we study biharmonic Reeb curves in 3-dimensional Kenmotsu manifold. Moreover, we apply biharmonic Reeb curves in special 3dimensional Kenmotsu manifold K. Finally, we characterize Bertrand mate of
the biharmonic Reeb curves in terms of their curvature and torsion in special
3-dimensional Kenmotsu manifold K.
2
Preliminaries
Let M 2n+1 (φ, ξ, η, g) be an almost contact Riemannian manifold with 1-form η,
the associated vector field ξ, (1, 1)-tensor field φ and the associated Riemannian
metric g. It is well known that [2]
φξ = 0, η (ξ) = 1, η (φX) = 0,
φ2 (X) = −X + η (X) ξ,
g (X, ξ) = η (X) ,
g (φX, φY ) = g (X, Y ) − η (X) η (Y ) ,
(2.1)
(2.2)
(2.3)
(2.4)
for any vector fields X, Y on M . Moreover,
(∇X φ) Y = −η (Y ) φ (X) − g (X, φY ) ξ,
X, Y ∈ χ (M ) ,
∇X ξ = X − η (X) ξ,
(2.5)
(2.6)
where ∇ denotes the Riemannian connection of g, then (M, φ, ξ, η, g) is called
an Kenmotsu manifold [2].
In Kenmotsu manifolds the following relations hold [2]:
(∇X η) Y
η (R (X, Y ) Z)
R (X, Y ) ξ
R (ξ, X) Y
R (ξ, X) ξ
=
=
=
=
=
g (φX, φY ) , 2.7
η (Y ) g (X, Z) − η (X) g (Y, Z) , 2.8
η (X) Y − η (Y ) X, 2.9
η (Y ) X − g (X, Y ) ξ, 2.10
X − η (X) ξ, 2.11
where R is the Riemannian curvature tensor.
(1)
(2)
(3)
(4)
(5)
Bertrand Mate of Biharmonic Reeb Curves...
3
69
Biharmonic Reeb Curves in the 3-Dimensional
Kenmotsu Manifold
Let γ be a curve on the 3-dimensional Kenmotsu manifold parametrized by arc
length. Let {T, N, B} be the Frenet frame fields tangent to the 3-dimensional
Kenmotsu manifold along γ defined as follows:
T is the unit vector field γ 0 tangent to γ, N is the unit vector field in
the direction of ∇T T (normal to γ), and B is chosen so that {T, N, B} is
a positively oriented orthonormal basis. Then, we have the following Frenet
formulas:
∇T T = κN,
∇T N = −κT + τ B, 3.1
∇T B = −τ N,
(6)
where κ is the curvature of γ and τ its torsion and
g (T, T) = 1, g (N, N) = 1, g (B, B) = 1, 3.2
g (T, N) = g (T, B) = g (N, B) = 0.
(7)
Lemma 3.1. (see [13]) If γ is a biharmonic Reeb curve which are either
tangent or normal to the Reeb vector field 3-dimensional Kenmotsu manifold,
then γ is a helix.
We consider the special 3-dimensional manifold
K = (x, y, z) ∈ R3 : (x, y, z) 6= (0, 0, 0) ,
where (x, y, z) are the standard coordinates in R3 . The vector fields
e1 = z
∂
∂
∂
, e2 = z , e3 = −z
∂x
∂y
∂z
(3.3)
are linearly independent at each point of K. Let g be the Riemannian metric
defined by
g (e1 , e1 ) = g (e2 , e2 ) = g (e3 , e3 ) = 1, 3.4
g (e1 , e2 ) = g (e2 , e3 ) = g (e1 , e3 ) = 0.
(8)
The characterising properties of χ(K) are the following commutation relations:
[e1 , e2 ] = 0, [e1 , e3 ] = e1 , [e2 , e3 ] = e2 .
Let η be the 1-form defined by
η(Z) = g(Z, e3 ) for any Z ∈ χ(M )
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Talat Körpinar et al.
Let φ be the (1,1) tensor field defined by
φ(e1 ) = −e2 , φ(e2 ) = e1 , φ(e3 ) = 0.
Then using the linearity of and g we have
η(e3 ) = 1,
φ2 (Z) = −Z + η(Z)e3 ,
g (φZ, φW ) = g (Z, W ) − η(Z)η(W ),
for any Z, W ∈ χ(K). Thus for e3 = ξ, (φ, ξ, η, g) defines an almost contact
metric structure on K.
Now, we consider biharmonicity of curves in the special three-dimensional
Kenmotsu manifold K.
Theorem 3.4. (see [13]) Let γ : I −→ K be a unit speed biharmonic Reeb
curve which are either tangent or normal to the Reeb vector field 3-dimensional
Kenmotsu manifold K. Then, the parametric equations of γ are
κ
κ
C1 sin5 ϕ
− cos ϕs
e
(
cos(
s + σ)
x (s) = 2
4
2
κ + sin ϕ cos2 ϕ
sin ϕ
sin2 ϕ
κ
+ cos ϕ sin( 2 s + σ)) + C2 ,
sin ϕ
5
C1 sin ϕ
κ
y (s) = 2
e− cos ϕs (− cos ϕ cos( 2 s + σ)3.14
4
2
κ + sin ϕ cos ϕ
sin ϕ
κ
κ
+ 2 sin( 2 s + σ)) + C3 ,
sin ϕ
sin ϕ
− cos ϕs
z (s) = C1 e
,
(9)
where C, C1 , C2 , C3 are constants of integration.
4
Bertrand Mate of Biharmonic Reeb Curves
in the Special Three-Dimensional Kenmotsu
Manifold K
A curve γ : I −→ K with κ 6= 0 is called a Bertrand curve if there exist a
curve γB : I −→ K such that the principal normal lines of γ and γB at s ∈ I
are equal. In this case γB is called a Bertrand mate of γ.
71
Bertrand Mate of Biharmonic Reeb Curves...
On the other hand, let γ : I −→ K be a Bertrand curve parametrized by
arc length. A Bertrand mate of γ is as follows:
γB (s) = γ (s) + λN (s) ,
∀s ∈ I,
(4.1)
where λ is constant.
Theorem 4.1. Let γ : I −→ K be a biharmonic curve parametrized by arc
length. If γB is a Bertrand mate of γ, then the parametric equations of γB
are
κ
λ sin ϕ
κ
κ
xB (s) =
cos( 2 s + σ) + cos ϕ sin( 2 s + σ) C̄1 s + C̄2
2
κ
sin ϕ
sin ϕ
sin ϕ
3
C1 sin ϕ − cos ϕs
κ
κ
+
e
(− cos σ cos( 2 s) + sin σ sin( 2 s)) + C2 , 4.2(10)
κ
sin ϕ
sin ϕ
κ
κ
κ
λ sin ϕ
− 2 sin( 2 s + σ) + cos ϕ cos( 2 s + σ) C̄1 s + C̄2
yB (s) =
κ
sin ϕ
sin ϕ
sin ϕ
3
C1 sin ϕ − cos ϕs
κ
κ
e
(sin σ cos( 2 s + σ) + cos σ sin( 2 s + σ)) + C3 ,
κ
sin ϕ
sin ϕ
λ
zB (s) =
C̄1 s + C̄2 + C1 e− cos ϕs ,
κ
where σ, C̄1 , C̄2 , C1 , C2 , C3 are constants of integration.
Proof. Assume that T is
T =T1 e1 + T2 e2 + T3 e3 ,
(4.3)
where T1 , T2 , T3 are differentiable functions on I.
From [13], we obtain
T = sin ϕ sin(
κ
κ
s + σ)e2 + cos ϕe3 .
2 s + σ)e1 + sin ϕ cos(
sin ϕ
sin2 ϕ
(4.4)
Using (3.3) in (4.4), we obtain
T = (z sin ϕ sin(
κ
κ
s + σ), −z cos ϕ).
2 s + σ), z sin ϕ cos(
sin ϕ
sin2 ϕ
(4.5)
Because, by making use of (3.3), we have
∇T T = (T10 + T1 T3 ) e1 + (T20 + T2 T3 ) e2 + T30 e3 .
From (3.1) and (4.4), we get
(4.6)
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Talat Körpinar et al.
κ
κ
κ
∇T T = sin ϕ
cos
s + σ + cos ϕ sin
s + σ e1 4.7(11)
sin2 ϕ
sin2 ϕ
sin2 ϕ
κ
κ
κ
+ sin ϕ − 2 sin
s + σ + cos ϕ cos
s+C
e2 .
sin ϕ
sin2 ϕ
sin2 ϕ
Then, by using Frenet formulas (3.1), we get
1
∇T T
κ
κ
κ
κ
1
cos( 2 s + σ) + cos ϕ sin( 2 s + σ) e1 4.8(12)
[sin ϕ
=
κ
sin2 ϕ
sin ϕ
sin ϕ
κ
κ
κ
+ sin ϕ − 2 sin( 2 s + σ) + cos ϕ cos( 2 s + σ) e2 ].
sin ϕ
sin ϕ
sin ϕ
N =
Finally, we substitute (3.5) and (4.8) into (4.1), we get (4.2). The proof is
completed.
Corollary 4.2. Let γ : I −→ K be a biharmonic curve parametrized by arc
length. If γB is a Bertrand mate of γ, then the parametric equations of γB in
terms of τ are
√
√
√
λ sin ϕ
1 − τ2
1 − τ2
1 − τ2
C̄
s
+
C̄
cos(
s
+
σ)
+
cos
ϕ
sin(
s
+
σ)
xB (s) = √
1
2
sin2 ϕ
sin2 ϕ
sin2 ϕ
1 − τ2
√
√
C1 sin3 ϕ − cos ϕs
1 − τ2
1 − τ2
+√
e
(− cos σ cos(
s)
+
sin
σ
sin(
s)) + C2 ,
sin2 ϕ
sin2 ϕ
1 − τ2
√
√
√
1 − τ2
1 − τ2
1 − τ2
λ sin ϕ
−
C̄
s
+
C̄
yB (s) = √
sin(
s
+
σ)
+
cos
ϕ
cos(
s
+
σ)
1
2
sin2 ϕ
sin2 ϕ
sin2 ϕ
1 − τ2
√
√
C1 sin3 ϕ − cos ϕs
1 − τ2
1 − τ2
√
e
(sin σ cos(
s
+
σ)
+
cos
σ
sin(
s + σ)) + C3 ,
sin2 ϕ
sin2 ϕ
1 − τ2
λ
zB (s) = √
C̄1 s + C̄2 + C1 e− cos ϕs ,
1 − τ2
where σ, C̄1 , C̄2 , C1 , C2 , C3 are constants of integration.
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