FACTA UNIVERSITATIS (NIS)
Ser. Math. Inform. Vol. 27 No 3 (2012), 337–344
ON M2 SURFACES OF BIHARMONIC B-GENERAL HELICESACCORDING TO BISHOP FRAME IN HEISENBERG GROUP
Heis3
Talat Korpinar and Essin Turhan
Abstract. In this paper, we study M2 surfaces of biharmonic B-general helices accord-ing to Bishop frame in the Heisenberg group Heis3. Finally, we characterize the M2
surfaces of biharmonic B-general helices in terms of Bishop frame in the Heisenberggroup Heis3.
1. Introduction
Harmonic maps f : (M, g) −→ (N,h) between manifolds are the critical points ofthe energy
E (f) =
∫M
e (f) vg,(1.1)
where vg is the volume form on (M, g) and
e (f) (x) :=1
2∥df (x)∥2T∗M⊗f−1TN
is the energy density of f at the point x ∈M .
Critical points of the energy functional are called harmonic maps.
In this paper, we study M2 surfaces of biharmonic B-general helices accordingto Bishop frame in the Heisenberg group Heis3. Finally, we characterize the M2
surfaces of biharmonic B-general helices in terms of Bishop frame in the Heisenberggroup Heis3.
Received September 5, 2012.; Accepted September 21, 2012.2010 Mathematics Subject Classification. Primary 14H55; Secondary 14J29, 30F15
337
338 Talat Korpinar and Essin Turhan
2. The Heisenberg Group Heis3
Heisenberg group Heis3 can be seen as the space R3 endowed with the followingmultipilcation:
(x, y, z)(x, y, z) = (x+ x, y + y, z + z − 1
2xy +
1
2xy).(2.1)
Heis3 is a three-dimensional, connected, simply connected and 2-step nilpotent Liegroup.
The Riemannian metric g is given by
g = dx2 + dy2 + (dz − xdy)2.
The Lie algebra of Heis3 has an orthonormal basis
e1 =∂
∂x, e2 =
∂
∂y+ x
∂
∂z, e3 =
∂
∂z,(2.2)
for which we have the Lie products
[e1, e2] = e3, [e2, e3] = [e3, e1] = 0
withg(e1, e1) = g(e2, e2) = g(e3, e3) = 1.
3. Biharmonic B-General Helices with Bishop Frame In TheHeisenberg Group Heis3
Let γ : I −→ Heis3 be a non geodesic curve on the Heisenberg group Heis3 pa-rameterized by arc length. Let {T,N,B} be the Frenet frame fields tangent to theHeisenberg group Heis3 along γ defined as follows:
T is the unit vector field γ′ tangent to γ, N is the unit vector field in thedirection of ∇TT (normal to γ), and B is chosen so that {T,N,B} is a positivelyoriented orthonormal basis. Then, we have the following Frenet formulas:
∇TT = κN,
∇TN = −κT+ τB,(3.1)
∇TB = −τN,
where κ is the curvature of γ and τ is 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.
On M2 Surfaces of Biharmonic B General Helices... 339
In the rest of the paper, we suppose everywhere κ = 0 and τ = 0.
The Bishop frame or parallel transport frame is an alternative approach todefining a moving frame that is well defined even when the curve has a vanishingsecond derivative. The Bishop frame is expressed as
∇TT = k1M1 + k2M2,
∇TM1 = −k1T,(3.3)
∇TM2 = −k2T,
where
g (T,T) = 1, g (M1,M1) = 1, g (M2,M2) = 1,(3.4)
g (T,M1) = g (T,M2) = g (M1,M2) = 0.
Here, we shall call the set {T,M1,M2} as Bishop trihedra, k1 and k2 as Bishopcurvatures, where θ (s) = arctan k2
k1, τ(s) = θ′ (s) and κ(s) =
√k22 + k21. Thus,
Bishop curvatures are defined by
k1 = κ(s) cos θ (s) ,(3.5)
k2 = κ(s) sin θ (s) .
With respect to the orthonormal basis {e1, e2, e3} we can write
T = T 1e1 + T 2e2 + T 3e3,
M1 = M11 e1 +M2
1 e2 +M31 e3,(3.6)
M2 = M12 e1 +M2
2 e2 +M32 e3.
Theorem 3.1. γ : I −→ Heis3 is a biharmonic curve with Bishop frame if andonly if
k21 + k22 = constant = C = 0,
k′′1 − Ck1 = k1
[1
4−(M3
2
)2]− k2M31M
32 ,(3.7)
k′′2 − Ck2 = k1M31M
32 + k2
[1
4−(M3
1
)2].
To separate a general helix according to Bishop frame from that of Frenet- Serretframe, in the rest of the paper, we shall use notation for the curve defined above asB-general helix, [10].
340 Talat Korpinar and Essin Turhan
4. M2 Surface of Biharmonic B-General Helices with Bishop Frame InThe Heisenberg Group Heis3
The M2 surface of γB is a ruled surface
E (s, u) = γB (s) + uM2 (s) .(4.1)
Lemma 4.2. Let γB : I −→ Heis3 be a unit speed biharmonic B-general helixwith non-zero natural curvatures. Then the M2 surface of γB is
E (s, u) = [sin θ
(k21+k2
2
sin2 θ− cos θ)
12
sin[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0]
+u cos θ cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0] + ζ2]e1
+[− sin θ
(k21+k2
2
sin2 θ− cos θ)
12
cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0]
+u cos θ sin[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0] + ζ3]e2(4.2)
+[−[sin θ
(k21+k2
2
sin2 θ− cos θ)
12
sin[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0] + ζ2]
[− sin θ
(k21+k2
2
sin2 θ− cos θ)
12
cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0] + ζ3]
+ (cos θ) s+sin2 θ
(k21+k2
2
sin2 θ− cos θ)
12
(s
2−
sin 2[(k21+k2
2
sin2 θ− cos θ)
12 s+ ζ0]
4(k21+k2
2
sin2 θ− cos θ)
12
− ζ1 sin θ
(k21+k2
2
sin2 θ− cos θ)
12
cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0]− u sin θ + ζ4]e3,
where ζ0, ζ1, ζ2, ζ3, ζ4 are constants of integration.
Proof. Using orthonormal basis (2.2) and (3.7), we obtain
T = (sin θ cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0], sin θ sin[(
k21 + k22sin2 θ
− cos θ)12 s+ ζ0],
cos θ +sin2 θ
(k21+k2
2
sin2 θ− cos θ)
12
sin2[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0](4.3)
+ζ1 sin θ sin[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0]),
where ζ1 is the constant of integration.
On M2 Surfaces of Biharmonic B General Helices... 341
T = sin θ cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0]e1 + sin θ sin[(
k21 + k22sin2 θ
− cos θ)12 s+ ζ0]e2
+cos θe3.(4.4)
On the other hand, using Bishop formulas (3.3) and (2.1), we have
M2 = cos θ cos[(k21 + k22sin2 θ
−cos θ)12 s+ζ0]e1+cos θ sin[(
k21 + k22sin2 θ
−cos θ)12 s+ζ0]e2−sin θe3.
(4.5)
Using the above equation, we have (4.2), and the theorem is proved.
We need the following lemma.
Lemma 4.2. Let γB : I −→ Heis3 be a unit speed biharmonic B-general helixwith non-zero natural curvatures. Then the M2 surface of γB are
xE (s, u) = [sin θ
(k21+k2
2
sin2 θ− cos θ)
12
sin[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0]
+u cos θ cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0] + ζ2],
yE (s, u) = [− sin θ
(k21+k2
2
sin2 θ− cos θ)
12
cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0]
+u cos θ sin[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0] + ζ3],
zE (s, u) = [sin θ
(k21+k2
2
sin2 θ− cos θ)
12
sin[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0]
+u cos θ cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0] + ζ2]
[− sin θ
(k21+k2
2
sin2 θ− cos θ)
12
cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0]
+u cos θ sin[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0]] + ζ3]
+[−[sin θ
(k21+k2
2
sin2 θ− cos θ)
12
sin[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0] + ζ2]
[− sin θ
(k21+k2
2
sin2 θ− cos θ)
12
cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0] + ζ3]
+ (cos θ) s+sin2 θ
(k21+k2
2
sin2 θ− cos θ)
12
(s
2−
sin 2[(k21+k2
2
sin2 θ− cos θ)
12 s+ ζ0]
4(k21+k2
2
sin2 θ− cos θ)
12
342 Talat Korpinar and Essin Turhan
− ζ1 sin θ
(k21+k2
2
sin2 θ− cos θ)
12
cos[(k21 + k22sin2 θ
− cos θ)12 s+ ζ0]− u sin θ + ζ4],
where ζ0, ζ1, ζ2, ζ3, ζ4 are constants of integration.
Proof. Using the orthonormal basis we easily have the above system. Hence,the proof is completed.
-1
0
1
2
3
-5
0
5
-5
0
5
Fig. 4.1: The first illustration.
On M2 Surfaces of Biharmonic B General Helices... 343
-1
0
1
2
3
-2
0
2
4
6
-5
0
5
Fig. 4.2: The second illustration
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344 Talat Korpinar and Essin Turhan
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Talat KORPINAR
Fırat University, Department of Mathematics
23119, Elazıg, TURKEY
Essin TURHAN
Fırat University, Department of Mathematics
23119, Elazıg, TURKEY