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Minimal surfaces in some 4-dimensional general (α,β)-spaces
J. Math. Anal. Appl. 444 (2016) 1027–1044
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Minimal surfaces in some 4-dimensional general (α, β)-spaces ✩ Zhong-Hua Hou ∗ , Yong-Nan Liu Institute of Mathematics, Dalian University of Technology, Dalian, PR China
a r t i c l e
i n f o
Article history: Received 1 April 2016 Available online 12 July 2016 Submitted by J. Xiao Keywords: Minimal submanifolds General (α, β)-metrics Finsler manifolds Mean curvature vector
a b s t r a c t In this paper, we study minimal submanifolds in some general (α, β)-spaces. We firstly establish the equations for minimal submanifolds in a general (α, β)-space. Then we construct some kinds of minimal surfaces in some 4-dimensional general (α, β)-space. © 2016 Elsevier Inc. All rights reserved.
1. Introduction By using the Busemann–Hausdorff (briefly BH) volume form, Z. Shen [6] introduced the notions of mean curvature and normal curvature for Finsler submanifolds. Later, by using the Holmes–Thompson (briefly HT) volume form, Q. He and Y. B. Shen [4] introduced another notion of mean curvature. Based on the BH-volume form, M. Souza, J. Spruck and K. Tenenblat [8,9] considered a Bernstein type theorem for minimal surfaces of rotation in a Randers–Minkowski space. Based on the HT-volume form, Q. He and W. Yang [7] constructed a minimal rotational hypersurface in a Minkowski (α, β)-space. N. Cui and Y. B. Shen [2] determined the BH-minimal and HT-minimal rotational hypersurfaces generated by plane curves rotating around the axis in the direction of β in Minkowski (α, β)-space. C. Yu and H. Zhu [1] defined a new class of Finsler metrics of the form F = αφ(x, β/α) and called it a general (α, β)-metrics, where φ(x, s) is a C ∞ function, α is a Riemannian metric and β is a 1-form. It is obvious that the navigation expression of a Randers metric F = ( λα2 + β 2 − β)/λ can be viewed as a general (α, β)-metric, where α is a Riemannian metric, β is a 1-form and λ = 1 − β2α . S. T. Yin, Q. He and D. H. Xie in [3] constructed a minimal rotational hypersurface in a non-Minkowski general (α, β)-metric. In this paper, we want to construct some kinds of minimal surfaces in some 4-dimensional non-Minkowski 4 , F ). In Sections 2 and 3, some basic concepts of general (α, β)-metrics and that general (α, β) space (M ✩
This work was supported in part by NSFC (No. 61473059).
* Corresponding author. E-mail addresses: [email protected] (Z.-H. Hou), [email protected] (Y.-N. Liu). http://dx.doi.org/10.1016/j.jmaa.2016.07.004 0022-247X/© 2016 Elsevier Inc. All rights reserved.
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4 , F) are introduced. In Section 4, we construct some kinds of rotational of minimal submanifolds in (M 4 minimal surfaces in (M , F ). In Section 5, we construct some kinds of minimal surfaces produced by a one-parameter subgroup and study the asymptotic behaviour of one kind of them. Finally, we construct a 4 , F) in Section 6. kind of ruled minimal hypersurfaces in (M 2. The (α, β) metric on smooth manifolds Let M be an n-dimensional smooth manifold and π : TM → M be the tangent bundle of M . Let (U, {xi }) be a local coordinate of M . Then (π −1 (U ), {xi , y j }) is the natural local coordinate of M , where Y = y j ∂xj ∈ Tx M for any x ∈ U . We denote ∂h/∂y i , ∂ 2 h/∂y i ∂y j , · · · , by hyi , hyi yj , · · · , for any smooth function h over TM , and use the following convention for index ranges unless otherwise stated: 1 ≤ i, j, k, · · · ≤ n,
1 ≤ A, B, C, · · · ≤ n + p,
n + 1 ≤ a, b, c, · · · ≤ n + p.
A Finsler metric on M is a function F : TM → [0, ∞) with the following properties: (1) F is smooth on TM \ {0}; (2) F (x, λy) = λF (x, y) for all λ > 0; (3) the induced quadratic form g is positive-definite, where g := gij dxi ⊗ dxj ,
1 2 [F ]yi yj . 2
gij =
(2.1)
The projection π : TM → M gives rise to the pull-back bundle π ∗ TM together with its dual bundle π T ∗ M over TM \ {0}. There is a global section ω = [F ]yi dxi in π ∗ T ∗ M which is called the Hilbert form. Its dual L = Li ∂xi with Li = y i /F in π ∗ TM is called the distinguished vector field. For any x ∈ M , let SM = x∈M Sx M with Sx M = {[y]|y ∈ Tx M } be the projective sphere bundle. Denote by gˆ the pull-back of the Sasaki metric from T (TM \ {0}) to T (SM ). Then the volume element dVSM of SM with respect to gˆ is expressed as (cf. [3]) ∗
dVSM = Ωdτ ∧ dx,
(2.2)
where Ω = F −n det(gij ), dx = dx1 ∧ · · · ∧ dxn and dτ =
n i ∧ · · · ∧ dy n . (−1)i−1 y i dy 1 ∧ · · · ∧ dy
(2.3)
i=1
The n-dimensional volume form of (M, F ) is defined to be dVM = σ(x)dx,
σ(x) =
1
Ωdτ,
cn−1
(2.4)
Sx M
where cn−1 denotes the volume of the unit Euclidean (n − 1)-sphere S n−1 . Let φ(x, s) be a positive C ∞ function with respect to x ∈ M and |s| < b0 for some 0 < b0 ≤ ∞. Let α = aij (x)y i y j be a Riemannian metric and β = bi y i be a 1-form with b := βα < b0 . Define F = αφ(x, β/α). Then F is a Finsler metric if and only if φ − sφs + (b2 − s2 )φss > 0,
(2.5)
for all x ∈ M and |s| ≤ b < b0 . Such a kind of Finsler metric is called the general (α, β)-metric on M (cf. [9]). By direct computations, we have (cf. [1]),
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gij = ρaij + ρ0 bi bj + ρ1 (bi αyi + bj αyj ) − sρ1 αyi αyj ,
(2.6)
det(gij ) = φn H(x, s) det(aij ),
(2.7)
g ij = ρ−1 [aij + ηbi bj + η0 α−1 (bi y j + bj y i ) + η1 α−2 y i y j ],
(2.8)
⎧ (g ij ) = (gij )−1 , (aij ) = (aij )−1 , bi = aij bj , ⎪ ⎪ ⎪ ⎪ ⎨ σ = sφ + (b2 − s2 )φs , ρ = φ(φ − sφs ), ρ0 = (φφs )s , ρ1 = ρs , ⎪ ⎪ ⎪ φ ρ σρ ⎪ ⎩ H(x, s) = φ3−n ρn−2 σs , η = − ss , η0 = − s , η1 = − 2 s . σs φσs φ σs
(2.9)
where
Let bi|j denote the coefficients of the covariant derivative of β with respect to α and write ⎧ rij = 12 (bi|j + bj|i ), sij = 12 (bi|j − bj|i ), r00 = rij y i y j , ⎪ ⎪ ⎨ ri = bj rji , si = bj sji , r0 = ri y i , s0 = si y i , ⎪ ⎪ ⎩ i r = aij rj , si = aij sj , r = bi ri , si0 = aij sjk y k .
(2.10)
3. Immersed submanifolds of general (α, β)-spaces , F ) be Finsler manifolds. Let f : (M, F ) → (M , F) be an immersion and dfx : Tx M → Let (M, F ) and (M be the differential of f at x ∈ M . Then f is called isometric if F (x, y) = F(f (x), dfx (y)) for any Tf (x) M (x, y) ∈ TM \ {0}. , { and (U, {xi }) be the local coordinate For isometric immersion f , let (U xi }) be a local coordinate of M ) ⊂ U . Then of M such that f −1 (U f (x) = { xA },
dfx (y) = df (y i ∂xi ) = yB ∂xB = { y B },
(3.1)
where x A = f A (x), yB = fiB y i and fiB = ∂f B /∂xi for any i and B. Then gij (x, y) = gAB ( x, y)fiA fjB .
(3.2)
) with respect to g (cf. [3,5]). Denote Let (π ∗ TM )⊥ be the orthogonal complement of π ∗ TM in π ∗ (f −1 T M h=
hA F2
∂xA ,
A i j A , hA = fij y y − fkA Gk + G
A fij =
∂2f A , ∂xi ∂xj
(3.3)
α are the geodesic coefficients of F and F respectively. Then h ∈ (π ∗ TM )⊥ , which is called where Gk and G the normal curvature of f . The mean curvature form of f is defined by μ=
h A Ωdτ d xA , cn−1 σ F2 1
(3.4)
Sx M
where hA = gAB hB . , F) is called minimal if any compact domain of M is the An isometric immersion f : (M, F ) → (M critical point of its volume functional with respect to any variation of f . , F ) be an isometric immersion. Then f is minimal if and only if Lemma 3.1. (cf. [3]) Let f : (M, F ) → (M μ = 0.
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metric F = α α), where α , F ) be a Finsler space with the general ( Let (M α, β) φ( x, β/ = A and β = bA y . Let f : (M, F ) → (M , F ) be an isometric immersion. Then F = f ∗ F = αφ(f (x), β/α),
aAB yA yB
(3.5)
where α = aij y i y j with aij = aAB fiA fjB and β = bi y i with bi = bA fiA . Let β = bA ∂xA and β⊥ be the projection of β into the normal bundle TM ⊥ with respect to α . Yin et al. [8] discussed the relation between the local orthonormal frame field of TM ⊥ with respect to α and the local orthonormal frame field of (π ∗ TM )⊥ with respect to F. And proved the following result: n+p , F) is an isometric immersion where F is the Lemma 3.2. (cf. [8]) Suppose that f : (M n , F ) → (M n+p general ( α, β) metric. Let {na }a=n+1 be a local orthonormal frame field of TM ⊥ such that nn+p is parallel to β⊥ . Denote a )β + F η0 ( a )L n a = χa na + ηβ(n α−1 β)(n
(3.6)
a )2 ], { is the distinguished where χa = 1/ ρ[1 + ηβ(n ρ, η, η0 } are defined as in (2.9) with respect to F , and L n+p ∗ ⊥ vector field. Then { na } is a local orthonormal frame of (π TM ) with respect to F . a=n+1
By applying Lemma 3.2, they obtained a description of Finslerian minimal immersions: n+p , F) is an isometric immersion where F is the Lemma 3.3. (cf. [8]) Suppose that f : (M n , F ) → (M n+p B general ( α, β) metric. Let {na = na ∂xB }a=n+1 be a local orthonormal frame field of TM ⊥ such that nn+p is parallel to β⊥ . Then f is minimal if and only if aBD nD a Sx M
B i j B ) ρH(fij y y +G dτ = 0, a )2 )φ2 αn+2 (1 + ηβ(n
(3.7)
for all a, where {φ, ρ, H} are defined as in (2.9) with respect to F . metric space β) 4. Minimal submanifolds of a general (α, n+p , F), where Consider the isometric immersion f : (M n , F ) → (M F =
α2 + β2 − β λ, λ
= 1 − β 2 > 0, λ α
(4.1)
is the navigation expression of a Randers metric. Then the induced metric is F = (γ − β)/λ,
γ=
2 + β2. λα
(4.2)
By direct computation, we get η = −1,
ρ=
α2 φ2 , Fγ
α n+1 + b2 ) F H(x, s) = (λ , α γ
where b2 = β2α . Substituting (4.3) into (3.7), we have
(4.3)
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metric (4.1). If α n+p , F) with general ( Lemma 4.1. (cf. [8]) Let (M n , F ) be a submanifold in (M α, β) is a Riemannian metric with constant sectional curvature and β is the Killing vector field, then f : (M n , F ) → n+p , F) is minimal if and only if (M
aBD nD a
B i j yy − fij
{γ=1}
sB (1 − β)2 2(1 − β) B − s0 dτ = 0 2 λ λ
(4.4)
for every a. 4 = {( x1 , x 2 , x 3 , x 4 ) ∈ R4 |( x1 )2 + ( x2 )2 + ( x3 )2 + ( x4 )2 < 1/k 2 } (where k is a given constant) Let M equipped with the following general ( α, β) metric: F =
α2 + β2 − β /λ, λ
α =
4
( y A )2
1/2 ,
β = k( x2 y1 − x 1 y2 + x 4 y3 − x 3 y4 ),
(4.5)
A=1
2 > 0. It clear that F is non-Minkowskian so that β is a Killing vector field. We want to = 1 − β and λ α 4 , F). construct minimal surfaces in (M Firstly, denote U1 = (cos v, sin v, 0, 0),
U2 = (− sin v, cos v, 0, 0),
(4.6)
U3 = (0, 0, cos v, sin v),
U4 = (0, 0, − sin v, cos v).
(4.7)
Then we have U1 = U2 ,
U2 = −U1 ,
U3 = U4 ,
U4 = −U3 .
(4.8)
v) = ( 4 , F ) defined by Let S1 : X(u, x1 , x 2 , x 3 , x 4 ) be a rotation surface in (M ( x1 , x 2 , x 3 , x 4 ) = [h(u) cos u]U1 + [h(u) sin u]U3 ,
(4.9)
for (u, v) ∈ Ω1 , where h(u) is a function to be determined that satisfies k2 h2 < 1. Then u = (h cos u) U1 + (h sin u) U3 , X
v = (h cos u)U2 + (h sin u)U4 , X
(4.10)
and the tangent vector field Y on S1 is expressed as u + y 2 X v , Y = ( y 1 , y2 , y3 , y4 ) = y 1 X
(4.11)
for arbitrary (y 1 , y 2 ). Moreover, we have uu = (h cos u) U1 + (h sin u) U3 , X
vv = −(h cos u)U1 − (h sin u)U3 . X
(4.12)
By direct computation, we obtain ⎧ 2 2 2 ⎪ ⎪ ⎪λ = 1−b = 1−k h , ⎨ ∗α α=X = [h2 + (h )2 ](y 1 )2 + h2 (y 2 )2 , ⎪ ⎪ ⎪ ⎩β = X ∗ β = k( 1 y2 + x 4 y3 − x 3 y4 ) = −kh2 y 2 . x2 y1 − x
(4.13)
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Put R = h2 + (h )2 and − 12 cos θ, y 1 = (λR)
y 2 = h−1 sin θ,
where θ ∈ [0, 2π]. Then γ=
1 )2 + h2 (y 2 )2 = 1, 2 + β 2 = (λR)(y λα
dτ = y 1 dy 2 − y 2 dy 1 = dθ/(h λR).
By direct computation, we have {γ=1}
2π dτ = , h λR
π , (y ) dτ = hλR λR
1 2
{γ=1}
(y 2 )2 dτ = {γ=1}
π . h3 λR
(4.14)
Denote W = (W 1 , W 2 , W 3 , W 4 ), where WA = {γ=1}
A 2 A (y 2 )2 − s (1 + β ) + 2β sA dτ, A (y 1 )2 + X X uu vv 2 0 λ λ
(4.15)
for 1 ≤ A ≤ 4. In this case, (4.4) is equivalent to
W, n3 = 0,
W, n4 = 0.
(4.16)
Let us compute W A . It follows from (2.10) that sD = aDA sBAbB = sBDbB ,
sD aDA sAB yB = sDB yB . 0 =
After direct computations, we obtain ( s1 , s2 , s3 , s4 ) = k2 [(h cos u)U1 + (h sin u)U3 ],
(4.17)
( s10 , s20 , s30 , s40 ) = ky 2 [(h cos u)U1 + (h sin u)U3 ] − ky 1 [(h cos u) U2 + (h sin u) U4 ].
(4.18)
Substituting (4.12), (4.14), (4.17) and (4.18) into (4.15), we obtain W = (W 1 , W 2 , W 3 , W 4 ) = Φ1 U1 + Ψ1 U3 ,
(4.19)
where cos u 2 1 2k + 2 , 3 R)1/2 hR h (λ λ (h sin u) sin u 2 π 1 − Ψ1 = 2k + 2 . 3 R)1/2 hR h (λ λ Φ1 =
π
(h cos u)
−
It is easy to check that n3 = (h cos u) U3 − (h sin u) U1 , are the orthogonal normal vector fields to S1 .
n4 = (sin u)U2 − (cos u)U4 ,
(4.20)
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Substituting (4.19) and (4.20) into (4.16), we get W, n4 = 0 and
W, n3 = (k 2 h2 − 1)hh + 3(h )2 + (k2 h2 + 2)h2 = 0.
(4.21)
Therefore we obtain the following − β/ λ, where α α2 + β2 /λ 4 , F ) be a general (α, β) space with F = λ Theorem 4.1. Let (M and β are defined by (4.5). Then the parameterized rotation surface S1 with v) = (h(u) cos u cos v, h(u) cos u sin v, h(u) sin u cos v, h(u) sin u sin v) X(u, 4 , F) is minimal if and only if h satisfies (4.21). in (M We proceed to solve equation (4.21). Let p = (h )2 and l = h2 . Then h =
dh dh dh dh dp = = h =h . du dh du dh dl
It follows that (4.21) becomes (k2 l − 1)l
dp + 3p + (k 2 l + 2)l = 0, dl
or equivalently,
3 d(p + l) + (p + l) = 0. dl l(k2 l − 1)
By a direct computation, we obtain p + l = C1 exp
3 −3 1 k2 l dl = C − dl = C exp 3 , 1 1 l(k2 l − 1) l k2 l − 1 k2 l − 1
where C1 is an integration constant. It follows that 3 h2 dh C1 h6 − h2 (k2 h2 − 1)3 2 =± = h = ± C1 − h , du k2 h2 − 1 (k2 h2 − 1)3 from which we have u = D1 ±
1 h
(k2 h2 − 1)3 dh C1 h4 − (k2 h2 − 1)3
(4.22)
where C1 and D1 are suitable integration constants. It follows from (4.22) that it is suitable to consider h as the parameter and u as the function of h. So we obtain the following 4 − β/ λ, where α α2 + β2 /λ Theorem 4.2. Let (M , F ) be a general (α, β) space with F = λ and β are defined by (4.5). Then the parameterized rotation surface S1 with X(h, v) = (h cos u(h) cos v, h cos u(h) sin v, h sin u(h) cos v, h sin u(h) sin v) 4 , F) is minimal if and only if in (M u(h) = D1 ±
1 h
(k2 h2 − 1)3 dh C1 h4 − (k2 h2 − 1)3
1 , C1 and D1 are suitable integration constants. where (h, v) ∈ Ω
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5. Minimal surfaces generated by one-parameter subgroups space (M 4 , F) generated by a one-parameter In this section, we study minimal surfaces in a general ( α, β) 4 . subgroup of SO(4, R) acting on a curve in M It is known that the Lie algebra of SO(4, R) is so(4, R) = {A ∈ M (4, R)|A + AT = 0}. For any A ∈ so(4, R), {eAt |t ∈ R} becomes a one-parameter subgroup of SO(4, R). 4 , F) generated by {eAt |t ∈ R} In the sequel, we proceed to construct two kinds of minimal surfaces in (M 4 acting on a curve in M . At first, let θ be an arbitrary constant and ⎛
0 ⎜ −1 ⎜ A=⎜ ⎝ − cos θ − sin θ
1 0 sin θ − cos θ
cos θ − sin θ 0 −1
⎞ sin θ cos θ ⎟ ⎟ ⎟, 1 ⎠ 0
⎛
1 ⎜ 0 ⎜ B=⎜ ⎝ sin θ − cos θ
0 1 cos θ sin θ
sin θ cos θ 1 0
⎞ − cos θ sin θ ⎟ ⎟ ⎟. 0 ⎠ 1
Then A ∈ so(4, R) and A2m+1 = (−1)m 22m A,
A2m = (−1)m 22m−1 B,
for any m ≥ 1. It follows that eAt =I4 +
∞
∞ t2m+1 t2m 2m 1 A2m+1 + A = (A sin 2t + B cos 2t + C), (2m + 1)! (2m)! 2 m=0 m=1
(5.1)
where C = 2I4 − B. Let ϕ be an arbitrary constant and U = (1, 0, sin ϕ, − cos ϕ),
V = (0, 1, − cos ϕ, − sin ϕ).
(5.2)
By direct computation, we have ⎧
ϕ − θ
ϕ − θ
ϕ − θ ⎪ u1 , UB = 2 cos u2 , UC = 2 cos u3 , ⎨ UA = 2 cos 2 2 2
ϕ − θ
ϕ − θ
ϕ − θ ⎪ ⎩ u1 , VB = 2 sin u2 , VC = 2 sin u3 , VA = 2 sin 2 2 2 where
ϕ − θ
ϕ − θ
ϕ + θ
ϕ + θ − sin , cos , cos , sin , 2 2 2 2
ϕ − θ
ϕ − θ
ϕ + θ
ϕ + θ u2 = cos , sin , sin , − cos , 2 2 2 2
ϕ − θ
ϕ + θ
ϕ + θ ϕ − θ u3 = sin , − cos , cos , sin . 2 2 2 2 u1 =
It easy to check that {u1 , u2 , u3 , u4 } forms an orthogonal basis of R, where
ϕ − θ
ϕ − θ
ϕ + θ
ϕ + θ u4 = cos , sin , − sin , cos . 2 2 2 2
(5.3)
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4 , F ) is a general (α, β) space where Suppose that (M 4 = {( x1 , x 2 , x 3 , x 4 ) ∈ R4 |( x1 )2 + ( x2 )2 < 1/k2 }, M λ, α2 + β2 − β)/ F = ( λ
α =
4
( y A )2
1/2 ,
(5.4)
β = k( x2 y1 − x 1 y2 ),
(5.5)
A=1
2 > 0. = 1 − ||β|| where k is a given constant and λ α Let γ(s) = x(s)U + y(s)V , s ∈ I, be a plane curve in span{U, V } ⊂ R4 where x(s) and y(s) are smooth functions satisfying k2 (x2 + y 2 ) < 1. Consider surface S2 in R4 defined by for (s, t) ∈ (I, R),
r(s, t) = γ(s)eAt + htu4 ,
where h is an arbitrary constant. It follows from (5.1) and (5.3) that r(s, t) =
1 1 x(s)(UA sin 2t + UB cos 2t + UC ) + y(s)(VA sin 2t + VB cos 2t + VC ) + (ht)u4 2 2
= P (s)(sin 2t)u1 + P (s)(cos 2t)u2 + Q(s)u3 + (ht)u4 ,
(5.6)
where P (s) = x(s) cos
ϕ − θ 2
+ y(s) sin
ϕ − θ 2
,
Q(s) = x(s) sin
ϕ − θ 2
− y(s) cos
ϕ − θ . 2
Without loss of generality, we suppose that u1 =
√
2(1, 0, 0, 0), u2 =
√
2(0, 1, 0, 0), u3 =
√
2(0, 0, 1, 0), u4 =
√
2(0, 0, 0, 1).
Denote that v1 = (sin 2t, cos 2t, 0, 0), v2 = (− cos 2t, sin 2t, 0, 0), v3 = (0, 0, 1, 0), v4 = (0, 0, 0, 1). Then we get r(s, t) =
√
2(P (s)v1 + Q(s)v3 + htv4 ).
(5.7)
Moreover, we have s = X ss = X
√ √
2(P v1 + Q v3 ), 2(P v1 + Q v3 ),
t = X
√
2(−2P v2 + hv4 ), √ tt = −4 2P v1 . X
(5.8) (5.9)
It follows that the tangent vector field Y on S2 is expressed by Y =
y1 , y2 , y3 , y4
s + y 2 X t . = y1 X
(5.10)
By direct computation, we have = 1 − 2k 2 P 2 , λ
α=
√ 2 [(P )2 + (Q )2 ](y 1 )2 + (4P 2 + h2 )(y 2 )2 ,
Denote S = (P )2 + (Q )2 and T = (1 − 2k 2 P 2 )h2 + 4P 2 . Then
β = 4kP 2 y 2 .
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√ 1 )2 + T (y 2 )2 . 2 + β 2 = 2 λS(y γ = λα In order that γ = 1 we suppose that −1/2 cos σ, y 1 = (2λS)
y 2 = (2T )−1/2 sin σ.
), and Then dτ = y 1 dy 2 − y 2 dy 1 = dσ/(2 λST
π (y ) dτ = , 3 S 3 T 4 λ
1 2
{γ=1}
{γ=1}
π dτ = , λST
{γ=1}
π (y 2 )2 dτ = . 3 4 λST
(5.11)
By direct computation, we obtain √ √ √ ( s10 , s20 , s30 , s40 ) = −k 2 2 P y 2 v1 + P y 1 v2 .
√ ( s1 , s2 , s3 , s4 ) = k2 2P v1 ,
(5.12)
Substituting (5.9), (5.11) and (5.12) into (4.15), we obtain (W 1 , W 2 , W 3 , W 4 ) = Φ2 v1 + Ψ2 v3 ,
(5.13)
where √ P k2 P 2π P − − Φ2 = , 3 ST 4S λT λ λ
√ 2π Q . Ψ2 = 3 S 3 T 4 λ
By directly checking, we can see that n3 = P v3 − Q v1 ,
n4 = 2P v4 + hv2
(5.14)
are the orthogonal normal vector fields to surface S2 . Substituting (5.13) and (5.14) into (4.16), we get W, n4 = 0 and
W, n3 =
P Q P Q − P Q 1 − + k2 = 0. 2 2 2 2 2 2 2 2 2 4(P + Q ) 1 − 2k P h − 2k h P + 4P
(5.15)
4 , F) be a general (α, β) space defined by (5.4) and (5.5). Then Theorem 5.1. Let (M t) = S2 : X(s,
√
2(P (s) sin 2t, P (s) cos 2t, Q(s), ht)
4 , F) is minimal if and only if h satisfies (5.15). in (M Let us solve (5.15). Denote R = dQ/dP . Then (5.15) becomes
1 3k2 R 2 − k2 h2 − + d(P 2 ). dR = 1 + R2 R h2 + 2(2 − k2 h2 )P 2 1 − 2k 2 P 2
Solving (5.16), we have 1+ or equivalently,
1 1 + R2 h2 + 2(2 − k2 h2 )P 2 = = , 2 2 R R C22 (1 − 2k2 P 2 )3
(5.16)
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R2 =
C22 (1 − 2k2 P 2 )3 , h2 + 2(2 − k2 h2 )P 2 − C22 (1 − 2k2 P 2 )3
1037
(5.17)
where C2 is an integration constant. From (5.17), we obtain
Q = D2 +
C2 (1 − 2k2 P 2 )3/2 h2 + 2(2 − k2 h2 )P 2 − C22 (1 − 2k2 P 2 )3
dP,
(5.18)
where D2 is an integration constant. 4 , F) be a general (α, β) space defined by (5.4) and (5.5). Then Theorem 5.2. Let (M S2 : X(P, t) =
√
2(P sin 2t, P cos 2t, Q(P ), ht)
4 , F) is minimal if and only if in (M Q = D2 +
C2 (1 − 2k2 P 2 )3/2 h2 + 2(2 − k2 h2 )P 2 − C22 (1 − 2k2 P 2 )3
dP,
where C2 and D2 are integration constants. 4 , F). Let Now we construct the second kind of minimal surfaces in (M ⎛
⎞ 1 a b 0 −b a ⎟ ⎟ ⎟, b 0 1⎠ −a −1 0
0 ⎜ −1 ⎜ A=⎜ ⎝ −a −b
⎛
0 ⎜ −1 ⎜ B=⎜ ⎝ 0 0
1 0 0 0 0 0 0 −1
⎞ 0 0⎟ ⎟ ⎟ ∈ so(4, R) 1⎠ 0
where a and b are arbitrary constants satisfying 0 < a2 + b2 = 1. By using the Mathematical Induction Method, we can prove that (−1)m 2m+1 (u − v 2m+1 )A − uv(u2m − v 2m )B , 2r (−1)m 2m+2 − v 2m+2 )BA + uv(u2m+1 − v 2m+1 )I4 , A2m+2 = (u 2r √ for every m ≥ 0, where r = a2 + b2 = 1, u = 1 + r and v = 1 − r. It follows that A2m+1 =
eAt =
1 (sin ut − sin vt)A + (u sin vt − v sin ut)B 2r
+ (cos vt − cos ut)BA + (u cos vt − v cos ut)I4 .
(5.19)
Choose a parameterized curve C : γ(s) = (−b, −a, 1, 0)h(s) + (b, a, 1, 0)es
(5.20)
4 , where h(s) is a smooth function to be determined. in M 4 . Then Let S3 : X = X(s, t) = γ(s)eAt be a parameterized surface in M 1 2 3 4 X(s, t) = x ,x ,x ,x = η1 Z1 + η2 Z2 , where a = r cos θ, b = r sin θ, and
(5.21)
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1038
η1 (s) =
Z1 = − sin(ut − θ),
1 [vh(s) + ues ], 2
1 η2 (s) = [uh(s) + ves ], 2
Z2 =
cos(ut − θ), cos ut, sin ut ,
sin(vt − θ), − cos(vt − θ), cos vt, sin vt .
(5.22)
Denote Z3 = cos(ut − θ), sin(ut − θ), sin ut, − cos ut , Z4 = cos(vt − θ), sin(vt − θ), − sin vt, cos vt .
(5.23)
It is directly to check that {Z1 , Z2 , Z3 , Z4 } forms an orthogonal frame field of R4 . Moreover (Z1 )t = −uZ3 ,
(Z2 )t = vZ4 ,
(Z4 )t = −vZ2 .
(Z3 )t = uZ1 ,
(5.24)
Taking partial derivatives to (5.21) with respect to s and t respectively, we have Xs = η1 Z1 + η2 Z2 ,
Xt = −(uη1 )Z3 + (vη2 )Z4 ,
(5.25)
where we using prime to indicate to take derivative with respect to s. Moreover, Xss = η1 Z1 + η2 Z2 ,
Xtt = −(u2 η1 )Z1 − (v 2 η2 )Z2 .
(5.26)
n4 = −(vη2 )Z3 − (uη1 )Z4 .
(5.27)
Denote n3 = η2 Z1 − η1 Z2 ,
Then {Xs , Xt , n3 , n4 } forms another orthogonal frame field of R4 . It is clear that the tangent vector field Y on S3 is expressed as Y = y1 , y2 , y3 , y4 = y 1 Xs + y 2 Xt . By direct computation, we have = 1 − 2k 2 (η 2 + η 2 ), λ 1 2
α=
2S(y 1 )2 + 2(u2 η12 + v 2 η22 )(y 2 )2 ,
β = −2k(uη12 + vη22 )y 2 ,
where S = (η1 )2 + (η2 )2 . Note that γ=
2 + β2 = λα
1 )2 + 2T (y 2 )2 , 2λS(y
(5.28)
where T = (u + rζ2 )η12 + (v − rζ1 )η22 and ζ1 = v + 4k2 rη12 ,
ζ2 = u − 4k2 rη22 .
(5.29)
In order that γ = 1 we suppose that −1/2 cos ϕ, y 1 = (2λS)
y 2 = (2T )−1/2 sin ϕ.
), and Then dτ = y 1 dy 2 − y 2 dy 1 = dϕ/(2 λST π π dτ = (y 1 )2 dτ = , , 3 S 3 T 4 λST λ {γ=1} {γ=1}
π (y 2 )2 dτ = . 3 4 λST {γ=1}
(5.30)
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It is easy to check that ( s1 , s2 , s3 , s4 ) = k2 ( x1 , x 2 , x 3 , x 4 ) = k2 (η1 Z1 + η2 Z2 ), ( s10 , s20 , s30 , s40 ) = k( y 2 , − y 1 , y4 , − y 3 ) = k(y 1 U1 + y 2 U2 ),
(5.31)
where U1 = η1 Z3 − η2 Z4 ,
U2 = (uη1 )Z1 + (vη2 )Z2 .
Substituting (5.26), (5.30) and (5.31) into (4.15), we have W = (W1 , W2 , W3 , W4 ) = Φ3 Z1 + Ψ3 Z2 ,
(5.32)
where
η 4k2 η1 π η1 ζ22 1 − Φ3 = − , 3 ST S λ λT 4 λ
η 4k2 η2 π η2 ζ12 2 − − Ψ3 = . 3 ST S λ λT 4 λ
(5.33)
Substituting (5.32) into (4.16), we get
W, n3 = 2(Φ3 η2 − Ψ3 η1 ),
W, n4 = 0.
(5.34)
It follows from (5.33) and (5.34) that S3 is minimal if and only if 0=
4k2 1 1 2 η1 η2 − η2 η1 + η2 η1 − η1 η2 + η2 η1 ζ1 − η1 η2 ζ22 . S λ λT
(5.35)
u v It is clear that (5.35) holds for η1 = 0 or η2 = 0, from which we have h(s) = − es or h(s) = − es . It v u follows that the parametric function of C in (5.20) becomes u+v v−u s a, ,0 e , v v v
u + v u + v u − v γ(s) = b, a, , 0 es . u u u
γ(s) =
u + v
b,
or
(5.36) (5.37)
And the parametric function of S3 becomes v − u s e sin(vt − θ), − cos(vt − θ), cos vt, sin vt , or (5.38) v u − v s e − sin(ut − θ), cos(ut − θ), cos ut, sin ut . X(s, t) = (5.39) u If η1 = 0 (resp. η2 = 0), then η2 = − 2r/v es (resp. η1 = 2r/u es ). It is easy to check that if (5.35) holds, then η1 = 0 (resp. η2 = 0). In either cases, γ(s) is an open half straight line and S3 is a ruled surface described by (5.38) or (5.39). In the sequel, we suppose that η1 η2 η1 η2 = 0. Using (5.29), we obtain X(s, t) =
ζ1 − v = 4k2 rη12 > 0,
u − ζ2 = 4k 2 rη22 > 0,
> 0, ζ2 − ζ1 = 2rλ
4k 2 rT = 2rζ1 ζ2 + uv(ζ1 − ζ2 ) > 0. In order that (5.35) has regular solution, ζ1 and ζ2 have to satisfy (5.40) simultaneously.
(5.40)
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Let Ω = {(ζ1 , ζ2 )|v < ζ1 < ζ2 < u} and Γ(ζ1 , ζ2 ) = 2rζ1 ζ2 + uv(ζ1 − ζ2 ). Then it is directly to check that Γ(ζ1 , ζ2 ) > 0 over Ω. On the other hand, it is easy to see that dζ2 −8k 2 rη2 dη2 η2 dη2 η2 η2 = = − = − = 0. ζ˙2 := dζ1 8k 2 rη1 dη1 η1 dη1 η1 η1
(5.41)
η 2 η2 η2 (ζ1 − v) ˙ ζ2 , η2 η1 − η1 η2 = η2 η1 1 + 12 − η = η 1 + 2 1 η2 η1 η1 (u − ζ2 )
(5.42)
η 2 (η )2 (u − ζ ) (η )2 η 2 (η )2 η22 2 + (ζ˙2 )2 . S = (η1 )2 1 + 2 2 = 1 21 + (ζ˙2 )2 = 1 21 2 (η1 ) η2 η1 η2 (ζ1 − v)
(5.43)
Hence, it follows that
and
Therefore dζ˙2 ζ˙2 (u − ζ2 ) 1 η22 + ζ˙2 + =− 2 ζ¨2 = (η2 η1 − η1 η2 ), dζ1 η2 (ζ1 − v) η2 (η1 )3 η12 or equivalently, η2 η1 − η1 η2 =
! η12 (η1 )3 ζ˙2 (u − ζ2 ) + ζ˙2 . (8k 2 r)ζ¨2 + 2 η2 η2 (ζ1 − v)
(5.44)
Substituting (5.40), (5.42), (5.43) and (5.44) into (5.35), we get η12 (η1 )3 ¨ 1 (ζ1 − v) ˙ (u − ζ2 ) ζ˙2 (u − ζ2 ) + ζ˙2 + + (ζ˙2 )2 2ζ2 + 2 2 1+ ζ2 η2 4k rη2 (ζ1 − v) (u − ζ2 ) (ζ1 − v) rλ ! 1 (ζ1 − v) ˙ 2 (u − ζ2 ) + ζ2 ζ2 + (ζ˙2 )2 . ζ12 + 2 (u − ζ ) (ζ − v) 2 1 4k rλT
0 = (4k2 r)
It follows from the assumption that the above equality turns into 1 (ζ1 − v) ˙ (u − ζ2 ) ζ˙2 (u − ζ2 ) + ζ˙2 + + (ζ˙2 )2 1+ ζ2 2 2 4k rη2 (ζ1 − v) (u − ζ2 ) (ζ1 − v) rλ 1 (ζ1 − v) ˙ 2 (u − ζ2 ) + + (ζ˙2 )2 . ζ12 + ζ2 ζ 2 (u − ζ2 ) 2 (ζ1 − v) 4k rλT
0 = 2ζ¨2 +
(5.45)
Substituting (5.29), (5.40) into (5.45) and sorting, we obtain
2 2(u − ζ2 ) rζ 2 1 rζ 2 ! ˙ + 1+ 1 + 1+ 2 ζ2 (ζ2 − ζ1 )(ζ1 − v) Γ ζ1 − v (ζ2 − ζ1 ) Γ
1 2 2(ζ1 − v) rζ 2 ! ˙ 2 rζ 2 + 1+ 1 (ζ2 ) + 1 + 2 (ζ˙2 )3 . u − ζ2 (ζ2 − ζ1 ) Γ (ζ2 − ζ1 )(u − ζ2 ) Γ
0 = 2ζ¨2 + + Denote
(5.46)
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2(u − ζ2 ) rζ 2 1+ 1 , (ζ2 − ζ1 )(ζ1 − v) Γ
2(ζ1 − v) rζ 2 χ3 = 1+ 2 , (ζ2 − ζ1 )(u − ζ2 ) Γ
χ0 =
2 1 rζ 2 + 1+ 2 , ζ1 − v (ζ2 − ζ1 ) Γ
1 2 rζ 2 + χ2 = 1+ 1 . u − ζ2 (ζ2 − ζ1 ) Γ
χ1 =
1041
(5.47) (5.48)
Then (5.46) turns into −2ζ¨2 = χ0 + χ1 ζ˙2 + χ2 (ζ˙2 )2 + χ3 (ζ˙2 )3 . (k)
Denote ζ1
(5.49)
= dk ζ1 /dζ2k for any k ∈ N. Then (1)
ζ1
=
1 , ζ˙2
(2)
ζ1
=−
ζ¨2 . ˙ (ζ2 )3
(5.50)
Substituting (5.50) into (5.49), we obtain (2)
2ζ1
(1)
(1)
(1)
= χ3 + χ2 ζ1 + χ1 (ζ1 )2 + χ0 (ζ1 )3 .
(5.51)
From (5.49) and (5.51), we can see that the relation between ζ1 and ζ2 is complicated. We could not obtain the explicit solutions to either (5.49) or (5.51). But we could discuss the asymptotic behaviour of the solutions of (5.49). Note that ζ2 turns to u when ζ1 turns to u and that ζ1 turns to v when ζ2 turns to v, because of v < ζ1 < ζ2 < u. At first, let us consider the asymptotic behaviour of the solutions of (5.49) when ζ1 turns to u. Denote ζ1 = ζ1 − u. Suppose that the Taylor expansion of ζ2 with respect to ζ1 is ζ2 = u + aζ1k + o(ζ1k ). We want to find the first k such that a = 0. By direct computation, we have ζ˙2 = akζ1k−1 + o(ζ1k−1 ),
ζ¨2 = ak(k − 1)ζ1k−2 + o(ζ1k−2 ),
(5.52)
and 4r k−1 + o(ζ1k−1 ), χ1 = −3ζ1−1 + o(ζ1−1 ), ζ 3a 1 1 6r χ2 = − ζ1−k + o(ζ1−k ), χ3 = ζ1−k−1 + o(ζ1−k−1 ). a a
χ0 =
(5.53)
Substituting (5.52) and (5.53) into (5.49), we obtain ak(k − 5)ζ1k−2 + o(ζ1k−2 ) ≡ 0, from which we get k = 5. Therefore ζ2 − u = O[(ζ1 − u)5 ]. Taking the same discussion to the solution of (5.51), we have ζ1 − v = O[(ζ2 − v)5 ], or equivalently ζ2 − v = O[(ζ1 − v)1/5 ]. To summarize the above discussion, we obtain the following conclusions: 4 , F ) be a general (α, β) space defined by (5.4) and (5.5) and S3 be a parameterized Theorem 5.3. Let (M 4 surface in (M , F ) defined by (5.21) and (5.22). (1) If η1 η2 η1 η2 ≡ 0, then S3 is a minimal surface defined by (5.38) or (5.39); (2) If η1 η2 η1 η2 = 0, then S3 is minimal if and only if ζ1 and ζ2 given by (5.29) satisfy (5.49) (or equivalently (5.51)). Furthermore, we have ζ2 − u = O[(ζ1 − u)5 ] as ζ1 → u, or ζ2 − v = O[(ζ1 − v)1/5 ] as ζ1 → v.
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4 , F ) 6. A kind of minimal hypersurfaces in (M From Theorem 5.3 we can see that the generating curve γ of S3 is difficult to get. However, we note that
W, n4 ≡ 0 along S3 . So it is natural to consider the property of the ruled hypersurface Σ3 generated by S3 and the vector field n4 over S3 . 4 , F) defined by Let Σ3 be a ruled parameterized hypersurface in (M t, p) = X(s, t) + pn3 = η1 Z1 + η2 Z2 , X(s,
(6.1)
where η1 = η1 + pη2 and η2 = η2 − pη1 . Then s = ( X η1 )Z1 + ( η2 )Z2 ,
t = −(u X η1 )Z3 + (v η2 )Z4 ,
p = η Z1 − η Z2 . X 2 1
The tangent vector field Y on Σ3 is expressed as s y 1 + X t y 2 + X p y 3 Y = ( y 1 , y2 , y3 , y4 ) = X for any (y 1 , y 2 , y 3 ). Moreover, the normal vector field to Σ3 is 4 = −(v η2 )Z3 − (u η1 )Z4 . n
(6.2)
By direct computations, we can see that = 1 − 2k 2 ( η12 + η22 ), λ α = 2[( η1 )2 + ( η2 )2 ](y 1 )2 + 2[(u η1 )2 + (v η2 )2 ](y 2 )2 + 2[(η1 )2 + (η2 )2 ](y 3 )2 ,
(6.3)
β = −2k[u η12 + v η22 ]y 2 . In order that γ =
2 + β 2 = 1, we suppose that λα y 1 = σ1 sin ϕ cos ψ,
y 2 = σ2 sin ϕ sin ψ,
y 3 = σ3 cos ϕ,
for all ϕ ∈ [0, π], ψ ∈ [0, 2π], where η )2 + ( 1/σ12 = 2λ[( η2 )2 ], 1
1/σ22 = 2[u2 η12 + v 2 η22 − 8k 2 r2 η12 η22 ],
)2 + (η )2 ]. 1/σ32 = 2λ[(η 1 2
Then we have dτ = σ1 σ2 σ3 sin ϕdϕ ∧ dψ and
dτ = 4πσ1 σ2 σ3 , {γ=1}
(y 1 )2 dτ =
4 π(σ1 )3 σ2 σ3 , 3
(y 3 )2 dτ =
4 πσ1 σ2 (σ3 )3 . 3
{γ=1}
(y 2 )2 dτ = {γ=1}
4 πσ1 (σ2 )3 σ3 , 3
{γ=1}
(6.4)
After direct computation, we obtain ( s1 , s2 , s3 , s4 ) = k2 ( x1 , x 2 , x 3 , x 4 ) = k2 ( η1 Z1 + η2 Z2 ), ( s10 , s20 , s30 , s40 ) = k( y 2 , − y 1 , y4 , − y 3 ) = k(U1 y 1 + U2 y 2 + U3 y 3 ),
(6.5)
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where U1 = ( η1 )Z3 − ( η2 )Z4 ,
U2 = (u η1 )Z1 + (v η2 )Z2 ,
U3 = η2 Z3 + η1 Z4 .
It is easy to see that ss = η Z1 + η Z2 , X 1 2
tt = −(u2 η1 )Z1 − (v 2 η2 )Z2 , X
pp = 0. X
(6.6)
Substituting (6.2) to (6.6) into (4.15) we deduce that W = (W 1 , W 2 , W 3 , W 4 ) = 4πσ1 σ2 σ3 (ΦZ1 + ΨZ2 ),
(6.7)
where Φ = η1 (σ1 )2 −
u − 4k 2 η2 r 2 2k 2 η1 2 2 − η (σ ) , 1 2 [1 − 2k 2 ( η12 + η22 )]2 1 − 2k2 ( η12 + η22 )
Ψ = η2 (σ1 )2 −
v + 4k 2 η2 r 2 2k 2 η2 1 2 − η (σ ) . 2 2 [1 − 2k 2 ( η12 + η22 )]2 1 − 2k 2 ( η12 + η22 )
4 ≡ 0. Therefore we have the following From (6.2) and (6.7) we can see that W, n − β/ λ, where α α2 + β2 /λ and β are λ 4 , F ) defined by (6.1). Then Σ3 is defined by (4.5). Let Σ3 be a ruled parameterized hypersurface in (M minimal. 4 , F ) be a general (α, β) space with F = Theorem 6.1. Let (M
In the sequel, we construct an explicit example of Σ3 . Suppose that (η1 )2 + (η2 )2 = ρ2 ,
(6.8)
where ρ is a given constant. From (6.8), we have (1 + r2 )(h )2 + 2(1 − r2 )es h + (1 + r2 )e2s = 2ρ2 , from which we get h = xes ± y
1 − z 2 e2s ,
(6.9)
where 1 − r2 x=− , 1 + r2
y=
2ρ2 , 1 + r2
z=
2r2 . ρ2 (1 + r2 )
(6.10)
1 − z 2 e2s d(zes ). zes
(6.11)
Taking integration on both sides of (6.9), we have h = xes ± y Denote sin t = zes . Then
√ 1 − z 2 e2s ds = xes ± y
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√
cos2 t 1 cos t 1 − z 2 e2s t s d(sin t) = dt = cos t + ln tan2 + C3 d(ze ) = s ze sin t sin t 2 2 1 = cos t + [ln(1 − cos t) − ln(1 + cos t)] + C3 2 1 ln(1 − 1 − z 2 e2s ) − ln(1 + 1 − z 2 e2s ) + C3 . = 1 − z 2 e2s + 2
(6.12)
Substituting (6.12) into (6.11), we obtain ! 1 1 − z 2 e2s + (6.13) ln(1 − 1 − z 2 e2s ) − ln(1 + 1 − z 2 e2s ) + C3 . 2 − β/ λ, where α α2 + β2 /λ 4 , F ) be a general (α, β) space with F = λ Theorem 6.2. Let (M and β are 3 4 , F) defined by (6.1) with h(s) given defined by (4.5). Let Σ be a ruled parameterized hypersurface in (M 3 by (6.13). Then Σ is minimal. h(s) = xes ± y
Acknowledgments The authors would like to express their hearty gratitude to referee for the valuable suggestions and constructive comments. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
S.S. Chern, Z. Shen, Riemann–Finsler Geometry, World Scientific, 2005. N. Cui, Y.B. Shen, Minimal rotational hypersurface in Minkowski (α, β)-space, Geom. Dedicata 151 (2011) 27–39. Q. He, Y.B. Shen, On the mean curvature of Finsler submanifolds, Chinese J. Contemp. Math. 27C (2006) 431–442. Q. He, W. Yang, Volume forms and minimal surfaces of rotation in Finsler space with (α, β)-metrics, Int. J. Math. 21 (2010) 1401–1411. Z. Shen, On Finsler geometry of submanifolds, Math. Ann. 311 (1998) 549–576. M. Souza, J. Spruck, K. Tenenblat, A Bernstein type theorem on a Randers space, Math. Ann. 329 (2004) 291–305. M. Souza, K. Tenenblat, Minimal surfaces of rotation in Finsler space with a Randers metric, Math. Ann. 325 (2003) 625–642. S.T. Yin, Q. He, D.H. Xie, Minimal submanifolds in general (α, β)-spaces, Ann. Polon. Math. 108 (1) (2013) 43–59. C. Yu, H. Zhu, On a new class of Finsler metrics, Differential Geom. Appl. 29 (2011) 244–254.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Minimal surfaces in some 4-dimensional general (α, β)-spaces ✩ Zhong-Hua Hou ∗ , Yong-Nan Liu Institute of Mathematics, Dalian University of Technology, Dalian, PR China
a r t i c l e
i n f o
Article history: Received 1 April 2016 Available online 12 July 2016 Submitted by J. Xiao Keywords: Minimal submanifolds General (α, β)-metrics Finsler manifolds Mean curvature vector
a b s t r a c t In this paper, we study minimal submanifolds in some general (α, β)-spaces. We firstly establish the equations for minimal submanifolds in a general (α, β)-space. Then we construct some kinds of minimal surfaces in some 4-dimensional general (α, β)-space. © 2016 Elsevier Inc. All rights reserved.
1. Introduction By using the Busemann–Hausdorff (briefly BH) volume form, Z. Shen [6] introduced the notions of mean curvature and normal curvature for Finsler submanifolds. Later, by using the Holmes–Thompson (briefly HT) volume form, Q. He and Y. B. Shen [4] introduced another notion of mean curvature. Based on the BH-volume form, M. Souza, J. Spruck and K. Tenenblat [8,9] considered a Bernstein type theorem for minimal surfaces of rotation in a Randers–Minkowski space. Based on the HT-volume form, Q. He and W. Yang [7] constructed a minimal rotational hypersurface in a Minkowski (α, β)-space. N. Cui and Y. B. Shen [2] determined the BH-minimal and HT-minimal rotational hypersurfaces generated by plane curves rotating around the axis in the direction of β in Minkowski (α, β)-space. C. Yu and H. Zhu [1] defined a new class of Finsler metrics of the form F = αφ(x, β/α) and called it a general (α, β)-metrics, where φ(x, s) is a C ∞ function, α is a Riemannian metric and β is a 1-form. It is obvious that the navigation expression of a Randers metric F = ( λα2 + β 2 − β)/λ can be viewed as a general (α, β)-metric, where α is a Riemannian metric, β is a 1-form and λ = 1 − β2α . S. T. Yin, Q. He and D. H. Xie in [3] constructed a minimal rotational hypersurface in a non-Minkowski general (α, β)-metric. In this paper, we want to construct some kinds of minimal surfaces in some 4-dimensional non-Minkowski 4 , F ). In Sections 2 and 3, some basic concepts of general (α, β)-metrics and that general (α, β) space (M ✩
This work was supported in part by NSFC (No. 61473059).
* Corresponding author. E-mail addresses: [email protected] (Z.-H. Hou), [email protected] (Y.-N. Liu). http://dx.doi.org/10.1016/j.jmaa.2016.07.004 0022-247X/© 2016 Elsevier Inc. All rights reserved.
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4 , F) are introduced. In Section 4, we construct some kinds of rotational of minimal submanifolds in (M 4 minimal surfaces in (M , F ). In Section 5, we construct some kinds of minimal surfaces produced by a one-parameter subgroup and study the asymptotic behaviour of one kind of them. Finally, we construct a 4 , F) in Section 6. kind of ruled minimal hypersurfaces in (M 2. The (α, β) metric on smooth manifolds Let M be an n-dimensional smooth manifold and π : TM → M be the tangent bundle of M . Let (U, {xi }) be a local coordinate of M . Then (π −1 (U ), {xi , y j }) is the natural local coordinate of M , where Y = y j ∂xj ∈ Tx M for any x ∈ U . We denote ∂h/∂y i , ∂ 2 h/∂y i ∂y j , · · · , by hyi , hyi yj , · · · , for any smooth function h over TM , and use the following convention for index ranges unless otherwise stated: 1 ≤ i, j, k, · · · ≤ n,
1 ≤ A, B, C, · · · ≤ n + p,
n + 1 ≤ a, b, c, · · · ≤ n + p.
A Finsler metric on M is a function F : TM → [0, ∞) with the following properties: (1) F is smooth on TM \ {0}; (2) F (x, λy) = λF (x, y) for all λ > 0; (3) the induced quadratic form g is positive-definite, where g := gij dxi ⊗ dxj ,
1 2 [F ]yi yj . 2
gij =
(2.1)
The projection π : TM → M gives rise to the pull-back bundle π ∗ TM together with its dual bundle π T ∗ M over TM \ {0}. There is a global section ω = [F ]yi dxi in π ∗ T ∗ M which is called the Hilbert form. Its dual L = Li ∂xi with Li = y i /F in π ∗ TM is called the distinguished vector field. For any x ∈ M , let SM = x∈M Sx M with Sx M = {[y]|y ∈ Tx M } be the projective sphere bundle. Denote by gˆ the pull-back of the Sasaki metric from T (TM \ {0}) to T (SM ). Then the volume element dVSM of SM with respect to gˆ is expressed as (cf. [3]) ∗
dVSM = Ωdτ ∧ dx,
(2.2)
where Ω = F −n det(gij ), dx = dx1 ∧ · · · ∧ dxn and dτ =
n i ∧ · · · ∧ dy n . (−1)i−1 y i dy 1 ∧ · · · ∧ dy
(2.3)
i=1
The n-dimensional volume form of (M, F ) is defined to be dVM = σ(x)dx,
σ(x) =
1
Ωdτ,
cn−1
(2.4)
Sx M
where cn−1 denotes the volume of the unit Euclidean (n − 1)-sphere S n−1 . Let φ(x, s) be a positive C ∞ function with respect to x ∈ M and |s| < b0 for some 0 < b0 ≤ ∞. Let α = aij (x)y i y j be a Riemannian metric and β = bi y i be a 1-form with b := βα < b0 . Define F = αφ(x, β/α). Then F is a Finsler metric if and only if φ − sφs + (b2 − s2 )φss > 0,
(2.5)
for all x ∈ M and |s| ≤ b < b0 . Such a kind of Finsler metric is called the general (α, β)-metric on M (cf. [9]). By direct computations, we have (cf. [1]),
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gij = ρaij + ρ0 bi bj + ρ1 (bi αyi + bj αyj ) − sρ1 αyi αyj ,
(2.6)
det(gij ) = φn H(x, s) det(aij ),
(2.7)
g ij = ρ−1 [aij + ηbi bj + η0 α−1 (bi y j + bj y i ) + η1 α−2 y i y j ],
(2.8)
⎧ (g ij ) = (gij )−1 , (aij ) = (aij )−1 , bi = aij bj , ⎪ ⎪ ⎪ ⎪ ⎨ σ = sφ + (b2 − s2 )φs , ρ = φ(φ − sφs ), ρ0 = (φφs )s , ρ1 = ρs , ⎪ ⎪ ⎪ φ ρ σρ ⎪ ⎩ H(x, s) = φ3−n ρn−2 σs , η = − ss , η0 = − s , η1 = − 2 s . σs φσs φ σs
(2.9)
where
Let bi|j denote the coefficients of the covariant derivative of β with respect to α and write ⎧ rij = 12 (bi|j + bj|i ), sij = 12 (bi|j − bj|i ), r00 = rij y i y j , ⎪ ⎪ ⎨ ri = bj rji , si = bj sji , r0 = ri y i , s0 = si y i , ⎪ ⎪ ⎩ i r = aij rj , si = aij sj , r = bi ri , si0 = aij sjk y k .
(2.10)
3. Immersed submanifolds of general (α, β)-spaces , F ) be Finsler manifolds. Let f : (M, F ) → (M , F) be an immersion and dfx : Tx M → Let (M, F ) and (M be the differential of f at x ∈ M . Then f is called isometric if F (x, y) = F(f (x), dfx (y)) for any Tf (x) M (x, y) ∈ TM \ {0}. , { and (U, {xi }) be the local coordinate For isometric immersion f , let (U xi }) be a local coordinate of M ) ⊂ U . Then of M such that f −1 (U f (x) = { xA },
dfx (y) = df (y i ∂xi ) = yB ∂xB = { y B },
(3.1)
where x A = f A (x), yB = fiB y i and fiB = ∂f B /∂xi for any i and B. Then gij (x, y) = gAB ( x, y)fiA fjB .
(3.2)
) with respect to g (cf. [3,5]). Denote Let (π ∗ TM )⊥ be the orthogonal complement of π ∗ TM in π ∗ (f −1 T M h=
hA F2
∂xA ,
A i j A , hA = fij y y − fkA Gk + G
A fij =
∂2f A , ∂xi ∂xj
(3.3)
α are the geodesic coefficients of F and F respectively. Then h ∈ (π ∗ TM )⊥ , which is called where Gk and G the normal curvature of f . The mean curvature form of f is defined by μ=
h A Ωdτ d xA , cn−1 σ F2 1
(3.4)
Sx M
where hA = gAB hB . , F) is called minimal if any compact domain of M is the An isometric immersion f : (M, F ) → (M critical point of its volume functional with respect to any variation of f . , F ) be an isometric immersion. Then f is minimal if and only if Lemma 3.1. (cf. [3]) Let f : (M, F ) → (M μ = 0.
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metric F = α α), where α , F ) be a Finsler space with the general ( Let (M α, β) φ( x, β/ = A and β = bA y . Let f : (M, F ) → (M , F ) be an isometric immersion. Then F = f ∗ F = αφ(f (x), β/α),
aAB yA yB
(3.5)
where α = aij y i y j with aij = aAB fiA fjB and β = bi y i with bi = bA fiA . Let β = bA ∂xA and β⊥ be the projection of β into the normal bundle TM ⊥ with respect to α . Yin et al. [8] discussed the relation between the local orthonormal frame field of TM ⊥ with respect to α and the local orthonormal frame field of (π ∗ TM )⊥ with respect to F. And proved the following result: n+p , F) is an isometric immersion where F is the Lemma 3.2. (cf. [8]) Suppose that f : (M n , F ) → (M n+p general ( α, β) metric. Let {na }a=n+1 be a local orthonormal frame field of TM ⊥ such that nn+p is parallel to β⊥ . Denote a )β + F η0 ( a )L n a = χa na + ηβ(n α−1 β)(n
(3.6)
a )2 ], { is the distinguished where χa = 1/ ρ[1 + ηβ(n ρ, η, η0 } are defined as in (2.9) with respect to F , and L n+p ∗ ⊥ vector field. Then { na } is a local orthonormal frame of (π TM ) with respect to F . a=n+1
By applying Lemma 3.2, they obtained a description of Finslerian minimal immersions: n+p , F) is an isometric immersion where F is the Lemma 3.3. (cf. [8]) Suppose that f : (M n , F ) → (M n+p B general ( α, β) metric. Let {na = na ∂xB }a=n+1 be a local orthonormal frame field of TM ⊥ such that nn+p is parallel to β⊥ . Then f is minimal if and only if aBD nD a Sx M
B i j B ) ρH(fij y y +G dτ = 0, a )2 )φ2 αn+2 (1 + ηβ(n
(3.7)
for all a, where {φ, ρ, H} are defined as in (2.9) with respect to F . metric space β) 4. Minimal submanifolds of a general (α, n+p , F), where Consider the isometric immersion f : (M n , F ) → (M F =
α2 + β2 − β λ, λ
= 1 − β 2 > 0, λ α
(4.1)
is the navigation expression of a Randers metric. Then the induced metric is F = (γ − β)/λ,
γ=
2 + β2. λα
(4.2)
By direct computation, we get η = −1,
ρ=
α2 φ2 , Fγ
α n+1 + b2 ) F H(x, s) = (λ , α γ
where b2 = β2α . Substituting (4.3) into (3.7), we have
(4.3)
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metric (4.1). If α n+p , F) with general ( Lemma 4.1. (cf. [8]) Let (M n , F ) be a submanifold in (M α, β) is a Riemannian metric with constant sectional curvature and β is the Killing vector field, then f : (M n , F ) → n+p , F) is minimal if and only if (M
aBD nD a
B i j yy − fij
{γ=1}
sB (1 − β)2 2(1 − β) B − s0 dτ = 0 2 λ λ
(4.4)
for every a. 4 = {( x1 , x 2 , x 3 , x 4 ) ∈ R4 |( x1 )2 + ( x2 )2 + ( x3 )2 + ( x4 )2 < 1/k 2 } (where k is a given constant) Let M equipped with the following general ( α, β) metric: F =
α2 + β2 − β /λ, λ
α =
4
( y A )2
1/2 ,
β = k( x2 y1 − x 1 y2 + x 4 y3 − x 3 y4 ),
(4.5)
A=1
2 > 0. It clear that F is non-Minkowskian so that β is a Killing vector field. We want to = 1 − β and λ α 4 , F). construct minimal surfaces in (M Firstly, denote U1 = (cos v, sin v, 0, 0),
U2 = (− sin v, cos v, 0, 0),
(4.6)
U3 = (0, 0, cos v, sin v),
U4 = (0, 0, − sin v, cos v).
(4.7)
Then we have U1 = U2 ,
U2 = −U1 ,
U3 = U4 ,
U4 = −U3 .
(4.8)
v) = ( 4 , F ) defined by Let S1 : X(u, x1 , x 2 , x 3 , x 4 ) be a rotation surface in (M ( x1 , x 2 , x 3 , x 4 ) = [h(u) cos u]U1 + [h(u) sin u]U3 ,
(4.9)
for (u, v) ∈ Ω1 , where h(u) is a function to be determined that satisfies k2 h2 < 1. Then u = (h cos u) U1 + (h sin u) U3 , X
v = (h cos u)U2 + (h sin u)U4 , X
(4.10)
and the tangent vector field Y on S1 is expressed as u + y 2 X v , Y = ( y 1 , y2 , y3 , y4 ) = y 1 X
(4.11)
for arbitrary (y 1 , y 2 ). Moreover, we have uu = (h cos u) U1 + (h sin u) U3 , X
vv = −(h cos u)U1 − (h sin u)U3 . X
(4.12)
By direct computation, we obtain ⎧ 2 2 2 ⎪ ⎪ ⎪λ = 1−b = 1−k h , ⎨ ∗α α=X = [h2 + (h )2 ](y 1 )2 + h2 (y 2 )2 , ⎪ ⎪ ⎪ ⎩β = X ∗ β = k( 1 y2 + x 4 y3 − x 3 y4 ) = −kh2 y 2 . x2 y1 − x
(4.13)
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Put R = h2 + (h )2 and − 12 cos θ, y 1 = (λR)
y 2 = h−1 sin θ,
where θ ∈ [0, 2π]. Then γ=
1 )2 + h2 (y 2 )2 = 1, 2 + β 2 = (λR)(y λα
dτ = y 1 dy 2 − y 2 dy 1 = dθ/(h λR).
By direct computation, we have {γ=1}
2π dτ = , h λR
π , (y ) dτ = hλR λR
1 2
{γ=1}
(y 2 )2 dτ = {γ=1}
π . h3 λR
(4.14)
Denote W = (W 1 , W 2 , W 3 , W 4 ), where WA = {γ=1}
A 2 A (y 2 )2 − s (1 + β ) + 2β sA dτ, A (y 1 )2 + X X uu vv 2 0 λ λ
(4.15)
for 1 ≤ A ≤ 4. In this case, (4.4) is equivalent to
W, n3 = 0,
W, n4 = 0.
(4.16)
Let us compute W A . It follows from (2.10) that sD = aDA sBAbB = sBDbB ,
sD aDA sAB yB = sDB yB . 0 =
After direct computations, we obtain ( s1 , s2 , s3 , s4 ) = k2 [(h cos u)U1 + (h sin u)U3 ],
(4.17)
( s10 , s20 , s30 , s40 ) = ky 2 [(h cos u)U1 + (h sin u)U3 ] − ky 1 [(h cos u) U2 + (h sin u) U4 ].
(4.18)
Substituting (4.12), (4.14), (4.17) and (4.18) into (4.15), we obtain W = (W 1 , W 2 , W 3 , W 4 ) = Φ1 U1 + Ψ1 U3 ,
(4.19)
where cos u 2 1 2k + 2 , 3 R)1/2 hR h (λ λ (h sin u) sin u 2 π 1 − Ψ1 = 2k + 2 . 3 R)1/2 hR h (λ λ Φ1 =
π
(h cos u)
−
It is easy to check that n3 = (h cos u) U3 − (h sin u) U1 , are the orthogonal normal vector fields to S1 .
n4 = (sin u)U2 − (cos u)U4 ,
(4.20)
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Substituting (4.19) and (4.20) into (4.16), we get W, n4 = 0 and
W, n3 = (k 2 h2 − 1)hh + 3(h )2 + (k2 h2 + 2)h2 = 0.
(4.21)
Therefore we obtain the following − β/ λ, where α α2 + β2 /λ 4 , F ) be a general (α, β) space with F = λ Theorem 4.1. Let (M and β are defined by (4.5). Then the parameterized rotation surface S1 with v) = (h(u) cos u cos v, h(u) cos u sin v, h(u) sin u cos v, h(u) sin u sin v) X(u, 4 , F) is minimal if and only if h satisfies (4.21). in (M We proceed to solve equation (4.21). Let p = (h )2 and l = h2 . Then h =
dh dh dh dh dp = = h =h . du dh du dh dl
It follows that (4.21) becomes (k2 l − 1)l
dp + 3p + (k 2 l + 2)l = 0, dl
or equivalently,
3 d(p + l) + (p + l) = 0. dl l(k2 l − 1)
By a direct computation, we obtain p + l = C1 exp
3 −3 1 k2 l dl = C − dl = C exp 3 , 1 1 l(k2 l − 1) l k2 l − 1 k2 l − 1
where C1 is an integration constant. It follows that 3 h2 dh C1 h6 − h2 (k2 h2 − 1)3 2 =± = h = ± C1 − h , du k2 h2 − 1 (k2 h2 − 1)3 from which we have u = D1 ±
1 h
(k2 h2 − 1)3 dh C1 h4 − (k2 h2 − 1)3
(4.22)
where C1 and D1 are suitable integration constants. It follows from (4.22) that it is suitable to consider h as the parameter and u as the function of h. So we obtain the following 4 − β/ λ, where α α2 + β2 /λ Theorem 4.2. Let (M , F ) be a general (α, β) space with F = λ and β are defined by (4.5). Then the parameterized rotation surface S1 with X(h, v) = (h cos u(h) cos v, h cos u(h) sin v, h sin u(h) cos v, h sin u(h) sin v) 4 , F) is minimal if and only if in (M u(h) = D1 ±
1 h
(k2 h2 − 1)3 dh C1 h4 − (k2 h2 − 1)3
1 , C1 and D1 are suitable integration constants. where (h, v) ∈ Ω
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5. Minimal surfaces generated by one-parameter subgroups space (M 4 , F) generated by a one-parameter In this section, we study minimal surfaces in a general ( α, β) 4 . subgroup of SO(4, R) acting on a curve in M It is known that the Lie algebra of SO(4, R) is so(4, R) = {A ∈ M (4, R)|A + AT = 0}. For any A ∈ so(4, R), {eAt |t ∈ R} becomes a one-parameter subgroup of SO(4, R). 4 , F) generated by {eAt |t ∈ R} In the sequel, we proceed to construct two kinds of minimal surfaces in (M 4 acting on a curve in M . At first, let θ be an arbitrary constant and ⎛
0 ⎜ −1 ⎜ A=⎜ ⎝ − cos θ − sin θ
1 0 sin θ − cos θ
cos θ − sin θ 0 −1
⎞ sin θ cos θ ⎟ ⎟ ⎟, 1 ⎠ 0
⎛
1 ⎜ 0 ⎜ B=⎜ ⎝ sin θ − cos θ
0 1 cos θ sin θ
sin θ cos θ 1 0
⎞ − cos θ sin θ ⎟ ⎟ ⎟. 0 ⎠ 1
Then A ∈ so(4, R) and A2m+1 = (−1)m 22m A,
A2m = (−1)m 22m−1 B,
for any m ≥ 1. It follows that eAt =I4 +
∞
∞ t2m+1 t2m 2m 1 A2m+1 + A = (A sin 2t + B cos 2t + C), (2m + 1)! (2m)! 2 m=0 m=1
(5.1)
where C = 2I4 − B. Let ϕ be an arbitrary constant and U = (1, 0, sin ϕ, − cos ϕ),
V = (0, 1, − cos ϕ, − sin ϕ).
(5.2)
By direct computation, we have ⎧
ϕ − θ
ϕ − θ
ϕ − θ ⎪ u1 , UB = 2 cos u2 , UC = 2 cos u3 , ⎨ UA = 2 cos 2 2 2
ϕ − θ
ϕ − θ
ϕ − θ ⎪ ⎩ u1 , VB = 2 sin u2 , VC = 2 sin u3 , VA = 2 sin 2 2 2 where
ϕ − θ
ϕ − θ
ϕ + θ
ϕ + θ − sin , cos , cos , sin , 2 2 2 2
ϕ − θ
ϕ − θ
ϕ + θ
ϕ + θ u2 = cos , sin , sin , − cos , 2 2 2 2
ϕ − θ
ϕ + θ
ϕ + θ ϕ − θ u3 = sin , − cos , cos , sin . 2 2 2 2 u1 =
It easy to check that {u1 , u2 , u3 , u4 } forms an orthogonal basis of R, where
ϕ − θ
ϕ − θ
ϕ + θ
ϕ + θ u4 = cos , sin , − sin , cos . 2 2 2 2
(5.3)
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4 , F ) is a general (α, β) space where Suppose that (M 4 = {( x1 , x 2 , x 3 , x 4 ) ∈ R4 |( x1 )2 + ( x2 )2 < 1/k2 }, M λ, α2 + β2 − β)/ F = ( λ
α =
4
( y A )2
1/2 ,
(5.4)
β = k( x2 y1 − x 1 y2 ),
(5.5)
A=1
2 > 0. = 1 − ||β|| where k is a given constant and λ α Let γ(s) = x(s)U + y(s)V , s ∈ I, be a plane curve in span{U, V } ⊂ R4 where x(s) and y(s) are smooth functions satisfying k2 (x2 + y 2 ) < 1. Consider surface S2 in R4 defined by for (s, t) ∈ (I, R),
r(s, t) = γ(s)eAt + htu4 ,
where h is an arbitrary constant. It follows from (5.1) and (5.3) that r(s, t) =
1 1 x(s)(UA sin 2t + UB cos 2t + UC ) + y(s)(VA sin 2t + VB cos 2t + VC ) + (ht)u4 2 2
= P (s)(sin 2t)u1 + P (s)(cos 2t)u2 + Q(s)u3 + (ht)u4 ,
(5.6)
where P (s) = x(s) cos
ϕ − θ 2
+ y(s) sin
ϕ − θ 2
,
Q(s) = x(s) sin
ϕ − θ 2
− y(s) cos
ϕ − θ . 2
Without loss of generality, we suppose that u1 =
√
2(1, 0, 0, 0), u2 =
√
2(0, 1, 0, 0), u3 =
√
2(0, 0, 1, 0), u4 =
√
2(0, 0, 0, 1).
Denote that v1 = (sin 2t, cos 2t, 0, 0), v2 = (− cos 2t, sin 2t, 0, 0), v3 = (0, 0, 1, 0), v4 = (0, 0, 0, 1). Then we get r(s, t) =
√
2(P (s)v1 + Q(s)v3 + htv4 ).
(5.7)
Moreover, we have s = X ss = X
√ √
2(P v1 + Q v3 ), 2(P v1 + Q v3 ),
t = X
√
2(−2P v2 + hv4 ), √ tt = −4 2P v1 . X
(5.8) (5.9)
It follows that the tangent vector field Y on S2 is expressed by Y =
y1 , y2 , y3 , y4
s + y 2 X t . = y1 X
(5.10)
By direct computation, we have = 1 − 2k 2 P 2 , λ
α=
√ 2 [(P )2 + (Q )2 ](y 1 )2 + (4P 2 + h2 )(y 2 )2 ,
Denote S = (P )2 + (Q )2 and T = (1 − 2k 2 P 2 )h2 + 4P 2 . Then
β = 4kP 2 y 2 .
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√ 1 )2 + T (y 2 )2 . 2 + β 2 = 2 λS(y γ = λα In order that γ = 1 we suppose that −1/2 cos σ, y 1 = (2λS)
y 2 = (2T )−1/2 sin σ.
), and Then dτ = y 1 dy 2 − y 2 dy 1 = dσ/(2 λST
π (y ) dτ = , 3 S 3 T 4 λ
1 2
{γ=1}
{γ=1}
π dτ = , λST
{γ=1}
π (y 2 )2 dτ = . 3 4 λST
(5.11)
By direct computation, we obtain √ √ √ ( s10 , s20 , s30 , s40 ) = −k 2 2 P y 2 v1 + P y 1 v2 .
√ ( s1 , s2 , s3 , s4 ) = k2 2P v1 ,
(5.12)
Substituting (5.9), (5.11) and (5.12) into (4.15), we obtain (W 1 , W 2 , W 3 , W 4 ) = Φ2 v1 + Ψ2 v3 ,
(5.13)
where √ P k2 P 2π P − − Φ2 = , 3 ST 4S λT λ λ
√ 2π Q . Ψ2 = 3 S 3 T 4 λ
By directly checking, we can see that n3 = P v3 − Q v1 ,
n4 = 2P v4 + hv2
(5.14)
are the orthogonal normal vector fields to surface S2 . Substituting (5.13) and (5.14) into (4.16), we get W, n4 = 0 and
W, n3 =
P Q P Q − P Q 1 − + k2 = 0. 2 2 2 2 2 2 2 2 2 4(P + Q ) 1 − 2k P h − 2k h P + 4P
(5.15)
4 , F) be a general (α, β) space defined by (5.4) and (5.5). Then Theorem 5.1. Let (M t) = S2 : X(s,
√
2(P (s) sin 2t, P (s) cos 2t, Q(s), ht)
4 , F) is minimal if and only if h satisfies (5.15). in (M Let us solve (5.15). Denote R = dQ/dP . Then (5.15) becomes
1 3k2 R 2 − k2 h2 − + d(P 2 ). dR = 1 + R2 R h2 + 2(2 − k2 h2 )P 2 1 − 2k 2 P 2
Solving (5.16), we have 1+ or equivalently,
1 1 + R2 h2 + 2(2 − k2 h2 )P 2 = = , 2 2 R R C22 (1 − 2k2 P 2 )3
(5.16)
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R2 =
C22 (1 − 2k2 P 2 )3 , h2 + 2(2 − k2 h2 )P 2 − C22 (1 − 2k2 P 2 )3
1037
(5.17)
where C2 is an integration constant. From (5.17), we obtain
Q = D2 +
C2 (1 − 2k2 P 2 )3/2 h2 + 2(2 − k2 h2 )P 2 − C22 (1 − 2k2 P 2 )3
dP,
(5.18)
where D2 is an integration constant. 4 , F) be a general (α, β) space defined by (5.4) and (5.5). Then Theorem 5.2. Let (M S2 : X(P, t) =
√
2(P sin 2t, P cos 2t, Q(P ), ht)
4 , F) is minimal if and only if in (M Q = D2 +
C2 (1 − 2k2 P 2 )3/2 h2 + 2(2 − k2 h2 )P 2 − C22 (1 − 2k2 P 2 )3
dP,
where C2 and D2 are integration constants. 4 , F). Let Now we construct the second kind of minimal surfaces in (M ⎛
⎞ 1 a b 0 −b a ⎟ ⎟ ⎟, b 0 1⎠ −a −1 0
0 ⎜ −1 ⎜ A=⎜ ⎝ −a −b
⎛
0 ⎜ −1 ⎜ B=⎜ ⎝ 0 0
1 0 0 0 0 0 0 −1
⎞ 0 0⎟ ⎟ ⎟ ∈ so(4, R) 1⎠ 0
where a and b are arbitrary constants satisfying 0 < a2 + b2 = 1. By using the Mathematical Induction Method, we can prove that (−1)m 2m+1 (u − v 2m+1 )A − uv(u2m − v 2m )B , 2r (−1)m 2m+2 − v 2m+2 )BA + uv(u2m+1 − v 2m+1 )I4 , A2m+2 = (u 2r √ for every m ≥ 0, where r = a2 + b2 = 1, u = 1 + r and v = 1 − r. It follows that A2m+1 =
eAt =
1 (sin ut − sin vt)A + (u sin vt − v sin ut)B 2r
+ (cos vt − cos ut)BA + (u cos vt − v cos ut)I4 .
(5.19)
Choose a parameterized curve C : γ(s) = (−b, −a, 1, 0)h(s) + (b, a, 1, 0)es
(5.20)
4 , where h(s) is a smooth function to be determined. in M 4 . Then Let S3 : X = X(s, t) = γ(s)eAt be a parameterized surface in M 1 2 3 4 X(s, t) = x ,x ,x ,x = η1 Z1 + η2 Z2 , where a = r cos θ, b = r sin θ, and
(5.21)
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η1 (s) =
Z1 = − sin(ut − θ),
1 [vh(s) + ues ], 2
1 η2 (s) = [uh(s) + ves ], 2
Z2 =
cos(ut − θ), cos ut, sin ut ,
sin(vt − θ), − cos(vt − θ), cos vt, sin vt .
(5.22)
Denote Z3 = cos(ut − θ), sin(ut − θ), sin ut, − cos ut , Z4 = cos(vt − θ), sin(vt − θ), − sin vt, cos vt .
(5.23)
It is directly to check that {Z1 , Z2 , Z3 , Z4 } forms an orthogonal frame field of R4 . Moreover (Z1 )t = −uZ3 ,
(Z2 )t = vZ4 ,
(Z4 )t = −vZ2 .
(Z3 )t = uZ1 ,
(5.24)
Taking partial derivatives to (5.21) with respect to s and t respectively, we have Xs = η1 Z1 + η2 Z2 ,
Xt = −(uη1 )Z3 + (vη2 )Z4 ,
(5.25)
where we using prime to indicate to take derivative with respect to s. Moreover, Xss = η1 Z1 + η2 Z2 ,
Xtt = −(u2 η1 )Z1 − (v 2 η2 )Z2 .
(5.26)
n4 = −(vη2 )Z3 − (uη1 )Z4 .
(5.27)
Denote n3 = η2 Z1 − η1 Z2 ,
Then {Xs , Xt , n3 , n4 } forms another orthogonal frame field of R4 . It is clear that the tangent vector field Y on S3 is expressed as Y = y1 , y2 , y3 , y4 = y 1 Xs + y 2 Xt . By direct computation, we have = 1 − 2k 2 (η 2 + η 2 ), λ 1 2
α=
2S(y 1 )2 + 2(u2 η12 + v 2 η22 )(y 2 )2 ,
β = −2k(uη12 + vη22 )y 2 ,
where S = (η1 )2 + (η2 )2 . Note that γ=
2 + β2 = λα
1 )2 + 2T (y 2 )2 , 2λS(y
(5.28)
where T = (u + rζ2 )η12 + (v − rζ1 )η22 and ζ1 = v + 4k2 rη12 ,
ζ2 = u − 4k2 rη22 .
(5.29)
In order that γ = 1 we suppose that −1/2 cos ϕ, y 1 = (2λS)
y 2 = (2T )−1/2 sin ϕ.
), and Then dτ = y 1 dy 2 − y 2 dy 1 = dϕ/(2 λST π π dτ = (y 1 )2 dτ = , , 3 S 3 T 4 λST λ {γ=1} {γ=1}
π (y 2 )2 dτ = . 3 4 λST {γ=1}
(5.30)
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It is easy to check that ( s1 , s2 , s3 , s4 ) = k2 ( x1 , x 2 , x 3 , x 4 ) = k2 (η1 Z1 + η2 Z2 ), ( s10 , s20 , s30 , s40 ) = k( y 2 , − y 1 , y4 , − y 3 ) = k(y 1 U1 + y 2 U2 ),
(5.31)
where U1 = η1 Z3 − η2 Z4 ,
U2 = (uη1 )Z1 + (vη2 )Z2 .
Substituting (5.26), (5.30) and (5.31) into (4.15), we have W = (W1 , W2 , W3 , W4 ) = Φ3 Z1 + Ψ3 Z2 ,
(5.32)
where
η 4k2 η1 π η1 ζ22 1 − Φ3 = − , 3 ST S λ λT 4 λ
η 4k2 η2 π η2 ζ12 2 − − Ψ3 = . 3 ST S λ λT 4 λ
(5.33)
Substituting (5.32) into (4.16), we get
W, n3 = 2(Φ3 η2 − Ψ3 η1 ),
W, n4 = 0.
(5.34)
It follows from (5.33) and (5.34) that S3 is minimal if and only if 0=
4k2 1 1 2 η1 η2 − η2 η1 + η2 η1 − η1 η2 + η2 η1 ζ1 − η1 η2 ζ22 . S λ λT
(5.35)
u v It is clear that (5.35) holds for η1 = 0 or η2 = 0, from which we have h(s) = − es or h(s) = − es . It v u follows that the parametric function of C in (5.20) becomes u+v v−u s a, ,0 e , v v v
u + v u + v u − v γ(s) = b, a, , 0 es . u u u
γ(s) =
u + v
b,
or
(5.36) (5.37)
And the parametric function of S3 becomes v − u s e sin(vt − θ), − cos(vt − θ), cos vt, sin vt , or (5.38) v u − v s e − sin(ut − θ), cos(ut − θ), cos ut, sin ut . X(s, t) = (5.39) u If η1 = 0 (resp. η2 = 0), then η2 = − 2r/v es (resp. η1 = 2r/u es ). It is easy to check that if (5.35) holds, then η1 = 0 (resp. η2 = 0). In either cases, γ(s) is an open half straight line and S3 is a ruled surface described by (5.38) or (5.39). In the sequel, we suppose that η1 η2 η1 η2 = 0. Using (5.29), we obtain X(s, t) =
ζ1 − v = 4k2 rη12 > 0,
u − ζ2 = 4k 2 rη22 > 0,
> 0, ζ2 − ζ1 = 2rλ
4k 2 rT = 2rζ1 ζ2 + uv(ζ1 − ζ2 ) > 0. In order that (5.35) has regular solution, ζ1 and ζ2 have to satisfy (5.40) simultaneously.
(5.40)
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Let Ω = {(ζ1 , ζ2 )|v < ζ1 < ζ2 < u} and Γ(ζ1 , ζ2 ) = 2rζ1 ζ2 + uv(ζ1 − ζ2 ). Then it is directly to check that Γ(ζ1 , ζ2 ) > 0 over Ω. On the other hand, it is easy to see that dζ2 −8k 2 rη2 dη2 η2 dη2 η2 η2 = = − = − = 0. ζ˙2 := dζ1 8k 2 rη1 dη1 η1 dη1 η1 η1
(5.41)
η 2 η2 η2 (ζ1 − v) ˙ ζ2 , η2 η1 − η1 η2 = η2 η1 1 + 12 − η = η 1 + 2 1 η2 η1 η1 (u − ζ2 )
(5.42)
η 2 (η )2 (u − ζ ) (η )2 η 2 (η )2 η22 2 + (ζ˙2 )2 . S = (η1 )2 1 + 2 2 = 1 21 + (ζ˙2 )2 = 1 21 2 (η1 ) η2 η1 η2 (ζ1 − v)
(5.43)
Hence, it follows that
and
Therefore dζ˙2 ζ˙2 (u − ζ2 ) 1 η22 + ζ˙2 + =− 2 ζ¨2 = (η2 η1 − η1 η2 ), dζ1 η2 (ζ1 − v) η2 (η1 )3 η12 or equivalently, η2 η1 − η1 η2 =
! η12 (η1 )3 ζ˙2 (u − ζ2 ) + ζ˙2 . (8k 2 r)ζ¨2 + 2 η2 η2 (ζ1 − v)
(5.44)
Substituting (5.40), (5.42), (5.43) and (5.44) into (5.35), we get η12 (η1 )3 ¨ 1 (ζ1 − v) ˙ (u − ζ2 ) ζ˙2 (u − ζ2 ) + ζ˙2 + + (ζ˙2 )2 2ζ2 + 2 2 1+ ζ2 η2 4k rη2 (ζ1 − v) (u − ζ2 ) (ζ1 − v) rλ ! 1 (ζ1 − v) ˙ 2 (u − ζ2 ) + ζ2 ζ2 + (ζ˙2 )2 . ζ12 + 2 (u − ζ ) (ζ − v) 2 1 4k rλT
0 = (4k2 r)
It follows from the assumption that the above equality turns into 1 (ζ1 − v) ˙ (u − ζ2 ) ζ˙2 (u − ζ2 ) + ζ˙2 + + (ζ˙2 )2 1+ ζ2 2 2 4k rη2 (ζ1 − v) (u − ζ2 ) (ζ1 − v) rλ 1 (ζ1 − v) ˙ 2 (u − ζ2 ) + + (ζ˙2 )2 . ζ12 + ζ2 ζ 2 (u − ζ2 ) 2 (ζ1 − v) 4k rλT
0 = 2ζ¨2 +
(5.45)
Substituting (5.29), (5.40) into (5.45) and sorting, we obtain
2 2(u − ζ2 ) rζ 2 1 rζ 2 ! ˙ + 1+ 1 + 1+ 2 ζ2 (ζ2 − ζ1 )(ζ1 − v) Γ ζ1 − v (ζ2 − ζ1 ) Γ
1 2 2(ζ1 − v) rζ 2 ! ˙ 2 rζ 2 + 1+ 1 (ζ2 ) + 1 + 2 (ζ˙2 )3 . u − ζ2 (ζ2 − ζ1 ) Γ (ζ2 − ζ1 )(u − ζ2 ) Γ
0 = 2ζ¨2 + + Denote
(5.46)
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2(u − ζ2 ) rζ 2 1+ 1 , (ζ2 − ζ1 )(ζ1 − v) Γ
2(ζ1 − v) rζ 2 χ3 = 1+ 2 , (ζ2 − ζ1 )(u − ζ2 ) Γ
χ0 =
2 1 rζ 2 + 1+ 2 , ζ1 − v (ζ2 − ζ1 ) Γ
1 2 rζ 2 + χ2 = 1+ 1 . u − ζ2 (ζ2 − ζ1 ) Γ
χ1 =
1041
(5.47) (5.48)
Then (5.46) turns into −2ζ¨2 = χ0 + χ1 ζ˙2 + χ2 (ζ˙2 )2 + χ3 (ζ˙2 )3 . (k)
Denote ζ1
(5.49)
= dk ζ1 /dζ2k for any k ∈ N. Then (1)
ζ1
=
1 , ζ˙2
(2)
ζ1
=−
ζ¨2 . ˙ (ζ2 )3
(5.50)
Substituting (5.50) into (5.49), we obtain (2)
2ζ1
(1)
(1)
(1)
= χ3 + χ2 ζ1 + χ1 (ζ1 )2 + χ0 (ζ1 )3 .
(5.51)
From (5.49) and (5.51), we can see that the relation between ζ1 and ζ2 is complicated. We could not obtain the explicit solutions to either (5.49) or (5.51). But we could discuss the asymptotic behaviour of the solutions of (5.49). Note that ζ2 turns to u when ζ1 turns to u and that ζ1 turns to v when ζ2 turns to v, because of v < ζ1 < ζ2 < u. At first, let us consider the asymptotic behaviour of the solutions of (5.49) when ζ1 turns to u. Denote ζ1 = ζ1 − u. Suppose that the Taylor expansion of ζ2 with respect to ζ1 is ζ2 = u + aζ1k + o(ζ1k ). We want to find the first k such that a = 0. By direct computation, we have ζ˙2 = akζ1k−1 + o(ζ1k−1 ),
ζ¨2 = ak(k − 1)ζ1k−2 + o(ζ1k−2 ),
(5.52)
and 4r k−1 + o(ζ1k−1 ), χ1 = −3ζ1−1 + o(ζ1−1 ), ζ 3a 1 1 6r χ2 = − ζ1−k + o(ζ1−k ), χ3 = ζ1−k−1 + o(ζ1−k−1 ). a a
χ0 =
(5.53)
Substituting (5.52) and (5.53) into (5.49), we obtain ak(k − 5)ζ1k−2 + o(ζ1k−2 ) ≡ 0, from which we get k = 5. Therefore ζ2 − u = O[(ζ1 − u)5 ]. Taking the same discussion to the solution of (5.51), we have ζ1 − v = O[(ζ2 − v)5 ], or equivalently ζ2 − v = O[(ζ1 − v)1/5 ]. To summarize the above discussion, we obtain the following conclusions: 4 , F ) be a general (α, β) space defined by (5.4) and (5.5) and S3 be a parameterized Theorem 5.3. Let (M 4 surface in (M , F ) defined by (5.21) and (5.22). (1) If η1 η2 η1 η2 ≡ 0, then S3 is a minimal surface defined by (5.38) or (5.39); (2) If η1 η2 η1 η2 = 0, then S3 is minimal if and only if ζ1 and ζ2 given by (5.29) satisfy (5.49) (or equivalently (5.51)). Furthermore, we have ζ2 − u = O[(ζ1 − u)5 ] as ζ1 → u, or ζ2 − v = O[(ζ1 − v)1/5 ] as ζ1 → v.
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4 , F ) 6. A kind of minimal hypersurfaces in (M From Theorem 5.3 we can see that the generating curve γ of S3 is difficult to get. However, we note that
W, n4 ≡ 0 along S3 . So it is natural to consider the property of the ruled hypersurface Σ3 generated by S3 and the vector field n4 over S3 . 4 , F) defined by Let Σ3 be a ruled parameterized hypersurface in (M t, p) = X(s, t) + pn3 = η1 Z1 + η2 Z2 , X(s,
(6.1)
where η1 = η1 + pη2 and η2 = η2 − pη1 . Then s = ( X η1 )Z1 + ( η2 )Z2 ,
t = −(u X η1 )Z3 + (v η2 )Z4 ,
p = η Z1 − η Z2 . X 2 1
The tangent vector field Y on Σ3 is expressed as s y 1 + X t y 2 + X p y 3 Y = ( y 1 , y2 , y3 , y4 ) = X for any (y 1 , y 2 , y 3 ). Moreover, the normal vector field to Σ3 is 4 = −(v η2 )Z3 − (u η1 )Z4 . n
(6.2)
By direct computations, we can see that = 1 − 2k 2 ( η12 + η22 ), λ α = 2[( η1 )2 + ( η2 )2 ](y 1 )2 + 2[(u η1 )2 + (v η2 )2 ](y 2 )2 + 2[(η1 )2 + (η2 )2 ](y 3 )2 ,
(6.3)
β = −2k[u η12 + v η22 ]y 2 . In order that γ =
2 + β 2 = 1, we suppose that λα y 1 = σ1 sin ϕ cos ψ,
y 2 = σ2 sin ϕ sin ψ,
y 3 = σ3 cos ϕ,
for all ϕ ∈ [0, π], ψ ∈ [0, 2π], where η )2 + ( 1/σ12 = 2λ[( η2 )2 ], 1
1/σ22 = 2[u2 η12 + v 2 η22 − 8k 2 r2 η12 η22 ],
)2 + (η )2 ]. 1/σ32 = 2λ[(η 1 2
Then we have dτ = σ1 σ2 σ3 sin ϕdϕ ∧ dψ and
dτ = 4πσ1 σ2 σ3 , {γ=1}
(y 1 )2 dτ =
4 π(σ1 )3 σ2 σ3 , 3
(y 3 )2 dτ =
4 πσ1 σ2 (σ3 )3 . 3
{γ=1}
(y 2 )2 dτ = {γ=1}
4 πσ1 (σ2 )3 σ3 , 3
{γ=1}
(6.4)
After direct computation, we obtain ( s1 , s2 , s3 , s4 ) = k2 ( x1 , x 2 , x 3 , x 4 ) = k2 ( η1 Z1 + η2 Z2 ), ( s10 , s20 , s30 , s40 ) = k( y 2 , − y 1 , y4 , − y 3 ) = k(U1 y 1 + U2 y 2 + U3 y 3 ),
(6.5)
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where U1 = ( η1 )Z3 − ( η2 )Z4 ,
U2 = (u η1 )Z1 + (v η2 )Z2 ,
U3 = η2 Z3 + η1 Z4 .
It is easy to see that ss = η Z1 + η Z2 , X 1 2
tt = −(u2 η1 )Z1 − (v 2 η2 )Z2 , X
pp = 0. X
(6.6)
Substituting (6.2) to (6.6) into (4.15) we deduce that W = (W 1 , W 2 , W 3 , W 4 ) = 4πσ1 σ2 σ3 (ΦZ1 + ΨZ2 ),
(6.7)
where Φ = η1 (σ1 )2 −
u − 4k 2 η2 r 2 2k 2 η1 2 2 − η (σ ) , 1 2 [1 − 2k 2 ( η12 + η22 )]2 1 − 2k2 ( η12 + η22 )
Ψ = η2 (σ1 )2 −
v + 4k 2 η2 r 2 2k 2 η2 1 2 − η (σ ) . 2 2 [1 − 2k 2 ( η12 + η22 )]2 1 − 2k 2 ( η12 + η22 )
4 ≡ 0. Therefore we have the following From (6.2) and (6.7) we can see that W, n − β/ λ, where α α2 + β2 /λ and β are λ 4 , F ) defined by (6.1). Then Σ3 is defined by (4.5). Let Σ3 be a ruled parameterized hypersurface in (M minimal. 4 , F ) be a general (α, β) space with F = Theorem 6.1. Let (M
In the sequel, we construct an explicit example of Σ3 . Suppose that (η1 )2 + (η2 )2 = ρ2 ,
(6.8)
where ρ is a given constant. From (6.8), we have (1 + r2 )(h )2 + 2(1 − r2 )es h + (1 + r2 )e2s = 2ρ2 , from which we get h = xes ± y
1 − z 2 e2s ,
(6.9)
where 1 − r2 x=− , 1 + r2
y=
2ρ2 , 1 + r2
z=
2r2 . ρ2 (1 + r2 )
(6.10)
1 − z 2 e2s d(zes ). zes
(6.11)
Taking integration on both sides of (6.9), we have h = xes ± y Denote sin t = zes . Then
√ 1 − z 2 e2s ds = xes ± y
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√
cos2 t 1 cos t 1 − z 2 e2s t s d(sin t) = dt = cos t + ln tan2 + C3 d(ze ) = s ze sin t sin t 2 2 1 = cos t + [ln(1 − cos t) − ln(1 + cos t)] + C3 2 1 ln(1 − 1 − z 2 e2s ) − ln(1 + 1 − z 2 e2s ) + C3 . = 1 − z 2 e2s + 2
(6.12)
Substituting (6.12) into (6.11), we obtain ! 1 1 − z 2 e2s + (6.13) ln(1 − 1 − z 2 e2s ) − ln(1 + 1 − z 2 e2s ) + C3 . 2 − β/ λ, where α α2 + β2 /λ 4 , F ) be a general (α, β) space with F = λ Theorem 6.2. Let (M and β are 3 4 , F) defined by (6.1) with h(s) given defined by (4.5). Let Σ be a ruled parameterized hypersurface in (M 3 by (6.13). Then Σ is minimal. h(s) = xes ± y
Acknowledgments The authors would like to express their hearty gratitude to referee for the valuable suggestions and constructive comments. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
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