Area Minimizing Unit Vector Fields on Antipodally Punctured Unit 2-Sphere
Fabiano G.B. Brito, Jackeline Conrado, Icaro Gonçalves, Adriana V. Nicoli
AAREA MINIMIZING UNIT VECTOR FIELDS ON ANTIPODALLYPUNCTURED UNIT 2-SPHERE
FABIANO G. B. BRITO , JACKELINE CONRADO , ICARO GONC¸ ALVES , AND ADRIANA V. NICOLI Abstract.
We provide a lower value for the volume of a unit vector field tangent to anantipodally Euclidean sphere S depending on the length of an ellipse determined by the indexesof its singularities. In memory of Amine Fawaz Introduction and main results
Inspired by [1] and [4], we establish sharp lower bounds for the total area of unit vectorfields on antipodally punctured Euclidean sphere S , and these values depend on the indexes oftheir singularities. Theorem.
Let (cid:126)v be a unit vector field defined on M = S \ { N, S } . If k = max { I (cid:126)v ( N ) , I (cid:126)v ( S ) } ,then vol( (cid:126)v ) ≥ πL ( ε k ) , where L ( ε k ) is the length of the ellipse x k + y ( k − = 1 with k > and I (cid:126)v ( P ) stands for thePoincar´e index of (cid:126)v around P . This is a natural extension of the theorem proved in [2] by P. Chacon, D. Johnson andthe first author. In [2], a general lower-bound for area of unit vector fields in S \{ N, S } isestablished. It turns out to be the area of north-south vector field with both indexes equal to1. In this context, our theorem provides certain lower bounds for each class of index.We also exhibit minimizing vector fields (cid:126)v k within each index class. These fields have areasgiven essentially by the length of ellipses depending just on the indexes in N and S . The second and fourth authors was financed by Coordena¸c˜ao de Aperfei¸coamento de Pessoal de N´ıvel Superior-Brasil (CAPES) - Finance Code 001. a r X i v : . [ m a t h . DG ] F e b FABIANO G. B. BRITO , JACKELINE CONRADO , ICARO GONC¸ ALVES , AND ADRIANA V. NICOLI Preliminaries and the Proof of Theorem
Let M = S \ { N, S } be the Euclidean sphere in which two antipodal points N and S are removed. Denote by g the usual metric of S induced from R , and by ∇ the Levi-Civitaconnection associated to g . Consider the oriented orthonormal local frame { e , e } on M , where e is tangent to the meridians and e to the parallels. Let (cid:126)v be a unit vector field tangent to M and consider another oriented orthonormal local frame (cid:8) (cid:126)v ⊥ , (cid:126)v (cid:9) on M and its dual basis { ω , ω } compatible with the orientation of { e , e } .In dimension 2, the volume of (cid:126)v is given byvol( (cid:126)v ) = (cid:90) S (cid:112) γ + δ ν, (1)where γ = g ( ∇ (cid:126)v (cid:126)v, (cid:126)v ⊥ ) and δ = g ( ∇ v ⊥ (cid:126)v ⊥ , (cid:126)v ) are the geodesic curvatures associated to (cid:126)v and (cid:126)v ⊥ ,respectively.Let S α be the parallel of S at latitude α ∈ ( − π , π ) and S β be the meridian of S atlongitude β ∈ (0 , π ). Proposition.
Let θ ∈ [0 , π/ be the oriented angle from e to (cid:126)v . If (cid:126)v = (cos θ ) e + (sin θ ) e and (cid:126)v ⊥ = ( − sin θ ) e + (cos θ ) e , then γ + δ = 1 + (tan α + θ ) + θ , where θ = dθ ( e ) , θ = dθ ( e ) .Proof. We have(2) γ = g (cid:0) ∇ (cid:126)v (cid:126)v, (cid:126)v ⊥ (cid:1) = g (cid:0) ∇ (cos θ ) e +(sin θ ) e [(cos θ ) e + (sin θ ) e ] , (cid:126)v ⊥ (cid:1) = g (cid:0) ∇ (cos θ ) e (cos θ ) e , (cid:126)v ⊥ (cid:1) + g (cid:0) ∇ (sin θ ) e (cos θ ) e , (cid:126)v ⊥ (cid:1) + g (cid:0) ∇ (cos θ ) e (sin θ ) e , (cid:126)v ⊥ (cid:1) + g (cid:0) ∇ (sin θ ) e (sin θ ) e , (cid:126)v ⊥ (cid:1) and(3) δ = g (cid:0) ∇ (cid:126)v ⊥ (cid:126)v ⊥ , (cid:126)v (cid:1) = g (cid:0) ∇ ( − sin θ ) e +(cos θ ) e [ − (sin θ ) e + (cos θ ) e ] , (cid:126)v (cid:1) = g (cid:0) ∇ ( − sin θ ) e ( − sin θ ) e , (cid:126)v (cid:1) + g (cid:0) ∇ (cos θ ) e ( − sin θ ) e , (cid:126)v (cid:1) + g (cid:0) ∇ ( − sin θ ) e (cos θ ) e , (cid:126)v (cid:1) + g (cid:0) ∇ (cos θ ) e (cos θ ) e , (cid:126)v (cid:1) . REA MINIMIZING UNIT VECTOR FIELDS ON ANTIPODALLY PUNCTURED UNIT 2-SPHERE 3
We write γ and δ as the following sums γ = A + B + C + D and δ = A (cid:48) + B (cid:48) + C (cid:48) + D (cid:48) , with A = g (cid:0) ∇ (cos θ ) e (cos θ ) e , (cid:126)v ⊥ (cid:1) , B = g (cid:0) ∇ (sin θ ) e (cos θ ) e , (cid:126)v ⊥ (cid:1) ,C = g (cid:0) ∇ (cos θ ) e (sin θ ) e , (cid:126)v ⊥ (cid:1) , D = g (cid:0) ∇ (sin θ ) e (sin θ ) e , (cid:126)v ⊥ (cid:1) and A (cid:48) = g (cid:0) ∇ ( − sin θ ) e ( − sin θ ) e , (cid:126)v (cid:1) , B (cid:48) = g (cid:0) ∇ (cos θ ) e ( − sin θ ) e , (cid:126)v (cid:1) C (cid:48) = g (cid:0) ∇ ( − sin θ ) e (cos θ ) e , (cid:126)v (cid:1) , D (cid:48) = g (cid:0) ∇ (cos θ ) e (cos θ ) e , (cid:126)v (cid:1) . First observe that tan α = g (cid:0) ∇ e e , e (cid:1) and ∇ e e = 0. By an elementary computation weobtain A = sin θ (cos θ ) θ + cos θ tan α, B = (sin θ ) θ ,C = (cos θ ) θ + sin θ cos θ tan α, D = (sin θ cos θ ) θ and A (cid:48) = (sin θ cos θ ) θ + sin θ tan α, B (cid:48) = ( − cos θ ) θ ,C (cid:48) = (sin θ ) θ + sin θ cos θ tan α, D (cid:48) = − (sin θ cos θ ) θ . Moreover, γ = (cos θ tan α + sin θ cos θ tan α ) + (sin θ (cos θ ) θ + (cos θ ) θ )+ ((sin θ ) θ + (sin θ cos θ ) θ )= cos θ tan α + (cos θ ) θ + (sin θ ) θ = cos θ (tan α + θ ) + (sin θ ) θ and δ = (sin θ tan α + sin θ cos θ tan α ) + ((sin θ cos θ ) θ + (sin θ ) θ )+ (( − cos θ ) θ − (sin θ cos θ ) θ )= sin θ tan α + (sin θ ) θ − (cos θ ) θ = sin θ (tan α + θ ) − (cos θ ) θ . Finally,(4) γ = cos θ (tan α + θ ) + (sin θ ) θ , (5) δ = sin θ (tan α + θ ) − (cos θ ) θ . FABIANO G. B. BRITO , JACKELINE CONRADO , ICARO GONC¸ ALVES , AND ADRIANA V. NICOLI From equations (4) and (5), we find1 + γ + δ = 1 + (cos θ (tan α + θ ) + (sin θ ) θ ) + (sin θ (tan α + θ ) − (cos θ ) θ ) = 1 + cos θ (tan α + θ ) + (sin θ ) θ + sin θ (tan α + θ ) + (cos θ ) θ = 1 + (tan α + θ ) + θ . Therefore, 1 + γ + δ = 1 + (tan α + θ ) + θ . (cid:3) Our proposition allows us to rewrite the volume functional as an integral depending on thelatitude α and the derivatives of θ (6) vol( (cid:126)v ) = (cid:90) M (cid:113) α + θ ) + θ . Proof of Theorem.
Given ϕ such that 0 ≤ ϕ ≤ π , α + θ ) + θ = (cid:18) cos ϕ + sin ϕ (cid:20)(cid:113) (tan α + θ ) + θ (cid:21)(cid:19) + (cid:18) − sin ϕ + cos ϕ (cid:20)(cid:113) (tan α + θ ) + θ (cid:21)(cid:19) . Hence, vol( (cid:126)v ) = (cid:90) M (cid:115)(cid:18) cos ϕ + sin ϕ (cid:20)(cid:113) (tan α + θ ) + θ (cid:21)(cid:19) + (cid:18) − sin ϕ + cos ϕ (cid:20)(cid:113) (tan α + θ ) + θ (cid:21)(cid:19) . Remember that 1 + (tan α + θ ) + θ ≥ α + θ ) , implies (cid:113) α + θ ) + θ ≥ (cid:113) α + θ ) . From the general inequality, √ a + b ≥ | a cos ϕ + b sin ϕ | , for any a , b , ϕ ∈ R , we have (cid:113) α + θ ) ≥ | cos ϕ + sin ϕ (tan( α ) + θ ) | . Therefore,(7) vol( (cid:126)v ) ≥ (cid:82) M (cid:113)(cid:0) cos ϕ + sin ϕ | tan α + θ | (cid:1) + (cid:0) − sin ϕ + cos ϕ | tan α + θ | (cid:1) ≥ (cid:82) M cos ϕ + sin ϕ | tan α + θ | . This inequality is valid for all ϕ such that 0 ≤ ϕ ≤ π .As a next step one consider the following conditions: REA MINIMIZING UNIT VECTOR FIELDS ON ANTIPODALLY PUNCTURED UNIT 2-SPHERE 5 i) ϕ k ( α ) = arctan (cid:18) tan α + k − α (cid:19) ;ii) tan (cid:0) ϕ k ( α ) (cid:1) = tan α + k − α . Replacing these conditions in equation (7) we find(8) vol( (cid:126)v ) ≥ (cid:90) M (cid:0) cos (cid:0) ϕ k ( α ) (cid:1) + sin (cid:0) ϕ k ( α ) (cid:1)(cid:1) | tan α + θ | ν. Condition (i) provides thatcos (cid:0) ϕ k ( α ) (cid:1) = cos α (cid:112) k − + 2( k −
1) sin α , − π ≤ α ≤ π , sin (cid:0) ϕ k ( α ) (cid:1) = k − α (cid:112) k − + 2( k −
1) sin α , − π ≤ α ≤ π . Thus, as the second part of the inequation (8) is equal to (9) lim α →− π (cid:34)(cid:90) π α (cid:90) π (cid:32) cos α (cid:112) k − + 2( k −
1) sin α + k − α (cid:112) k − + 2( k −
1) sin α | tan α + θ | (cid:33)(cid:35) dβdα. Remember that Cartan’s connection form ω is given by ω = δω + γω , where { ω , ω } is dual basis of { (cid:126)v ⊥ , (cid:126)v } . Let i ∗ : S α (cid:44) → S be the inclusion map, and e =sin θ(cid:126)v ⊥ + cos θ(cid:126)v , then i ∗ ( ω )( e ) = δ sin θ + γ cos θ. From equations (4) and (5), we have i ∗ ( ω )( e ) = sin θ (cid:2) sin θ (cid:0) tan α + θ (cid:1) − cos θ ( θ ) (cid:3) + cos θ (cid:2) cos θ (cid:0) tan α + θ (cid:1) + sin θ ( θ ) (cid:3) = tan α + θ . Thus, from (9)(10) vol( (cid:126)v ) ≥ lim α →− π (cid:32)(cid:90) π α (cid:32)(cid:90) π (cid:32) cos α + (cid:0) ( k −
1) + sin α (cid:1) i ∗ ( ω )( e ) (cid:112) k − + 2( k −
1) sin α (cid:33) dβ (cid:33) dα (cid:33) . FABIANO G. B. BRITO , JACKELINE CONRADO , ICARO GONC¸ ALVES , AND ADRIANA V. NICOLI In order to compute the integral of i ∗ ω over the parallel of S at constant latitude α , we followthe same arguments in the proof Theorem 1.1 of [2]. Denote by ω the connection form ω and S α = { ( x, y, z ∈ R ; z ≥ sin α, α ≤ α ≤ π } . The 2-form dω is given by dω = ω ∧ ω . A simple application of Stokes’ theorem implies that (cid:90) S α dω = 2 π (cid:0) I N ( (cid:126)v ) (cid:1) − (cid:90) S α i ∗ ω . Suppose that I N ( (cid:126)v ) = sup { I N ( (cid:126)v ) , I S ( (cid:126)v ) } = k , we obtain(11) (cid:90) S α i ∗ ( ω )( e ) dβ = 2 πk − Area (cid:0) S α (cid:1) = 2 πk − π (cid:0) − sin α (cid:1) = 2 π (cid:0) k − α (cid:1) . From inequation (10),vol( (cid:126)v ) ≥ lim α →− π (cid:18)(cid:82) π α (cid:18)(cid:82) π (cid:18) cos α + (cid:0) ( k − α (cid:1) i ∗ ( ω )( e ) √ k − +2( k −
1) sin α (cid:19) dβ (cid:19) dα (cid:19) = lim α →− π (cid:82) π α (cid:18) cos α √ k − +2( k −
1) sin α (cid:82) π dβ + (cid:0) ( k − α (cid:1) √ k − +2( k −
1) sin α (cid:82) S α i ∗ ω (cid:19) dα = lim α →− π (cid:82) π α (cid:32) π cos α √ k − +2( k −
1) sin α + π (cid:0) ( k − α (cid:1) √ k − +2( k −
1) sin α (cid:33) dα, where the last inequality is obtained from (11). Therefore,vol( (cid:126)v ) ≥ π lim α →− π (cid:90) π α (cid:32) cos α + (cid:0) ( k −
1) + sin α (cid:1) (cid:112) k − + 2( k −
1) sin α (cid:33) dα. Analogously, vol( (cid:126)v ) ≥ π lim α →− π (cid:90) π α (cid:16)(cid:112) k − + 2( k −
1) sin α (cid:17) dα. A trigonometrical identity give usvol( (cid:126)v ) ≥ π (cid:90) π − π (cid:114) ( k − + 4( k −
1) sin (cid:16) α π (cid:17) dα. Assume that t = α + π , then(12) vol( (cid:126)v ) ≥ π (cid:90) π (cid:113) ( k − + 4( k −
1) sin tdt. REA MINIMIZING UNIT VECTOR FIELDS ON ANTIPODALLY PUNCTURED UNIT 2-SPHERE 7
Consider k > ε k given by x k + y ( k − = 1 . Let µ be a parametrization for ε k defined by µ ( t ) = ( k cos t, ( k −
2) sin t ). Its length is(13) L ( ε k ) = 4 (cid:90) π (cid:18)(cid:113) ( k − + 4( k −
1) sin t (cid:19) dt. Therefore, vol( (cid:126)v ) ≥ πL ( ε k ) . (cid:3) Area-minimizing vector fields (cid:126)v k on M Now we are going to exhibit a family of unit vector fields attaining the lower value fromthe main Theorem. At the end of this Section, we also show a geometric interpretation of theareas of these vector fields.Our previous computations implyvol( (cid:126)v ) = (cid:82) [1 + θ + (tan α + θ ) ] ≥ (cid:82) [1 + (tan α + θ ) ] ≥ (cid:82) | cos ϕ + sin ϕ (tan α + θ ) | . Assuming vol( (cid:126)v ) = πL ( ε k ), we have θ = 0 and cos ϕ (tan α + θ ) = sin ϕ , where ϕ ∈ R . Then θ = tan ϕ − tan α . Finally ϕ = ϕ ( α ) = arctan (cid:0) tan α + k − α (cid:1) , which implies θ = k − α .We may summarize this discution in a definition including p (cid:55)→ e ( p ), which is a minimalvector field tangent to the meridians of S \{ N, S } , see [1] and [2]. Definition.
Let k be a positive integer, k (cid:54) = 2 and define:(1) (cid:126)v ( p ) = (cid:126)e ( p ) , if k = 1 ;(2) (cid:126)v k ( p ) = cos θ ( p ) (cid:126)e ( p ) + sin θ ( p ) (cid:126)e ( p ) , if k > , where θ : S \ { N, S } → R satisfy θ ( p ) = k − (cid:112) x + y . FABIANO G. B. BRITO , JACKELINE CONRADO , ICARO GONC¸ ALVES , AND ADRIANA V. NICOLI Notice that θ has constant variation along the parallel x + y = cos α , with α ∈ (cid:0) − π , π (cid:1) constant, and this includes the case where k = 1.If we use spherical coordinates ( β, α ) so that p = (cos α cos β, cos α sin β, sin α ), we can saythat the vector (cid:126)v k spins at a constant speed of rotation along the parallel α . Moreover, (cid:126)v k givesexactly k − α parallel, with respect to the referential { (cid:126)e , (cid:126)e } , and itgives k turns with respect to a fixed polar referential, in this case, θ ( p ) = k − α .In the case of a vector field with two singularities at N and S with indexes 4 and − k = 4vol( (cid:126)v ) ≥ π (cid:90) π √ sin t dt Figure 1.
Singularity of index4 in north pole
Figure 2.
Singularity of index − (cid:126)v k , its image (cid:126)v k ( S \{ N, S } ) is a surface in T S \{ N, S } .Using the well-known Pappus-Guldin’s Theorem, its area equals the area of a suitable ellipsoidof revolution, see the Figure 3. REA MINIMIZING UNIT VECTOR FIELDS ON ANTIPODALLY PUNCTURED UNIT 2-SPHERE 9
Figure 3.
Elippsoid of revolution
Acknowledgment
We would like to thank professor Giovanni Nunes for carefully reading and commenting aprevious version of this manuscript.
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Email address : [email protected], [email protected] Dpto. de Matem´atica, Instituto de Matem´atica e Estat´ıstica, Universidade de S˜ao Paulo,R. do Mat˜ao 1010, S˜ao Paulo-SP, 05508-900, Brazil.
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