Surfaces making constant angle with certain vector fields in 3-spaces

Surfaces making constant angle with certainvector fields in 3-spacesAna-Irina NistorUniversidad de Granada - May 30, 2012A.I.Nistor (KUL) CAS Granada, May 31, 2012 1 / 43

OutlineOutline1 Introduction2 Constant angle with R direction in Product SpacesCAS in M 2 × RCPD in M 2 × RCAS&CPD in Lorentzian Product spaces M 2 × R 13 Constant angle with the position vectorConstant slope surfaces in E 34 Constant angle with a Killing vector field in E 35 CAS in Heisenberg group6 CAS in Solvable Lie groupsA.I.Nistor (KUL) CAS Granada, May 31, 2012 2 / 43

IntroductionConstant Angle SurfacesA constant angle surface (CAS in short) is an oriented surface for whichits normal makes a constant angle with a fixed direction, which is chosenin each case as a preferred direction in the ambient space:1 R direction in M 2 × R, M 2 × R 12 position vector in E 3 and E 3 13 Killing vector field in E 34 Killing vector field e 3 = ∂ z in Nil 35 left invariant vector field in G(µ 1 , µ 2 ), in particular inSol 3 = G(−1, 1)A.I.Nistor (KUL) CAS Granada, May 31, 2012 3 / 43

IntroductionConstant Angle Surfaces and Principal DirectionsA property of CAS:When the ambient is of the form M 2 × R, a favored direction is R. It isknown that for a constant angle surface in E 3 , S 2 × R or in H 2 × R, theprojection of ∂ ∂t(where t is the global parameter on R) onto the tangentplane of the immersed surface, denoted by T , is a principal direction 1 withthe corresponding principal curvature 2 identically zero.Study surfaces endowed with a principal direction T which will be called acanonical principal direction (CPD in short).1 eigenvector of the shape operator2 eigenvalue of the shape operatorA.I.Nistor (KUL) CAS Granada, May 31, 2012 3 / 43

Constant angle with R direction in Product SpacesResults - CAS in M 2 × RCAS in M 2 × RF. Dillen, J. Fastenakels, J. Van der Veken, L. Vrancken, Constantangle suraces in S 2 × R, Monatsh. Math. 152 (2)(2007), 89–96.F. Dillen, M.I. Munteanu, Constant Angle Surfaces in H 2 × R, Bull.Braz. Math. Soc. 40 (1) (2009) 1, 85–97.M.I. Munteanu, N., A new approach on constant angle surfaces in E 3 ,Turkish J. Math. 33 (2) (2009), 169–178.A.I.Nistor (KUL) CAS Granada, May 31, 2012 4 / 43

Constant angle with R direction in Product SpacesResults - CPD in M 2 × RCPD in M 2 × RF. Dillen, J. Fastenakels, J. Van der Veken, Surfaces in S 2 × R with acanonical principal direction, Ann. Glob. Anal. Geom., 35(2009) 4,381–396.F. Dillen, M.I. Munteanu, N., Surfaces in H 2 × R with a canonicalprincipal direction, Taiwanese J. Math., 15 (2011) 5, 2265-2289.M.I. Munteanu, N., Complete classification of surfaces with acanonical principal direction in the Euclidean space E 3 , Cent. Eur. J.Math., 9(2011)2, 378–389.A.I.Nistor (KUL) CAS Granada, May 31, 2012 5 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1CAS&CPD in M 2 × R 1 - Preliminaries• M 2 × R 1 the product of M 2 (c) - a 2-dimensional space form of constantsectional curvature c and R 1• 〈 , 〉 = 〈 , 〉 M 2 − dt 2 - the Lorentzian metric• the surface M is given by L : M → M 2 × R 1• ξ is a δ-unit normal vector field to M, 〈ξ, ξ〉 = δ ∈ {−1, 1}.δ = −1, ξ is timelike iff M is spacelikeδ = 1, ξ is spacelike iff M is timelike• ∂ t := (∂/∂t), t ∈ R 1 is a unitary timelike vector field on M 2 × R 1 ,∂ t = T + ΘξΘ = cosh θ (M is spacelike) or Θ = sinh θ (M is timelike).• for every tangent vector field X on M:∇ X T = ΘSX , (1)X (Θ) = −δ〈SX , T 〉. (2)A.I.Nistor (KUL) CAS Granada, May 31, 2012 6 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1Preliminaries - more about anglesRecall that ∂ t is timelike in (M 2 × R 1 , 〈 , 〉).If M is a spacelike surface, then ξ is timelike, and we will consider it withthe same time-orientation as ∂ t , i.e. future-directed, 〈∂ t , ξ〉 < 0.• The angle θ between two unitary timelike vectors was defined for thefirst time in 1984, as:cosh θ = −〈∂ t , ξ〉.If M is a timelike surface, then ξ is spacelike.• The angle θ between a timelike and a spacelike unitary vectors wasdefined for the first time in 2005, as:sinh θ = 〈∂ t , ξ〉.G. Birman, K. Nomizu, Trigonometry in Lorentzian geometry, Amer.Math. Monthly, 91(1984)9, 543–549.E. Ne˘sović, M.Petrović - Torga˘sev, L. Verstraelen, Curves in Lorentzianspaces, Boll. Unione Mat. Ital., serie 8, 8-B(2005)3, 685–696.A.I.Nistor (KUL) CAS Granada, May 31, 2012 7 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1Spacelike surfaces in S 2 × R 1 and H 2 × R 1• constant angle spacelike surfaces in Minkowski 3−space (when c = 0)have been classified inR. López, M.I. Munteanu, Constant angle surfaces in Minkowskispace, Bull. Belg. Math. Soc. - Simon Stevin , 18 (2011) 2, 271–286.• we extend this study to constant angle spacelike surfaces in S 2 × R 1(c = 1) and H 2 × R 1 (c = −1) .• we study canonical principal directions for spacelike surfaces in theambient space M 2 × R 1 .A.I.Nistor (KUL) CAS Granada, May 31, 2012 8 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1Spacelike surfaces in S 2 × R 1 and H 2 × R 1Theorem (Fu, N.)Let M be a spacelike surface immersed into Lorentzian product spaceM 2 (c) × R 1 , with c = −1, 1. Then, M is a constant angle spacelike surfaceif and only if the immersion L is locally given by:(a) If c = 1, then L : M → S 2 × R 1 ↩→ R 4 1 ,L(u, v) = ( cos(u cosh θ)f (v) + sin(u cosh θ)f (v) × f ′ (v), −u sinh θ ) ,where f is a unit speed curve in S 2 .(b) If c = −1, then L : M → H 2 × R 1 ↩→ R 4 2 ,L(u, v) = ( cosh(u cosh θ)f (v)+sinh(u cosh θ)f (v)⊠f ′ (v), −u sinh θ ) ,where f is a unit speed curve in H 2 .In both cases θ ≠ 0 denotes the constant hyperbolic angle.If θ = 0, then ∂ t is a normal vector field and M is an open part of thespacelike surface: • S 2 × {t 0 } for c = 1• H 2 × {t 0 } for c = −1.A.I.Nistor (KUL) CAS Granada, May 31, 2012 9 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1Spacelike surfaces in S 2 × R 1 and H 2 × R 1Theorem (Fu, N.)Let L : M → M 2 (c) × R 1 , c = −1, 1 be a spacelike surface and let θ be thehyperbolic angle function. Then, T is a canonical principal direction for Mif and only if M is parameterized as:(a) If c = 1, then(L : M → S 2 × R 1 ↩→ R 4 1 , )L(u, v) = cos φ(u) f (v) + sin φ(u) N f (v), χ(u) , where f is aregular curve on S 2 and N f (v) =f (v)×f ′ (v)√〈f ′ (v),f ′ (v)〉is the normal of f .(b) If c = −1,(then L : M → H 2 × R 1 ↩→ R 4 2 , )L(u, v) = cosh φ(u) f (v) + sinh φ(u) N f (v), χ(u) , where f is aregular curve on H 2 and N f (v) =f (v)⊠f ′ (v)√〈f ′ (v),f ′ (v)〉is the normal of f .Moreover, φ(u) =∫ ucosh θ(τ)dτ and χ(u) = −∫ usinh θ(τ)dτ.A.I.Nistor (KUL) CAS Granada, May 31, 2012 10 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1CPD for spacelike surfaces in E 3 1Theorem (N.)Let L : M → E 3 1 be a spacelike surface isometrically immersed in E3 1 andlet θ(p) ≠ 0 be the hyperbolic angle function. Then, M has a canonicalprincipal direction if and only if M is parametrized by:(c.1) L(u, v) =∫( ) ucos v, sin v, 0 cosh θ(τ)dτ − ( 0, 0, 1 ) ∫ usinh θ(τ)dτ + γ(v),where ( ∫ ∫γ(v) = ψ(v) sin v dv, −)ψ(v) cos v dv, 0 , ψ ∈ C ∞ (M),(c.2) L(u, v) =(cos v0 , sin v 0 , 0 ) ∫ ucosh θ(τ)dτ − ( 0, 0, 1 ) ∫ usinh θ(τ)dτ + vγ 0 ,()where γ 0 = − sin v 0 , cos v 0 , 0 , and v 0 is a real constant.A.I.Nistor (KUL) CAS Granada, May 31, 2012 11 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1CPD for spacelike surfaces in E 3 1 with H = 0Theorem (N.)The only maximal spacelike surfaces in E 3 1 with a canonical principal directionare the catenoids of 1st kind, L : M → E 3 1 ,(√L(u, v) = u 2 − m 2 cos v, √ u 2 − m 2 sin v, m ln(u + √ )u 2 − m 2 ) ,m ∈ R ∗ .RemarkUnder the assumption of flatness, we obtain the generalized cylinders fromcase (c.2) of the classification theorem.A.I.Nistor (KUL) CAS Granada, May 31, 2012 12 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1The catenoid of 1st kindThe catenoid of 1st kind maybe obtained rotating thecurve( ( t) )m sinhm − ln m , 0, taround the Oz axis.Figure: m = 1, t ∈ [−3, 3], v ∈ [0, 2π]A.I.Nistor (KUL) CAS Granada, May 31, 2012 13 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1Timelike surfaces in S 2 × R 1 and H 2 × R 1• constant angle timelike surfaces in Minkowski 3−space (when c = 0)have been classified inF. Güler, G. Şaffak, E. Kasap, Timelike constant angle surfaces inMinkowski space R 3 1 , Int. J. Contemp. Math. Sci., 6(2011)44,2189–2200.• we study constant angle timelike surfaces in S 2 × R 1 (c = 1) andH 2 × R 1 (c = −1) .• we study canonical principal directions for timelike surfaces in theambient space M 2 × R 1 .A.I.Nistor (KUL) CAS Granada, May 31, 2012 14 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1Timelike CAS S 2 × R 1 and H 2 × R 1Theorem (Fu, N.)Let L : M → M 2 (c) × R 1 be a timelike surface immersed into Lorentzianproduct space M 2 (c) × R 1 . Then M is a constant angle timelike surface ifand only if the immersion L is locally given by:(a) If c = 1, then L : M → S 2 × R 1 ,L(u, v) = ( cos(u sinh θ)f (v) + sin(u sinh θ)f (v) × f ′ (v), u cosh θ ) ,where f is a unit speed curve in S 2 ,(b) If c = −1, then L : M → H 2 × R 1 ,L(u, v) = ( cosh(u sinh θ)f (v) + sinh(u sinh θ)f (v) ⊠ f ′ (v), u cosh θ ) ,where f is a unit speed curve in H 2 .In both cases θ ≠ 0 denotes the constant hyperbolic angle.If θ = 0, then ∂ t is a tangent vector field, has no normal component, andM is an open part of γ × R 1 , where γ ∈ M 2 (c), c ∈ {−1, 1}.A.I.Nistor (KUL) CAS Granada, May 31, 2012 15 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1CPD for timelike surfaces in M 2 × R 1Theorem (Fu, N.)Let L : M → M 2 (c) × R 1 be a timelike surface and let θ be the hyperbolicangle function. Then, T is a canonical principal direction for M if and onlyif M is parametrized as:(a) If c = 1, then ( L : M → S 2 × R 1 ,)L(u, v) = cos χ(u) f (v) + sin χ(u) N f (v), φ(u) , where f is aregular curve on S 2 and N f (v) =f (v)×f ′ (v)√〈f ′ (v),f ′ (v)〉is the normal of f .(b) If c = −1, ( then L : M → H 2 × R 1 ,)L(u, v) = cosh χ(u) f (v) + sinh χ(u) N f (v), φ(u) , where f is aregular curve on H 2 and N f (v) =f (v)⊠f ′ (v)√〈f ′ (v),f ′ (v)〉is the normal of f .Moreover, φ(u) =∫ ucosh θ(τ)dτ and χ(u) =∫ usinh θ(τ)dτ.A.I.Nistor (KUL) CAS Granada, May 31, 2012 16 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1CPD for timelike surfaces in M 2 × R 1Theorem (Fu, N.)Let L : M → M 2 (c) × R 1 be a timelike surface and let θ be the hyperbolicangle function. Then, T is a canonical principal direction for M if and onlyif M is parametrized as:(c) If c = 0, then L : M → E 3 (1 ,)(c.1) L(u, v) = χ(u) cos v, χ(u) sin v, φ(u) + γ(v),( ∫∫)where γ(v) = − ψ(v) sin v dv, ψ(v) cos v dv, 0 ,and ψ is a(smooth function,)(c.2) L(u, v) = χ(u) cos v 0 , χ(u) sin v 0 , φ(u) + γ 0 v,where γ 0 = ( − sin v 0 , cos v 0 , 0 ) , and v 0 is a real constant.Moreover, φ(u) =∫ ucosh θ(τ)dτ and χ(u) =∫ usinh θ(τ)dτ.A.I.Nistor (KUL) CAS Granada, May 31, 2012 16 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1CPD for timelike surfaces in E 3 1 - minimalityCorollary (Fu, N.)The only flat timelike surfaces M immersed in E 3 1 endowed with a canonicalprincipal direction are given by the cylindrical surfaces parametrized in case(c.2) of previous Theorem.Theorem (Fu, N.)The only minimal timelike surfaces M immersed in E 3 1 endowed with acanonical principal direction are given by the catenoids of 3rd kind parametrizedas:L(t, v) =(m cos t m cos v, m cos t )m sin v, t , m ∈ R ∗ .A.I.Nistor (KUL) CAS Granada, May 31, 2012 17 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1The catenoid of 3rd kind...... may be obtained rotating thecurve:(m cos t m , 0, t ), m ∈ R ∗ ,around the Oz axis.Figure: m = 1, t ∈ [0, 2π], v ∈ [0, 2π]A.I.Nistor (KUL) CAS Granada, May 31, 2012 18 / 43

Constant angle with R direction in Product Spaces CAS&CPD in Lorentzian Product spaces M 2 × R 1PreprintsThe problems of- CAS for spacelike and timelike surfaces in S 2 × R 1 and H 2 × R 1- CPD for spacelike and timelike surfaces in M 2 × R 1are studied in :Y. Fu, N. Constant angle property and canonical principal directionsfor surfaces in M 2 (c) × R 1 , preprint 2012.- CPD for spacelike surfaces in E 3 1 :N., A note on spacelike surfaces in Minkowski 3-space, preprint 2011.A.I.Nistor (KUL) CAS Granada, May 31, 2012 19 / 43

Constant angle with the position vector Constant slope surfaces in E 3Constant angle with the position vectorLogarithmic spirals =⇒ constant slope surfacesLogarithmic spiral: planar curve having the property that the angle θbetween its tangent and the radial direction at every point is constant.A.I.Nistor (KUL) CAS Granada, May 31, 2012 20 / 43

Constant angle with the position vector Constant slope surfaces in E 3Constant angle with the position vectorIn other words, the logarithmic spiral is the curve whose tangent makes aconstant angle θ with the position vector in every point.Question - surfaces:Passing from curves to surfaces, find all surfaces in the Euclidean3-space making a constant angle with the position vector.A.I.Nistor (KUL) CAS Granada, May 31, 2012 20 / 43

Constant angle with the position vector Constant slope surfaces in E 3Constant angle with the position vectorAnswer: constant slope surfacesTheorem (Munteanu)Let r : M −→ E 3 be an isometric immersion. Then M is of constant slope ifand only if either it is an open part of the Euclidean 2-sphere centeredin the origin, or it can be parameterized byr(u, v) = u sin θ ( cos ϕ(u) f (v) + sin ϕ(u) f (v) × f ′ (v) )where θ is a constant (angle) different from 0, ϕ(u) = cot θ log u and f isa unit speed curve on the Euclidean sphere S 2 .θ = 0: the position vector is normal to the surface, M is an open partof the Euclidean 2-sphere;θ = π 2: M is a cone with the vertex in origin, or a plane passingthrough origin.A.I.Nistor (KUL) CAS Granada, May 31, 2012 20 / 43

ExamplesConstant angle with the position vector Constant slope surfaces in E 3θ = π 5θ = π 15f (v) = (cos v, sin v, 0) f (v) = (cos 2 v, cos v sin v, sin v)parametric lines: blue: logarithmic spiralblack: the spherical curve f .A.I.Nistor (KUL) CAS Granada, May 31, 2012 21 / 43

Constant angle with the position vector Constant slope surfaces in E 3Constant slope surfacesM.I. Munteanu, From Golden Spirals to Constant Slope Surfaces, J.Math.Phys., 51 (2010) 7, 073507:1–9.Y. Fu, D. Yang, On constant slope spacelike surfaces in 3-dimensionalMinkowski space, J. Math. Analysis Appl., 385 (2012) 1, 208–220.Y. Fu, X. Wang, Classification of Timelike Constant Slope Surfaces in3-Dimensional Minkowski Space, Res. Math. 2012.A.I.Nistor (KUL) CAS Granada, May 31, 2012 22 / 43

Constant angle with a Killing vector field in E 3PreliminariesConstant angle with a Killing vector field -Preliminaries• E 3 = (R 3 , 〈 , 〉),• ◦ ∇ - Levi-Civita connection corresponding to 〈 , 〉 in E 3 ,• V is Killing iff it satisfies the Killing equation:for any vector fields X , Y in R 3 .• The set〈 ◦ ∇X V , Y 〉 + 〈 ◦ ∇Y V , X 〉 = 0,{∂ x , ∂ y , ∂ z , −y∂ x + x∂ y , z∂ y − y∂ z , z∂ x − x∂ z }gives a basis of Killing vector fields in E 3 .A.I.Nistor (KUL) CAS Granada, May 31, 2012 23 / 43

Constant angle with a Killing vector field in E 3Classification results for curvesCurves - constant angle with a Killing field• If ˜γ is a straight line, then γ is a helix.W.l.o.g. the line can be taken to be (parallel with) one of the coordinateaxes, and this is an integral curve of a Killing vector vector field in E 3 .Motivated by this remark, a natural question appears:• which curves make a constant angle with a Killing vector field in E 3 ?A.I.Nistor (KUL) CAS Granada, May 31, 2012 24 / 43

Constant angle with a Killing vector field in E 3Classification results for curvesCurves - constant angle with a Killing field• If ˜γ is a straight line, then γ is a helix.W.l.o.g. the line can be taken to be (parallel with) one of the coordinateaxes, and this is an integral curve of a Killing vector vector field in E 3 .Motivated by this remark, a natural question appears:• which curves make a constant angle with a Killing vector field in E 3 ?Recall that we have a basis of Killing vector fields in E 3 :{∂ x , ∂ y , ∂ z , −y∂ x + x∂ y , z∂ y − y∂ z , z∂ x − x∂ z }A.I.Nistor (KUL) CAS Granada, May 31, 2012 24 / 43

Constant angle with a Killing vector field in E 3Classification results for curvesCurves - constant angle with a Killing field• If ˜γ is a straight line, then γ is a helix.W.l.o.g. the line can be taken to be (parallel with) one of the coordinateaxes, and this is an integral curve of a Killing vector vector field in E 3 .Motivated by this remark, a natural question appears:• which curves make a constant angle with a Killing vector field in E 3 ?Recall that we have a basis of Killing vector fields in E 3 :{∂ x , ∂ y , ∂ z , −y∂ x + x∂ y , z∂ y − y∂ z , z∂ x − x∂ z }A.I.Nistor (KUL) CAS Granada, May 31, 2012 24 / 43

Constant angle with a Killing vector field in E 3Classification results for curvesCurves - constant angle with a Killing field• If ˜γ is a straight line, then γ is a helix.W.l.o.g. the line can be taken to be (parallel with) one of the coordinateaxes, and this is an integral curve of a Killing vector vector field in E 3 .Motivated by this remark, a natural question appears:• which curves make a constant angle with a Killing vector field in E 3 ?Recall that we have a basis of Killing vector fields in E 3 :{∂ x , ∂ y , ∂ z , −y∂ x + x∂ y , z∂ y − y∂ z , z∂ x − x∂ z }• which curves make a constant angle with V in E 3 ?A.I.Nistor (KUL) CAS Granada, May 31, 2012 24 / 43

Constant angle with a Killing vector field in E 3Classification results for curvesPlane curves - constant angle with VTheorem (Munteanu, N.)A curve in the xy-plane makes constant angle θ with the Killing vector fieldV = −y∂ x + x∂ y if and only of it is given by one of the following cases:(a) or a straight line passing through the origin,(b) either the circle S 1 (r 0 ) centred in the origin and of radius r 0 ,(c) or the logarithmic spiral ρ(φ) = e tan θ (φ−φ 0) .Sketch of proof.For the curve p.a.l. γ(s) = (ρ(s) cos φ(s), ρ(s) sin φ(s)), the constantangle condition becomes: ρ(s)φ ′ (s) = cos θ.• if θ = π 2 : ρ(s) ≠ 0 and φ(s) = φ 0, (a),• if θ = 0: ρ(s) = ρ 0 and φ(s) = s ρ 0+ φ 0 , (b),• if θ ≠ 0: ρ(s) = s sin θ + s 0 and φ(s) = cot θ ln(s sin θ + s 0 ) + φ 0 , (c).A.I.Nistor (KUL) CAS Granada, May 31, 2012 25 / 43

Constant angle with a Killing vector field in E 3Classification results for curvesSpace curves - constant angle with VIn cylindrical coordinates: γ(s) =Theorem (Munteanu, N.)()ρ(s) cos φ(s), ρ(s) sin φ(s), z(s)A curve γ in the Euclidean space E 3 \Oz makes a constant angle θ with theKilling vector field V = −y∂ x + x∂ y if and only if, is given, in cylindricalcoordinates (ρ, φ, z), up to vertical translations and rotations around z-axis,by: ρ(s) = ρ 0 + sin θ∫ s∫ sz(s) = sin θ sin ω(ζ)dζ,∫cos ω(ζ)dζ, φ(s) = cos θwhere ρ 0 ∈ R and ω is a smooth function on I ⊂ R.sdζρ(ζ) ,A.I.Nistor (KUL) CAS Granada, May 31, 2012 26 / 43

Constant angle with a Killing vector field in E 3Classification results for curvesExamples for different values of ωΘ⩵Π 20, Ωs⩵40.060.040.000.020.050.10 0.05 0.050.020.050.100.040.000.050.050.060.100.050.000.08Figure: Space curve making constant angle with V (left) and its projection(right): ω = ω 0A.I.Nistor (KUL) CAS Granada, May 31, 2012 27 / 43

Constant angle with a Killing vector field in E 3Classification results for curvesExamples for different values of ω0.05Θ⩵Π 20,Ωs⩵3s5110 70.000.050.05510 6110 7 510 6 510 6 110 70.00510 60.00 0.050.05110 7Figure: Space curve making constant angle with V (left) and its projection(right): ω = ms + nA.I.Nistor (KUL) CAS Granada, May 31, 2012 27 / 43

Constant angle with a Killing vector field in E 3Classification results for curvesExamples for different values of ωΘ⩵Π 20, Ωs⩵arccoss0.0150.10.0100.0050.00.015 0.010 0.005 0.005 0.010 0.015 0.0200.0050.10.050.0100.050.000.000.050.050.015Figure: Space curve making constant angle with V (left) and its projection(right): ω = arccos(s)A.I.Nistor (KUL) CAS Granada, May 31, 2012 27 / 43

Constant angle with a Killing vector field in E 3Surfaces - PreliminariesSurfaces making constant angle with a Killing vector field• V = −y∂ x + x∂ y must be non-null, thus the surface M lies in E 3 \ Oz;• g the metric on M and by ∇ the associated Levi-Civita connection,• N denotes the unit normal to the surface M;• denote ∠(V , N) := θ - constant angle;If θ = π 2, then M is a surface of revolution.If θ = 0, then we obtain half-planes having z−axis as boundary.• projecting V on the tangent plane to M:V = T + µ cos θξ,where ξ is the unit normal to M, T is the tangent part, with‖T ‖ = µ sin θ and µ = ‖V ‖.• choose an orthonormal basis {e 1 , e 2 } on the tangent plane to M s.t.e 1 =T‖T ‖ and e 2 ⊥ e 1 .• If follows that V = µ(sin θe 1 + cos θξ).A.I.Nistor (KUL) CAS Granada, May 31, 2012 28 / 43

Constant angle with a Killing vector field in E 3Surfaces - Basic formulasSurfaces making constant angle with a Killing vector fieldFor an arbitrary vector field X in E 3 , we have◦∇X V = k × X , (3)where k = (0, 0, 1) and × stands for the usual cross product in E 3 .Consider {e 1 , e 2 , k, ξ} in a point on M and define the angles:∠(ξ, k) := ϕ, ∠(e 1 , k) := η ∠(e 2 , k) := ψ,which are not independent, cos ϕ = − sin θ sin ψ cos η = cos θ sin ψ.We decompose k × e 1 and k × e 2 in the basis {e 1 , e 2 , ξ},k ×e 1 = − sin θ sin ψ e 2 −cos ψξ, k ×e 2 = sin θ sin ψ e 1 +cos θ sin ψξ. (4)If X is tangent to M, then◦()∇X V = X (µ) sin θe 1 + cos θξ()+ µ sin θ ∇ X e 1 + h(X , e 1 )− µ cos θAX .(5)A.I.Nistor (KUL) CAS Granada, May 31, 2012 29 / 43

Constant angle with a Killing vector field in E 3Surfaces - Basic formulasSurfaces making constant angle with a Killing vector fieldFrom (3), (5) and (4) we gete 1 (µ) = − cos θ cos ψ, e 2 (µ) = sin ψ.As a consequence, we obtain the shape operator:()sin θ cos ψ−S =µ0, (6)0 λwhere λ is a smooth function on M, and the Levi-Civita connection:∇ e1 e 1 = − sin ψµ e 2, ∇ e1 e 2 = sin ψµ e 1,∇ e2 e 1 = λ cotanθ e 2 , ∇ e2 e 2 = −λ cotanθ e 1 .From (6) we see that e 1 and e 2 are principal directions on M.A.I.Nistor (KUL) CAS Granada, May 31, 2012 30 / 43

Constant angle with a Killing vector field in E 3Surfaces - Basic formulasSurfaces making constant angle with a Killing vector fieldThen, using the expressions of the Levi-Civita connection we may computethe Lie bracket of e 1 and e 2 :[e 1 , e 2 ] = sin ψµ e 1 − λ cotanθ e 2 . (7)Consequently, a compatibility condition is found computing [e 1 , e 2 ](µ) intwo ways:−cos ψ e 1 (ψ)+cos θ sin ψ e 2 (ψ) =cos θ sin ψ cos ψµ+λ cotanθ sin ψ. (8)A.I.Nistor (KUL) CAS Granada, May 31, 2012 31 / 43

Constant angle with a Killing vector field in E 3Surfaces making constant angle with a Killing vector fieldCoordinates (x, y, z) ↦→ (ρ cos φ, ρ sin φ, z)From now on we use cylindrical coordinates, such that the parametrizationof the surface M may be thought as()F : D ⊂ R 2 −→ E 3 \ Oz, (u, v) ↦→ ρ(u, v), φ(u, v), z(u, v) . (9)The Euclidean metric in E 3 becomes a warped metric〈 , 〉 = dρ 2 + dz 2 + ρ 2 dφ 2Note that the Killing vector field V coincides with ∂ φ .The basis {e 1 , e 2 , ξ} may be expressed in terms of the new coordinates as:e 1 = − cos θ cos ψ ∂ ρ + sin θµ ∂ φ + cos θ sin ψ ∂ z ,e 2 = sin ψ ∂ ρ + cos ψ ∂ z , (10)ξ = sin θ cos ψ ∂ ρ + cos θµ ∂ φ − sin θ sin ψ ∂ z .A.I.Nistor (KUL) CAS Granada, May 31, 2012 32 / 43

Constant angle with a Killing vector field in E 3Surfaces making constant angle with a Killing vector fieldClassification theoremTheorem (Munteanu, N.)Let M be a surface isometrically immersed in E 3 \Oz and consider the Killingvector field V = −y∂ x + x∂ y . Then M makes a constant angle θ with V ifand only if is one of the following surfaces, up to vertical translations androtations about z−axis:(a) a half-plane with z−axis as boundary,(b) a rotational surface around z−axis,A.I.Nistor (KUL) CAS Granada, May 31, 2012 33 / 43

Constant angle with a Killing vector field in E 3Surfaces making constant angle with a Killing vector fieldClassification theoremTheorem (Munteanu, N.)Let M be a surface isometrically immersed in E 3 \Oz and consider the Killingvector field V = −y∂ x + x∂ y . Then M makes a constant angle θ with V ifand only if is one of the following surfaces, up to vertical translations androtations about z−axis:(c) a right cylinder over a logarithmic spiral given by :F (u, z) = ( u cos θ, log(cu − tan θ ), z ) ,c ∈ R ∗For θ = π 3 ; and c = 3we get this figure:A.I.Nistor (KUL) CAS Granada, May 31, 2012 33 / 43

Constant angle with a Killing vector field in E 3Classification theoremSurfaces making constant angle with a Killing vector fieldTheorem (Munteanu, N.)Let M be a surface isometrically immersed in E 3 \Oz and consider the Killingvector field V = −y∂ x + x∂ y . Then M makes a constant angle θ with V ifand only if is one of the following surfaces, up to vertical translations androtations about z−axis:(d) the Dini’s surface defined in cylindrical coordinates (ρ, φ, z) by(F (u, v) = −cos θ sin(cu) cv tan θ(, − − tan θ log tan cu )c cos θ2where c is a nonzero real constant.v −cos θ cos(cu)c,), (11)A.I.Nistor (KUL) CAS Granada, May 31, 2012 33 / 43

Constant angle with a Killing vector field in E 3Dini’s surface - parametrization...Surfaces making constant angle with a Killing vector field... from cylindrical back to Euclidean coordinatesF (u, v) =cos θ sin(cu)− cosccos θ sin(cu)− sincv −cos θ cos(cu)ccv tan θ−cos θcv tan θ−cos θ − tan θ log tan cu ,2 − tan θ log tan cu ,2( )cv tan θcu ↦→ u 1 and −cos θ− tan θ log tan cu 2↦→ u 2x(u 1 , u 2 ) = − cos θ sin u c 1 cos u 2 ,y(u 1 , u 2 ) = − cos θ sin u c 1 sin u 2 ,z(u 1 , u 2 ) = − cos θc cos u 1 + log tan u 12¡¡−cos θc tan θ u 2.Figure: θ = π 3 , c = √ 32A.I.Nistor (KUL) CAS Granada, May 31, 2012 34 / 43

Constant angle with a Killing vector field in E 3More over Dini’s surfaceSurfaces making constant angle with a Killing vector field• This surface is named after Ulisse Dini (1845 -1918), who obtained it studying helicoidal surfaces.• Dini’s surface is a helicoidal surface with axis Oz:F (ρ, φ) = ( ρ cos φ, ρ sin φ, hφ + Λ(ρ) ) ,where (Λ ◦ ρ)(u) = − cos θc(log ( tan cu ) )2 + cos(cu)and the pitch equals to h = − cos θc tan θ .• It may be obtained twisting the pseudosphere ofradius cos θc.• It has constant negative Gaussian curvaturedepending on the constant angle θ,K = −c 2 tan 2 θ, c ∈ R ∗ .Figure: θ = π 3 , c = √ 32A.I.Nistor (KUL) CAS Granada, May 31, 2012 35 / 43

Constant angle with a Killing vector field in E 3Surfaces making constant angle with a Killing vector fieldFinal remarksProposition (Munteanu, N.)The parametric curves of Dini’s surface are circular helices(v-param) andspherical curves(u−param).Corollary (Munteanu, N.)Looking backward, the u−parameter curves make the constant angle π 2 −θ with the Killing vector field V and the affine function ω (appearing inTheorem of space curves) is given by ω(s) = cs, c ∈ R ∗ .Let M make a constant angle with the Killing vector field V , and:M is totally geodesic iff it is a vertical plane with the boundary Oz;M is minimal not totally geodesic iff it is a catenoid about z−axis;M is flat iff it is a vertical plane with the boundary z−axis, a flatrotational surface or a right cylinder over a logarithmic spiral.A.I.Nistor (KUL) CAS Granada, May 31, 2012 36 / 43

Constant angle with a Killing vector field in E 3Final remarksSurfaces making constant angle with a Killing vector fieldFigure: PseudosphereFigure: Dini’s surfaceSource: http://virtualmathmuseum.org/galleryS.htmlA.I.Nistor (KUL) CAS Granada, May 31, 2012 37 / 43

ArticlesConstant angle with a Killing vector field in E 3ArticlesThe problems of- curves making constant angle with a rotational Killing vector field in E 3- surfaces making constant angle with a rotational Killing vector field in E 3are studied in :M.I. Munteanu, N. Surfaces in E 3 making constant angle with Killingvector fields, Internat. J. Math., 23 (2012) 6, 1250023:1-16.A.I.Nistor (KUL) CAS Granada, May 31, 2012 38 / 43

CAS in Heisenberg groupCAS in Heisenberg group Nil 3R 3 : (x, y, z) ∗ (x, y, z) =(x + x, y + y, z + z + xy 2 − xy ).2Remark that the mapping⎧⎛⎞⎫ ⎛⎨ 1 a bR 3 ⎬→ ⎝ 0 1 c ⎠⎩0 0 1 ∣ a, b, c ∈ R ⎭ : (x, y, z) ↦→ ⎝1 x z + xy 20 1 y0 0 1⎞⎠is an isomorphism between (R 3 , ∗) and a subgroup of GL(3, R).For every τ ≠ 0: left-invariant Riemannian metric on (R 3 , ∗)() 2g = dx 2 + dy 2 + 4τ 2 dz + . After the change of coordinatesy dx−x dy2(x, y, 2τz) ↦→ (x, y, z), g = dx 2 + dy 2 + (dz + τ(y dx − x dy)) 2Some authors: only if τ = 1 2 .Nil 3 = (R 3 , ∗) with g.A.I.Nistor (KUL) CAS Granada, May 31, 2012 39 / 43

CAS in Heisenberg groupCAS in Heisenberg group Nil 3The following vector fields form a left-invariant orthonormal frame on Nil 3 :e 1 = ∂ x − τy∂ z , e 2 = ∂ y + τx∂ z , e 3 = ∂ z .The geometry of Nil 3 can be described in terms of this frame.The Killing vector field e 3Nil 3 .plays an important role in the geometry ofDefinitionWe say that a surface in the Heisenberg group Nil 3 is a constant anglesurface if the angle θ between the unit normal and the direction e 3 is thesame at every point.- cannot have θ = 0 - contradiction with: [e 1 , e 2 ] = 2τe 3 , [e 2 , e 3 ] = 0[e 3 , e 1 ] = 0, since τ ≠ 0.A.I.Nistor (KUL) CAS Granada, May 31, 2012 40 / 43

CAS in Heisenberg groupCAS in Heisenberg group Nil 3The following vector fields form a left-invariant orthonormal frame on Nil 3 :e 1 = ∂ x − τy∂ z , e 2 = ∂ y + τx∂ z , e 3 = ∂ z .The geometry of Nil 3 can be described in terms of this frame.The Killing vector field e 3Nil 3 .plays an important role in the geometry ofDefinitionWe say that a surface in the Heisenberg group Nil 3 is a constant anglesurface if the angle θ between the unit normal and the direction e 3 is thesame at every point.- cannot have θ = 0 - contradiction with: [e 1 , e 2 ] = 2τe 3 , [e 2 , e 3 ] = 0[e 3 , e 1 ] = 0, since τ ≠ 0.A.I.Nistor (KUL) CAS Granada, May 31, 2012 40 / 43

CAS in Heisenberg groupCAS in Heisenberg group Nil 3The following vector fields form a left-invariant orthonormal frame on Nil 3 :e 1 = ∂ x − τy∂ z , e 2 = ∂ y + τx∂ z , e 3 = ∂ z .The geometry of Nil 3 can be described in terms of this frame.The Killing vector field e 3Nil 3 .plays an important role in the geometry ofDefinitionWe say that a surface in the Heisenberg group Nil 3 is a constant anglesurface if the angle θ between the unit normal and the direction e 3 is thesame at every point.- cannot have θ = 0 - contradiction with: [e 1 , e 2 ] = 2τe 3 , [e 2 , e 3 ] = 0[e 3 , e 1 ] = 0, since τ ≠ 0.A.I.Nistor (KUL) CAS Granada, May 31, 2012 40 / 43

CAS in Heisenberg groupCAS in Heisenberg group Nil 3Theorem (Fastenakels, Munteanu, Van der Veken)Let M be a constant angle surface in the Heisenberg group Nil 3 . Then Mis isometric to an open part of one of the following types of surfaces:(i) a Hopf-cylinder,(ii) a surface given byr(u, v) =(12τ tan θ sin u + f 1(v), − 12τ tan θ cos u + f 2(v),− 14τ tan2 θ u − 1 2 tan θ cos uf 1(v) − 1 2 tan θ sin uf 2(v) − τf 3 (v)with (f ′1 )2 + (f ′2 )2 = sin 2 θ and f ′3 (v) = f ′1 (v)f 2(v) − f 1 (v)f ′2 (v).)J. Fastenakels, M.I. Munteanu, J. Van der Veken, Constant AngleSurfaces in the Heisenberg group, Acta Math. Sinica(Engl. Ser.), 27(2011) 4, 747–756.A.I.Nistor (KUL) CAS Granada, May 31, 2012 41 / 43

CAS in Solvable Lie groupsCAS in Solvable Lie groups... to be continued...A.I.Nistor (KUL) CAS Granada, May 31, 2012 42 / 43

The endThank you for attention !A.I.Nistor (KUL) CAS Granada, May 31, 2012 43 / 43