thehyperbolic space Umbilical routes along geodesics and hypercycles in Differential Geometry and its Applications

the hyperbolic space

Maciej Czarnecki

WydziałMatematykiiInformatyki,KatedraGeometrii,UniwersytetŁódzki,Łódź,Poland

a r t i cl e i n f o a b s t r a c t

Articlehistory:

Received25June2018 Receivedinrevisedform21 November2018

Availableonlinexxxx

CommunicatedbyE.Garcia-Rio

MSC:

53C12 53A30 53A35

Keywords:

Foliation Umbilical Hypercycle Geodesic Umbilicalroute

Givenageodesiclineγ inthehyperbolicspaceHⁿ weformulatea necessaryand sufficient condition for a function along this geodesic which measure the mean curvatureoftotallyumbilicalleavesofafoliationorthogonaltoγ.Thenweextend theresulttoγ beingahypercyclei.e.ageodesiconahypersurfaceequidistantfrom thetotallygeodesicone.

©2019ElsevierB.V.Allrightsreserved.

0. Introduction

In the geometric theory of foliations, a question on foliations with totally umbilical leaves comes just after that ontotally geodesic foliations. Thelast onefor compact or finite volume manifolds hasdefinite and negative answer (see [8] for some history). In [13] Langevin and Walczak proved that on a closed manifoldofconstantnon-zerocurvaturethere arenototallyumbilicalfoliations.Foropenmanifoldsthere are geometrical classifications of totally geodesic foliations in the hyperbolic space by Ferus ([10]) and Browne([4]).

The question on totally umbilicalroutes along curves inthe real hyperbolic space Hⁿ was formulated in [8]. In [1] we announced a solution for geodesics, now extended to transversals which are horocycles or hypercycles. More general result was obtained by the author and Langevin – see Remark 4.3 for a mention.

E-mailaddress:maczar@math.uni.lodz.pl.

https://doi.org/10.1016/j.difgeo.2019.02.001 0926-2245/©2019ElsevierB.V.Allrightsreserved.

ThemostgeneralresultofthepaperisTheorem3.1givingnecessaryandsufficientconditionforafunction alongarc-lengthparametrizedhypercycletogeneratetotallyumbilicalfoliations.Namely,ifthehypercycle has (constant) geodesic curvature cos ϕ then this condition states that the mean curvature of leaves h starts with at least−sin ϕ, ends with at mostsin ϕ andbetween itsmodified hyperbolic arcus tangentis a(sin ϕ)-Lipschitzfunction.The conditionismorevisible incase ofageodesictransversal(Theorem 2.2):

ath◦ h is1-Lipschitz.Respectiveinequalitiesfordifferentiableh appear inTheorem3.5 andTheorem 2.6.

Thelastoneformulatesash^≥ h²− 1 andshowsthatclosetoleavesofsmallmeancurvaturethereismore roomforchangeofh.

1. Umbilicalhypersurfacesofthe hyperbolicspace

Umbilicity is a standard notion in Riemannian geometry and one of the easiest which is conformally invariant.

A point on a submanifold of aRiemannian manifold is called umbilical if all eigenvalues of the shape operator at this point are equal. Inthis casethe mean curvature at the points is this common value (or its opposite,dependingonorientation).Consequently, asubmanifoldistotally umbilical ifconsists onlyof umbilicalpointsandatotally umbilicalfoliation onaRiemannianmanifoldisafoliationwithalltheleaves totallyumbilical.

For the real n-dimensionalhyperbolic space Hⁿ consider its half-spacemodel i.e. theset Π^n,+ ={x ∈ Rⁿ | xn> 0} endowed withtheRiemannianmetric

g(X, Y )_x= 1 x²_nX, Y  where .,. denotesthestandardEuclidean innerproduct.

Thehyperbolic distanceinthehalf-spaceisgivenbytheformula(cf.[2])

d(x, y) = 2 ath

ˆx − ˆy²+ (x_n− yn)²

ˆx − ˆy²+ (xn+ yn)² where x = (xˆ 1,. . . ,x_n−1) andanalogouslyfory.Here

ath = (tanh)⁻¹ : t→ ln

1 + t 1− t. IntheparticularcaseΠ^2,+⊂ C,

d(z, w) = 2 ath

z− w z− ¯w

,

especiallyd(ai,bi)=ln^a_bfora,b> 0.

Everyisometryofthehalf-spacemodelisaconformaldiffeomorphismΠ^n,+ ontoitselfi.e.acomposition ofahorizontaltranslation,adilation,aninversioninasphereorthogonaltotheidealboundaryoridentity, and anorthogonaltransformationinthefirstn− 1 variables(cf. [2]).

Inparticular, foranytwogeodesiclinesthere isanisometrysendingonetoanother.

Hⁿ isanHadamardmanifoldsointhepurelymetricway(cf.[3])wecoulddefinetheidealboundaryand horospheres.Inthehalf-spacemodel,theidealboundary isatopological(n−1) sphere

Rⁿ⁻¹× {0}

∪{∞}.

A horosphere isaspheretangenttoRⁿ⁻¹× {0} (withouttangencypoint)orahyperplane paralleltoit.

Fig. 1. Hyperspheres and horospheres.

Totallygeodesichypersurfaces areopenhemi-spheresoropenhalf-hyperplanesorthogonaltoRⁿ⁻¹×{0}.

Aconnectedcomponentofasetequidistantfrom atotallygeodesic hypersurfaceiscalledahypersphere.In thehalf-spacemodel,ahypersphereisapartofasphereorhyperplanetransverselyintersectingRⁿ⁻¹× {0}

includedinΠ^n,+.(SeeFig.1.)

A horosphere has natural orientation inside it. For a hypersphere S we induce its orientation from orientationofthecorrespondingtotallygeodesichypersurfaceS₀ fromwhichS isequidistantinsuchaway thatahyperballbounded byS sharestheidealboundarywiththehyperbolichalf-spaceboundedbyS0. Definition1.1.Weusethecommonnamegeneralizedhypersphere foracomplete hypersurfaceinHⁿwhich iseitherhorosphere,hypersphereoratotallygeodesichypersurfaceandattachto suchageneralizedhyper- sphere itsangleofintersection withtheidealboundary.

Thusahorosphereis0-or-π-hypersphere(dependingonitsend)whileatotallygeodesichypersurfaceis a ^π₂-hypersphere.

Proposition 1.2.Let ageneralizedhypersphere beorientedinside (i.e. “down” inthehalf-space model)and makesexternal angleβ withtheidealboundary.Then itsmeancurvature isconstant andequals−cos β.

Ahorosphere of end∞ is ofconstant meancurvature 1.

Proof. UsingChristoffelsymbols for aconformal changeofRiemannianmetric([9])onecancalculatethe secondfundamentalform forhyperplanesinΠ^n,+.

AssumethatthehypersphereisrepresentedbyapartofhyperplaneS makingangleβ with Rⁿ⁻¹× {0}

andv itsunitnormalvector i.e.v = (v1,. . . ,vn)⊥ S andv= 1. Thenobviouslyvn =±cos β.

Lużyńczyk in [14] observed that the shape operator of S ∩ Π^n,+ is simply vn· Id so with the given orientationh=−cos β. 2

Proposition1.3.AconnectedcompleteunboundedhypersurfaceofHⁿistotallyumbilicaliffitisageneralized hypersphere.

Proof. Itisclassical(cf.[16])thatanytotallyumbilicalhypersurfaceofRⁿ iscontainedinasphereor ina hyperplane.

The half-spacemodel ofHⁿ is conformallyequivalent to Rⁿ. Conformal diffeomorphisms preserve um- bilicityhenceallconnectedcompletetotallyumbilicalhypersurfacesinthehalf-spacemodelarenonempty intersections ofΠⁿ⁺byasphereor ahyperplane.

Amongthem there aremetricsphereswhich arebounded so anyunboundedcomplete umbilicalhyper- surface is the cross-section of a sphere or hyperplane not disjoint with ideal boundary i.e. a generalized hypersphere.Moredetailedexplanationcanbe foundin[5]. 2

Definition 1.4. A ϕ-hypercycle is ageodesic line onaϕ-hypersphere, ϕ∈ [0,π]. In thehalf-spacemodela ϕ-hypercycle isacross-sectionofaϕ-hyperspherewitha2-dimensionalplanethroughitscenterorsimply open raymakingangleϕ withtheidealboundary.

Example 1.5. A generalized ϕ-hypercycle has constant geodesic curvature equal |cos ϕ| (cf. [6]). In Π^2,+

hypercycles (atthesametimehyperspheres,n= 2)are

1. geodesic (ϕ= ^π₂)of ideal ends 0 and ∞ being positive imaginary half-axis iR+ parametrized by arc- lengthast→ ie^t;

2. ϕ-hypercycle Eϕ = e^iϕR+, ϕ ∈  0,^π₂

, of ideal ends 0 and ∞ parametrized by arc-length as t →

e^{t sin ϕ+iϕ};

3. horocycle(ϕ= π)withtheidealend∞ having arc-lengthparametrizationst→ t+ ia witha> 0.

2. Umbilicalroutesalonggeodesics

In thissection we shallstudy totally umbilicalfoliationsofHⁿ orthogonalto agiven geodesicline.We shall provide a Lipschitz-type condition (Theorem 2.2) and a differentiable condition (Theorem 2.6) for changeofthemeancurvatureofleavesalongthegeodesic.

Definition 2.1. Let γ : R→ Hⁿ be an arc-lengthparametrized curve. Wesay thata realfunction h is an umbilical route along γ if the family (Lt) of generalized hyperspheres orthogonal to γ and having mean curvature h(t) atγ(t) couldbe extendedtoatotallyumbilicalfoliationofHⁿ.

Incodimension1 therealhyperbolic spaceistheonlycarryingnon-trivial umbilical routes.

Inanopenballthereisnofoliationtangenttotheboundaryandtotallyumbilicalcompletehypersurfaces inotherconstantcurvaturespaces,Rⁿ andSⁿ,arefullspheresorhyperplanes.Ontheotherhand,anytwo nonparallel hyperplanesinRⁿ intersect.Thus inSⁿ there isno umbilicalroute andinRⁿ theonly oneis constantlyequal0 alongastraightline.

In nonconstant curvature even very regular symmetric spaces like the complex hyperbolic space CHⁿ havenototallyumbilicalhypersurfaces.

We start with a very simple case of umbilical route along a geodesic where the situation is clear and formulaearepredictable.

Theorem 2.2. Let F be a transversely C⁰ codimension 1 totally umbilical foliation of Hⁿ orthogonal to an arc-length parametrized geodesic line γ. If for any t ∈ R the mean curvature of the leaf (taken with orientation oppositetoγ)throughγ(t) ish(t) then |h|≤ 1 and therearet₋,t+∈ [−∞,+∞] suchthat

(i) h|(−∞,t−]≡ −1,

(ii) the function (ath◦ h)|(t₋,t+) is 1-Lipschitz, (iii) h|[t+,+∞)≡ 1.

(1)

Conversely,if h:R→ [−1,1] is acontinuous functionsatisfying (1) thenh isan umbilicalroutealong any geodesic linein Hⁿ.

Fortheproofwe needelementarylemmas.

Lemma2.3.Lets> 0,β ∈ [0,π) andC beacircleofcenterC∈ C andradiusR orthogonaltotheimaginary axis iR at pointi· s andmakingexternal angleβ withtherealaxis.

Fig. 2. Hypercycles orthogonal to a geodesic.

Then

R = s

1 + cos β, C = i s cos β 1 + cos β andthepoint(s)of theintersection C ∩ R areoftheform

a_± =±s tanβ 2.

Proof. SinceC isorthogonalto iR itscenterC∈ iR andC = s− R (Fig.2). Atapointa∈ C ∩ R radius isperpendicularto thetangent.Ifβ isacute(other casesaresimilar)then 0aC =^π₂ − β and

s− R

R = sinπ 2− β

which implies R = s 1 + cos β. ThuswehaveC and

a =±R sin β = ±s sin β

1 + cos β =±s tanβ

2. 2

Lemma2.4.Let0< s1< s2,β1,β2∈ [0,π) andC1,C2⊂ C becirclesorthogonaltotheimaginaryaxisiR at pointsis1,is2 andmeeting thereal axisR atangles β1,β2,respectively.

ThenC1 andC2 do notintersect in theupper half-planeΠ^2,+ iff cot^β₂²

cot^β₂¹ ≤ s₂ s1

providedthat β1,β2∈ (0,π) or β1= β2= 0.

Proof. For agiven C1 whichintersectsR transversally (β1 > 0)theonly situationof C1∩ C2∩ Π^2,+=∅ is thata2− ≤ a1− anda1+≤ a2+. HenceweobtaintheinequalitybyLemma2.3. 2

Corollary 2.5. LetS bea hyperspherewhich isof constant distanceδ from atotally geodesic hypersurface.

If β∈ 0,^π₂

issuchthat themeancurvature ofS is equalto−cos β thencos β = tanh δ.

Proof. Afterahorizontaltranslation andadilation (bothare hyperbolic isometries)we mayassumethat S is containedinaspherecenteredontheimaginary axis andintersects this axisat pointis, s> 1,while

the corresponding totally geodesichypersurfacemeets iR atthe pointi. Then by definitionof hyperbolic distance inΠ^n,+ wehaveδ = ln s.

Now wehaveequalityinLemma2.4 andthesecondangleis ^π₂ so e^δ= cot^β₂.Hence

δ = ln cotβ

2 = ln1 + cos β

sin β = ln 1 + cos β

1− cos²β = ath(cos β).

In[11] we couldfindthisformulaintheequivalent formcot β = sinh δ. 2 Now wearepreparedfor

ProofofTheorem2.2. AgeodesicsphereinHⁿ cannotserveasaleafofacodimension1 foliationbecause itsinteriorhasnonzeroEulercharacteristicandcarriesnofoliationtangenttotheboundary.Thustheonly possible leavesoftotallyumbilicalfoliationsonHⁿ aregeneralizedhyperspheresand infact|h|≤ 1.

To prove (i) observe that if some leaf L_γ(t) is a horosphere “centered” at the begin γ(−∞) then all precedingleavesmustbehorospheresofthesame“centre”–thereisnoroomforotheridealboundary.This argument worksinproofof(iii)as well.

We useaconformaltransformation ofΠ^n,+ whichisthen hyperbolic isometryto putthegeodesic γ as then-thhalf-axis

A_n,+={x1= . . . = x_n−1 = 0, x_n > 0}

oriented“up”.Anygeneralizedsphererepresentingageneralizedhypersphereorthogonaltoγ hasitscenter ontheA_n,+.

Consider sectionof Π^n,+ by any2-dimensional plane P containingthen-th axis and orthogonalto the idealboundary.ThenP∩Π^n,+isisometrictoΠ^2,+andF ∩P isgeneralizedhypercyclefoliationorthogonal to γ = P∩ An,+.

In Π^2,+ weparametrizethegeodesic γ byarc-length,namely γ(t)= ie^t. Fixt1 and forany t2> t1 use criterionfromLemma2.4tohaveL_γ(t1)∩Lγ(t2)=∅.ByProposition1.2themeancurvatureofL_γ(ti)equals h(t_i)=−cos β(t_i),i= 1,2,thus

e^t2

e^t1 ≥ cot^β(t₂²⁾ cot^β(t₂¹⁾ =

1−h(t2) 1−(h(t2))²

1−h(t1) 1−(h(t1))²

=

1+h(t2) 1−h(t2)

−1

1+h(t1) 1−h(t1)

−1 = e^{−ath(h(t2))} e^{−ath(h(t1))}

whichisexactly(ii).

Now assumethat h iscontinuous, bounded by 1 andsatisfy (1). Conditions (i) and (iii) imply proper foliationinsidethelastofinitialhorospheresandthefirstoffinishingones.From(ii)weknowthatgeneralized hyperspheresofgiven meancurvaturearepairwisedisjoint.

Since everyleaf dividesHⁿ into two hyperbolic half-spacesand h iscontinuous, there is aleafthrough everypointbetweeninitialandfinishinghorospheresandfamilyofhypersphereswithgivenh alongγ covers alltheHⁿ. 2

If weassumethatafoliationis transverselydifferentiable thenthe conditiononumbilicalroute iseven simpler.

Theorem2.6.Forfunctionh ofmeancurvatureofleavesofatransverselyC¹totallyumbilicalcodimension 1 foliationofHⁿalongarc-lengthparametrizedgeodesic(transversalorientation oppositetothegeodesic)there are t₋,t₊∈ [−∞,+∞] such that

Fig. 3. “Rising sun” foliation inH²of common ends 0 and∞.

(i) h|(−∞,t−]≡ −1, (ii) h^ ≥ h²− 1, (iii) h|[t+,+∞)≡ 1.

(2)

Conversely, if h : R → [−1,1] is a C¹-function satisfying (2) then h is an umbilical route along any geodesic lineinHⁿ.

Proof. Forany ε> 0 (1)(ii)gives

−ath(h(t + ε))− ath(h(t))

ε ≤ 1

Aslimitinε→ 0 weobtain

h^

1− h² ≥ −1. 2

Remark2.7.Thecondition(2)(ii)onthederivativeofh couldbereformulatedintermsofintersectionangle as

β^≥ − sin β.

Example2.8.

1. Totallygeodesicfoliationwithh≡ 0 isrepresentedbyconcentricspheres.

2. Horosphericalfoliationwithh≡ −1 isrepresentedbyspherestangentatonepoint.

3. A function h = −tanh is extremal for estimation (2)(ii). A foliation defined by h is represented by sphereshaving an(n− 2)-dimensionalsphere⊂ Rⁿ⁻¹× {0}∪ {∞} incommon.Thisisso-calledpencil of spheres definedbylinearcombinationsofequations oftwointersectingspheres.

Indimension2 such afoliationlookslike “risingsun”(Fig. 3).

Proposition 2.9. For a given (even non-discrete) family of generalized hyperspheres pairwise disjoint and orthogonaltoagivengeodesic thereisatotally umbilicalfoliationofwholeHⁿ containingthesehypercycles asleaves.

Proof. Let γ be an arc-length parametrized geodesicline inHⁿ and (L_tα)_α∈A be afamily of generalized hypersphereswhicharepairwisedisjoint.AssumethatL_tα containsγ(t_α) andhasmeancurvatureequalto h(t_α).

We shall construct an umbilical route along γ. First add to the set X = {tα | α ∈ A} set X^ of its accumulation points. In X^ define values of h as limits. Observe thatfunction h on X∪ X^ is such that ath◦ h is1-Lipschitz.Thiscomesfrom disjointnessofthefamilyandLemma2.3.

The set R\ (X ∪ X^) is open so it is a unionof disjoint open intervals. On everyof its closure define h as linearfunction. If the interval is infinite then h is constant. Hence we obtaina continuous function h:R→ [−1,1] suchthatthefunctionath◦ h is1-Lipschitz.

In fact, if interval [a,b] such that(a,b) ⊂ R\ (X ∪ X^) thenwe useconvexity of ath if h(a),h(b) ≥ 0 (respectively, convexity of −ath if h(a),h(b) ≤ 0). Otherwise, there is c ∈ [a,b] such that h(c) = 0 and theargumentaboveworksforintervals[a,c] and[c,b] byaddingtotallygeodesichypersurfaceas L_c tothe family.

BytheTheorem2.2functionh isthenumbilicalroutealongγ. 2

Remark2.10.Theaboveconstruction couldbemodifiedtoobtainasmoothumbilicalrouteifthefamilyof hyperspheresisdiscrete.

On theother hand, omittingassumption onorthogonality tosome geodesicline makes Proposition2.9 false.Itis enoughtotaketwo disjoint totallygeodesichypersurfacesand athirdwhich isdisjoint butnot separatingthem.

3. Umbilicalroutesalonghypercycles

Now weshall consider totallyumbilicalfoliationsorthogonal toahypercycle. Similarlyto thecaseof a geodesic weshall provide aLipschitz-typeand differential condition.Theyare somethingtechnical butin termsofintersectionangle(Remark3.6)inequalityiseasiertoapply.

Theorem3.1.Let0< ϕ<^π₂.AssumethatF isatransverselyC⁰codimension1 totallyumbilicalfoliationof Hⁿ orthogonaltoanarc-lengthparametrizedϕ-hypercycle γ.Ifforanyt∈ R themeancurvatureoftheleaf (takenwithorientation oppositetoγ)throughγ(t) ish(t) then|h|≤ sin ϕ andtherearet₋,t+∈ [−∞,+∞]

such that

(i) h|(−∞,t−] ≡ − sin ϕ,

(ii) t→ ln sin ϕ− h(t) h(t) cos ϕ +

1− (h(t))² sin ϕ is a (sin ϕ)-Lipschitz function on (t₋, t+), (iii) h|[t+,+∞)≡ sin ϕ.

(3)

Conversely,if h:R→ [−sin ϕ,sin ϕ] isacontinuousfunctionsatisfying (3) thenh isan umbilicalroute along any ϕ-hypercycle inHⁿ.

Weshallmodify Lemmas2.3and 2.4.DenotebyE_ϕ theopenraye^iϕR+⊂ C.

Lemma 3.2. Let ϕ∈ 0,^π₂

,s> 0, β ∈ [0,π),and β = ^π₂ + ϕ. Assumethat C ⊂ C isa circleof centerC and radiusR meetingorthogonally Eϕ atonlyone pointse^iϕ andmeeting therealaxis R atangleβ.

Then β∈_π

2 − ϕ,^π₂+ ϕ ,

R = s sin ϕ

sin ϕ + cos β, C = s cos ϕ cos β

sin ϕ + cos β + is sin ϕ cos β sin ϕ + cos β and thepoint(s)of theintersectionC ∩ R areof theform

a_∓= s cos(ϕ± β) sin ϕ + cos β.

Fig. 4. Hypercycles orthogonal to a hypercycle.

Proof. SinceC ⊥ Eϕ,C = ce^iϕ forsomec∈ R (Fig.4). Thus

R =ce^iϕ− se^iϕ=|s − c| = s − c

becauseifc= s+ R then(s+ 2R)e^iϕwouldbethesecondpointofintersectionEϕ∩ C.ThetoppointofC is

C + iR = (s− R)e^iϕ+ iR = (s− R) cos ϕ + i(s sin ϕ + (1 − sin ϕ)R).

ShiftinghorizontallybytherealpartofC totheleft wereducethesituationtoLemma2.3.Now

R = s sin ϕ + (1− sin ϕ)R 1 + cos β hence

R = s sin ϕ

sin ϕ + cos β and C = (s− R)e^iϕ hencetherealandimaginarypartofC are

C = s cos ϕ cos β

sin ϕ + cos β, C = s sin ϕ cos β sin ϕ + cos β Moreover,C intersectstherealaxis atpoints

C ± (C + R) tanβ

2 = s cos ϕ cos β sin ϕ + cos β

±

s sin ϕ cos β

sin ϕ + cos β + s sin ϕ sin ϕ + cos β

 sin β 1 + cos β

= s cos(ϕ∓ β) sin ϕ + cos β.

ObservethatC cutsE_ϕ atonlyonepointiff|C|≤ R i.e.|cos β|≤ sin ϕ so ^π₂− ϕ≤ β < ^π₂ + ϕ.Butthis meansthattheleft andrighthandpointsofC ∩ R arerespectively

a₋= s cos(ϕ + β)

sin ϕ + cos β ≤ 0, a+= s cos(ϕ− β)

sin ϕ + cos β > 0. 2 Lemma 3.3. Let ϕ∈

0,^π₂

, 0< s₁< s₂,and β₁,β₂ ∈_π

2− ϕ,^π₂ + ϕ

.Assume that C1,C2 ⊂ C are circles orthogonal toE_ϕ atpointss₁e^iϕ,s₂e^iϕ andmeetingthereal axis R atanglesβ₁,β₂,respectively.

Then C1 and C2 donot intersect intheupperhalf-planeΠ^2,+ iff

sin ϕ+cos β2

cos(ϕ+β2) sin ϕ+cos β1

cos(ϕ+β1)

≤s₂ s1

.

Proof. Under these assumptions circles C1 and C2 do not intersect in Π^2,+ iff C1 is inside C2 (including internaltangency)ortheyintersectonthesideof−s1e^iϕnot“toohigh”i.e.upperintersectionpointisstill underoronR.

SincecentersofC1andC2 lieonthelinespanned byE_ϕ,pointsC1∩ C2aresymmetricinthisline.Thus iftheleftoneisunderthereal axisthenthesameistruefortheright one.

This impliesthatthesecond conditionisequivalent to therequestthattheleft handpoint ofC1∩ R is to therightofthelefthandpointofC2∩ R.NowitisenoughtouseLemma3.2. 2

Remark3.4. If β = ^π₂ + ϕ the role ofcircle meetingR atthis angle playsastraightline and theirunique common pointis _{cos β}^s .

If C1 issuchaline thenC2 hasnoroomto bend andmust aline parallel toC1.Similarly,ifC2 is aline meetingR atangle ^π₂ − ϕ thenC1 C2.

Proof of Theorem 3.1. As in the proof of Theorem 2.2 we conclude that the only possible leaves are generalizedhypercycles.EveryofthemisdiffeomorphictoRⁿ⁻¹anddividesHⁿintotwopartsdiffeomorphic to Rⁿ.FromthetopologicalpointofviewthefoliationisaproductRⁿ⁻¹× R soanytransversalmeetsany leafat mostonce.

LikeinTheorem2.2wereducethesituationtothedimension2 withAn,+beingthegeodesicfromwhich theϕ-hypercycleisequidistant.Lemma3.2implies|h|≤ sin ϕ.(i)and(iii)areexplainedinRemark3.4.

To prove(ii)recallthatE_ϕhasarc-lengthparametrization γ(t) = e^{t sin ϕ+iϕ}, t∈ R and use Lemma3.3.Infact,forgivent1< t2wehave

e^t2−t1sin ϕ

= e^{t2sin ϕ} e^{t1sin ϕ} ≥

sin ϕ−h(t2)

−h(t2) cos ϕ−

1−(h(t2))² sin ϕ sin ϕ−h(t1)

−h(t1) cos ϕ−

1−(h(t1))² sin ϕ

Now applyinglogarithmto bothsidesweseethatthefunction

t→ ln sin ϕ− h(t) h(t) cos ϕ +

1− (h(t))² sin ϕ is (sin ϕ)-Lipschitz.

For theconverse, argument from Theorem 2.2 works similarly but the foliation orthogonal to one hy- percycle does not fillall the Hⁿ. Anyway leaves of such afoliation have alimit (on both sides) which is anumbilicalhypersurface.Adomain boundedbyahyperspherecanbe easilyfoliatedaddingleavesofthe samemeancurvature(Fig. 5). 2

Fig. 5. Extension of umbilical foliation orthogonal to a hypercycle.

Differentiationof(3)(ii)leadsto

Theorem3.5.Thefunctionh ofmeancurvatureofleavesoftotallyumbilicaltransversallyC¹codimension 1 foliation of Hⁿ along arc-length parametrized ϕ-hypercycle satisfies |h| ≤ sin ϕ and there are t₋,t+ ∈ [−∞,+∞] such that

(i) h|(−∞,t−]≡ − sin ϕ, (ii) h^(t)≥ (h(t)−sin ϕ)

h(t) cos ϕ+

1−(h(t))² sin ϕ

1−(h(t))² 1−h sin ϕ+

1−(h(t))²cos ϕ on (t₋, t+), (iii) h|[t+,+∞)≡ sin ϕ.

(4)

Conversely,if h:R→ [−sin ϕ,sin ϕ] isaC¹-function satisfying (4) thenh is an umbilical route along any ϕ-hypercycle inHⁿ.

Remark3.6.TheconditionfromTheorem3.5 looksshorterintermsofintersectionangle

β^≥ (sin ϕ + cos β) cos(ϕ + β) 1 + sin(ϕ + β) . Indeed,itisenoughtoreformulateinequality

sin ϕ + cos β

− cos(ϕ + β)



≤ sin ϕ.

AnyhorocyclecouldbetransformedintoalineparalleltoRⁿ⁻¹× {0} byinversioncenteredattheendof thehorocycle.Thusanysphereorthogonaltoitintersectsthehorocycleintwopoints.Theonlygeneralized hyperspheres orthogonal to the horocycle and meetingit once are 0-hyperspheres represented byvertical hyperplanes.This motivatesthefollowing

Corollary3.7. The onlyumbilicalroute along ahorocycleish≡ 0.

Example3.8.

1. Afamilyofdisjointtotallygeodesichypersurfacesorthogonaltoahypercycleatanyofitspointfoliates wholetheHⁿ.

2. Constantcurvaturefoliationwithh≡ sin ϕ isrepresentedbyparallelhyperplaneswhichareorthogonal to(cos ϕ)-hypercycles.

3. Meancurvatureofleavesofafoliationbyconcentricspheres(withthecenteroutsideofΠ^n,+)orthogonal to someϕ-hypercyclevaries over (−sin ϕ,sin ϕ) along thehypercycle butonremaining domain ofHⁿ couldincludeevenhorospheres.

4. Finalremarks

4.1.In[10] Ferusclassified totallygeodesiccodimension1 foliationsofHⁿasthosewithorthogonaltransver- salofgeodesiccurvature≤ 1.

At any point of such a curve in Hⁿ one can find a generalized hypercycle in contact of order 2. If curvature of orthogonal transversal exceeded 1 then even totally geodesic leaves would intersect.In such sense,itsuffices toconsider onlygeneralizedhypercycles forlocal studyoftotallyumbilicalfoliations.

4.2.Ifk_g(p) and k_n(p) denoterespectivelythenormof thesecondfundamentalform oftheleafthroughp and geodesiccurvatureofanorthogonaltransversalthenk_g²+ k_n² ≤ 1 incaseoftotallyumbilicalfoliations along hypercycles.

Thisisaveryspecialcaseof Hadamardfoliations(cf.[7])forwhichthisestimationisprobablyalsotrue.

4.3.TotallyumbilicalfoliationsofHⁿcouldbedescribedinapurelyconformalway.Infact,theidealbound- ary and totally umbilicalleaves are represented by generalized spheres and the mean curvature depends onlyonangleof intersection.

This motivates descriptionofsuch objectsinthe spaceof spheres–deSitter spacewhich isquadricin the Lorentz space (a comprehensive study of this theory can be found in [12] and [15]). The author and Langevin gave alocal classificationbased on boosted time cones inthe de Sitterspace and deduce some global factsoncurvatureoforthogonaltransversals.

4.4.Inthepaperwerestrictedtohypercyclicorthogonaltransversalsasthemostsimilartototallyumbilical higher-dimensionalsubmanifolds.

We could define abi-umbilicalfoliation as totallyumbilical foliationwith a/allorthogonal transversals being totallyumbilical.Theclassificationofbi-umbilical foliationsonHⁿ maybeofsomeinterestevenfor codimension2 inH⁴.

References

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[13]R.Langevin,P.G.Walczak,Conformalgeometryoffoliations,Geom.Dedic.132(2008)135–178.

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[15]J.O’Hara,EnergyofKnotsandConformalGeometry,WorldScientific,2003.

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