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Research ArticleInvariant Surfaces under Hyperbolic Translations inHyperbolic Space
Mahmut Mak1 and Baki KarlJLa2
1 Department of Mathematics Faculty of Arts and Sciences Ahi Evran University 40100 Kırsehir Turkey2Department of Mathematics Faculty of Arts and Sciences Gazi University 06500 Ankara Turkey
Correspondence should be addressed to Mahmut Mak mahmutmakgmailcom
Received 23 May 2014 Accepted 19 September 2014 Published 20 November 2014
Academic Editor Henggui Zhang
Copyright copy 2014 M Mak and B Karlıga This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
We consider hyperbolic rotation (1198660) hyperbolic translation (119866
1) and horocyclic rotation (119866
2) groups in H3 which is called
Minkowski model of hyperbolic spaceThen we investigate extrinsic differential geometry of invariant surfaces under subgroups of119866
0in H3 Also we give explicit parametrization of these invariant surfaces with respect to constant hyperbolic curvature of profile
curves Finally we obtain some corollaries for flat andminimal invariant surfaces which are associated with de Sitter and hyperbolicshape operator in H3
1 Introduction
Hyperbolic space has five analytic models which are isomet-rically equivalent to each other [1 2] In this study we chooseMinkowski model of hyperbolic space which is denoted byH3 In a different point of view we may consider invariantsurface as rotational surface In this sense rotational surfacesin different ambient spaceswere studied bymany authors Forinstance in [3] Carmo and Dajczer define rotational hyper-surfaces with constant mean curvature (cmc) in hyperbolic119899-spaceThey also give a local parametrization of this surfacein terms of the cmc under some special conditions In [4]Mori studied elliptic spherical and parabolic type rotationalsurfaces with cmc in H3 In [5] the total classification of thetimelike and spacelike hyperbolic rotation surfaces is givenin terms of cmc in 3-dimensional de Sitter space S3
1 As a
general form explicit parametrizations of rotational surfaceswith cmc are given in Minkowski 119899-space by [6]
This paper is organized as follows In Section 2 we givebriefly the notions of H-point H-line H-plane and H-distance in hyperbolic geometry ofH3Throughout this workthe prefix ldquoH-rdquo is used is the sense of belonging to hyperbolicspace It is well known that H-isometry is a map which ispreserved H-distance in H3 The set of H-isometries is a
group which is identified with restriction of isometries ofMinkowski 4-space R4
1to H3 Let the group of H-isometries
be denoted by 119866 in H3 We consider subgroups 1198660 119866
1 and
1198662of 119866 with respect to leaving fixed timelike spacelike
and lightlike planes of R4
1 respectively Then we give the
notions of H-rotation H-translation and horocyclic rotationwhich are one-parameter actions of 119866 in H3 Moreover weobtain some properties of H-isometries There exist threekinds of totally umbilical surfaces which are called H-sphereequidistant surface and horosphere in H3 We obtain aclassification ofH-isometries by the subgroups119866
0119866
1 and119866
2
with respect to leaving fixed equidistant surfaces H-spheresand horospheres in H3 respectively In Section 3 we givethe basic theory of extrinsic differential geometry of curvesand surfaces in H3 In Section 4 we investigate surfaceswhich are invariant under a subgroup of H-translations inH3 Moreover in the sense of de Sitter and hyperbolic shapeoperator in H3 we study extrinsic differential geometry ofthese invariant surfaces by using notations in [7 8] We givea relation between one of the principal curvatures of theinvariant surface and hyperbolic curvature of profile curve ofthe invariant surface inH3 In a different viewpoint we obtainexplicit parametrization of some invariant surfaces in terms
Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2014 Article ID 838564 12 pageshttpdxdoiorg1011552014838564
2 Journal of Applied Mathematics
of constant hyperbolic curvature of profile curve Moreoverwe give some geometric results with respect to constanthyperbolic curvature of profile curve for flat and minimalinvariant surfaces in H3 Finally we give a classificationtheorem for the totally umbilical invariant surfaces in H3
2 Isometries of H3
In [9] Reynold give a brief introduction to hyperbolicgeometry of hyperbolic plane H2 Also he described explicitdescriptions of the hyperbolic metric and the isometries ofthe hyperbolic plane In this section we consider hyperbolicgeometry in H3 We especially determine isometry groups ofH3with respect to causal character of hyperplanes ofR4
1 then
these isometry groups are classified in terms of leaving thosetotally umbilic surfaces of H3 fixed
Let R4
1denote the 4-dimensional Minkowski space that
is the real vector space R4 endowed with the scalar product
⟨x y⟩ = minus11990901199100
+
3
sum
119894=1
119909119894119910119894
(1)
for all x = (1199090 119909
1 119909
2 119909
3) y = (119910
0 119910
1 119910
2 119910
3) isin R4
Let e0 e
1 e
2 e
3 be pseudo-orthonormal basis forR4
1 Then
⟨e119894 e
119895⟩ = 120575
119894119895120576119895for signatures 120576
0= minus1 120576
1= 120576
2= 120576
3= 1 The
function
119902 R4
1997888rarr R 119902 (x) = ⟨x x⟩ (2)
is called the associated quadratic form of ⟨sdot sdot⟩A vector k isin R4
1is called spacelike timelike and lightlike
if ⟨k k⟩ gt 0 (or k = 0) ⟨k k⟩ lt 0 and ⟨k k⟩ = 0respectively The Lorentzian norm of a vector k is defined byk = radic|⟨k k⟩|
The sets
H3
= x isin R4
1⟨x x⟩ = minus1 119909
0ge 1
S3
1= x isin R
4
1⟨x x⟩ = 1
LC+
= x isin R4
1| ⟨x x⟩ = 0 119909
0gt 0
(3)
are calledMinkowskimodel of hyperbolic space de Sitter spaceand future light cone respectively
Let 119875 be a vector subspace of R4
1 Then 119875 is said to be
timelike spacelike and lightlike if and only if 119875 contains atimelike vector and every nonzero vector in 119875 is spacelikeotherwise respectively
Nowwe give basic notions for hyperbolic geometry inH3From now on we use the prefix ldquoH-rdquo instead of ldquohyperbolicrdquofor brevity
An H-point is intersection 1198800
cap H3 such that 1198800is 1-
dimensional timelike subspace of R4
1and is called 119860
1198800 An
H-line is intersection 1198801
cap H3 such that 1198801is 2-dimensional
timelike subspace of R4
1and is called 119897
1198801 An H-plane is
intersection 1198802
cap H3 such that 1198802is 3-dimensional timelike
subspace of R4
1and is called 119863
1198802
H-coordinate axes 1198970119895
are denoted by intersections 1198970119895
=
1198810119895
capH3 such that1198810119895
= Spe0 e
119895 for 119895 = 1 2 3H-coordinate
planes 119863119894119895are denoted by intersections 119863
119894119895= 119882
0119894119895cap H3 such
that 1198820119894119895
= Spe0 e
119894 e
119895 for 119894 119895 = 1 2 3 H-upper (H-lower)
half-spaces of 119863119894119895are defined by intersections H3 and upper
(lower) half-space of 1198820119894119895
A hyperplane in R4
1is defined by HP(k 119888) = x isin R4
1|
⟨x k⟩ = 119888 for a pseudo-normal k isin R4
1and a real number 119888 If
k is spacelike timelike or lightlike HP(k 119888) is called timelikespacelike or lightlike respectively
Three kinds of totally umbilic surfaces haveH3 which aregiven by intersections of H3 and hyperplanes HP(k 119888) in R4
1
A surface HP(k 119888)capH3 is calledH-sphere equidistant surfaceand horosphere if HP(k 119888) is spacelike timelike and lightlikerespectively
We now give the existence and uniqueness of any H-lineor H-plane in H3 Any given two distinct points determineunique 2-plane through origin inR4
1and three distinct points
determine unique 3-plane through origin in R4
1 So the
following propositions are clear
Proposition 1 Any given two distinct H-points lie on a uniqueH-line in H3
Proposition 2 Any given three distinct H-points lie on aunique H-plane in H3
Also we say that H-line segments 119897119860119861 H-ray 119897997888997888rarr
119860119861are
determined by two different H-points119860 and119861 in natural way
Definition 3 Let 120574 [119886 119887] sub R rarr 119897119860119861
If we take any hyperbolic space curve instead of 119897119860119861
inDefinition 3 thenH-arc length of any hyperbolic space curveis calculated by formula (4) in the same way Moreover H-distance between H-points 119860 and 119861 is given by
If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed timelike plane 119881
01of R4
1is given by
T01
= R01
120579J1198981J1198993 119898 119899 = 0 1 (12)
such that
R01
120579=
[[[
[
1 0 0 0
0 1 0 0
0 0 cos 120579 minus sin 120579
0 0 sin 120579 cos 120579
]]]
]
(13)
for all 120579 isin R In other cases if 119879(11988102
) = 11988102and 119879(119881
03) = 119881
03
then general forms of H-isometries that leave fixed timelikeplanes 119881
02and 119881
03of R4
1are given by
T02
= R02
120579J1198981J1198992
T03
= R03
120579J1198982J1198993
119898 119899 = 0 1
(14)
such that
R02
120579=
[[[
[
1 0 0 0
0 cos 120579 0 sin 120579
0 0 1 0
0 minus sin 120579 0 cos 120579
]]]
]
R03
120579=
[[[
[
1 0 0 0
0 cos 120579 minus sin 120579 0
0 sin 120579 cos 120579 0
0 0 0 1
]]]
]
(15)
for all 120579 isin R respectively Thus we say that the group 1198660is
union of disjoint subgroups of 119866+
0and 119866
minus
0such that
119866+
0= R01
11989811205791R02
11989821205792R03
11989831205793| 119898
119894isin Z 120579
119895isin R
119866minus
0= R01
11989811205791R02
11989821205792R03
11989831205793J11989841J11989852J11989863
|
119898119894isin Z 120579
119895isin R 119898
4+ 119898
5+ 119898
6equiv 1 (mod 2)
(16)
that is 1198660
= 119866+
0cup 119866
minus
0
We suppose that 119879 isin 1198661 Then 119879 leaves fixed spacelike
planes 119881119894119895
= Spe119894 e
119895 for 119894 119895 = 1 2 3 of R4
1 That is 119879(119881
119894119895) =
119881119894119895If 119879(119881
23) = 119881
23for 119894 = 2 119895 = 3 then entries of T must
be 11990502
= 0 11990512
= 0 11990522
= 1 and 11990532
= 0 and 11990503
= 0 11990513
= 011990523
= 0 and 11990533
= 1 By using (7) and (8) we have 11990520
= 011990530
= 0 11990521
= 0 and 11990531
= 0 and the following equationsystem
minus1199052
00+ 119905
2
10= minus1 minus119905
2
01+ 119905
2
11= 1
minus11990500
11990501
+ 11990510
11990511
= 0 (11990500
11990511
minus 11990501
11990510
) 11990522
11990533
= plusmn1
(17)
If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed spacelike plane 119881
Now we give corollaries about some properties of H-isometries and transition relation betweenH-coordinate axes1198970119895with H-coordinate planes 119863
119894119895
Corollary 9 Any H-coordinate axis is converted to each otherby suitable H-rotation That is
R0119895
120579l01119903
=
l02119903
119895 = 3 120579 =120587
2
l03119903
119895 = 2 120579 =3120587
2
(39)
Corollary 10 AH-plane consists in suitableH-coordinate axisand H-rotation Namely
R0119895
120601l0119896119903
=
D23
(119903120601) 119895 = 1 119896 = 2
D13
(119903120601) 119895 = 2 119896 = 3
D12
(119903120601) 119895 = 3 119896 = 1
(40)
6 Journal of Applied Mathematics
Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is
R0119895
120579D12
(119903120601)=
D23
(119903120601) 119895 = 3 120579 =
3120587
2
D13
(119903120601) 119895 = 1 120579 =
120587
2
(41)
Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely
R0119897
minus120595R0119896
minus120601R0119895
minus120579H012
120582R0119895
120579R0119896
120601R0119897
120595
=
H013
120582 119895 = 1 120579 =
120587
2 120601 = 0 120595 = 0
H031
120582 119895 = 1 119896 = 2 120579 =
120587
2 120601 =
120587
2 120595 = 0
H032
120582 119895 = 1 119896 = 2 119897 = 3 120579 =
120587
2 120601 =
120587
2 120595 =
120587
2
H023
120582 119895 = 3 119896 = 2 120579 = minus
120587
2 120601 = minus
120587
2 120595 = 0
H021
120582 119895 = 3 120579 = minus
120587
2 120601 = 0 120595 = 0
(42)
After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed
Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878
1015840
for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3
Theorem 14 An H-sphere is invariant under H-translation inH3
Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3
cap HP(k minus119896) for 119896 gt 0 So
HP (k minus119896) = x isin R4
1|
minus V01199090
+ V11199091
+ V21199092
+ V31199093
= minus119896 119896 gt 0
(43)
Moreover for w = kk isin H3 and 119888 = 119896k we have
HP (w minus119888) = x isin R4
1| ⟨xw⟩ = minus119888 (44)
Since w isin H3
w = (cosh 1199040 sinh 119904
0cos120601
0 sinh 119904
0sin120601
0cos 120579
0
sinh 1199040sin120601
0sin 120579
0)
(45)
for any hyperbolic polar coordinates 1199040
isin [0 infin) 1206010
isin [0 120587]
and 1205790
isin [0 2120587] If we apply H-isometry T = L01minus1199040
R03
minus1206010R01
minus1205790isin
119866 then we have unit timelike vector e0such that
119879 (w) = e0 (46)
However unit timelike normal vector e0is invariant under
1198710119895
119904 That is
1198710119895
119904(e
0) = (cosh 119904) e
0 119895 = 1 2 3 (47)
For this reason if is an H-sphere which is generated fromspacelike hyperplane
HP (e0 minus119888) = x isin R
4
1| 119909
0= 119888 119888 ge 1 (48)
then we have
1198710119895
119904() = (49)
by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)
The following theorems also can be proved using similarmethod
Theorem 15 An equidistant surface is invariant under H-rotation in H3
Theorem 16 A horosphere is invariant under horocyclicrotation in H3
Finally we give the following corollary
Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866
0 119866
1 and 119866
2in H3
respectively
3 Differential Geometry of Curves andSurfaces in H3
In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]
TheLorentzian vector product of vectors x1 x2 x3 is givenby
0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3
1 Then the following maps 119864 119880 rarr 119878
3
1
119864(119906) = 120578(119906) and 119871plusmn
119880 rarr 119871119862+ 119871
plusmn(119906) = x(119906) plusmn 120578(119906)
are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906
0) can be identified with
identity mapping 119868119879119901119872
on the tangent space 119879119901119872 at x(119906
0) =
119901 isin 119872 We have that minus119889119871plusmn
= minus119868119879119901119872
plusmn (minus119889119864)For any given x(119906
0) = 119901 isin 119872 the linear transforms 119860
119901=
minus119889119864(1199060) 119879
119901119872 rarr 119879
119901119872 and 119878
plusmn
119901= minus119889119871
plusmn(119906
0) 119879
119901119872 rarr
119879119901119872 are called de Sitter shape operator and hyperbolic shape
operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and
119878plusmn
119901are denoted by 119896
119894(119901) and 119896
plusmn
119894(119901) for 119894 = 1 2 respectively
Obviously 119860119901and 119878
plusmn
119901have same eigenvectors Also the
eigenvalues satisfy
119896plusmn
119894(119901) = minus1 plusmn 119896
119894(119901) 119894 = 1 2 (55)
where 119896119894(119901) and 119896
plusmn
119894(119901) are called de Sitter principal curvature
and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872
respectivelyThe de Sitter Gauss curvature and the de Sitter mean
curvature of 119872 are given by
119870119889
(1199060) = det119860
119901= 119896
1(119901) 119896
2(119901)
119867119889
(1199060) =
1
2Tr119860
119901=
1198961
(119901) + 1198962
(119901)
2
(56)
at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss
curvature and the hyperbolic mean curvature of 119872 are givenby
119870plusmn
ℎ(119906
0) = det 119878
plusmn
119901= 119896
plusmn
1(119901) 119896
plusmn
2(119901)
119867plusmn
ℎ(119906
0) =
1
2Tr 119878
plusmn
119901=
119896plusmn
1(119901) + 119896
plusmn
2(119901)
2
(57)
at x(1199060) = 119901 respectively Evidently we have the following
relations119870
plusmn
ℎ= 1 ∓ 2119867
119889+ 119870
119889
119867plusmn
ℎ= minus1 plusmn 119867
119889
(58)
We say that a point x(1199060) = 119901 isin 119872 is an umbilical point
if 1198961(119901) = 119896
2(119901) Also 119872 is totally umbilical if all points
on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])
Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification
(1) Supposing that 1198962
= 1
(a) if 119896 = 0 and 1198962
lt 1 then M is a part of anequidistant surface
(b) if 119896 = 0 and 1198962
gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)
(2) If 1198962
= 1 then M is a part of horosphere
4 1198661-Invariant Surfaces in H3
In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces
Let 119872 = x(119880) be a regular surface via embedding x
119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860
the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4
1H3 and 119872 respectively Then the Gauss
and Weingarten explicit formulas for 119872 in H3 are given by
2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since
120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that
120578 (119904 119905) = ((12057231205721015840
4minus 120572
1015840
31205724) cosh 119904 (120572
31205721015840
4minus 120572
1015840
31205724) sinh 119904
12057211205721015840
4minus 120572
1015840
11205724 120572
1015840
11205723
minus 12057211205721015840
3)
(67)
and it is clear that
⟨120578 120597119904⟩ equiv ⟨120578 120597
119905⟩ equiv 0 (68)
for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A
119901= [
119886 119888
119887 119889] at any x(119904 119905) = 119901 isin 119872 where
119886 =
⟨minus119863120597119904
120578 120597119904⟩
⟨120597119904 120597
119904⟩
=
⟨119863119909119904
119909119904 120578⟩
⟨119909119904 119909
119904⟩
119887 =
⟨minus119863120597119904
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119904
119909119905 120578⟩
⟨119909119905 119909
119905⟩
119888 = 119887
119889 =
⟨minus119863120597119905
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119905
119909119905 120578⟩
⟨119909119905 119909
119905⟩
(69)
After basic calculations the de Sitter principal curvatures of119872 are
1198961
=1205721015840
31205724
minus 12057231205721015840
4
1205721
(70)
1198962
= 12057210158401015840
1(120572
1015840
31205724
minus 12057231205721015840
4) + 120572
10158401015840
3(120572
11205721015840
4minus 120572
1015840
11205724)
+ 12057210158401015840
4(120572
1015840
11205723
minus 12057211205721015840
3)
(71)
Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581
ℎ 120591
ℎ in H3
Proposition 22 Thebinormal vector of the profile curve of1198661-
invariant surface 119872 is constant in H3
Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591
ℎ= 0 Moreover by
(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)
119863te = 0 This completes the proof
From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582
0 120582
1 120582
2 120582
3) such that 120582
119894
is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872
Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896
2=
1205821120581ℎ
Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581
ℎe = 120572 and 120572
1015840and 120572
10158401015840 Also by (50)and (71) we obtain that 120572 and 120572
1015840and 120572
10158401015840= (0 119896
2 0 0) Thus it
follows that 120581ℎe = (0 119896
2 0 0) For this reason we have that
1198962
= 1205821120581ℎ
As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are
119870119889
(119901) =1205721015840
31205724
minus 12057231205721015840
4
1205721
1205821120581ℎ (72)
119867119889
(119901) =
(1205721015840
31205724
minus 12057231205721015840
4) + 120582
1120581ℎ1205721
21205721
(73)
Journal of Applied Mathematics 9
at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are
119870plusmn
ℎ(119901) =
(1205721
∓ (1205721015840
31205724
minus 12057231205721015840
4)) (1 ∓ 120582
1120581ℎ)
1205721
(74)
119867plusmn
ℎ(119901) =
1205721
(minus2 plusmn 1205821120581ℎ) plusmn (120572
1015840
31205724
minus 12057231205721015840
4)
21205721
(75)
at any x(119904 119905) = 119901 respectively
Proposition 24 Let 120572 119868 rarr 11986323
sub 1198673 120572(119905) =
(1205721(119905) 0 120572
3(119905) 120572
4(119905)) be unit speed regular profile curve of119866
1-
invariant surface 119872 Then its components are given by
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(76)
Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations
minus1205721
(119905)2
+ 1205723
(119905)2
+ 1205724
(119905)2
= minus1 (77)
minus1205721015840
1(119905)
2+ 120572
1015840
3(119905)
2+ 120572
1015840
4(119905)
2= 1 (78)
for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that
1205723 (119905) = radic120572
1 (119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
(79)
such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that
1205931015840(119905)
2=
1205721 (119905)
2minus 120572
1015840
1(119905)
2minus 1
(1205721
(119905)2
minus 1)2
(80)
Finally by (80) we have that
120593 (119905) = plusmn int
119905
0
radic1205721
(119906)2
minus 1205721015840
1(119906)
2minus 1
1205721 (119906)
2minus 1
119889119906 (81)
such that 1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1 gt 0 for all 119905 isin 119868 Without loss
of generality when we choose positive of signature of 120593 thiscompletes the proof
Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3
Now we will give some results which are obtained by (72)and (74)
Corollary 26 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205723
= 0 or 1205724
= 0 then 119872 is a part of de Sitter flatsurface
(ii) if 120581ℎ
= 0 then 119872 is a part of de Sitter flat surface
(iii) if 1205821
= 0 then 119872 is a part of de Sitter flat surface
Corollary 27 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 If 1205723
= 1205831205724such that 120583 isin R then 119872 is a
de Sitter flat surface
Theorem 28 Let 120572 be the profile curve of1198661-invariant surface
119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ
=
plusmn11205821
Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn
ℎ=
0 By (74) it follows that
1205721
(119905) ∓ (1205721015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (82)
or
1 ∓ 1205821120581ℎ
(119905) = 0 (83)
for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572
1(119905) ∓
(minusradic1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1) = 0 Hence it follows that
1205721015840
1(119905)
2+ 1 = 0 (84)
There is no real solution of (84) This means that the onlyone solution is 120581
ℎ= plusmn1120582
1by (83) On the other hand if we
assume that 120581ℎ
= plusmn11205821 then the proof is clear
Corollary 29 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205821
= 1 and 120581ℎ
= 1 then 119872 is 119870+
ℎ-flat surface which
is generated from horocyle
(ii) if 1205821
= minus1 and 120581ℎ
= 1 then 119872 is 119870minus
ℎ-flat surface which
is generated from horocyle
Now we will give theorem and corollaries for 1198661-
invariant surface which satisfy minimal condition in H3 by(73) and (75)
Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 inH3 Then 119872 is de
10 Journal of Applied Mathematics
Sitter minimal surface if and only if the parametrization of 120572 isgiven by
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(85)
with the condition 120581ℎ
isin (minus11205821 1120582
1) such that 119888
1is an
arbitrary constant
Proof Suppose that 119872 is de Sitter minimal surface that is119867
119889equiv 0 By (73) it follows that
(1205721015840
3(119905) 120572
4 (119905) minus 1205723 (119905) 120572
1015840
4(119905)) + 120582
1120581ℎ1205721 (119905) = 0 (86)
for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation
1205721015840
1(119905)
2minus (1 minus 120582
2
11205812
ℎ) 120572
1 (119905)2
+ 1 = 0 (87)
There exists only one real solution of (87) under the condition1 minus 120582
2
11205812
ℎgt 0 Moreover 120582
1must not be zero by Corollary 26
So that we obtain
120581ℎ
isin (minus1
1205821
1
1205821
) (88)
Hence the solution is
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
(89)
where 1198881is an arbitrary constant under the condition (88)
Finally the parametrization of 120572 is given explicitly byProposition 24
On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867
119889equiv 0
Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 in H3 Then 119872 is
hyperbolic minimal surface if and only if the parametrizationof 120572 is given by
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721
(119905)2
minus 1205721015840
1(119905)
2minus 1
1205721 (119905)
2minus 1
119889119906
(90)
with the condition 1minus(2∓1205821120581ℎ)2
gt 0 such that 1198881is an arbitrary
constant
Proof Suppose that 119872 is hyperbolic minimal surface that is119867
plusmn
ℎequiv 0 By (75) it follows that
(minus2 plusmn 1205821120581ℎ) 120572
1(119905) plusmn (120572
1015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (91)
If Proposition 24 is applied to (91) we have that
1205721015840
1(119905)
2+ ((minus2 plusmn 120582
1120581ℎ)2
minus 1) 1205721
(119905)2
+ 1 = 0 (92)
There exists only one real solution of (92) under the condition
(minus2 plusmn 1205821120581ℎ)2
minus 1 lt 0 (93)
Thus the solution is
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
(94)
where 1198881is an arbitrary constant under the condition (93)
However 1205821must not be zero by Corollary 26 So that if 119872
is119867+
ℎ-minimal surface (119867minus
ℎ-minimal surface) thenwe obtain
120581ℎ
isin (11205821 3120582
1) (120581
ℎisin (minus3120582
1 minus1120582
1)) by (93) Finally the
parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of
119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867
plusmn
ℎequiv 0
Now we will give classification theorem for totally umbil-ical 119866
1-invariant surfaces
Journal of Applied Mathematics 11
Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-
invariant surface 119872 inH3 Then the hyperbolic curvature of 120572
is constant
Proof Let 119872 be totally umbilical 1198661-invariant surface By
Theorem 23 and (70) we may assume that 1198961(119901) = 119896
2(119901) =
1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following
Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905
= x119905119904in
(96) then it follows that 12058211205811015840
ℎ(119905) = 0 for all 119905 isin 119868 Moreover
1205821must not be zero by Corollary 26Thus 120581
ℎis constant
Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following
classification
(1) Supposing that 1205852
= 1
(a) if 120585 = 0 and 1205852
lt 1 then 119872 is a part of anequidistant surface
(b) if 120585 = 0 and 1205852
gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane
(2) If 1205852
= 1 then 119872 is a part of horosphere
where 120585 = 1205821120581ℎis a constant
Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and
Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860
119901= 120585119868
2
where 1198682is identity matrix Finally the proof is complete by
Lemma 20
Now we will give some examples of 1198661-invariant surface
in H3 Let the Poincare ball model of hyperbolic space begiven by
B3
= (1199091 119909
2 119909
3) isin R
3|
3
sum
119894=1
1199092
119894lt 1 (97)
with the hyperbolic metric 1198891199042
= 4(1198891199092
1+119889119909
2
2+119889119909
2
3)(1minus119909
2
1minus
1199092
2minus 119909
2
3) Then it is well known that stereographic projection
of H3 is given by
Φ H3
997888rarr B3
Φ (1199090 119909
1 119909
2 119909
3) = (
1199091
1 + 1199090
1199092
1 + 1199090
1199093
1 + 1199090
)
(98)
We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3
Example 34 The 1198661-invariant surface which is generated
from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ
= radic2 is drawn in Figure 1(a)
Example 35 Let the profile curve of 119872 be given by
120572 (119905) = (radic2 +1
2(minus1 + radic2) 119905
2 0 1 minus
1
2(minus1 + radic2) 119905
2 119905)
(99)
such that hyperbolic curvature 120581ℎ
= 1 Then 119872 is hyperbolicflat 119866
1-invariant surface which is generated from horocycle
in H3 (see Figure 1(b))
Example 36 The 1198661-invariant surface which is generated
from
120572 (119905) = (1
3(minus1 + 4 cosh
radic3119905
2) 0
2
3(minus1 + cosh
radic3119905
2)
2
radic3sinh
radic3119905
2)
(100)
with hyperbolic curvature 120581ℎ
= 12 is drawn in Figure 1(c)
Example 37 Let the profile curve of 119872 be given by
120572 (119905)
= (2 cosh 119905
radic3minus sinh 119905
radic3 0 cosh 119905
radic3minus 2 sinh 119905
radic3 radic2)
(101)
such that hyperbolic curvature 120581ℎ
= radic2radic3 Then 119872 istotally umbilical 119866
1-invariant surface with 120585
2= 23 in H3
(see Figure 1(d))
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
12 Journal of Applied Mathematics
(a) (b)
(c) (d)
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993
of constant hyperbolic curvature of profile curve Moreoverwe give some geometric results with respect to constanthyperbolic curvature of profile curve for flat and minimalinvariant surfaces in H3 Finally we give a classificationtheorem for the totally umbilical invariant surfaces in H3
2 Isometries of H3
In [9] Reynold give a brief introduction to hyperbolicgeometry of hyperbolic plane H2 Also he described explicitdescriptions of the hyperbolic metric and the isometries ofthe hyperbolic plane In this section we consider hyperbolicgeometry in H3 We especially determine isometry groups ofH3with respect to causal character of hyperplanes ofR4
1 then
these isometry groups are classified in terms of leaving thosetotally umbilic surfaces of H3 fixed
Let R4
1denote the 4-dimensional Minkowski space that
is the real vector space R4 endowed with the scalar product
⟨x y⟩ = minus11990901199100
+
3
sum
119894=1
119909119894119910119894
(1)
for all x = (1199090 119909
1 119909
2 119909
3) y = (119910
0 119910
1 119910
2 119910
3) isin R4
Let e0 e
1 e
2 e
3 be pseudo-orthonormal basis forR4
1 Then
⟨e119894 e
119895⟩ = 120575
119894119895120576119895for signatures 120576
0= minus1 120576
1= 120576
2= 120576
3= 1 The
function
119902 R4
1997888rarr R 119902 (x) = ⟨x x⟩ (2)
is called the associated quadratic form of ⟨sdot sdot⟩A vector k isin R4
1is called spacelike timelike and lightlike
if ⟨k k⟩ gt 0 (or k = 0) ⟨k k⟩ lt 0 and ⟨k k⟩ = 0respectively The Lorentzian norm of a vector k is defined byk = radic|⟨k k⟩|
The sets
H3
= x isin R4
1⟨x x⟩ = minus1 119909
0ge 1
S3
1= x isin R
4
1⟨x x⟩ = 1
LC+
= x isin R4
1| ⟨x x⟩ = 0 119909
0gt 0
(3)
are calledMinkowskimodel of hyperbolic space de Sitter spaceand future light cone respectively
Let 119875 be a vector subspace of R4
1 Then 119875 is said to be
timelike spacelike and lightlike if and only if 119875 contains atimelike vector and every nonzero vector in 119875 is spacelikeotherwise respectively
Nowwe give basic notions for hyperbolic geometry inH3From now on we use the prefix ldquoH-rdquo instead of ldquohyperbolicrdquofor brevity
An H-point is intersection 1198800
cap H3 such that 1198800is 1-
dimensional timelike subspace of R4
1and is called 119860
1198800 An
H-line is intersection 1198801
cap H3 such that 1198801is 2-dimensional
timelike subspace of R4
1and is called 119897
1198801 An H-plane is
intersection 1198802
cap H3 such that 1198802is 3-dimensional timelike
subspace of R4
1and is called 119863
1198802
H-coordinate axes 1198970119895
are denoted by intersections 1198970119895
=
1198810119895
capH3 such that1198810119895
= Spe0 e
119895 for 119895 = 1 2 3H-coordinate
planes 119863119894119895are denoted by intersections 119863
119894119895= 119882
0119894119895cap H3 such
that 1198820119894119895
= Spe0 e
119894 e
119895 for 119894 119895 = 1 2 3 H-upper (H-lower)
half-spaces of 119863119894119895are defined by intersections H3 and upper
(lower) half-space of 1198820119894119895
A hyperplane in R4
1is defined by HP(k 119888) = x isin R4
1|
⟨x k⟩ = 119888 for a pseudo-normal k isin R4
1and a real number 119888 If
k is spacelike timelike or lightlike HP(k 119888) is called timelikespacelike or lightlike respectively
Three kinds of totally umbilic surfaces haveH3 which aregiven by intersections of H3 and hyperplanes HP(k 119888) in R4
1
A surface HP(k 119888)capH3 is calledH-sphere equidistant surfaceand horosphere if HP(k 119888) is spacelike timelike and lightlikerespectively
We now give the existence and uniqueness of any H-lineor H-plane in H3 Any given two distinct points determineunique 2-plane through origin inR4
1and three distinct points
determine unique 3-plane through origin in R4
1 So the
following propositions are clear
Proposition 1 Any given two distinct H-points lie on a uniqueH-line in H3
Proposition 2 Any given three distinct H-points lie on aunique H-plane in H3
Also we say that H-line segments 119897119860119861 H-ray 119897997888997888rarr
119860119861are
determined by two different H-points119860 and119861 in natural way
Definition 3 Let 120574 [119886 119887] sub R rarr 119897119860119861
If we take any hyperbolic space curve instead of 119897119860119861
inDefinition 3 thenH-arc length of any hyperbolic space curveis calculated by formula (4) in the same way Moreover H-distance between H-points 119860 and 119861 is given by
If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed timelike plane 119881
01of R4
1is given by
T01
= R01
120579J1198981J1198993 119898 119899 = 0 1 (12)
such that
R01
120579=
[[[
[
1 0 0 0
0 1 0 0
0 0 cos 120579 minus sin 120579
0 0 sin 120579 cos 120579
]]]
]
(13)
for all 120579 isin R In other cases if 119879(11988102
) = 11988102and 119879(119881
03) = 119881
03
then general forms of H-isometries that leave fixed timelikeplanes 119881
02and 119881
03of R4
1are given by
T02
= R02
120579J1198981J1198992
T03
= R03
120579J1198982J1198993
119898 119899 = 0 1
(14)
such that
R02
120579=
[[[
[
1 0 0 0
0 cos 120579 0 sin 120579
0 0 1 0
0 minus sin 120579 0 cos 120579
]]]
]
R03
120579=
[[[
[
1 0 0 0
0 cos 120579 minus sin 120579 0
0 sin 120579 cos 120579 0
0 0 0 1
]]]
]
(15)
for all 120579 isin R respectively Thus we say that the group 1198660is
union of disjoint subgroups of 119866+
0and 119866
minus
0such that
119866+
0= R01
11989811205791R02
11989821205792R03
11989831205793| 119898
119894isin Z 120579
119895isin R
119866minus
0= R01
11989811205791R02
11989821205792R03
11989831205793J11989841J11989852J11989863
|
119898119894isin Z 120579
119895isin R 119898
4+ 119898
5+ 119898
6equiv 1 (mod 2)
(16)
that is 1198660
= 119866+
0cup 119866
minus
0
We suppose that 119879 isin 1198661 Then 119879 leaves fixed spacelike
planes 119881119894119895
= Spe119894 e
119895 for 119894 119895 = 1 2 3 of R4
1 That is 119879(119881
119894119895) =
119881119894119895If 119879(119881
23) = 119881
23for 119894 = 2 119895 = 3 then entries of T must
be 11990502
= 0 11990512
= 0 11990522
= 1 and 11990532
= 0 and 11990503
= 0 11990513
= 011990523
= 0 and 11990533
= 1 By using (7) and (8) we have 11990520
= 011990530
= 0 11990521
= 0 and 11990531
= 0 and the following equationsystem
minus1199052
00+ 119905
2
10= minus1 minus119905
2
01+ 119905
2
11= 1
minus11990500
11990501
+ 11990510
11990511
= 0 (11990500
11990511
minus 11990501
11990510
) 11990522
11990533
= plusmn1
(17)
If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed spacelike plane 119881
Now we give corollaries about some properties of H-isometries and transition relation betweenH-coordinate axes1198970119895with H-coordinate planes 119863
119894119895
Corollary 9 Any H-coordinate axis is converted to each otherby suitable H-rotation That is
R0119895
120579l01119903
=
l02119903
119895 = 3 120579 =120587
2
l03119903
119895 = 2 120579 =3120587
2
(39)
Corollary 10 AH-plane consists in suitableH-coordinate axisand H-rotation Namely
R0119895
120601l0119896119903
=
D23
(119903120601) 119895 = 1 119896 = 2
D13
(119903120601) 119895 = 2 119896 = 3
D12
(119903120601) 119895 = 3 119896 = 1
(40)
6 Journal of Applied Mathematics
Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is
R0119895
120579D12
(119903120601)=
D23
(119903120601) 119895 = 3 120579 =
3120587
2
D13
(119903120601) 119895 = 1 120579 =
120587
2
(41)
Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely
R0119897
minus120595R0119896
minus120601R0119895
minus120579H012
120582R0119895
120579R0119896
120601R0119897
120595
=
H013
120582 119895 = 1 120579 =
120587
2 120601 = 0 120595 = 0
H031
120582 119895 = 1 119896 = 2 120579 =
120587
2 120601 =
120587
2 120595 = 0
H032
120582 119895 = 1 119896 = 2 119897 = 3 120579 =
120587
2 120601 =
120587
2 120595 =
120587
2
H023
120582 119895 = 3 119896 = 2 120579 = minus
120587
2 120601 = minus
120587
2 120595 = 0
H021
120582 119895 = 3 120579 = minus
120587
2 120601 = 0 120595 = 0
(42)
After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed
Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878
1015840
for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3
Theorem 14 An H-sphere is invariant under H-translation inH3
Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3
cap HP(k minus119896) for 119896 gt 0 So
HP (k minus119896) = x isin R4
1|
minus V01199090
+ V11199091
+ V21199092
+ V31199093
= minus119896 119896 gt 0
(43)
Moreover for w = kk isin H3 and 119888 = 119896k we have
HP (w minus119888) = x isin R4
1| ⟨xw⟩ = minus119888 (44)
Since w isin H3
w = (cosh 1199040 sinh 119904
0cos120601
0 sinh 119904
0sin120601
0cos 120579
0
sinh 1199040sin120601
0sin 120579
0)
(45)
for any hyperbolic polar coordinates 1199040
isin [0 infin) 1206010
isin [0 120587]
and 1205790
isin [0 2120587] If we apply H-isometry T = L01minus1199040
R03
minus1206010R01
minus1205790isin
119866 then we have unit timelike vector e0such that
119879 (w) = e0 (46)
However unit timelike normal vector e0is invariant under
1198710119895
119904 That is
1198710119895
119904(e
0) = (cosh 119904) e
0 119895 = 1 2 3 (47)
For this reason if is an H-sphere which is generated fromspacelike hyperplane
HP (e0 minus119888) = x isin R
4
1| 119909
0= 119888 119888 ge 1 (48)
then we have
1198710119895
119904() = (49)
by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)
The following theorems also can be proved using similarmethod
Theorem 15 An equidistant surface is invariant under H-rotation in H3
Theorem 16 A horosphere is invariant under horocyclicrotation in H3
Finally we give the following corollary
Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866
0 119866
1 and 119866
2in H3
respectively
3 Differential Geometry of Curves andSurfaces in H3
In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]
TheLorentzian vector product of vectors x1 x2 x3 is givenby
0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3
1 Then the following maps 119864 119880 rarr 119878
3
1
119864(119906) = 120578(119906) and 119871plusmn
119880 rarr 119871119862+ 119871
plusmn(119906) = x(119906) plusmn 120578(119906)
are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906
0) can be identified with
identity mapping 119868119879119901119872
on the tangent space 119879119901119872 at x(119906
0) =
119901 isin 119872 We have that minus119889119871plusmn
= minus119868119879119901119872
plusmn (minus119889119864)For any given x(119906
0) = 119901 isin 119872 the linear transforms 119860
119901=
minus119889119864(1199060) 119879
119901119872 rarr 119879
119901119872 and 119878
plusmn
119901= minus119889119871
plusmn(119906
0) 119879
119901119872 rarr
119879119901119872 are called de Sitter shape operator and hyperbolic shape
operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and
119878plusmn
119901are denoted by 119896
119894(119901) and 119896
plusmn
119894(119901) for 119894 = 1 2 respectively
Obviously 119860119901and 119878
plusmn
119901have same eigenvectors Also the
eigenvalues satisfy
119896plusmn
119894(119901) = minus1 plusmn 119896
119894(119901) 119894 = 1 2 (55)
where 119896119894(119901) and 119896
plusmn
119894(119901) are called de Sitter principal curvature
and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872
respectivelyThe de Sitter Gauss curvature and the de Sitter mean
curvature of 119872 are given by
119870119889
(1199060) = det119860
119901= 119896
1(119901) 119896
2(119901)
119867119889
(1199060) =
1
2Tr119860
119901=
1198961
(119901) + 1198962
(119901)
2
(56)
at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss
curvature and the hyperbolic mean curvature of 119872 are givenby
119870plusmn
ℎ(119906
0) = det 119878
plusmn
119901= 119896
plusmn
1(119901) 119896
plusmn
2(119901)
119867plusmn
ℎ(119906
0) =
1
2Tr 119878
plusmn
119901=
119896plusmn
1(119901) + 119896
plusmn
2(119901)
2
(57)
at x(1199060) = 119901 respectively Evidently we have the following
relations119870
plusmn
ℎ= 1 ∓ 2119867
119889+ 119870
119889
119867plusmn
ℎ= minus1 plusmn 119867
119889
(58)
We say that a point x(1199060) = 119901 isin 119872 is an umbilical point
if 1198961(119901) = 119896
2(119901) Also 119872 is totally umbilical if all points
on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])
Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification
(1) Supposing that 1198962
= 1
(a) if 119896 = 0 and 1198962
lt 1 then M is a part of anequidistant surface
(b) if 119896 = 0 and 1198962
gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)
(2) If 1198962
= 1 then M is a part of horosphere
4 1198661-Invariant Surfaces in H3
In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces
Let 119872 = x(119880) be a regular surface via embedding x
119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860
the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4
1H3 and 119872 respectively Then the Gauss
and Weingarten explicit formulas for 119872 in H3 are given by
2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since
120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that
120578 (119904 119905) = ((12057231205721015840
4minus 120572
1015840
31205724) cosh 119904 (120572
31205721015840
4minus 120572
1015840
31205724) sinh 119904
12057211205721015840
4minus 120572
1015840
11205724 120572
1015840
11205723
minus 12057211205721015840
3)
(67)
and it is clear that
⟨120578 120597119904⟩ equiv ⟨120578 120597
119905⟩ equiv 0 (68)
for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A
119901= [
119886 119888
119887 119889] at any x(119904 119905) = 119901 isin 119872 where
119886 =
⟨minus119863120597119904
120578 120597119904⟩
⟨120597119904 120597
119904⟩
=
⟨119863119909119904
119909119904 120578⟩
⟨119909119904 119909
119904⟩
119887 =
⟨minus119863120597119904
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119904
119909119905 120578⟩
⟨119909119905 119909
119905⟩
119888 = 119887
119889 =
⟨minus119863120597119905
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119905
119909119905 120578⟩
⟨119909119905 119909
119905⟩
(69)
After basic calculations the de Sitter principal curvatures of119872 are
1198961
=1205721015840
31205724
minus 12057231205721015840
4
1205721
(70)
1198962
= 12057210158401015840
1(120572
1015840
31205724
minus 12057231205721015840
4) + 120572
10158401015840
3(120572
11205721015840
4minus 120572
1015840
11205724)
+ 12057210158401015840
4(120572
1015840
11205723
minus 12057211205721015840
3)
(71)
Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581
ℎ 120591
ℎ in H3
Proposition 22 Thebinormal vector of the profile curve of1198661-
invariant surface 119872 is constant in H3
Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591
ℎ= 0 Moreover by
(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)
119863te = 0 This completes the proof
From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582
0 120582
1 120582
2 120582
3) such that 120582
119894
is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872
Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896
2=
1205821120581ℎ
Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581
ℎe = 120572 and 120572
1015840and 120572
10158401015840 Also by (50)and (71) we obtain that 120572 and 120572
1015840and 120572
10158401015840= (0 119896
2 0 0) Thus it
follows that 120581ℎe = (0 119896
2 0 0) For this reason we have that
1198962
= 1205821120581ℎ
As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are
119870119889
(119901) =1205721015840
31205724
minus 12057231205721015840
4
1205721
1205821120581ℎ (72)
119867119889
(119901) =
(1205721015840
31205724
minus 12057231205721015840
4) + 120582
1120581ℎ1205721
21205721
(73)
Journal of Applied Mathematics 9
at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are
119870plusmn
ℎ(119901) =
(1205721
∓ (1205721015840
31205724
minus 12057231205721015840
4)) (1 ∓ 120582
1120581ℎ)
1205721
(74)
119867plusmn
ℎ(119901) =
1205721
(minus2 plusmn 1205821120581ℎ) plusmn (120572
1015840
31205724
minus 12057231205721015840
4)
21205721
(75)
at any x(119904 119905) = 119901 respectively
Proposition 24 Let 120572 119868 rarr 11986323
sub 1198673 120572(119905) =
(1205721(119905) 0 120572
3(119905) 120572
4(119905)) be unit speed regular profile curve of119866
1-
invariant surface 119872 Then its components are given by
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(76)
Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations
minus1205721
(119905)2
+ 1205723
(119905)2
+ 1205724
(119905)2
= minus1 (77)
minus1205721015840
1(119905)
2+ 120572
1015840
3(119905)
2+ 120572
1015840
4(119905)
2= 1 (78)
for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that
1205723 (119905) = radic120572
1 (119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
(79)
such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that
1205931015840(119905)
2=
1205721 (119905)
2minus 120572
1015840
1(119905)
2minus 1
(1205721
(119905)2
minus 1)2
(80)
Finally by (80) we have that
120593 (119905) = plusmn int
119905
0
radic1205721
(119906)2
minus 1205721015840
1(119906)
2minus 1
1205721 (119906)
2minus 1
119889119906 (81)
such that 1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1 gt 0 for all 119905 isin 119868 Without loss
of generality when we choose positive of signature of 120593 thiscompletes the proof
Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3
Now we will give some results which are obtained by (72)and (74)
Corollary 26 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205723
= 0 or 1205724
= 0 then 119872 is a part of de Sitter flatsurface
(ii) if 120581ℎ
= 0 then 119872 is a part of de Sitter flat surface
(iii) if 1205821
= 0 then 119872 is a part of de Sitter flat surface
Corollary 27 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 If 1205723
= 1205831205724such that 120583 isin R then 119872 is a
de Sitter flat surface
Theorem 28 Let 120572 be the profile curve of1198661-invariant surface
119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ
=
plusmn11205821
Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn
ℎ=
0 By (74) it follows that
1205721
(119905) ∓ (1205721015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (82)
or
1 ∓ 1205821120581ℎ
(119905) = 0 (83)
for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572
1(119905) ∓
(minusradic1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1) = 0 Hence it follows that
1205721015840
1(119905)
2+ 1 = 0 (84)
There is no real solution of (84) This means that the onlyone solution is 120581
ℎ= plusmn1120582
1by (83) On the other hand if we
assume that 120581ℎ
= plusmn11205821 then the proof is clear
Corollary 29 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205821
= 1 and 120581ℎ
= 1 then 119872 is 119870+
ℎ-flat surface which
is generated from horocyle
(ii) if 1205821
= minus1 and 120581ℎ
= 1 then 119872 is 119870minus
ℎ-flat surface which
is generated from horocyle
Now we will give theorem and corollaries for 1198661-
invariant surface which satisfy minimal condition in H3 by(73) and (75)
Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 inH3 Then 119872 is de
10 Journal of Applied Mathematics
Sitter minimal surface if and only if the parametrization of 120572 isgiven by
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(85)
with the condition 120581ℎ
isin (minus11205821 1120582
1) such that 119888
1is an
arbitrary constant
Proof Suppose that 119872 is de Sitter minimal surface that is119867
119889equiv 0 By (73) it follows that
(1205721015840
3(119905) 120572
4 (119905) minus 1205723 (119905) 120572
1015840
4(119905)) + 120582
1120581ℎ1205721 (119905) = 0 (86)
for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation
1205721015840
1(119905)
2minus (1 minus 120582
2
11205812
ℎ) 120572
1 (119905)2
+ 1 = 0 (87)
There exists only one real solution of (87) under the condition1 minus 120582
2
11205812
ℎgt 0 Moreover 120582
1must not be zero by Corollary 26
So that we obtain
120581ℎ
isin (minus1
1205821
1
1205821
) (88)
Hence the solution is
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
(89)
where 1198881is an arbitrary constant under the condition (88)
Finally the parametrization of 120572 is given explicitly byProposition 24
On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867
119889equiv 0
Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 in H3 Then 119872 is
hyperbolic minimal surface if and only if the parametrizationof 120572 is given by
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721
(119905)2
minus 1205721015840
1(119905)
2minus 1
1205721 (119905)
2minus 1
119889119906
(90)
with the condition 1minus(2∓1205821120581ℎ)2
gt 0 such that 1198881is an arbitrary
constant
Proof Suppose that 119872 is hyperbolic minimal surface that is119867
plusmn
ℎequiv 0 By (75) it follows that
(minus2 plusmn 1205821120581ℎ) 120572
1(119905) plusmn (120572
1015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (91)
If Proposition 24 is applied to (91) we have that
1205721015840
1(119905)
2+ ((minus2 plusmn 120582
1120581ℎ)2
minus 1) 1205721
(119905)2
+ 1 = 0 (92)
There exists only one real solution of (92) under the condition
(minus2 plusmn 1205821120581ℎ)2
minus 1 lt 0 (93)
Thus the solution is
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
(94)
where 1198881is an arbitrary constant under the condition (93)
However 1205821must not be zero by Corollary 26 So that if 119872
is119867+
ℎ-minimal surface (119867minus
ℎ-minimal surface) thenwe obtain
120581ℎ
isin (11205821 3120582
1) (120581
ℎisin (minus3120582
1 minus1120582
1)) by (93) Finally the
parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of
119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867
plusmn
ℎequiv 0
Now we will give classification theorem for totally umbil-ical 119866
1-invariant surfaces
Journal of Applied Mathematics 11
Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-
invariant surface 119872 inH3 Then the hyperbolic curvature of 120572
is constant
Proof Let 119872 be totally umbilical 1198661-invariant surface By
Theorem 23 and (70) we may assume that 1198961(119901) = 119896
2(119901) =
1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following
Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905
= x119905119904in
(96) then it follows that 12058211205811015840
ℎ(119905) = 0 for all 119905 isin 119868 Moreover
1205821must not be zero by Corollary 26Thus 120581
ℎis constant
Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following
classification
(1) Supposing that 1205852
= 1
(a) if 120585 = 0 and 1205852
lt 1 then 119872 is a part of anequidistant surface
(b) if 120585 = 0 and 1205852
gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane
(2) If 1205852
= 1 then 119872 is a part of horosphere
where 120585 = 1205821120581ℎis a constant
Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and
Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860
119901= 120585119868
2
where 1198682is identity matrix Finally the proof is complete by
Lemma 20
Now we will give some examples of 1198661-invariant surface
in H3 Let the Poincare ball model of hyperbolic space begiven by
B3
= (1199091 119909
2 119909
3) isin R
3|
3
sum
119894=1
1199092
119894lt 1 (97)
with the hyperbolic metric 1198891199042
= 4(1198891199092
1+119889119909
2
2+119889119909
2
3)(1minus119909
2
1minus
1199092
2minus 119909
2
3) Then it is well known that stereographic projection
of H3 is given by
Φ H3
997888rarr B3
Φ (1199090 119909
1 119909
2 119909
3) = (
1199091
1 + 1199090
1199092
1 + 1199090
1199093
1 + 1199090
)
(98)
We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3
Example 34 The 1198661-invariant surface which is generated
from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ
= radic2 is drawn in Figure 1(a)
Example 35 Let the profile curve of 119872 be given by
120572 (119905) = (radic2 +1
2(minus1 + radic2) 119905
2 0 1 minus
1
2(minus1 + radic2) 119905
2 119905)
(99)
such that hyperbolic curvature 120581ℎ
= 1 Then 119872 is hyperbolicflat 119866
1-invariant surface which is generated from horocycle
in H3 (see Figure 1(b))
Example 36 The 1198661-invariant surface which is generated
from
120572 (119905) = (1
3(minus1 + 4 cosh
radic3119905
2) 0
2
3(minus1 + cosh
radic3119905
2)
2
radic3sinh
radic3119905
2)
(100)
with hyperbolic curvature 120581ℎ
= 12 is drawn in Figure 1(c)
Example 37 Let the profile curve of 119872 be given by
120572 (119905)
= (2 cosh 119905
radic3minus sinh 119905
radic3 0 cosh 119905
radic3minus 2 sinh 119905
radic3 radic2)
(101)
such that hyperbolic curvature 120581ℎ
= radic2radic3 Then 119872 istotally umbilical 119866
1-invariant surface with 120585
2= 23 in H3
(see Figure 1(d))
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
12 Journal of Applied Mathematics
(a) (b)
(c) (d)
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993
If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed timelike plane 119881
01of R4
1is given by
T01
= R01
120579J1198981J1198993 119898 119899 = 0 1 (12)
such that
R01
120579=
[[[
[
1 0 0 0
0 1 0 0
0 0 cos 120579 minus sin 120579
0 0 sin 120579 cos 120579
]]]
]
(13)
for all 120579 isin R In other cases if 119879(11988102
) = 11988102and 119879(119881
03) = 119881
03
then general forms of H-isometries that leave fixed timelikeplanes 119881
02and 119881
03of R4
1are given by
T02
= R02
120579J1198981J1198992
T03
= R03
120579J1198982J1198993
119898 119899 = 0 1
(14)
such that
R02
120579=
[[[
[
1 0 0 0
0 cos 120579 0 sin 120579
0 0 1 0
0 minus sin 120579 0 cos 120579
]]]
]
R03
120579=
[[[
[
1 0 0 0
0 cos 120579 minus sin 120579 0
0 sin 120579 cos 120579 0
0 0 0 1
]]]
]
(15)
for all 120579 isin R respectively Thus we say that the group 1198660is
union of disjoint subgroups of 119866+
0and 119866
minus
0such that
119866+
0= R01
11989811205791R02
11989821205792R03
11989831205793| 119898
119894isin Z 120579
119895isin R
119866minus
0= R01
11989811205791R02
11989821205792R03
11989831205793J11989841J11989852J11989863
|
119898119894isin Z 120579
119895isin R 119898
4+ 119898
5+ 119898
6equiv 1 (mod 2)
(16)
that is 1198660
= 119866+
0cup 119866
minus
0
We suppose that 119879 isin 1198661 Then 119879 leaves fixed spacelike
planes 119881119894119895
= Spe119894 e
119895 for 119894 119895 = 1 2 3 of R4
1 That is 119879(119881
119894119895) =
119881119894119895If 119879(119881
23) = 119881
23for 119894 = 2 119895 = 3 then entries of T must
be 11990502
= 0 11990512
= 0 11990522
= 1 and 11990532
= 0 and 11990503
= 0 11990513
= 011990523
= 0 and 11990533
= 1 By using (7) and (8) we have 11990520
= 011990530
= 0 11990521
= 0 and 11990531
= 0 and the following equationsystem
minus1199052
00+ 119905
2
10= minus1 minus119905
2
01+ 119905
2
11= 1
minus11990500
11990501
+ 11990510
11990511
= 0 (11990500
11990511
minus 11990501
11990510
) 11990522
11990533
= plusmn1
(17)
If the above system is solved under time orientation preserv-ing and sign cases then general form of H-isometries thatleave fixed spacelike plane 119881
Now we give corollaries about some properties of H-isometries and transition relation betweenH-coordinate axes1198970119895with H-coordinate planes 119863
119894119895
Corollary 9 Any H-coordinate axis is converted to each otherby suitable H-rotation That is
R0119895
120579l01119903
=
l02119903
119895 = 3 120579 =120587
2
l03119903
119895 = 2 120579 =3120587
2
(39)
Corollary 10 AH-plane consists in suitableH-coordinate axisand H-rotation Namely
R0119895
120601l0119896119903
=
D23
(119903120601) 119895 = 1 119896 = 2
D13
(119903120601) 119895 = 2 119896 = 3
D12
(119903120601) 119895 = 3 119896 = 1
(40)
6 Journal of Applied Mathematics
Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is
R0119895
120579D12
(119903120601)=
D23
(119903120601) 119895 = 3 120579 =
3120587
2
D13
(119903120601) 119895 = 1 120579 =
120587
2
(41)
Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely
R0119897
minus120595R0119896
minus120601R0119895
minus120579H012
120582R0119895
120579R0119896
120601R0119897
120595
=
H013
120582 119895 = 1 120579 =
120587
2 120601 = 0 120595 = 0
H031
120582 119895 = 1 119896 = 2 120579 =
120587
2 120601 =
120587
2 120595 = 0
H032
120582 119895 = 1 119896 = 2 119897 = 3 120579 =
120587
2 120601 =
120587
2 120595 =
120587
2
H023
120582 119895 = 3 119896 = 2 120579 = minus
120587
2 120601 = minus
120587
2 120595 = 0
H021
120582 119895 = 3 120579 = minus
120587
2 120601 = 0 120595 = 0
(42)
After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed
Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878
1015840
for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3
Theorem 14 An H-sphere is invariant under H-translation inH3
Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3
cap HP(k minus119896) for 119896 gt 0 So
HP (k minus119896) = x isin R4
1|
minus V01199090
+ V11199091
+ V21199092
+ V31199093
= minus119896 119896 gt 0
(43)
Moreover for w = kk isin H3 and 119888 = 119896k we have
HP (w minus119888) = x isin R4
1| ⟨xw⟩ = minus119888 (44)
Since w isin H3
w = (cosh 1199040 sinh 119904
0cos120601
0 sinh 119904
0sin120601
0cos 120579
0
sinh 1199040sin120601
0sin 120579
0)
(45)
for any hyperbolic polar coordinates 1199040
isin [0 infin) 1206010
isin [0 120587]
and 1205790
isin [0 2120587] If we apply H-isometry T = L01minus1199040
R03
minus1206010R01
minus1205790isin
119866 then we have unit timelike vector e0such that
119879 (w) = e0 (46)
However unit timelike normal vector e0is invariant under
1198710119895
119904 That is
1198710119895
119904(e
0) = (cosh 119904) e
0 119895 = 1 2 3 (47)
For this reason if is an H-sphere which is generated fromspacelike hyperplane
HP (e0 minus119888) = x isin R
4
1| 119909
0= 119888 119888 ge 1 (48)
then we have
1198710119895
119904() = (49)
by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)
The following theorems also can be proved using similarmethod
Theorem 15 An equidistant surface is invariant under H-rotation in H3
Theorem 16 A horosphere is invariant under horocyclicrotation in H3
Finally we give the following corollary
Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866
0 119866
1 and 119866
2in H3
respectively
3 Differential Geometry of Curves andSurfaces in H3
In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]
TheLorentzian vector product of vectors x1 x2 x3 is givenby
0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3
1 Then the following maps 119864 119880 rarr 119878
3
1
119864(119906) = 120578(119906) and 119871plusmn
119880 rarr 119871119862+ 119871
plusmn(119906) = x(119906) plusmn 120578(119906)
are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906
0) can be identified with
identity mapping 119868119879119901119872
on the tangent space 119879119901119872 at x(119906
0) =
119901 isin 119872 We have that minus119889119871plusmn
= minus119868119879119901119872
plusmn (minus119889119864)For any given x(119906
0) = 119901 isin 119872 the linear transforms 119860
119901=
minus119889119864(1199060) 119879
119901119872 rarr 119879
119901119872 and 119878
plusmn
119901= minus119889119871
plusmn(119906
0) 119879
119901119872 rarr
119879119901119872 are called de Sitter shape operator and hyperbolic shape
operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and
119878plusmn
119901are denoted by 119896
119894(119901) and 119896
plusmn
119894(119901) for 119894 = 1 2 respectively
Obviously 119860119901and 119878
plusmn
119901have same eigenvectors Also the
eigenvalues satisfy
119896plusmn
119894(119901) = minus1 plusmn 119896
119894(119901) 119894 = 1 2 (55)
where 119896119894(119901) and 119896
plusmn
119894(119901) are called de Sitter principal curvature
and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872
respectivelyThe de Sitter Gauss curvature and the de Sitter mean
curvature of 119872 are given by
119870119889
(1199060) = det119860
119901= 119896
1(119901) 119896
2(119901)
119867119889
(1199060) =
1
2Tr119860
119901=
1198961
(119901) + 1198962
(119901)
2
(56)
at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss
curvature and the hyperbolic mean curvature of 119872 are givenby
119870plusmn
ℎ(119906
0) = det 119878
plusmn
119901= 119896
plusmn
1(119901) 119896
plusmn
2(119901)
119867plusmn
ℎ(119906
0) =
1
2Tr 119878
plusmn
119901=
119896plusmn
1(119901) + 119896
plusmn
2(119901)
2
(57)
at x(1199060) = 119901 respectively Evidently we have the following
relations119870
plusmn
ℎ= 1 ∓ 2119867
119889+ 119870
119889
119867plusmn
ℎ= minus1 plusmn 119867
119889
(58)
We say that a point x(1199060) = 119901 isin 119872 is an umbilical point
if 1198961(119901) = 119896
2(119901) Also 119872 is totally umbilical if all points
on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])
Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification
(1) Supposing that 1198962
= 1
(a) if 119896 = 0 and 1198962
lt 1 then M is a part of anequidistant surface
(b) if 119896 = 0 and 1198962
gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)
(2) If 1198962
= 1 then M is a part of horosphere
4 1198661-Invariant Surfaces in H3
In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces
Let 119872 = x(119880) be a regular surface via embedding x
119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860
the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4
1H3 and 119872 respectively Then the Gauss
and Weingarten explicit formulas for 119872 in H3 are given by
2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since
120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that
120578 (119904 119905) = ((12057231205721015840
4minus 120572
1015840
31205724) cosh 119904 (120572
31205721015840
4minus 120572
1015840
31205724) sinh 119904
12057211205721015840
4minus 120572
1015840
11205724 120572
1015840
11205723
minus 12057211205721015840
3)
(67)
and it is clear that
⟨120578 120597119904⟩ equiv ⟨120578 120597
119905⟩ equiv 0 (68)
for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A
119901= [
119886 119888
119887 119889] at any x(119904 119905) = 119901 isin 119872 where
119886 =
⟨minus119863120597119904
120578 120597119904⟩
⟨120597119904 120597
119904⟩
=
⟨119863119909119904
119909119904 120578⟩
⟨119909119904 119909
119904⟩
119887 =
⟨minus119863120597119904
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119904
119909119905 120578⟩
⟨119909119905 119909
119905⟩
119888 = 119887
119889 =
⟨minus119863120597119905
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119905
119909119905 120578⟩
⟨119909119905 119909
119905⟩
(69)
After basic calculations the de Sitter principal curvatures of119872 are
1198961
=1205721015840
31205724
minus 12057231205721015840
4
1205721
(70)
1198962
= 12057210158401015840
1(120572
1015840
31205724
minus 12057231205721015840
4) + 120572
10158401015840
3(120572
11205721015840
4minus 120572
1015840
11205724)
+ 12057210158401015840
4(120572
1015840
11205723
minus 12057211205721015840
3)
(71)
Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581
ℎ 120591
ℎ in H3
Proposition 22 Thebinormal vector of the profile curve of1198661-
invariant surface 119872 is constant in H3
Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591
ℎ= 0 Moreover by
(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)
119863te = 0 This completes the proof
From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582
0 120582
1 120582
2 120582
3) such that 120582
119894
is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872
Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896
2=
1205821120581ℎ
Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581
ℎe = 120572 and 120572
1015840and 120572
10158401015840 Also by (50)and (71) we obtain that 120572 and 120572
1015840and 120572
10158401015840= (0 119896
2 0 0) Thus it
follows that 120581ℎe = (0 119896
2 0 0) For this reason we have that
1198962
= 1205821120581ℎ
As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are
119870119889
(119901) =1205721015840
31205724
minus 12057231205721015840
4
1205721
1205821120581ℎ (72)
119867119889
(119901) =
(1205721015840
31205724
minus 12057231205721015840
4) + 120582
1120581ℎ1205721
21205721
(73)
Journal of Applied Mathematics 9
at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are
119870plusmn
ℎ(119901) =
(1205721
∓ (1205721015840
31205724
minus 12057231205721015840
4)) (1 ∓ 120582
1120581ℎ)
1205721
(74)
119867plusmn
ℎ(119901) =
1205721
(minus2 plusmn 1205821120581ℎ) plusmn (120572
1015840
31205724
minus 12057231205721015840
4)
21205721
(75)
at any x(119904 119905) = 119901 respectively
Proposition 24 Let 120572 119868 rarr 11986323
sub 1198673 120572(119905) =
(1205721(119905) 0 120572
3(119905) 120572
4(119905)) be unit speed regular profile curve of119866
1-
invariant surface 119872 Then its components are given by
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(76)
Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations
minus1205721
(119905)2
+ 1205723
(119905)2
+ 1205724
(119905)2
= minus1 (77)
minus1205721015840
1(119905)
2+ 120572
1015840
3(119905)
2+ 120572
1015840
4(119905)
2= 1 (78)
for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that
1205723 (119905) = radic120572
1 (119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
(79)
such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that
1205931015840(119905)
2=
1205721 (119905)
2minus 120572
1015840
1(119905)
2minus 1
(1205721
(119905)2
minus 1)2
(80)
Finally by (80) we have that
120593 (119905) = plusmn int
119905
0
radic1205721
(119906)2
minus 1205721015840
1(119906)
2minus 1
1205721 (119906)
2minus 1
119889119906 (81)
such that 1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1 gt 0 for all 119905 isin 119868 Without loss
of generality when we choose positive of signature of 120593 thiscompletes the proof
Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3
Now we will give some results which are obtained by (72)and (74)
Corollary 26 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205723
= 0 or 1205724
= 0 then 119872 is a part of de Sitter flatsurface
(ii) if 120581ℎ
= 0 then 119872 is a part of de Sitter flat surface
(iii) if 1205821
= 0 then 119872 is a part of de Sitter flat surface
Corollary 27 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 If 1205723
= 1205831205724such that 120583 isin R then 119872 is a
de Sitter flat surface
Theorem 28 Let 120572 be the profile curve of1198661-invariant surface
119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ
=
plusmn11205821
Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn
ℎ=
0 By (74) it follows that
1205721
(119905) ∓ (1205721015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (82)
or
1 ∓ 1205821120581ℎ
(119905) = 0 (83)
for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572
1(119905) ∓
(minusradic1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1) = 0 Hence it follows that
1205721015840
1(119905)
2+ 1 = 0 (84)
There is no real solution of (84) This means that the onlyone solution is 120581
ℎ= plusmn1120582
1by (83) On the other hand if we
assume that 120581ℎ
= plusmn11205821 then the proof is clear
Corollary 29 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205821
= 1 and 120581ℎ
= 1 then 119872 is 119870+
ℎ-flat surface which
is generated from horocyle
(ii) if 1205821
= minus1 and 120581ℎ
= 1 then 119872 is 119870minus
ℎ-flat surface which
is generated from horocyle
Now we will give theorem and corollaries for 1198661-
invariant surface which satisfy minimal condition in H3 by(73) and (75)
Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 inH3 Then 119872 is de
10 Journal of Applied Mathematics
Sitter minimal surface if and only if the parametrization of 120572 isgiven by
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(85)
with the condition 120581ℎ
isin (minus11205821 1120582
1) such that 119888
1is an
arbitrary constant
Proof Suppose that 119872 is de Sitter minimal surface that is119867
119889equiv 0 By (73) it follows that
(1205721015840
3(119905) 120572
4 (119905) minus 1205723 (119905) 120572
1015840
4(119905)) + 120582
1120581ℎ1205721 (119905) = 0 (86)
for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation
1205721015840
1(119905)
2minus (1 minus 120582
2
11205812
ℎ) 120572
1 (119905)2
+ 1 = 0 (87)
There exists only one real solution of (87) under the condition1 minus 120582
2
11205812
ℎgt 0 Moreover 120582
1must not be zero by Corollary 26
So that we obtain
120581ℎ
isin (minus1
1205821
1
1205821
) (88)
Hence the solution is
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
(89)
where 1198881is an arbitrary constant under the condition (88)
Finally the parametrization of 120572 is given explicitly byProposition 24
On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867
119889equiv 0
Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 in H3 Then 119872 is
hyperbolic minimal surface if and only if the parametrizationof 120572 is given by
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721
(119905)2
minus 1205721015840
1(119905)
2minus 1
1205721 (119905)
2minus 1
119889119906
(90)
with the condition 1minus(2∓1205821120581ℎ)2
gt 0 such that 1198881is an arbitrary
constant
Proof Suppose that 119872 is hyperbolic minimal surface that is119867
plusmn
ℎequiv 0 By (75) it follows that
(minus2 plusmn 1205821120581ℎ) 120572
1(119905) plusmn (120572
1015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (91)
If Proposition 24 is applied to (91) we have that
1205721015840
1(119905)
2+ ((minus2 plusmn 120582
1120581ℎ)2
minus 1) 1205721
(119905)2
+ 1 = 0 (92)
There exists only one real solution of (92) under the condition
(minus2 plusmn 1205821120581ℎ)2
minus 1 lt 0 (93)
Thus the solution is
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
(94)
where 1198881is an arbitrary constant under the condition (93)
However 1205821must not be zero by Corollary 26 So that if 119872
is119867+
ℎ-minimal surface (119867minus
ℎ-minimal surface) thenwe obtain
120581ℎ
isin (11205821 3120582
1) (120581
ℎisin (minus3120582
1 minus1120582
1)) by (93) Finally the
parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of
119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867
plusmn
ℎequiv 0
Now we will give classification theorem for totally umbil-ical 119866
1-invariant surfaces
Journal of Applied Mathematics 11
Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-
invariant surface 119872 inH3 Then the hyperbolic curvature of 120572
is constant
Proof Let 119872 be totally umbilical 1198661-invariant surface By
Theorem 23 and (70) we may assume that 1198961(119901) = 119896
2(119901) =
1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following
Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905
= x119905119904in
(96) then it follows that 12058211205811015840
ℎ(119905) = 0 for all 119905 isin 119868 Moreover
1205821must not be zero by Corollary 26Thus 120581
ℎis constant
Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following
classification
(1) Supposing that 1205852
= 1
(a) if 120585 = 0 and 1205852
lt 1 then 119872 is a part of anequidistant surface
(b) if 120585 = 0 and 1205852
gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane
(2) If 1205852
= 1 then 119872 is a part of horosphere
where 120585 = 1205821120581ℎis a constant
Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and
Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860
119901= 120585119868
2
where 1198682is identity matrix Finally the proof is complete by
Lemma 20
Now we will give some examples of 1198661-invariant surface
in H3 Let the Poincare ball model of hyperbolic space begiven by
B3
= (1199091 119909
2 119909
3) isin R
3|
3
sum
119894=1
1199092
119894lt 1 (97)
with the hyperbolic metric 1198891199042
= 4(1198891199092
1+119889119909
2
2+119889119909
2
3)(1minus119909
2
1minus
1199092
2minus 119909
2
3) Then it is well known that stereographic projection
of H3 is given by
Φ H3
997888rarr B3
Φ (1199090 119909
1 119909
2 119909
3) = (
1199091
1 + 1199090
1199092
1 + 1199090
1199093
1 + 1199090
)
(98)
We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3
Example 34 The 1198661-invariant surface which is generated
from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ
= radic2 is drawn in Figure 1(a)
Example 35 Let the profile curve of 119872 be given by
120572 (119905) = (radic2 +1
2(minus1 + radic2) 119905
2 0 1 minus
1
2(minus1 + radic2) 119905
2 119905)
(99)
such that hyperbolic curvature 120581ℎ
= 1 Then 119872 is hyperbolicflat 119866
1-invariant surface which is generated from horocycle
in H3 (see Figure 1(b))
Example 36 The 1198661-invariant surface which is generated
from
120572 (119905) = (1
3(minus1 + 4 cosh
radic3119905
2) 0
2
3(minus1 + cosh
radic3119905
2)
2
radic3sinh
radic3119905
2)
(100)
with hyperbolic curvature 120581ℎ
= 12 is drawn in Figure 1(c)
Example 37 Let the profile curve of 119872 be given by
120572 (119905)
= (2 cosh 119905
radic3minus sinh 119905
radic3 0 cosh 119905
radic3minus 2 sinh 119905
radic3 radic2)
(101)
such that hyperbolic curvature 120581ℎ
= radic2radic3 Then 119872 istotally umbilical 119866
1-invariant surface with 120585
2= 23 in H3
(see Figure 1(d))
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
12 Journal of Applied Mathematics
(a) (b)
(c) (d)
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993
Now we give corollaries about some properties of H-isometries and transition relation betweenH-coordinate axes1198970119895with H-coordinate planes 119863
119894119895
Corollary 9 Any H-coordinate axis is converted to each otherby suitable H-rotation That is
R0119895
120579l01119903
=
l02119903
119895 = 3 120579 =120587
2
l03119903
119895 = 2 120579 =3120587
2
(39)
Corollary 10 AH-plane consists in suitableH-coordinate axisand H-rotation Namely
R0119895
120601l0119896119903
=
D23
(119903120601) 119895 = 1 119896 = 2
D13
(119903120601) 119895 = 2 119896 = 3
D12
(119903120601) 119895 = 3 119896 = 1
(40)
6 Journal of Applied Mathematics
Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is
R0119895
120579D12
(119903120601)=
D23
(119903120601) 119895 = 3 120579 =
3120587
2
D13
(119903120601) 119895 = 1 120579 =
120587
2
(41)
Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely
R0119897
minus120595R0119896
minus120601R0119895
minus120579H012
120582R0119895
120579R0119896
120601R0119897
120595
=
H013
120582 119895 = 1 120579 =
120587
2 120601 = 0 120595 = 0
H031
120582 119895 = 1 119896 = 2 120579 =
120587
2 120601 =
120587
2 120595 = 0
H032
120582 119895 = 1 119896 = 2 119897 = 3 120579 =
120587
2 120601 =
120587
2 120595 =
120587
2
H023
120582 119895 = 3 119896 = 2 120579 = minus
120587
2 120601 = minus
120587
2 120595 = 0
H021
120582 119895 = 3 120579 = minus
120587
2 120601 = 0 120595 = 0
(42)
After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed
Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878
1015840
for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3
Theorem 14 An H-sphere is invariant under H-translation inH3
Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3
cap HP(k minus119896) for 119896 gt 0 So
HP (k minus119896) = x isin R4
1|
minus V01199090
+ V11199091
+ V21199092
+ V31199093
= minus119896 119896 gt 0
(43)
Moreover for w = kk isin H3 and 119888 = 119896k we have
HP (w minus119888) = x isin R4
1| ⟨xw⟩ = minus119888 (44)
Since w isin H3
w = (cosh 1199040 sinh 119904
0cos120601
0 sinh 119904
0sin120601
0cos 120579
0
sinh 1199040sin120601
0sin 120579
0)
(45)
for any hyperbolic polar coordinates 1199040
isin [0 infin) 1206010
isin [0 120587]
and 1205790
isin [0 2120587] If we apply H-isometry T = L01minus1199040
R03
minus1206010R01
minus1205790isin
119866 then we have unit timelike vector e0such that
119879 (w) = e0 (46)
However unit timelike normal vector e0is invariant under
1198710119895
119904 That is
1198710119895
119904(e
0) = (cosh 119904) e
0 119895 = 1 2 3 (47)
For this reason if is an H-sphere which is generated fromspacelike hyperplane
HP (e0 minus119888) = x isin R
4
1| 119909
0= 119888 119888 ge 1 (48)
then we have
1198710119895
119904() = (49)
by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)
The following theorems also can be proved using similarmethod
Theorem 15 An equidistant surface is invariant under H-rotation in H3
Theorem 16 A horosphere is invariant under horocyclicrotation in H3
Finally we give the following corollary
Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866
0 119866
1 and 119866
2in H3
respectively
3 Differential Geometry of Curves andSurfaces in H3
In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]
TheLorentzian vector product of vectors x1 x2 x3 is givenby
0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3
1 Then the following maps 119864 119880 rarr 119878
3
1
119864(119906) = 120578(119906) and 119871plusmn
119880 rarr 119871119862+ 119871
plusmn(119906) = x(119906) plusmn 120578(119906)
are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906
0) can be identified with
identity mapping 119868119879119901119872
on the tangent space 119879119901119872 at x(119906
0) =
119901 isin 119872 We have that minus119889119871plusmn
= minus119868119879119901119872
plusmn (minus119889119864)For any given x(119906
0) = 119901 isin 119872 the linear transforms 119860
119901=
minus119889119864(1199060) 119879
119901119872 rarr 119879
119901119872 and 119878
plusmn
119901= minus119889119871
plusmn(119906
0) 119879
119901119872 rarr
119879119901119872 are called de Sitter shape operator and hyperbolic shape
operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and
119878plusmn
119901are denoted by 119896
119894(119901) and 119896
plusmn
119894(119901) for 119894 = 1 2 respectively
Obviously 119860119901and 119878
plusmn
119901have same eigenvectors Also the
eigenvalues satisfy
119896plusmn
119894(119901) = minus1 plusmn 119896
119894(119901) 119894 = 1 2 (55)
where 119896119894(119901) and 119896
plusmn
119894(119901) are called de Sitter principal curvature
and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872
respectivelyThe de Sitter Gauss curvature and the de Sitter mean
curvature of 119872 are given by
119870119889
(1199060) = det119860
119901= 119896
1(119901) 119896
2(119901)
119867119889
(1199060) =
1
2Tr119860
119901=
1198961
(119901) + 1198962
(119901)
2
(56)
at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss
curvature and the hyperbolic mean curvature of 119872 are givenby
119870plusmn
ℎ(119906
0) = det 119878
plusmn
119901= 119896
plusmn
1(119901) 119896
plusmn
2(119901)
119867plusmn
ℎ(119906
0) =
1
2Tr 119878
plusmn
119901=
119896plusmn
1(119901) + 119896
plusmn
2(119901)
2
(57)
at x(1199060) = 119901 respectively Evidently we have the following
relations119870
plusmn
ℎ= 1 ∓ 2119867
119889+ 119870
119889
119867plusmn
ℎ= minus1 plusmn 119867
119889
(58)
We say that a point x(1199060) = 119901 isin 119872 is an umbilical point
if 1198961(119901) = 119896
2(119901) Also 119872 is totally umbilical if all points
on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])
Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification
(1) Supposing that 1198962
= 1
(a) if 119896 = 0 and 1198962
lt 1 then M is a part of anequidistant surface
(b) if 119896 = 0 and 1198962
gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)
(2) If 1198962
= 1 then M is a part of horosphere
4 1198661-Invariant Surfaces in H3
In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces
Let 119872 = x(119880) be a regular surface via embedding x
119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860
the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4
1H3 and 119872 respectively Then the Gauss
and Weingarten explicit formulas for 119872 in H3 are given by
2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since
120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that
120578 (119904 119905) = ((12057231205721015840
4minus 120572
1015840
31205724) cosh 119904 (120572
31205721015840
4minus 120572
1015840
31205724) sinh 119904
12057211205721015840
4minus 120572
1015840
11205724 120572
1015840
11205723
minus 12057211205721015840
3)
(67)
and it is clear that
⟨120578 120597119904⟩ equiv ⟨120578 120597
119905⟩ equiv 0 (68)
for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A
119901= [
119886 119888
119887 119889] at any x(119904 119905) = 119901 isin 119872 where
119886 =
⟨minus119863120597119904
120578 120597119904⟩
⟨120597119904 120597
119904⟩
=
⟨119863119909119904
119909119904 120578⟩
⟨119909119904 119909
119904⟩
119887 =
⟨minus119863120597119904
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119904
119909119905 120578⟩
⟨119909119905 119909
119905⟩
119888 = 119887
119889 =
⟨minus119863120597119905
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119905
119909119905 120578⟩
⟨119909119905 119909
119905⟩
(69)
After basic calculations the de Sitter principal curvatures of119872 are
1198961
=1205721015840
31205724
minus 12057231205721015840
4
1205721
(70)
1198962
= 12057210158401015840
1(120572
1015840
31205724
minus 12057231205721015840
4) + 120572
10158401015840
3(120572
11205721015840
4minus 120572
1015840
11205724)
+ 12057210158401015840
4(120572
1015840
11205723
minus 12057211205721015840
3)
(71)
Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581
ℎ 120591
ℎ in H3
Proposition 22 Thebinormal vector of the profile curve of1198661-
invariant surface 119872 is constant in H3
Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591
ℎ= 0 Moreover by
(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)
119863te = 0 This completes the proof
From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582
0 120582
1 120582
2 120582
3) such that 120582
119894
is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872
Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896
2=
1205821120581ℎ
Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581
ℎe = 120572 and 120572
1015840and 120572
10158401015840 Also by (50)and (71) we obtain that 120572 and 120572
1015840and 120572
10158401015840= (0 119896
2 0 0) Thus it
follows that 120581ℎe = (0 119896
2 0 0) For this reason we have that
1198962
= 1205821120581ℎ
As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are
119870119889
(119901) =1205721015840
31205724
minus 12057231205721015840
4
1205721
1205821120581ℎ (72)
119867119889
(119901) =
(1205721015840
31205724
minus 12057231205721015840
4) + 120582
1120581ℎ1205721
21205721
(73)
Journal of Applied Mathematics 9
at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are
119870plusmn
ℎ(119901) =
(1205721
∓ (1205721015840
31205724
minus 12057231205721015840
4)) (1 ∓ 120582
1120581ℎ)
1205721
(74)
119867plusmn
ℎ(119901) =
1205721
(minus2 plusmn 1205821120581ℎ) plusmn (120572
1015840
31205724
minus 12057231205721015840
4)
21205721
(75)
at any x(119904 119905) = 119901 respectively
Proposition 24 Let 120572 119868 rarr 11986323
sub 1198673 120572(119905) =
(1205721(119905) 0 120572
3(119905) 120572
4(119905)) be unit speed regular profile curve of119866
1-
invariant surface 119872 Then its components are given by
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(76)
Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations
minus1205721
(119905)2
+ 1205723
(119905)2
+ 1205724
(119905)2
= minus1 (77)
minus1205721015840
1(119905)
2+ 120572
1015840
3(119905)
2+ 120572
1015840
4(119905)
2= 1 (78)
for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that
1205723 (119905) = radic120572
1 (119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
(79)
such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that
1205931015840(119905)
2=
1205721 (119905)
2minus 120572
1015840
1(119905)
2minus 1
(1205721
(119905)2
minus 1)2
(80)
Finally by (80) we have that
120593 (119905) = plusmn int
119905
0
radic1205721
(119906)2
minus 1205721015840
1(119906)
2minus 1
1205721 (119906)
2minus 1
119889119906 (81)
such that 1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1 gt 0 for all 119905 isin 119868 Without loss
of generality when we choose positive of signature of 120593 thiscompletes the proof
Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3
Now we will give some results which are obtained by (72)and (74)
Corollary 26 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205723
= 0 or 1205724
= 0 then 119872 is a part of de Sitter flatsurface
(ii) if 120581ℎ
= 0 then 119872 is a part of de Sitter flat surface
(iii) if 1205821
= 0 then 119872 is a part of de Sitter flat surface
Corollary 27 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 If 1205723
= 1205831205724such that 120583 isin R then 119872 is a
de Sitter flat surface
Theorem 28 Let 120572 be the profile curve of1198661-invariant surface
119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ
=
plusmn11205821
Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn
ℎ=
0 By (74) it follows that
1205721
(119905) ∓ (1205721015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (82)
or
1 ∓ 1205821120581ℎ
(119905) = 0 (83)
for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572
1(119905) ∓
(minusradic1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1) = 0 Hence it follows that
1205721015840
1(119905)
2+ 1 = 0 (84)
There is no real solution of (84) This means that the onlyone solution is 120581
ℎ= plusmn1120582
1by (83) On the other hand if we
assume that 120581ℎ
= plusmn11205821 then the proof is clear
Corollary 29 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205821
= 1 and 120581ℎ
= 1 then 119872 is 119870+
ℎ-flat surface which
is generated from horocyle
(ii) if 1205821
= minus1 and 120581ℎ
= 1 then 119872 is 119870minus
ℎ-flat surface which
is generated from horocyle
Now we will give theorem and corollaries for 1198661-
invariant surface which satisfy minimal condition in H3 by(73) and (75)
Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 inH3 Then 119872 is de
10 Journal of Applied Mathematics
Sitter minimal surface if and only if the parametrization of 120572 isgiven by
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(85)
with the condition 120581ℎ
isin (minus11205821 1120582
1) such that 119888
1is an
arbitrary constant
Proof Suppose that 119872 is de Sitter minimal surface that is119867
119889equiv 0 By (73) it follows that
(1205721015840
3(119905) 120572
4 (119905) minus 1205723 (119905) 120572
1015840
4(119905)) + 120582
1120581ℎ1205721 (119905) = 0 (86)
for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation
1205721015840
1(119905)
2minus (1 minus 120582
2
11205812
ℎ) 120572
1 (119905)2
+ 1 = 0 (87)
There exists only one real solution of (87) under the condition1 minus 120582
2
11205812
ℎgt 0 Moreover 120582
1must not be zero by Corollary 26
So that we obtain
120581ℎ
isin (minus1
1205821
1
1205821
) (88)
Hence the solution is
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
(89)
where 1198881is an arbitrary constant under the condition (88)
Finally the parametrization of 120572 is given explicitly byProposition 24
On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867
119889equiv 0
Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 in H3 Then 119872 is
hyperbolic minimal surface if and only if the parametrizationof 120572 is given by
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721
(119905)2
minus 1205721015840
1(119905)
2minus 1
1205721 (119905)
2minus 1
119889119906
(90)
with the condition 1minus(2∓1205821120581ℎ)2
gt 0 such that 1198881is an arbitrary
constant
Proof Suppose that 119872 is hyperbolic minimal surface that is119867
plusmn
ℎequiv 0 By (75) it follows that
(minus2 plusmn 1205821120581ℎ) 120572
1(119905) plusmn (120572
1015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (91)
If Proposition 24 is applied to (91) we have that
1205721015840
1(119905)
2+ ((minus2 plusmn 120582
1120581ℎ)2
minus 1) 1205721
(119905)2
+ 1 = 0 (92)
There exists only one real solution of (92) under the condition
(minus2 plusmn 1205821120581ℎ)2
minus 1 lt 0 (93)
Thus the solution is
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
(94)
where 1198881is an arbitrary constant under the condition (93)
However 1205821must not be zero by Corollary 26 So that if 119872
is119867+
ℎ-minimal surface (119867minus
ℎ-minimal surface) thenwe obtain
120581ℎ
isin (11205821 3120582
1) (120581
ℎisin (minus3120582
1 minus1120582
1)) by (93) Finally the
parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of
119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867
plusmn
ℎequiv 0
Now we will give classification theorem for totally umbil-ical 119866
1-invariant surfaces
Journal of Applied Mathematics 11
Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-
invariant surface 119872 inH3 Then the hyperbolic curvature of 120572
is constant
Proof Let 119872 be totally umbilical 1198661-invariant surface By
Theorem 23 and (70) we may assume that 1198961(119901) = 119896
2(119901) =
1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following
Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905
= x119905119904in
(96) then it follows that 12058211205811015840
ℎ(119905) = 0 for all 119905 isin 119868 Moreover
1205821must not be zero by Corollary 26Thus 120581
ℎis constant
Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following
classification
(1) Supposing that 1205852
= 1
(a) if 120585 = 0 and 1205852
lt 1 then 119872 is a part of anequidistant surface
(b) if 120585 = 0 and 1205852
gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane
(2) If 1205852
= 1 then 119872 is a part of horosphere
where 120585 = 1205821120581ℎis a constant
Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and
Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860
119901= 120585119868
2
where 1198682is identity matrix Finally the proof is complete by
Lemma 20
Now we will give some examples of 1198661-invariant surface
in H3 Let the Poincare ball model of hyperbolic space begiven by
B3
= (1199091 119909
2 119909
3) isin R
3|
3
sum
119894=1
1199092
119894lt 1 (97)
with the hyperbolic metric 1198891199042
= 4(1198891199092
1+119889119909
2
2+119889119909
2
3)(1minus119909
2
1minus
1199092
2minus 119909
2
3) Then it is well known that stereographic projection
of H3 is given by
Φ H3
997888rarr B3
Φ (1199090 119909
1 119909
2 119909
3) = (
1199091
1 + 1199090
1199092
1 + 1199090
1199093
1 + 1199090
)
(98)
We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3
Example 34 The 1198661-invariant surface which is generated
from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ
= radic2 is drawn in Figure 1(a)
Example 35 Let the profile curve of 119872 be given by
120572 (119905) = (radic2 +1
2(minus1 + radic2) 119905
2 0 1 minus
1
2(minus1 + radic2) 119905
2 119905)
(99)
such that hyperbolic curvature 120581ℎ
= 1 Then 119872 is hyperbolicflat 119866
1-invariant surface which is generated from horocycle
in H3 (see Figure 1(b))
Example 36 The 1198661-invariant surface which is generated
from
120572 (119905) = (1
3(minus1 + 4 cosh
radic3119905
2) 0
2
3(minus1 + cosh
radic3119905
2)
2
radic3sinh
radic3119905
2)
(100)
with hyperbolic curvature 120581ℎ
= 12 is drawn in Figure 1(c)
Example 37 Let the profile curve of 119872 be given by
120572 (119905)
= (2 cosh 119905
radic3minus sinh 119905
radic3 0 cosh 119905
radic3minus 2 sinh 119905
radic3 radic2)
(101)
such that hyperbolic curvature 120581ℎ
= radic2radic3 Then 119872 istotally umbilical 119866
1-invariant surface with 120585
2= 23 in H3
(see Figure 1(d))
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
12 Journal of Applied Mathematics
(a) (b)
(c) (d)
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993
Now we give corollaries about some properties of H-isometries and transition relation betweenH-coordinate axes1198970119895with H-coordinate planes 119863
119894119895
Corollary 9 Any H-coordinate axis is converted to each otherby suitable H-rotation That is
R0119895
120579l01119903
=
l02119903
119895 = 3 120579 =120587
2
l03119903
119895 = 2 120579 =3120587
2
(39)
Corollary 10 AH-plane consists in suitableH-coordinate axisand H-rotation Namely
R0119895
120601l0119896119903
=
D23
(119903120601) 119895 = 1 119896 = 2
D13
(119903120601) 119895 = 2 119896 = 3
D12
(119903120601) 119895 = 3 119896 = 1
(40)
6 Journal of Applied Mathematics
Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is
R0119895
120579D12
(119903120601)=
D23
(119903120601) 119895 = 3 120579 =
3120587
2
D13
(119903120601) 119895 = 1 120579 =
120587
2
(41)
Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely
R0119897
minus120595R0119896
minus120601R0119895
minus120579H012
120582R0119895
120579R0119896
120601R0119897
120595
=
H013
120582 119895 = 1 120579 =
120587
2 120601 = 0 120595 = 0
H031
120582 119895 = 1 119896 = 2 120579 =
120587
2 120601 =
120587
2 120595 = 0
H032
120582 119895 = 1 119896 = 2 119897 = 3 120579 =
120587
2 120601 =
120587
2 120595 =
120587
2
H023
120582 119895 = 3 119896 = 2 120579 = minus
120587
2 120601 = minus
120587
2 120595 = 0
H021
120582 119895 = 3 120579 = minus
120587
2 120601 = 0 120595 = 0
(42)
After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed
Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878
1015840
for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3
Theorem 14 An H-sphere is invariant under H-translation inH3
Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3
cap HP(k minus119896) for 119896 gt 0 So
HP (k minus119896) = x isin R4
1|
minus V01199090
+ V11199091
+ V21199092
+ V31199093
= minus119896 119896 gt 0
(43)
Moreover for w = kk isin H3 and 119888 = 119896k we have
HP (w minus119888) = x isin R4
1| ⟨xw⟩ = minus119888 (44)
Since w isin H3
w = (cosh 1199040 sinh 119904
0cos120601
0 sinh 119904
0sin120601
0cos 120579
0
sinh 1199040sin120601
0sin 120579
0)
(45)
for any hyperbolic polar coordinates 1199040
isin [0 infin) 1206010
isin [0 120587]
and 1205790
isin [0 2120587] If we apply H-isometry T = L01minus1199040
R03
minus1206010R01
minus1205790isin
119866 then we have unit timelike vector e0such that
119879 (w) = e0 (46)
However unit timelike normal vector e0is invariant under
1198710119895
119904 That is
1198710119895
119904(e
0) = (cosh 119904) e
0 119895 = 1 2 3 (47)
For this reason if is an H-sphere which is generated fromspacelike hyperplane
HP (e0 minus119888) = x isin R
4
1| 119909
0= 119888 119888 ge 1 (48)
then we have
1198710119895
119904() = (49)
by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)
The following theorems also can be proved using similarmethod
Theorem 15 An equidistant surface is invariant under H-rotation in H3
Theorem 16 A horosphere is invariant under horocyclicrotation in H3
Finally we give the following corollary
Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866
0 119866
1 and 119866
2in H3
respectively
3 Differential Geometry of Curves andSurfaces in H3
In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]
TheLorentzian vector product of vectors x1 x2 x3 is givenby
0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3
1 Then the following maps 119864 119880 rarr 119878
3
1
119864(119906) = 120578(119906) and 119871plusmn
119880 rarr 119871119862+ 119871
plusmn(119906) = x(119906) plusmn 120578(119906)
are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906
0) can be identified with
identity mapping 119868119879119901119872
on the tangent space 119879119901119872 at x(119906
0) =
119901 isin 119872 We have that minus119889119871plusmn
= minus119868119879119901119872
plusmn (minus119889119864)For any given x(119906
0) = 119901 isin 119872 the linear transforms 119860
119901=
minus119889119864(1199060) 119879
119901119872 rarr 119879
119901119872 and 119878
plusmn
119901= minus119889119871
plusmn(119906
0) 119879
119901119872 rarr
119879119901119872 are called de Sitter shape operator and hyperbolic shape
operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and
119878plusmn
119901are denoted by 119896
119894(119901) and 119896
plusmn
119894(119901) for 119894 = 1 2 respectively
Obviously 119860119901and 119878
plusmn
119901have same eigenvectors Also the
eigenvalues satisfy
119896plusmn
119894(119901) = minus1 plusmn 119896
119894(119901) 119894 = 1 2 (55)
where 119896119894(119901) and 119896
plusmn
119894(119901) are called de Sitter principal curvature
and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872
respectivelyThe de Sitter Gauss curvature and the de Sitter mean
curvature of 119872 are given by
119870119889
(1199060) = det119860
119901= 119896
1(119901) 119896
2(119901)
119867119889
(1199060) =
1
2Tr119860
119901=
1198961
(119901) + 1198962
(119901)
2
(56)
at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss
curvature and the hyperbolic mean curvature of 119872 are givenby
119870plusmn
ℎ(119906
0) = det 119878
plusmn
119901= 119896
plusmn
1(119901) 119896
plusmn
2(119901)
119867plusmn
ℎ(119906
0) =
1
2Tr 119878
plusmn
119901=
119896plusmn
1(119901) + 119896
plusmn
2(119901)
2
(57)
at x(1199060) = 119901 respectively Evidently we have the following
relations119870
plusmn
ℎ= 1 ∓ 2119867
119889+ 119870
119889
119867plusmn
ℎ= minus1 plusmn 119867
119889
(58)
We say that a point x(1199060) = 119901 isin 119872 is an umbilical point
if 1198961(119901) = 119896
2(119901) Also 119872 is totally umbilical if all points
on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])
Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification
(1) Supposing that 1198962
= 1
(a) if 119896 = 0 and 1198962
lt 1 then M is a part of anequidistant surface
(b) if 119896 = 0 and 1198962
gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)
(2) If 1198962
= 1 then M is a part of horosphere
4 1198661-Invariant Surfaces in H3
In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces
Let 119872 = x(119880) be a regular surface via embedding x
119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860
the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4
1H3 and 119872 respectively Then the Gauss
and Weingarten explicit formulas for 119872 in H3 are given by
2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since
120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that
120578 (119904 119905) = ((12057231205721015840
4minus 120572
1015840
31205724) cosh 119904 (120572
31205721015840
4minus 120572
1015840
31205724) sinh 119904
12057211205721015840
4minus 120572
1015840
11205724 120572
1015840
11205723
minus 12057211205721015840
3)
(67)
and it is clear that
⟨120578 120597119904⟩ equiv ⟨120578 120597
119905⟩ equiv 0 (68)
for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A
119901= [
119886 119888
119887 119889] at any x(119904 119905) = 119901 isin 119872 where
119886 =
⟨minus119863120597119904
120578 120597119904⟩
⟨120597119904 120597
119904⟩
=
⟨119863119909119904
119909119904 120578⟩
⟨119909119904 119909
119904⟩
119887 =
⟨minus119863120597119904
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119904
119909119905 120578⟩
⟨119909119905 119909
119905⟩
119888 = 119887
119889 =
⟨minus119863120597119905
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119905
119909119905 120578⟩
⟨119909119905 119909
119905⟩
(69)
After basic calculations the de Sitter principal curvatures of119872 are
1198961
=1205721015840
31205724
minus 12057231205721015840
4
1205721
(70)
1198962
= 12057210158401015840
1(120572
1015840
31205724
minus 12057231205721015840
4) + 120572
10158401015840
3(120572
11205721015840
4minus 120572
1015840
11205724)
+ 12057210158401015840
4(120572
1015840
11205723
minus 12057211205721015840
3)
(71)
Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581
ℎ 120591
ℎ in H3
Proposition 22 Thebinormal vector of the profile curve of1198661-
invariant surface 119872 is constant in H3
Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591
ℎ= 0 Moreover by
(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)
119863te = 0 This completes the proof
From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582
0 120582
1 120582
2 120582
3) such that 120582
119894
is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872
Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896
2=
1205821120581ℎ
Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581
ℎe = 120572 and 120572
1015840and 120572
10158401015840 Also by (50)and (71) we obtain that 120572 and 120572
1015840and 120572
10158401015840= (0 119896
2 0 0) Thus it
follows that 120581ℎe = (0 119896
2 0 0) For this reason we have that
1198962
= 1205821120581ℎ
As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are
119870119889
(119901) =1205721015840
31205724
minus 12057231205721015840
4
1205721
1205821120581ℎ (72)
119867119889
(119901) =
(1205721015840
31205724
minus 12057231205721015840
4) + 120582
1120581ℎ1205721
21205721
(73)
Journal of Applied Mathematics 9
at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are
119870plusmn
ℎ(119901) =
(1205721
∓ (1205721015840
31205724
minus 12057231205721015840
4)) (1 ∓ 120582
1120581ℎ)
1205721
(74)
119867plusmn
ℎ(119901) =
1205721
(minus2 plusmn 1205821120581ℎ) plusmn (120572
1015840
31205724
minus 12057231205721015840
4)
21205721
(75)
at any x(119904 119905) = 119901 respectively
Proposition 24 Let 120572 119868 rarr 11986323
sub 1198673 120572(119905) =
(1205721(119905) 0 120572
3(119905) 120572
4(119905)) be unit speed regular profile curve of119866
1-
invariant surface 119872 Then its components are given by
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(76)
Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations
minus1205721
(119905)2
+ 1205723
(119905)2
+ 1205724
(119905)2
= minus1 (77)
minus1205721015840
1(119905)
2+ 120572
1015840
3(119905)
2+ 120572
1015840
4(119905)
2= 1 (78)
for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that
1205723 (119905) = radic120572
1 (119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
(79)
such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that
1205931015840(119905)
2=
1205721 (119905)
2minus 120572
1015840
1(119905)
2minus 1
(1205721
(119905)2
minus 1)2
(80)
Finally by (80) we have that
120593 (119905) = plusmn int
119905
0
radic1205721
(119906)2
minus 1205721015840
1(119906)
2minus 1
1205721 (119906)
2minus 1
119889119906 (81)
such that 1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1 gt 0 for all 119905 isin 119868 Without loss
of generality when we choose positive of signature of 120593 thiscompletes the proof
Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3
Now we will give some results which are obtained by (72)and (74)
Corollary 26 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205723
= 0 or 1205724
= 0 then 119872 is a part of de Sitter flatsurface
(ii) if 120581ℎ
= 0 then 119872 is a part of de Sitter flat surface
(iii) if 1205821
= 0 then 119872 is a part of de Sitter flat surface
Corollary 27 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 If 1205723
= 1205831205724such that 120583 isin R then 119872 is a
de Sitter flat surface
Theorem 28 Let 120572 be the profile curve of1198661-invariant surface
119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ
=
plusmn11205821
Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn
ℎ=
0 By (74) it follows that
1205721
(119905) ∓ (1205721015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (82)
or
1 ∓ 1205821120581ℎ
(119905) = 0 (83)
for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572
1(119905) ∓
(minusradic1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1) = 0 Hence it follows that
1205721015840
1(119905)
2+ 1 = 0 (84)
There is no real solution of (84) This means that the onlyone solution is 120581
ℎ= plusmn1120582
1by (83) On the other hand if we
assume that 120581ℎ
= plusmn11205821 then the proof is clear
Corollary 29 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205821
= 1 and 120581ℎ
= 1 then 119872 is 119870+
ℎ-flat surface which
is generated from horocyle
(ii) if 1205821
= minus1 and 120581ℎ
= 1 then 119872 is 119870minus
ℎ-flat surface which
is generated from horocyle
Now we will give theorem and corollaries for 1198661-
invariant surface which satisfy minimal condition in H3 by(73) and (75)
Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 inH3 Then 119872 is de
10 Journal of Applied Mathematics
Sitter minimal surface if and only if the parametrization of 120572 isgiven by
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(85)
with the condition 120581ℎ
isin (minus11205821 1120582
1) such that 119888
1is an
arbitrary constant
Proof Suppose that 119872 is de Sitter minimal surface that is119867
119889equiv 0 By (73) it follows that
(1205721015840
3(119905) 120572
4 (119905) minus 1205723 (119905) 120572
1015840
4(119905)) + 120582
1120581ℎ1205721 (119905) = 0 (86)
for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation
1205721015840
1(119905)
2minus (1 minus 120582
2
11205812
ℎ) 120572
1 (119905)2
+ 1 = 0 (87)
There exists only one real solution of (87) under the condition1 minus 120582
2
11205812
ℎgt 0 Moreover 120582
1must not be zero by Corollary 26
So that we obtain
120581ℎ
isin (minus1
1205821
1
1205821
) (88)
Hence the solution is
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
(89)
where 1198881is an arbitrary constant under the condition (88)
Finally the parametrization of 120572 is given explicitly byProposition 24
On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867
119889equiv 0
Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 in H3 Then 119872 is
hyperbolic minimal surface if and only if the parametrizationof 120572 is given by
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721
(119905)2
minus 1205721015840
1(119905)
2minus 1
1205721 (119905)
2minus 1
119889119906
(90)
with the condition 1minus(2∓1205821120581ℎ)2
gt 0 such that 1198881is an arbitrary
constant
Proof Suppose that 119872 is hyperbolic minimal surface that is119867
plusmn
ℎequiv 0 By (75) it follows that
(minus2 plusmn 1205821120581ℎ) 120572
1(119905) plusmn (120572
1015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (91)
If Proposition 24 is applied to (91) we have that
1205721015840
1(119905)
2+ ((minus2 plusmn 120582
1120581ℎ)2
minus 1) 1205721
(119905)2
+ 1 = 0 (92)
There exists only one real solution of (92) under the condition
(minus2 plusmn 1205821120581ℎ)2
minus 1 lt 0 (93)
Thus the solution is
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
(94)
where 1198881is an arbitrary constant under the condition (93)
However 1205821must not be zero by Corollary 26 So that if 119872
is119867+
ℎ-minimal surface (119867minus
ℎ-minimal surface) thenwe obtain
120581ℎ
isin (11205821 3120582
1) (120581
ℎisin (minus3120582
1 minus1120582
1)) by (93) Finally the
parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of
119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867
plusmn
ℎequiv 0
Now we will give classification theorem for totally umbil-ical 119866
1-invariant surfaces
Journal of Applied Mathematics 11
Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-
invariant surface 119872 inH3 Then the hyperbolic curvature of 120572
is constant
Proof Let 119872 be totally umbilical 1198661-invariant surface By
Theorem 23 and (70) we may assume that 1198961(119901) = 119896
2(119901) =
1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following
Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905
= x119905119904in
(96) then it follows that 12058211205811015840
ℎ(119905) = 0 for all 119905 isin 119868 Moreover
1205821must not be zero by Corollary 26Thus 120581
ℎis constant
Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following
classification
(1) Supposing that 1205852
= 1
(a) if 120585 = 0 and 1205852
lt 1 then 119872 is a part of anequidistant surface
(b) if 120585 = 0 and 1205852
gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane
(2) If 1205852
= 1 then 119872 is a part of horosphere
where 120585 = 1205821120581ℎis a constant
Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and
Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860
119901= 120585119868
2
where 1198682is identity matrix Finally the proof is complete by
Lemma 20
Now we will give some examples of 1198661-invariant surface
in H3 Let the Poincare ball model of hyperbolic space begiven by
B3
= (1199091 119909
2 119909
3) isin R
3|
3
sum
119894=1
1199092
119894lt 1 (97)
with the hyperbolic metric 1198891199042
= 4(1198891199092
1+119889119909
2
2+119889119909
2
3)(1minus119909
2
1minus
1199092
2minus 119909
2
3) Then it is well known that stereographic projection
of H3 is given by
Φ H3
997888rarr B3
Φ (1199090 119909
1 119909
2 119909
3) = (
1199091
1 + 1199090
1199092
1 + 1199090
1199093
1 + 1199090
)
(98)
We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3
Example 34 The 1198661-invariant surface which is generated
from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ
= radic2 is drawn in Figure 1(a)
Example 35 Let the profile curve of 119872 be given by
120572 (119905) = (radic2 +1
2(minus1 + radic2) 119905
2 0 1 minus
1
2(minus1 + radic2) 119905
2 119905)
(99)
such that hyperbolic curvature 120581ℎ
= 1 Then 119872 is hyperbolicflat 119866
1-invariant surface which is generated from horocycle
in H3 (see Figure 1(b))
Example 36 The 1198661-invariant surface which is generated
from
120572 (119905) = (1
3(minus1 + 4 cosh
radic3119905
2) 0
2
3(minus1 + cosh
radic3119905
2)
2
radic3sinh
radic3119905
2)
(100)
with hyperbolic curvature 120581ℎ
= 12 is drawn in Figure 1(c)
Example 37 Let the profile curve of 119872 be given by
120572 (119905)
= (2 cosh 119905
radic3minus sinh 119905
radic3 0 cosh 119905
radic3minus 2 sinh 119905
radic3 radic2)
(101)
such that hyperbolic curvature 120581ℎ
= radic2radic3 Then 119872 istotally umbilical 119866
1-invariant surface with 120585
2= 23 in H3
(see Figure 1(d))
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
12 Journal of Applied Mathematics
(a) (b)
(c) (d)
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993
Corollary 11 Any H-coordinate plane is converted to eachother by suitable H-rotation That is
R0119895
120579D12
(119903120601)=
D23
(119903120601) 119895 = 3 120579 =
3120587
2
D13
(119903120601) 119895 = 1 120579 =
120587
2
(41)
Corollary 12 Any horocyclic rotation is converted to eachother by suitable H-rotations Namely
R0119897
minus120595R0119896
minus120601R0119895
minus120579H012
120582R0119895
120579R0119896
120601R0119897
120595
=
H013
120582 119895 = 1 120579 =
120587
2 120601 = 0 120595 = 0
H031
120582 119895 = 1 119896 = 2 120579 =
120587
2 120601 =
120587
2 120595 = 0
H032
120582 119895 = 1 119896 = 2 119897 = 3 120579 =
120587
2 120601 =
120587
2 120595 =
120587
2
H023
120582 119895 = 3 119896 = 2 120579 = minus
120587
2 120601 = minus
120587
2 120595 = 0
H021
120582 119895 = 3 120579 = minus
120587
2 120601 = 0 120595 = 0
(42)
After the notion of congruent in H3 we will give adifferent classification theorem of H-isometries in terms ofleaving those totally umbilic surfaces of H3 fixed
Definition 13 Let 119878 and 1198781015840 be two subsets of H3 If 119879(119878) = 119878
1015840
for some 119879 isin 119866 then 119878 and 1198781015840 are called congruent in H3
Theorem 14 An H-sphere is invariant under H-translation inH3
Proof Suppose that 119872 is an H-sphere Then there exists aspacelike hyperplane HP(k minus119896) with timelike normal k suchthat 119872 = H3
cap HP(k minus119896) for 119896 gt 0 So
HP (k minus119896) = x isin R4
1|
minus V01199090
+ V11199091
+ V21199092
+ V31199093
= minus119896 119896 gt 0
(43)
Moreover for w = kk isin H3 and 119888 = 119896k we have
HP (w minus119888) = x isin R4
1| ⟨xw⟩ = minus119888 (44)
Since w isin H3
w = (cosh 1199040 sinh 119904
0cos120601
0 sinh 119904
0sin120601
0cos 120579
0
sinh 1199040sin120601
0sin 120579
0)
(45)
for any hyperbolic polar coordinates 1199040
isin [0 infin) 1206010
isin [0 120587]
and 1205790
isin [0 2120587] If we apply H-isometry T = L01minus1199040
R03
minus1206010R01
minus1205790isin
119866 then we have unit timelike vector e0such that
119879 (w) = e0 (46)
However unit timelike normal vector e0is invariant under
1198710119895
119904 That is
1198710119895
119904(e
0) = (cosh 119904) e
0 119895 = 1 2 3 (47)
For this reason if is an H-sphere which is generated fromspacelike hyperplane
HP (e0 minus119888) = x isin R
4
1| 119909
0= 119888 119888 ge 1 (48)
then we have
1198710119895
119904() = (49)
by (47) Therefore is invariant under H-translationsFinallyThe proof is completed since119872 and are congruentby (46)
The following theorems also can be proved using similarmethod
Theorem 15 An equidistant surface is invariant under H-rotation in H3
Theorem 16 A horosphere is invariant under horocyclicrotation in H3
Finally we give the following corollary
Corollary 17 Equidistant surfaces H-spheres and horo-spheres are invariant under groups 119866
0 119866
1 and 119866
2in H3
respectively
3 Differential Geometry of Curves andSurfaces in H3
In this section we give the basic theory of extrinsic differen-tial geometry of curves and surfaces in H3 Unless otherwisestated we use the notation in [7 8]
TheLorentzian vector product of vectors x1 x2 x3 is givenby
0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3
1 Then the following maps 119864 119880 rarr 119878
3
1
119864(119906) = 120578(119906) and 119871plusmn
119880 rarr 119871119862+ 119871
plusmn(119906) = x(119906) plusmn 120578(119906)
are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906
0) can be identified with
identity mapping 119868119879119901119872
on the tangent space 119879119901119872 at x(119906
0) =
119901 isin 119872 We have that minus119889119871plusmn
= minus119868119879119901119872
plusmn (minus119889119864)For any given x(119906
0) = 119901 isin 119872 the linear transforms 119860
119901=
minus119889119864(1199060) 119879
119901119872 rarr 119879
119901119872 and 119878
plusmn
119901= minus119889119871
plusmn(119906
0) 119879
119901119872 rarr
119879119901119872 are called de Sitter shape operator and hyperbolic shape
operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and
119878plusmn
119901are denoted by 119896
119894(119901) and 119896
plusmn
119894(119901) for 119894 = 1 2 respectively
Obviously 119860119901and 119878
plusmn
119901have same eigenvectors Also the
eigenvalues satisfy
119896plusmn
119894(119901) = minus1 plusmn 119896
119894(119901) 119894 = 1 2 (55)
where 119896119894(119901) and 119896
plusmn
119894(119901) are called de Sitter principal curvature
and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872
respectivelyThe de Sitter Gauss curvature and the de Sitter mean
curvature of 119872 are given by
119870119889
(1199060) = det119860
119901= 119896
1(119901) 119896
2(119901)
119867119889
(1199060) =
1
2Tr119860
119901=
1198961
(119901) + 1198962
(119901)
2
(56)
at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss
curvature and the hyperbolic mean curvature of 119872 are givenby
119870plusmn
ℎ(119906
0) = det 119878
plusmn
119901= 119896
plusmn
1(119901) 119896
plusmn
2(119901)
119867plusmn
ℎ(119906
0) =
1
2Tr 119878
plusmn
119901=
119896plusmn
1(119901) + 119896
plusmn
2(119901)
2
(57)
at x(1199060) = 119901 respectively Evidently we have the following
relations119870
plusmn
ℎ= 1 ∓ 2119867
119889+ 119870
119889
119867plusmn
ℎ= minus1 plusmn 119867
119889
(58)
We say that a point x(1199060) = 119901 isin 119872 is an umbilical point
if 1198961(119901) = 119896
2(119901) Also 119872 is totally umbilical if all points
on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])
Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification
(1) Supposing that 1198962
= 1
(a) if 119896 = 0 and 1198962
lt 1 then M is a part of anequidistant surface
(b) if 119896 = 0 and 1198962
gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)
(2) If 1198962
= 1 then M is a part of horosphere
4 1198661-Invariant Surfaces in H3
In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces
Let 119872 = x(119880) be a regular surface via embedding x
119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860
the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4
1H3 and 119872 respectively Then the Gauss
and Weingarten explicit formulas for 119872 in H3 are given by
2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since
120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that
120578 (119904 119905) = ((12057231205721015840
4minus 120572
1015840
31205724) cosh 119904 (120572
31205721015840
4minus 120572
1015840
31205724) sinh 119904
12057211205721015840
4minus 120572
1015840
11205724 120572
1015840
11205723
minus 12057211205721015840
3)
(67)
and it is clear that
⟨120578 120597119904⟩ equiv ⟨120578 120597
119905⟩ equiv 0 (68)
for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A
119901= [
119886 119888
119887 119889] at any x(119904 119905) = 119901 isin 119872 where
119886 =
⟨minus119863120597119904
120578 120597119904⟩
⟨120597119904 120597
119904⟩
=
⟨119863119909119904
119909119904 120578⟩
⟨119909119904 119909
119904⟩
119887 =
⟨minus119863120597119904
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119904
119909119905 120578⟩
⟨119909119905 119909
119905⟩
119888 = 119887
119889 =
⟨minus119863120597119905
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119905
119909119905 120578⟩
⟨119909119905 119909
119905⟩
(69)
After basic calculations the de Sitter principal curvatures of119872 are
1198961
=1205721015840
31205724
minus 12057231205721015840
4
1205721
(70)
1198962
= 12057210158401015840
1(120572
1015840
31205724
minus 12057231205721015840
4) + 120572
10158401015840
3(120572
11205721015840
4minus 120572
1015840
11205724)
+ 12057210158401015840
4(120572
1015840
11205723
minus 12057211205721015840
3)
(71)
Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581
ℎ 120591
ℎ in H3
Proposition 22 Thebinormal vector of the profile curve of1198661-
invariant surface 119872 is constant in H3
Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591
ℎ= 0 Moreover by
(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)
119863te = 0 This completes the proof
From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582
0 120582
1 120582
2 120582
3) such that 120582
119894
is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872
Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896
2=
1205821120581ℎ
Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581
ℎe = 120572 and 120572
1015840and 120572
10158401015840 Also by (50)and (71) we obtain that 120572 and 120572
1015840and 120572
10158401015840= (0 119896
2 0 0) Thus it
follows that 120581ℎe = (0 119896
2 0 0) For this reason we have that
1198962
= 1205821120581ℎ
As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are
119870119889
(119901) =1205721015840
31205724
minus 12057231205721015840
4
1205721
1205821120581ℎ (72)
119867119889
(119901) =
(1205721015840
31205724
minus 12057231205721015840
4) + 120582
1120581ℎ1205721
21205721
(73)
Journal of Applied Mathematics 9
at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are
119870plusmn
ℎ(119901) =
(1205721
∓ (1205721015840
31205724
minus 12057231205721015840
4)) (1 ∓ 120582
1120581ℎ)
1205721
(74)
119867plusmn
ℎ(119901) =
1205721
(minus2 plusmn 1205821120581ℎ) plusmn (120572
1015840
31205724
minus 12057231205721015840
4)
21205721
(75)
at any x(119904 119905) = 119901 respectively
Proposition 24 Let 120572 119868 rarr 11986323
sub 1198673 120572(119905) =
(1205721(119905) 0 120572
3(119905) 120572
4(119905)) be unit speed regular profile curve of119866
1-
invariant surface 119872 Then its components are given by
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(76)
Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations
minus1205721
(119905)2
+ 1205723
(119905)2
+ 1205724
(119905)2
= minus1 (77)
minus1205721015840
1(119905)
2+ 120572
1015840
3(119905)
2+ 120572
1015840
4(119905)
2= 1 (78)
for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that
1205723 (119905) = radic120572
1 (119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
(79)
such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that
1205931015840(119905)
2=
1205721 (119905)
2minus 120572
1015840
1(119905)
2minus 1
(1205721
(119905)2
minus 1)2
(80)
Finally by (80) we have that
120593 (119905) = plusmn int
119905
0
radic1205721
(119906)2
minus 1205721015840
1(119906)
2minus 1
1205721 (119906)
2minus 1
119889119906 (81)
such that 1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1 gt 0 for all 119905 isin 119868 Without loss
of generality when we choose positive of signature of 120593 thiscompletes the proof
Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3
Now we will give some results which are obtained by (72)and (74)
Corollary 26 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205723
= 0 or 1205724
= 0 then 119872 is a part of de Sitter flatsurface
(ii) if 120581ℎ
= 0 then 119872 is a part of de Sitter flat surface
(iii) if 1205821
= 0 then 119872 is a part of de Sitter flat surface
Corollary 27 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 If 1205723
= 1205831205724such that 120583 isin R then 119872 is a
de Sitter flat surface
Theorem 28 Let 120572 be the profile curve of1198661-invariant surface
119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ
=
plusmn11205821
Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn
ℎ=
0 By (74) it follows that
1205721
(119905) ∓ (1205721015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (82)
or
1 ∓ 1205821120581ℎ
(119905) = 0 (83)
for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572
1(119905) ∓
(minusradic1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1) = 0 Hence it follows that
1205721015840
1(119905)
2+ 1 = 0 (84)
There is no real solution of (84) This means that the onlyone solution is 120581
ℎ= plusmn1120582
1by (83) On the other hand if we
assume that 120581ℎ
= plusmn11205821 then the proof is clear
Corollary 29 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205821
= 1 and 120581ℎ
= 1 then 119872 is 119870+
ℎ-flat surface which
is generated from horocyle
(ii) if 1205821
= minus1 and 120581ℎ
= 1 then 119872 is 119870minus
ℎ-flat surface which
is generated from horocyle
Now we will give theorem and corollaries for 1198661-
invariant surface which satisfy minimal condition in H3 by(73) and (75)
Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 inH3 Then 119872 is de
10 Journal of Applied Mathematics
Sitter minimal surface if and only if the parametrization of 120572 isgiven by
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(85)
with the condition 120581ℎ
isin (minus11205821 1120582
1) such that 119888
1is an
arbitrary constant
Proof Suppose that 119872 is de Sitter minimal surface that is119867
119889equiv 0 By (73) it follows that
(1205721015840
3(119905) 120572
4 (119905) minus 1205723 (119905) 120572
1015840
4(119905)) + 120582
1120581ℎ1205721 (119905) = 0 (86)
for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation
1205721015840
1(119905)
2minus (1 minus 120582
2
11205812
ℎ) 120572
1 (119905)2
+ 1 = 0 (87)
There exists only one real solution of (87) under the condition1 minus 120582
2
11205812
ℎgt 0 Moreover 120582
1must not be zero by Corollary 26
So that we obtain
120581ℎ
isin (minus1
1205821
1
1205821
) (88)
Hence the solution is
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
(89)
where 1198881is an arbitrary constant under the condition (88)
Finally the parametrization of 120572 is given explicitly byProposition 24
On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867
119889equiv 0
Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 in H3 Then 119872 is
hyperbolic minimal surface if and only if the parametrizationof 120572 is given by
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721
(119905)2
minus 1205721015840
1(119905)
2minus 1
1205721 (119905)
2minus 1
119889119906
(90)
with the condition 1minus(2∓1205821120581ℎ)2
gt 0 such that 1198881is an arbitrary
constant
Proof Suppose that 119872 is hyperbolic minimal surface that is119867
plusmn
ℎequiv 0 By (75) it follows that
(minus2 plusmn 1205821120581ℎ) 120572
1(119905) plusmn (120572
1015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (91)
If Proposition 24 is applied to (91) we have that
1205721015840
1(119905)
2+ ((minus2 plusmn 120582
1120581ℎ)2
minus 1) 1205721
(119905)2
+ 1 = 0 (92)
There exists only one real solution of (92) under the condition
(minus2 plusmn 1205821120581ℎ)2
minus 1 lt 0 (93)
Thus the solution is
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
(94)
where 1198881is an arbitrary constant under the condition (93)
However 1205821must not be zero by Corollary 26 So that if 119872
is119867+
ℎ-minimal surface (119867minus
ℎ-minimal surface) thenwe obtain
120581ℎ
isin (11205821 3120582
1) (120581
ℎisin (minus3120582
1 minus1120582
1)) by (93) Finally the
parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of
119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867
plusmn
ℎequiv 0
Now we will give classification theorem for totally umbil-ical 119866
1-invariant surfaces
Journal of Applied Mathematics 11
Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-
invariant surface 119872 inH3 Then the hyperbolic curvature of 120572
is constant
Proof Let 119872 be totally umbilical 1198661-invariant surface By
Theorem 23 and (70) we may assume that 1198961(119901) = 119896
2(119901) =
1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following
Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905
= x119905119904in
(96) then it follows that 12058211205811015840
ℎ(119905) = 0 for all 119905 isin 119868 Moreover
1205821must not be zero by Corollary 26Thus 120581
ℎis constant
Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following
classification
(1) Supposing that 1205852
= 1
(a) if 120585 = 0 and 1205852
lt 1 then 119872 is a part of anequidistant surface
(b) if 120585 = 0 and 1205852
gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane
(2) If 1205852
= 1 then 119872 is a part of horosphere
where 120585 = 1205821120581ℎis a constant
Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and
Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860
119901= 120585119868
2
where 1198682is identity matrix Finally the proof is complete by
Lemma 20
Now we will give some examples of 1198661-invariant surface
in H3 Let the Poincare ball model of hyperbolic space begiven by
B3
= (1199091 119909
2 119909
3) isin R
3|
3
sum
119894=1
1199092
119894lt 1 (97)
with the hyperbolic metric 1198891199042
= 4(1198891199092
1+119889119909
2
2+119889119909
2
3)(1minus119909
2
1minus
1199092
2minus 119909
2
3) Then it is well known that stereographic projection
of H3 is given by
Φ H3
997888rarr B3
Φ (1199090 119909
1 119909
2 119909
3) = (
1199091
1 + 1199090
1199092
1 + 1199090
1199093
1 + 1199090
)
(98)
We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3
Example 34 The 1198661-invariant surface which is generated
from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ
= radic2 is drawn in Figure 1(a)
Example 35 Let the profile curve of 119872 be given by
120572 (119905) = (radic2 +1
2(minus1 + radic2) 119905
2 0 1 minus
1
2(minus1 + radic2) 119905
2 119905)
(99)
such that hyperbolic curvature 120581ℎ
= 1 Then 119872 is hyperbolicflat 119866
1-invariant surface which is generated from horocycle
in H3 (see Figure 1(b))
Example 36 The 1198661-invariant surface which is generated
from
120572 (119905) = (1
3(minus1 + 4 cosh
radic3119905
2) 0
2
3(minus1 + cosh
radic3119905
2)
2
radic3sinh
radic3119905
2)
(100)
with hyperbolic curvature 120581ℎ
= 12 is drawn in Figure 1(c)
Example 37 Let the profile curve of 119872 be given by
120572 (119905)
= (2 cosh 119905
radic3minus sinh 119905
radic3 0 cosh 119905
radic3minus 2 sinh 119905
radic3 radic2)
(101)
such that hyperbolic curvature 120581ℎ
= radic2radic3 Then 119872 istotally umbilical 119866
1-invariant surface with 120585
2= 23 in H3
(see Figure 1(d))
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
12 Journal of Applied Mathematics
(a) (b)
(c) (d)
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993
0 119894 = 1 2 We also regard 120578 as unit normal vector field along119872 in H3 Moreover x(119906) plusmn 120578(119906) is a lightlike vector sincex(119906) isin H3 120578(119906) isin S3
1 Then the following maps 119864 119880 rarr 119878
3
1
119864(119906) = 120578(119906) and 119871plusmn
119880 rarr 119871119862+ 119871
plusmn(119906) = x(119906) plusmn 120578(119906)
are called de Sitter Gauss map and light cone Gauss map ofx respectively [8] Under the identification of 119880 and 119872 viathe embedding x the derivative 119889x(119906
0) can be identified with
identity mapping 119868119879119901119872
on the tangent space 119879119901119872 at x(119906
0) =
119901 isin 119872 We have that minus119889119871plusmn
= minus119868119879119901119872
plusmn (minus119889119864)For any given x(119906
0) = 119901 isin 119872 the linear transforms 119860
119901=
minus119889119864(1199060) 119879
119901119872 rarr 119879
119901119872 and 119878
plusmn
119901= minus119889119871
plusmn(119906
0) 119879
119901119872 rarr
119879119901119872 are called de Sitter shape operator and hyperbolic shape
operator of x(119880) = 119872 respectivelyThe eigenvalues of119860119901and
119878plusmn
119901are denoted by 119896
119894(119901) and 119896
plusmn
119894(119901) for 119894 = 1 2 respectively
Obviously 119860119901and 119878
plusmn
119901have same eigenvectors Also the
eigenvalues satisfy
119896plusmn
119894(119901) = minus1 plusmn 119896
119894(119901) 119894 = 1 2 (55)
where 119896119894(119901) and 119896
plusmn
119894(119901) are called de Sitter principal curvature
and hyperbolic principal curvature of 119872 at x(1199060) = 119901 isin 119872
respectivelyThe de Sitter Gauss curvature and the de Sitter mean
curvature of 119872 are given by
119870119889
(1199060) = det119860
119901= 119896
1(119901) 119896
2(119901)
119867119889
(1199060) =
1
2Tr119860
119901=
1198961
(119901) + 1198962
(119901)
2
(56)
at x(1199060) = 119901 respectively Similarly The hyperbolic Gauss
curvature and the hyperbolic mean curvature of 119872 are givenby
119870plusmn
ℎ(119906
0) = det 119878
plusmn
119901= 119896
plusmn
1(119901) 119896
plusmn
2(119901)
119867plusmn
ℎ(119906
0) =
1
2Tr 119878
plusmn
119901=
119896plusmn
1(119901) + 119896
plusmn
2(119901)
2
(57)
at x(1199060) = 119901 respectively Evidently we have the following
relations119870
plusmn
ℎ= 1 ∓ 2119867
119889+ 119870
119889
119867plusmn
ℎ= minus1 plusmn 119867
119889
(58)
We say that a point x(1199060) = 119901 isin 119872 is an umbilical point
if 1198961(119901) = 119896
2(119901) Also 119872 is totally umbilical if all points
on 119872 are umbilical Now we give the following classificationtheoremof totally umbilical surfaces inH3 (cf [8 Proposition21])
Lemma 20 Suppose that119872 = x(119880) is totally umbilicalThen119896(119901) is a constant 119896 Under this condition one has the followingclassification
(1) Supposing that 1198962
= 1
(a) if 119896 = 0 and 1198962
lt 1 then M is a part of anequidistant surface
(b) if 119896 = 0 and 1198962
gt 1 then M is a part of a sphere(c) if 119896 = 0 then M is a part of a plane (H-plane)
(2) If 1198962
= 1 then M is a part of horosphere
4 1198661-Invariant Surfaces in H3
In this section we investigate surfaces which are invariantunder some one parameter subgroup of H-translations inH3Moreover we study extrinsic differential geometry of theseinvariant surfaces
Let 119872 = x(119880) be a regular surface via embedding x
119880 rarr H3 such that open subset 119880 sub R2 We denote by 119860
the shape operator of 119872 with respect to unit normal vectorfield 120578 in H3 Let us represent by 119863 119863 and 119863 the Levi-Civitaconnections of R4
1H3 and 119872 respectively Then the Gauss
and Weingarten explicit formulas for 119872 in H3 are given by
2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since
120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that
120578 (119904 119905) = ((12057231205721015840
4minus 120572
1015840
31205724) cosh 119904 (120572
31205721015840
4minus 120572
1015840
31205724) sinh 119904
12057211205721015840
4minus 120572
1015840
11205724 120572
1015840
11205723
minus 12057211205721015840
3)
(67)
and it is clear that
⟨120578 120597119904⟩ equiv ⟨120578 120597
119905⟩ equiv 0 (68)
for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A
119901= [
119886 119888
119887 119889] at any x(119904 119905) = 119901 isin 119872 where
119886 =
⟨minus119863120597119904
120578 120597119904⟩
⟨120597119904 120597
119904⟩
=
⟨119863119909119904
119909119904 120578⟩
⟨119909119904 119909
119904⟩
119887 =
⟨minus119863120597119904
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119904
119909119905 120578⟩
⟨119909119905 119909
119905⟩
119888 = 119887
119889 =
⟨minus119863120597119905
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119905
119909119905 120578⟩
⟨119909119905 119909
119905⟩
(69)
After basic calculations the de Sitter principal curvatures of119872 are
1198961
=1205721015840
31205724
minus 12057231205721015840
4
1205721
(70)
1198962
= 12057210158401015840
1(120572
1015840
31205724
minus 12057231205721015840
4) + 120572
10158401015840
3(120572
11205721015840
4minus 120572
1015840
11205724)
+ 12057210158401015840
4(120572
1015840
11205723
minus 12057211205721015840
3)
(71)
Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581
ℎ 120591
ℎ in H3
Proposition 22 Thebinormal vector of the profile curve of1198661-
invariant surface 119872 is constant in H3
Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591
ℎ= 0 Moreover by
(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)
119863te = 0 This completes the proof
From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582
0 120582
1 120582
2 120582
3) such that 120582
119894
is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872
Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896
2=
1205821120581ℎ
Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581
ℎe = 120572 and 120572
1015840and 120572
10158401015840 Also by (50)and (71) we obtain that 120572 and 120572
1015840and 120572
10158401015840= (0 119896
2 0 0) Thus it
follows that 120581ℎe = (0 119896
2 0 0) For this reason we have that
1198962
= 1205821120581ℎ
As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are
119870119889
(119901) =1205721015840
31205724
minus 12057231205721015840
4
1205721
1205821120581ℎ (72)
119867119889
(119901) =
(1205721015840
31205724
minus 12057231205721015840
4) + 120582
1120581ℎ1205721
21205721
(73)
Journal of Applied Mathematics 9
at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are
119870plusmn
ℎ(119901) =
(1205721
∓ (1205721015840
31205724
minus 12057231205721015840
4)) (1 ∓ 120582
1120581ℎ)
1205721
(74)
119867plusmn
ℎ(119901) =
1205721
(minus2 plusmn 1205821120581ℎ) plusmn (120572
1015840
31205724
minus 12057231205721015840
4)
21205721
(75)
at any x(119904 119905) = 119901 respectively
Proposition 24 Let 120572 119868 rarr 11986323
sub 1198673 120572(119905) =
(1205721(119905) 0 120572
3(119905) 120572
4(119905)) be unit speed regular profile curve of119866
1-
invariant surface 119872 Then its components are given by
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(76)
Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations
minus1205721
(119905)2
+ 1205723
(119905)2
+ 1205724
(119905)2
= minus1 (77)
minus1205721015840
1(119905)
2+ 120572
1015840
3(119905)
2+ 120572
1015840
4(119905)
2= 1 (78)
for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that
1205723 (119905) = radic120572
1 (119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
(79)
such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that
1205931015840(119905)
2=
1205721 (119905)
2minus 120572
1015840
1(119905)
2minus 1
(1205721
(119905)2
minus 1)2
(80)
Finally by (80) we have that
120593 (119905) = plusmn int
119905
0
radic1205721
(119906)2
minus 1205721015840
1(119906)
2minus 1
1205721 (119906)
2minus 1
119889119906 (81)
such that 1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1 gt 0 for all 119905 isin 119868 Without loss
of generality when we choose positive of signature of 120593 thiscompletes the proof
Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3
Now we will give some results which are obtained by (72)and (74)
Corollary 26 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205723
= 0 or 1205724
= 0 then 119872 is a part of de Sitter flatsurface
(ii) if 120581ℎ
= 0 then 119872 is a part of de Sitter flat surface
(iii) if 1205821
= 0 then 119872 is a part of de Sitter flat surface
Corollary 27 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 If 1205723
= 1205831205724such that 120583 isin R then 119872 is a
de Sitter flat surface
Theorem 28 Let 120572 be the profile curve of1198661-invariant surface
119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ
=
plusmn11205821
Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn
ℎ=
0 By (74) it follows that
1205721
(119905) ∓ (1205721015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (82)
or
1 ∓ 1205821120581ℎ
(119905) = 0 (83)
for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572
1(119905) ∓
(minusradic1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1) = 0 Hence it follows that
1205721015840
1(119905)
2+ 1 = 0 (84)
There is no real solution of (84) This means that the onlyone solution is 120581
ℎ= plusmn1120582
1by (83) On the other hand if we
assume that 120581ℎ
= plusmn11205821 then the proof is clear
Corollary 29 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205821
= 1 and 120581ℎ
= 1 then 119872 is 119870+
ℎ-flat surface which
is generated from horocyle
(ii) if 1205821
= minus1 and 120581ℎ
= 1 then 119872 is 119870minus
ℎ-flat surface which
is generated from horocyle
Now we will give theorem and corollaries for 1198661-
invariant surface which satisfy minimal condition in H3 by(73) and (75)
Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 inH3 Then 119872 is de
10 Journal of Applied Mathematics
Sitter minimal surface if and only if the parametrization of 120572 isgiven by
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(85)
with the condition 120581ℎ
isin (minus11205821 1120582
1) such that 119888
1is an
arbitrary constant
Proof Suppose that 119872 is de Sitter minimal surface that is119867
119889equiv 0 By (73) it follows that
(1205721015840
3(119905) 120572
4 (119905) minus 1205723 (119905) 120572
1015840
4(119905)) + 120582
1120581ℎ1205721 (119905) = 0 (86)
for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation
1205721015840
1(119905)
2minus (1 minus 120582
2
11205812
ℎ) 120572
1 (119905)2
+ 1 = 0 (87)
There exists only one real solution of (87) under the condition1 minus 120582
2
11205812
ℎgt 0 Moreover 120582
1must not be zero by Corollary 26
So that we obtain
120581ℎ
isin (minus1
1205821
1
1205821
) (88)
Hence the solution is
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
(89)
where 1198881is an arbitrary constant under the condition (88)
Finally the parametrization of 120572 is given explicitly byProposition 24
On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867
119889equiv 0
Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 in H3 Then 119872 is
hyperbolic minimal surface if and only if the parametrizationof 120572 is given by
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721
(119905)2
minus 1205721015840
1(119905)
2minus 1
1205721 (119905)
2minus 1
119889119906
(90)
with the condition 1minus(2∓1205821120581ℎ)2
gt 0 such that 1198881is an arbitrary
constant
Proof Suppose that 119872 is hyperbolic minimal surface that is119867
plusmn
ℎequiv 0 By (75) it follows that
(minus2 plusmn 1205821120581ℎ) 120572
1(119905) plusmn (120572
1015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (91)
If Proposition 24 is applied to (91) we have that
1205721015840
1(119905)
2+ ((minus2 plusmn 120582
1120581ℎ)2
minus 1) 1205721
(119905)2
+ 1 = 0 (92)
There exists only one real solution of (92) under the condition
(minus2 plusmn 1205821120581ℎ)2
minus 1 lt 0 (93)
Thus the solution is
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
(94)
where 1198881is an arbitrary constant under the condition (93)
However 1205821must not be zero by Corollary 26 So that if 119872
is119867+
ℎ-minimal surface (119867minus
ℎ-minimal surface) thenwe obtain
120581ℎ
isin (11205821 3120582
1) (120581
ℎisin (minus3120582
1 minus1120582
1)) by (93) Finally the
parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of
119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867
plusmn
ℎequiv 0
Now we will give classification theorem for totally umbil-ical 119866
1-invariant surfaces
Journal of Applied Mathematics 11
Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-
invariant surface 119872 inH3 Then the hyperbolic curvature of 120572
is constant
Proof Let 119872 be totally umbilical 1198661-invariant surface By
Theorem 23 and (70) we may assume that 1198961(119901) = 119896
2(119901) =
1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following
Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905
= x119905119904in
(96) then it follows that 12058211205811015840
ℎ(119905) = 0 for all 119905 isin 119868 Moreover
1205821must not be zero by Corollary 26Thus 120581
ℎis constant
Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following
classification
(1) Supposing that 1205852
= 1
(a) if 120585 = 0 and 1205852
lt 1 then 119872 is a part of anequidistant surface
(b) if 120585 = 0 and 1205852
gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane
(2) If 1205852
= 1 then 119872 is a part of horosphere
where 120585 = 1205821120581ℎis a constant
Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and
Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860
119901= 120585119868
2
where 1198682is identity matrix Finally the proof is complete by
Lemma 20
Now we will give some examples of 1198661-invariant surface
in H3 Let the Poincare ball model of hyperbolic space begiven by
B3
= (1199091 119909
2 119909
3) isin R
3|
3
sum
119894=1
1199092
119894lt 1 (97)
with the hyperbolic metric 1198891199042
= 4(1198891199092
1+119889119909
2
2+119889119909
2
3)(1minus119909
2
1minus
1199092
2minus 119909
2
3) Then it is well known that stereographic projection
of H3 is given by
Φ H3
997888rarr B3
Φ (1199090 119909
1 119909
2 119909
3) = (
1199091
1 + 1199090
1199092
1 + 1199090
1199093
1 + 1199090
)
(98)
We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3
Example 34 The 1198661-invariant surface which is generated
from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ
= radic2 is drawn in Figure 1(a)
Example 35 Let the profile curve of 119872 be given by
120572 (119905) = (radic2 +1
2(minus1 + radic2) 119905
2 0 1 minus
1
2(minus1 + radic2) 119905
2 119905)
(99)
such that hyperbolic curvature 120581ℎ
= 1 Then 119872 is hyperbolicflat 119866
1-invariant surface which is generated from horocycle
in H3 (see Figure 1(b))
Example 36 The 1198661-invariant surface which is generated
from
120572 (119905) = (1
3(minus1 + 4 cosh
radic3119905
2) 0
2
3(minus1 + cosh
radic3119905
2)
2
radic3sinh
radic3119905
2)
(100)
with hyperbolic curvature 120581ℎ
= 12 is drawn in Figure 1(c)
Example 37 Let the profile curve of 119872 be given by
120572 (119905)
= (2 cosh 119905
radic3minus sinh 119905
radic3 0 cosh 119905
radic3minus 2 sinh 119905
radic3 radic2)
(101)
such that hyperbolic curvature 120581ℎ
= radic2radic3 Then 119872 istotally umbilical 119866
1-invariant surface with 120585
2= 23 in H3
(see Figure 1(d))
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
12 Journal of Applied Mathematics
(a) (b)
(c) (d)
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993
2 and also 1205721(119905) gt 0 for all 119905 isin 119868 since
120572(119868) isin H3 If the unit normal vector of 119872 inH3 is denoted by120578(119904 119905) = 120596(119904 119905)120596(119904 119905) then we have that
120578 (119904 119905) = ((12057231205721015840
4minus 120572
1015840
31205724) cosh 119904 (120572
31205721015840
4minus 120572
1015840
31205724) sinh 119904
12057211205721015840
4minus 120572
1015840
11205724 120572
1015840
11205723
minus 12057211205721015840
3)
(67)
and it is clear that
⟨120578 120597119904⟩ equiv ⟨120578 120597
119905⟩ equiv 0 (68)
for all (119904 119905) isin 119880 From (59) and (60) the matrix of de Sittershape operator of119872with respect to orthogonal tangent frame120595 ofX(119872) is A
119901= [
119886 119888
119887 119889] at any x(119904 119905) = 119901 isin 119872 where
119886 =
⟨minus119863120597119904
120578 120597119904⟩
⟨120597119904 120597
119904⟩
=
⟨119863119909119904
119909119904 120578⟩
⟨119909119904 119909
119904⟩
119887 =
⟨minus119863120597119904
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119904
119909119905 120578⟩
⟨119909119905 119909
119905⟩
119888 = 119887
119889 =
⟨minus119863120597119905
120578 120597119905⟩
⟨120597119905 120597
119905⟩
=
⟨119863119909119905
119909119905 120578⟩
⟨119909119905 119909
119905⟩
(69)
After basic calculations the de Sitter principal curvatures of119872 are
1198961
=1205721015840
31205724
minus 12057231205721015840
4
1205721
(70)
1198962
= 12057210158401015840
1(120572
1015840
31205724
minus 12057231205721015840
4) + 120572
10158401015840
3(120572
11205721015840
4minus 120572
1015840
11205724)
+ 12057210158401015840
4(120572
1015840
11205723
minus 12057211205721015840
3)
(71)
Let Frenet-Serret apparatus of 119872 be denoted bytn e 120581
ℎ 120591
ℎ in H3
Proposition 22 Thebinormal vector of the profile curve of1198661-
invariant surface 119872 is constant in H3
Proof Let 120572 be the profile curve of 119872 By (61) we know that120572 is a hyperbolic plane curve that is 120591
ℎ= 0 Moreover by
(52) and (59) we have that 119863te = minus120591ℎn = 0 Hence by (52)
119863te = 0 This completes the proof
From now on let the binormal vector of the profilecurve of 119872 be given by e = (120582
0 120582
1 120582
2 120582
3) such that 120582
119894
is scalar for 119894 = 0 1 2 3 Now we will give the importantrelation between the one of de Sitter principal curvatures andhyperbolic curvature of the profile curve of 119872
Theorem 23 Let119872 be1198661-invariant surface inH3 Then 119896
2=
1205821120581ℎ
Proof Let the binormal vector of the profile curve of 119872 bedenoted by e In Section 3 from the definition of Serret-Frenet vectors we have that 120581
ℎe = 120572 and 120572
1015840and 120572
10158401015840 Also by (50)and (71) we obtain that 120572 and 120572
1015840and 120572
10158401015840= (0 119896
2 0 0) Thus it
follows that 120581ℎe = (0 119896
2 0 0) For this reason we have that
1198962
= 1205821120581ℎ
As a result of Theorem 23 the de Sitter Gauss curvatureand the de Sitter mean curvature of 119872 = x(119880) are
119870119889
(119901) =1205721015840
31205724
minus 12057231205721015840
4
1205721
1205821120581ℎ (72)
119867119889
(119901) =
(1205721015840
31205724
minus 12057231205721015840
4) + 120582
1120581ℎ1205721
21205721
(73)
Journal of Applied Mathematics 9
at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are
119870plusmn
ℎ(119901) =
(1205721
∓ (1205721015840
31205724
minus 12057231205721015840
4)) (1 ∓ 120582
1120581ℎ)
1205721
(74)
119867plusmn
ℎ(119901) =
1205721
(minus2 plusmn 1205821120581ℎ) plusmn (120572
1015840
31205724
minus 12057231205721015840
4)
21205721
(75)
at any x(119904 119905) = 119901 respectively
Proposition 24 Let 120572 119868 rarr 11986323
sub 1198673 120572(119905) =
(1205721(119905) 0 120572
3(119905) 120572
4(119905)) be unit speed regular profile curve of119866
1-
invariant surface 119872 Then its components are given by
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(76)
Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations
minus1205721
(119905)2
+ 1205723
(119905)2
+ 1205724
(119905)2
= minus1 (77)
minus1205721015840
1(119905)
2+ 120572
1015840
3(119905)
2+ 120572
1015840
4(119905)
2= 1 (78)
for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that
1205723 (119905) = radic120572
1 (119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
(79)
such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that
1205931015840(119905)
2=
1205721 (119905)
2minus 120572
1015840
1(119905)
2minus 1
(1205721
(119905)2
minus 1)2
(80)
Finally by (80) we have that
120593 (119905) = plusmn int
119905
0
radic1205721
(119906)2
minus 1205721015840
1(119906)
2minus 1
1205721 (119906)
2minus 1
119889119906 (81)
such that 1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1 gt 0 for all 119905 isin 119868 Without loss
of generality when we choose positive of signature of 120593 thiscompletes the proof
Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3
Now we will give some results which are obtained by (72)and (74)
Corollary 26 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205723
= 0 or 1205724
= 0 then 119872 is a part of de Sitter flatsurface
(ii) if 120581ℎ
= 0 then 119872 is a part of de Sitter flat surface
(iii) if 1205821
= 0 then 119872 is a part of de Sitter flat surface
Corollary 27 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 If 1205723
= 1205831205724such that 120583 isin R then 119872 is a
de Sitter flat surface
Theorem 28 Let 120572 be the profile curve of1198661-invariant surface
119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ
=
plusmn11205821
Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn
ℎ=
0 By (74) it follows that
1205721
(119905) ∓ (1205721015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (82)
or
1 ∓ 1205821120581ℎ
(119905) = 0 (83)
for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572
1(119905) ∓
(minusradic1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1) = 0 Hence it follows that
1205721015840
1(119905)
2+ 1 = 0 (84)
There is no real solution of (84) This means that the onlyone solution is 120581
ℎ= plusmn1120582
1by (83) On the other hand if we
assume that 120581ℎ
= plusmn11205821 then the proof is clear
Corollary 29 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205821
= 1 and 120581ℎ
= 1 then 119872 is 119870+
ℎ-flat surface which
is generated from horocyle
(ii) if 1205821
= minus1 and 120581ℎ
= 1 then 119872 is 119870minus
ℎ-flat surface which
is generated from horocyle
Now we will give theorem and corollaries for 1198661-
invariant surface which satisfy minimal condition in H3 by(73) and (75)
Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 inH3 Then 119872 is de
10 Journal of Applied Mathematics
Sitter minimal surface if and only if the parametrization of 120572 isgiven by
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(85)
with the condition 120581ℎ
isin (minus11205821 1120582
1) such that 119888
1is an
arbitrary constant
Proof Suppose that 119872 is de Sitter minimal surface that is119867
119889equiv 0 By (73) it follows that
(1205721015840
3(119905) 120572
4 (119905) minus 1205723 (119905) 120572
1015840
4(119905)) + 120582
1120581ℎ1205721 (119905) = 0 (86)
for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation
1205721015840
1(119905)
2minus (1 minus 120582
2
11205812
ℎ) 120572
1 (119905)2
+ 1 = 0 (87)
There exists only one real solution of (87) under the condition1 minus 120582
2
11205812
ℎgt 0 Moreover 120582
1must not be zero by Corollary 26
So that we obtain
120581ℎ
isin (minus1
1205821
1
1205821
) (88)
Hence the solution is
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
(89)
where 1198881is an arbitrary constant under the condition (88)
Finally the parametrization of 120572 is given explicitly byProposition 24
On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867
119889equiv 0
Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 in H3 Then 119872 is
hyperbolic minimal surface if and only if the parametrizationof 120572 is given by
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721
(119905)2
minus 1205721015840
1(119905)
2minus 1
1205721 (119905)
2minus 1
119889119906
(90)
with the condition 1minus(2∓1205821120581ℎ)2
gt 0 such that 1198881is an arbitrary
constant
Proof Suppose that 119872 is hyperbolic minimal surface that is119867
plusmn
ℎequiv 0 By (75) it follows that
(minus2 plusmn 1205821120581ℎ) 120572
1(119905) plusmn (120572
1015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (91)
If Proposition 24 is applied to (91) we have that
1205721015840
1(119905)
2+ ((minus2 plusmn 120582
1120581ℎ)2
minus 1) 1205721
(119905)2
+ 1 = 0 (92)
There exists only one real solution of (92) under the condition
(minus2 plusmn 1205821120581ℎ)2
minus 1 lt 0 (93)
Thus the solution is
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
(94)
where 1198881is an arbitrary constant under the condition (93)
However 1205821must not be zero by Corollary 26 So that if 119872
is119867+
ℎ-minimal surface (119867minus
ℎ-minimal surface) thenwe obtain
120581ℎ
isin (11205821 3120582
1) (120581
ℎisin (minus3120582
1 minus1120582
1)) by (93) Finally the
parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of
119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867
plusmn
ℎequiv 0
Now we will give classification theorem for totally umbil-ical 119866
1-invariant surfaces
Journal of Applied Mathematics 11
Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-
invariant surface 119872 inH3 Then the hyperbolic curvature of 120572
is constant
Proof Let 119872 be totally umbilical 1198661-invariant surface By
Theorem 23 and (70) we may assume that 1198961(119901) = 119896
2(119901) =
1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following
Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905
= x119905119904in
(96) then it follows that 12058211205811015840
ℎ(119905) = 0 for all 119905 isin 119868 Moreover
1205821must not be zero by Corollary 26Thus 120581
ℎis constant
Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following
classification
(1) Supposing that 1205852
= 1
(a) if 120585 = 0 and 1205852
lt 1 then 119872 is a part of anequidistant surface
(b) if 120585 = 0 and 1205852
gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane
(2) If 1205852
= 1 then 119872 is a part of horosphere
where 120585 = 1205821120581ℎis a constant
Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and
Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860
119901= 120585119868
2
where 1198682is identity matrix Finally the proof is complete by
Lemma 20
Now we will give some examples of 1198661-invariant surface
in H3 Let the Poincare ball model of hyperbolic space begiven by
B3
= (1199091 119909
2 119909
3) isin R
3|
3
sum
119894=1
1199092
119894lt 1 (97)
with the hyperbolic metric 1198891199042
= 4(1198891199092
1+119889119909
2
2+119889119909
2
3)(1minus119909
2
1minus
1199092
2minus 119909
2
3) Then it is well known that stereographic projection
of H3 is given by
Φ H3
997888rarr B3
Φ (1199090 119909
1 119909
2 119909
3) = (
1199091
1 + 1199090
1199092
1 + 1199090
1199093
1 + 1199090
)
(98)
We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3
Example 34 The 1198661-invariant surface which is generated
from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ
= radic2 is drawn in Figure 1(a)
Example 35 Let the profile curve of 119872 be given by
120572 (119905) = (radic2 +1
2(minus1 + radic2) 119905
2 0 1 minus
1
2(minus1 + radic2) 119905
2 119905)
(99)
such that hyperbolic curvature 120581ℎ
= 1 Then 119872 is hyperbolicflat 119866
1-invariant surface which is generated from horocycle
in H3 (see Figure 1(b))
Example 36 The 1198661-invariant surface which is generated
from
120572 (119905) = (1
3(minus1 + 4 cosh
radic3119905
2) 0
2
3(minus1 + cosh
radic3119905
2)
2
radic3sinh
radic3119905
2)
(100)
with hyperbolic curvature 120581ℎ
= 12 is drawn in Figure 1(c)
Example 37 Let the profile curve of 119872 be given by
120572 (119905)
= (2 cosh 119905
radic3minus sinh 119905
radic3 0 cosh 119905
radic3minus 2 sinh 119905
radic3 radic2)
(101)
such that hyperbolic curvature 120581ℎ
= radic2radic3 Then 119872 istotally umbilical 119866
1-invariant surface with 120585
2= 23 in H3
(see Figure 1(d))
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
12 Journal of Applied Mathematics
(a) (b)
(c) (d)
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993
at any x(119904 119905) = 119901 respectively Moreover if we apply (58)then the hyperbolicGauss curvature and the hyperbolicmeancurvature of 119872 are
119870plusmn
ℎ(119901) =
(1205721
∓ (1205721015840
31205724
minus 12057231205721015840
4)) (1 ∓ 120582
1120581ℎ)
1205721
(74)
119867plusmn
ℎ(119901) =
1205721
(minus2 plusmn 1205821120581ℎ) plusmn (120572
1015840
31205724
minus 12057231205721015840
4)
21205721
(75)
at any x(119904 119905) = 119901 respectively
Proposition 24 Let 120572 119868 rarr 11986323
sub 1198673 120572(119905) =
(1205721(119905) 0 120572
3(119905) 120572
4(119905)) be unit speed regular profile curve of119866
1-
invariant surface 119872 Then its components are given by
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(76)
Proof Suppose that the profile curve of 119872 is unit speed andregular So that it satisfies the following equations
minus1205721
(119905)2
+ 1205723
(119905)2
+ 1205724
(119905)2
= minus1 (77)
minus1205721015840
1(119905)
2+ 120572
1015840
3(119905)
2+ 120572
1015840
4(119905)
2= 1 (78)
for all 119905 isin 119868 By (77) and 1205721(119905) ge 1 we have that
1205723 (119905) = radic120572
1 (119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
(79)
such that 120593 is a differentiable function Moreover by (78) and(79) we obtain that
1205931015840(119905)
2=
1205721 (119905)
2minus 120572
1015840
1(119905)
2minus 1
(1205721
(119905)2
minus 1)2
(80)
Finally by (80) we have that
120593 (119905) = plusmn int
119905
0
radic1205721
(119906)2
minus 1205721015840
1(119906)
2minus 1
1205721 (119906)
2minus 1
119889119906 (81)
such that 1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1 gt 0 for all 119905 isin 119868 Without loss
of generality when we choose positive of signature of 120593 thiscompletes the proof
Remark 25 If 119872 is a de Sitter flat surface in H3 then we saythat 119872 is an H-plane in H3
Now we will give some results which are obtained by (72)and (74)
Corollary 26 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205723
= 0 or 1205724
= 0 then 119872 is a part of de Sitter flatsurface
(ii) if 120581ℎ
= 0 then 119872 is a part of de Sitter flat surface
(iii) if 1205821
= 0 then 119872 is a part of de Sitter flat surface
Corollary 27 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 If 1205723
= 1205831205724such that 120583 isin R then 119872 is a
de Sitter flat surface
Theorem 28 Let 120572 be the profile curve of1198661-invariant surface
119872 in H3 Then 119872 is hyperbolic flat surface if and only if 120581ℎ
=
plusmn11205821
Proof Suppose that119872 is hyperbolic flat surface that is119870plusmn
ℎ=
0 By (74) it follows that
1205721
(119905) ∓ (1205721015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (82)
or
1 ∓ 1205821120581ℎ
(119905) = 0 (83)
for all 119905 isin 119868 Firstly let us find solution of (82) IfProposition 24 is applied to (82) we have that 120572
1(119905) ∓
(minusradic1205721(119905)
2minus 120572
1015840
1(119905)
2minus 1) = 0 Hence it follows that
1205721015840
1(119905)
2+ 1 = 0 (84)
There is no real solution of (84) This means that the onlyone solution is 120581
ℎ= plusmn1120582
1by (83) On the other hand if we
assume that 120581ℎ
= plusmn11205821 then the proof is clear
Corollary 29 Let 120572 be the profile curve of 1198661-invariant
surface 119872 in H3 Then
(i) if 1205821
= 1 and 120581ℎ
= 1 then 119872 is 119870+
ℎ-flat surface which
is generated from horocyle
(ii) if 1205821
= minus1 and 120581ℎ
= 1 then 119872 is 119870minus
ℎ-flat surface which
is generated from horocyle
Now we will give theorem and corollaries for 1198661-
invariant surface which satisfy minimal condition in H3 by(73) and (75)
Theorem 30 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 inH3 Then 119872 is de
10 Journal of Applied Mathematics
Sitter minimal surface if and only if the parametrization of 120572 isgiven by
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(85)
with the condition 120581ℎ
isin (minus11205821 1120582
1) such that 119888
1is an
arbitrary constant
Proof Suppose that 119872 is de Sitter minimal surface that is119867
119889equiv 0 By (73) it follows that
(1205721015840
3(119905) 120572
4 (119905) minus 1205723 (119905) 120572
1015840
4(119905)) + 120582
1120581ℎ1205721 (119905) = 0 (86)
for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation
1205721015840
1(119905)
2minus (1 minus 120582
2
11205812
ℎ) 120572
1 (119905)2
+ 1 = 0 (87)
There exists only one real solution of (87) under the condition1 minus 120582
2
11205812
ℎgt 0 Moreover 120582
1must not be zero by Corollary 26
So that we obtain
120581ℎ
isin (minus1
1205821
1
1205821
) (88)
Hence the solution is
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
(89)
where 1198881is an arbitrary constant under the condition (88)
Finally the parametrization of 120572 is given explicitly byProposition 24
On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867
119889equiv 0
Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 in H3 Then 119872 is
hyperbolic minimal surface if and only if the parametrizationof 120572 is given by
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721
(119905)2
minus 1205721015840
1(119905)
2minus 1
1205721 (119905)
2minus 1
119889119906
(90)
with the condition 1minus(2∓1205821120581ℎ)2
gt 0 such that 1198881is an arbitrary
constant
Proof Suppose that 119872 is hyperbolic minimal surface that is119867
plusmn
ℎequiv 0 By (75) it follows that
(minus2 plusmn 1205821120581ℎ) 120572
1(119905) plusmn (120572
1015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (91)
If Proposition 24 is applied to (91) we have that
1205721015840
1(119905)
2+ ((minus2 plusmn 120582
1120581ℎ)2
minus 1) 1205721
(119905)2
+ 1 = 0 (92)
There exists only one real solution of (92) under the condition
(minus2 plusmn 1205821120581ℎ)2
minus 1 lt 0 (93)
Thus the solution is
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
(94)
where 1198881is an arbitrary constant under the condition (93)
However 1205821must not be zero by Corollary 26 So that if 119872
is119867+
ℎ-minimal surface (119867minus
ℎ-minimal surface) thenwe obtain
120581ℎ
isin (11205821 3120582
1) (120581
ℎisin (minus3120582
1 minus1120582
1)) by (93) Finally the
parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of
119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867
plusmn
ℎequiv 0
Now we will give classification theorem for totally umbil-ical 119866
1-invariant surfaces
Journal of Applied Mathematics 11
Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-
invariant surface 119872 inH3 Then the hyperbolic curvature of 120572
is constant
Proof Let 119872 be totally umbilical 1198661-invariant surface By
Theorem 23 and (70) we may assume that 1198961(119901) = 119896
2(119901) =
1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following
Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905
= x119905119904in
(96) then it follows that 12058211205811015840
ℎ(119905) = 0 for all 119905 isin 119868 Moreover
1205821must not be zero by Corollary 26Thus 120581
ℎis constant
Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following
classification
(1) Supposing that 1205852
= 1
(a) if 120585 = 0 and 1205852
lt 1 then 119872 is a part of anequidistant surface
(b) if 120585 = 0 and 1205852
gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane
(2) If 1205852
= 1 then 119872 is a part of horosphere
where 120585 = 1205821120581ℎis a constant
Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and
Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860
119901= 120585119868
2
where 1198682is identity matrix Finally the proof is complete by
Lemma 20
Now we will give some examples of 1198661-invariant surface
in H3 Let the Poincare ball model of hyperbolic space begiven by
B3
= (1199091 119909
2 119909
3) isin R
3|
3
sum
119894=1
1199092
119894lt 1 (97)
with the hyperbolic metric 1198891199042
= 4(1198891199092
1+119889119909
2
2+119889119909
2
3)(1minus119909
2
1minus
1199092
2minus 119909
2
3) Then it is well known that stereographic projection
of H3 is given by
Φ H3
997888rarr B3
Φ (1199090 119909
1 119909
2 119909
3) = (
1199091
1 + 1199090
1199092
1 + 1199090
1199093
1 + 1199090
)
(98)
We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3
Example 34 The 1198661-invariant surface which is generated
from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ
= radic2 is drawn in Figure 1(a)
Example 35 Let the profile curve of 119872 be given by
120572 (119905) = (radic2 +1
2(minus1 + radic2) 119905
2 0 1 minus
1
2(minus1 + radic2) 119905
2 119905)
(99)
such that hyperbolic curvature 120581ℎ
= 1 Then 119872 is hyperbolicflat 119866
1-invariant surface which is generated from horocycle
in H3 (see Figure 1(b))
Example 36 The 1198661-invariant surface which is generated
from
120572 (119905) = (1
3(minus1 + 4 cosh
radic3119905
2) 0
2
3(minus1 + cosh
radic3119905
2)
2
radic3sinh
radic3119905
2)
(100)
with hyperbolic curvature 120581ℎ
= 12 is drawn in Figure 1(c)
Example 37 Let the profile curve of 119872 be given by
120572 (119905)
= (2 cosh 119905
radic3minus sinh 119905
radic3 0 cosh 119905
radic3minus 2 sinh 119905
radic3 radic2)
(101)
such that hyperbolic curvature 120581ℎ
= radic2radic3 Then 119872 istotally umbilical 119866
1-invariant surface with 120585
2= 23 in H3
(see Figure 1(d))
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
12 Journal of Applied Mathematics
(a) (b)
(c) (d)
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993
Sitter minimal surface if and only if the parametrization of 120572 isgiven by
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724 (119905) = radic120572
1 (119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721 (119906)
2minus 120572
1015840
1(119906)
2minus 1
1205721
(119906)2
minus 1119889119906
(85)
with the condition 120581ℎ
isin (minus11205821 1120582
1) such that 119888
1is an
arbitrary constant
Proof Suppose that 119872 is de Sitter minimal surface that is119867
119889equiv 0 By (73) it follows that
(1205721015840
3(119905) 120572
4 (119905) minus 1205723 (119905) 120572
1015840
4(119905)) + 120582
1120581ℎ1205721 (119905) = 0 (86)
for all 119905 isin 119868 By using Proposition 24 we have the followingdifferential equation
1205721015840
1(119905)
2minus (1 minus 120582
2
11205812
ℎ) 120572
1 (119905)2
+ 1 = 0 (87)
There exists only one real solution of (87) under the condition1 minus 120582
2
11205812
ℎgt 0 Moreover 120582
1must not be zero by Corollary 26
So that we obtain
120581ℎ
isin (minus1
1205821
1
1205821
) (88)
Hence the solution is
1205721
(119905) = ((minus2 + 1205822
11205812
ℎ) cosh ((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ)
minus1205822
11205812
ℎsinh((119905 + 119888
1) radic1 minus 120582
2
11205812
ℎ))
times (minus2 (1 minus 1205822
11205812
ℎ))
minus1
(89)
where 1198881is an arbitrary constant under the condition (88)
Finally the parametrization of 120572 is given explicitly byProposition 24
On the other hand let the parametrization of profile curve120572 of 119872 be given by (85) under the condition (87) Then itsatisfies (86) It means that 119867
119889equiv 0
Theorem 31 Let 120572 be the profile curve with constant hyper-bolic curvature of 119866
1-invariant surface 119872 in H3 Then 119872 is
hyperbolic minimal surface if and only if the parametrizationof 120572 is given by
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
1205723
(119905) = radic1205721
(119905)2
minus 1 cos120593 (119905)
1205724
(119905) = radic1205721
(119905)2
minus 1 sin120593 (119905)
120593 (119905) = int
119905
0
radic1205721
(119905)2
minus 1205721015840
1(119905)
2minus 1
1205721 (119905)
2minus 1
119889119906
(90)
with the condition 1minus(2∓1205821120581ℎ)2
gt 0 such that 1198881is an arbitrary
constant
Proof Suppose that 119872 is hyperbolic minimal surface that is119867
plusmn
ℎequiv 0 By (75) it follows that
(minus2 plusmn 1205821120581ℎ) 120572
1(119905) plusmn (120572
1015840
3(119905) 120572
4(119905) minus 120572
3(119905) 120572
1015840
4(119905)) = 0 (91)
If Proposition 24 is applied to (91) we have that
1205721015840
1(119905)
2+ ((minus2 plusmn 120582
1120581ℎ)2
minus 1) 1205721
(119905)2
+ 1 = 0 (92)
There exists only one real solution of (92) under the condition
(minus2 plusmn 1205821120581ℎ)2
minus 1 lt 0 (93)
Thus the solution is
1205721
(119905)
= (((2 ∓ 1205821120581ℎ)2
minus 2) cosh ((119905 + 1198881) radic1 minus (2 ∓ 120582
1120581ℎ)2)
+ (2 minus 1205821120581ℎ)2 sinh((119905 + 119888
1) radic1 minus (2 ∓ 120582
1120581ℎ)2))
times (2 (minus1 + (2 ∓ 1205821120581ℎ)2))
minus1
(94)
where 1198881is an arbitrary constant under the condition (93)
However 1205821must not be zero by Corollary 26 So that if 119872
is119867+
ℎ-minimal surface (119867minus
ℎ-minimal surface) thenwe obtain
120581ℎ
isin (11205821 3120582
1) (120581
ℎisin (minus3120582
1 minus1120582
1)) by (93) Finally the
parametrization of 120572 is given explicitly by Proposition 24Conversely let the parametrization of profile curve 120572 of
119872 be given by (90) under the condition (93)Then it satisfies(91) It means that 119867
plusmn
ℎequiv 0
Now we will give classification theorem for totally umbil-ical 119866
1-invariant surfaces
Journal of Applied Mathematics 11
Theorem 32 Let 120572 be the profile curve of totally umbilical 1198661-
invariant surface 119872 inH3 Then the hyperbolic curvature of 120572
is constant
Proof Let 119872 be totally umbilical 1198661-invariant surface By
Theorem 23 and (70) we may assume that 1198961(119901) = 119896
2(119901) =
1205821120581ℎ(119905) for all x(119904 119905) = 119901 isin 119872 Then we have the following
Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905
= x119905119904in
(96) then it follows that 12058211205811015840
ℎ(119905) = 0 for all 119905 isin 119868 Moreover
1205821must not be zero by Corollary 26Thus 120581
ℎis constant
Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following
classification
(1) Supposing that 1205852
= 1
(a) if 120585 = 0 and 1205852
lt 1 then 119872 is a part of anequidistant surface
(b) if 120585 = 0 and 1205852
gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane
(2) If 1205852
= 1 then 119872 is a part of horosphere
where 120585 = 1205821120581ℎis a constant
Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and
Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860
119901= 120585119868
2
where 1198682is identity matrix Finally the proof is complete by
Lemma 20
Now we will give some examples of 1198661-invariant surface
in H3 Let the Poincare ball model of hyperbolic space begiven by
B3
= (1199091 119909
2 119909
3) isin R
3|
3
sum
119894=1
1199092
119894lt 1 (97)
with the hyperbolic metric 1198891199042
= 4(1198891199092
1+119889119909
2
2+119889119909
2
3)(1minus119909
2
1minus
1199092
2minus 119909
2
3) Then it is well known that stereographic projection
of H3 is given by
Φ H3
997888rarr B3
Φ (1199090 119909
1 119909
2 119909
3) = (
1199091
1 + 1199090
1199092
1 + 1199090
1199093
1 + 1199090
)
(98)
We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3
Example 34 The 1198661-invariant surface which is generated
from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ
= radic2 is drawn in Figure 1(a)
Example 35 Let the profile curve of 119872 be given by
120572 (119905) = (radic2 +1
2(minus1 + radic2) 119905
2 0 1 minus
1
2(minus1 + radic2) 119905
2 119905)
(99)
such that hyperbolic curvature 120581ℎ
= 1 Then 119872 is hyperbolicflat 119866
1-invariant surface which is generated from horocycle
in H3 (see Figure 1(b))
Example 36 The 1198661-invariant surface which is generated
from
120572 (119905) = (1
3(minus1 + 4 cosh
radic3119905
2) 0
2
3(minus1 + cosh
radic3119905
2)
2
radic3sinh
radic3119905
2)
(100)
with hyperbolic curvature 120581ℎ
= 12 is drawn in Figure 1(c)
Example 37 Let the profile curve of 119872 be given by
120572 (119905)
= (2 cosh 119905
radic3minus sinh 119905
radic3 0 cosh 119905
radic3minus 2 sinh 119905
radic3 radic2)
(101)
such that hyperbolic curvature 120581ℎ
= radic2radic3 Then 119872 istotally umbilical 119866
1-invariant surface with 120585
2= 23 in H3
(see Figure 1(d))
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
12 Journal of Applied Mathematics
(a) (b)
(c) (d)
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993
Also if we use the equations 119863x119905120578119904 = 119863x119904120578119905 and x119904119905
= x119905119904in
(96) then it follows that 12058211205811015840
ℎ(119905) = 0 for all 119905 isin 119868 Moreover
1205821must not be zero by Corollary 26Thus 120581
ℎis constant
Corollary 33 Let 120572 be the profile curve of totally umbilical1198661-invariant surface 119872 in H3 Then we have the following
classification
(1) Supposing that 1205852
= 1
(a) if 120585 = 0 and 1205852
lt 1 then 119872 is a part of anequidistant surface
(b) if 120585 = 0 and 1205852
gt 1 then 119872 is a part of a sphere(c) if 120585 = 0 then 119872 is a part of a H-plane
(2) If 1205852
= 1 then 119872 is a part of horosphere
where 120585 = 1205821120581ℎis a constant
Proof We suppose that 120585 = 1205821120581ℎ By Proposition 22 and
Theorem 32we have that 120585 is constantMoreover 120585 is de Sitterprincipal curvature of 119872 by Theorem 23 Since 119872 is totallyumbilical surface de Sitter shape operator of 119872 is 119860
119901= 120585119868
2
where 1198682is identity matrix Finally the proof is complete by
Lemma 20
Now we will give some examples of 1198661-invariant surface
in H3 Let the Poincare ball model of hyperbolic space begiven by
B3
= (1199091 119909
2 119909
3) isin R
3|
3
sum
119894=1
1199092
119894lt 1 (97)
with the hyperbolic metric 1198891199042
= 4(1198891199092
1+119889119909
2
2+119889119909
2
3)(1minus119909
2
1minus
1199092
2minus 119909
2
3) Then it is well known that stereographic projection
of H3 is given by
Φ H3
997888rarr B3
Φ (1199090 119909
1 119909
2 119909
3) = (
1199091
1 + 1199090
1199092
1 + 1199090
1199093
1 + 1199090
)
(98)
We can draw the pictures of surface x(119880) = 119872 by usingstereographic projection Φ That is Φ(119872) sub B3 such thatx(119880) = 119872 sub H3
Example 34 The 1198661-invariant surface which is generated
from 120572(119905) = (radic2 0 cos 119905 sin 119905) with hyperbolic curvature120581ℎ
= radic2 is drawn in Figure 1(a)
Example 35 Let the profile curve of 119872 be given by
120572 (119905) = (radic2 +1
2(minus1 + radic2) 119905
2 0 1 minus
1
2(minus1 + radic2) 119905
2 119905)
(99)
such that hyperbolic curvature 120581ℎ
= 1 Then 119872 is hyperbolicflat 119866
1-invariant surface which is generated from horocycle
in H3 (see Figure 1(b))
Example 36 The 1198661-invariant surface which is generated
from
120572 (119905) = (1
3(minus1 + 4 cosh
radic3119905
2) 0
2
3(minus1 + cosh
radic3119905
2)
2
radic3sinh
radic3119905
2)
(100)
with hyperbolic curvature 120581ℎ
= 12 is drawn in Figure 1(c)
Example 37 Let the profile curve of 119872 be given by
120572 (119905)
= (2 cosh 119905
radic3minus sinh 119905
radic3 0 cosh 119905
radic3minus 2 sinh 119905
radic3 radic2)
(101)
such that hyperbolic curvature 120581ℎ
= radic2radic3 Then 119872 istotally umbilical 119866
1-invariant surface with 120585
2= 23 in H3
(see Figure 1(d))
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
12 Journal of Applied Mathematics
(a) (b)
(c) (d)
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993
Figure 1 Examples of some 1198661-invariant surfaces
Acknowledgment
The research of Mahmut Mak was partially supported bythe grant of Ahi Evran University Scientific Research Project(PYOFEN400914014)
References
[1] J W Cannon W J Floyd R Kenyon and R W ParryldquoHyperbolic geometryrdquo MSRI Publications vol 31 pp 69ndash721997
[2] J G Ratcliffe Foundations of Hyperbolic Manifolds vol 149Springer New York NY USA 2nd edition 2006
[3] M do Carmo and M Dajczer ldquoRotation hypersurfaces inspaces of constant curvaturerdquo Transactions of the AmericanMathematical Society vol 277 no 2 pp 685ndash709 1983
[4] H Mori ldquoStable complete constant mean curvature surfaces in1198773 and 119867
3rdquo Transactions of the AmericanMathematical Societyvol 278 no 2 pp 671ndash687 1983
[5] H Liu and G Liu ldquoHyperbolic rotation surfaces of constantmean curvature in 3-de Sitter spacerdquo Bulletin of the Belgian
Mathematical Society Simon Stevin vol 7 no 3 pp 455ndash4662000
[6] U Dursun ldquoRotation hypersurfaces in Lorentz-Minkowskispace with constant mean curvaturerdquo Taiwanese Journal ofMathematics vol 14 no 2 pp 685ndash705 2010
[7] S Izumiya D Pei and T Sano ldquoSingularities of hyperbolicGauss mapsrdquo Proceedings of the London Mathematical Societyvol 86 no 2 pp 485ndash512 2003
[8] S Izumiya K Saji and M Takahashi ldquoHorospherical flatsurfaces in hyperbolic 3-spacerdquo Journal of the MathematicalSociety of Japan vol 62 no 3 pp 789ndash849 2010
[9] W F Reynolds ldquoHyperbolic geometry on a hyperboloidrdquo TheAmerican Mathematical Monthly vol 100 no 5 pp 442ndash4551993