Mathematical Combinatorics (International Book Series), vol. I, 2020

2020·2 MB·English

by  Mao, Linfan

ISBN 978-1-59973-654-9
VOLUME 1, 2020
MATHEMATICAL COMBINATORICS
(INTERNATIONAL BOOK SERIES)
Edited By Linfan MAO
THE MADIS OF CHINESE ACADEMY OF SCIENCES AND
ACADEMY OF MATHEMATICAL COMBINATORICS & APPLICATIONS, USA
March, 2020
Vol.1, 2020 ISBN 978-1-59973-654-9
MATHEMATICAL COMBINATORICS
(INTERNATIONAL BOOK SERIES)
Edited By Linfan MAO
(www.mathcombin.com)
The Madis of Chinese Academy of Sciences and
Academy of Mathematical Combinatorics & Applications, USA
March, 2020
Aims and Scope: The mathematical combinatorics is a subject that applying combinatorial
notiontoallmathematicsandallsciencesforunderstandingtherealityofthingsintheuniverse,
motivatedbyCC ConjectureofDr.LinfanMAOonmathematicalsciences. TheInternational
J.Mathematical Combinatorics (ISSN 1937-1055) is a fully refereed international journal,
sponsored by the MADIS of Chinese Academy of Sciences and published in USA quarterly,
which publishes original research papers and survey articles in all aspects of mathematical
combinatorics, Smarandache multi-spaces, Smarandache geometries, non-Euclidean geometry,
topology and their applications to other sciences. Topics in detail to be covered are:
Mathematical combinatorics;
Smarandachemulti-spacesandSmarandachegeometrieswithapplicationstoothersciences;
Topological graphs; Algebraic graphs; Random graphs; Combinatorial maps; Graph and
map enumeration; Combinatorial designs; Combinatorial enumeration;
Differential Geometry; Geometry on manifolds; Low Dimensional Topology; Differential
Topology; Topology of Manifolds;
Geometrical aspects of Mathematical Physics and Relations with Manifold Topology;
Mathematical theory on gravitational fields and parallel universes;
Applications of Combinatorics to mathematics and theoretical physics.
Generally,papersonapplicationsofcombinatoricstoothermathematicsandothersciences
are welcome by this journal.
It is also available from the below international databases:
Serials Group/Editorial Department of EBSCO Publishing
10 Estes St. Ipswich, MA 01938-2106, USA
Tel.: [phone], Ext. 2262 Fax: [phone]
http://www.ebsco.com/home/printsubs/priceproj.asp
and
Gale Directory of Publications and Broadcast Media, Gale, a part of Cengage Learning
27500 Drake Rd. Farmington Hills, MI 48331-3535, USA
Tel.: [phone], ext. 1326; 1-800-347-GALE Fax: [phone]
http://www.gale.com
IndexingandReviews: MathematicalReviews(USA),ZentralblattMath(Germany),Refer-
ativnyi Zhurnal (Russia), Mathematika (Russia), Directory of Open Access (DoAJ), EBSCO
(USA), International Scientific Indexing (ISI, impact factor 2.012), Institute for Scientific In-
formation (PA, USA), Library of Congress Subject Headings (USA).
Subscription A subscription can be ordered by an email directly to
Linfan Mao
The Editor-in-Chief of International Journal of Mathematical Combinatorics
Chinese Academy of Mathematics and System Science Beijing, 100190, P.R.China, and also the
President of Academy of Mathematical Combinatorics & Applications (AMCA), Colorado, USA
Email: [email protected]
Price: US$48.00
Editorial Board (4th)
Editor-in-Chief
Linfan MAO Shaofei Du
ChineseAcademyofMathematicsandSystem Capital Normal University, P.R.China
Science, P.R.China Email: [email protected]
and
Xiaodong Hu
Academy of Mathematical Combinatorics &
ChineseAcademyofMathematicsandSystem
Applications, Colorado, USA
Science, P.R.China
Email: [email protected]
Email: [email protected]
Deputy Editor-in-Chief
Yuanqiu Huang
Hunan Normal University, P.R.China
Guohua Song Email: [email protected]
Beijing University of Civil Engineering and
H.Iseri
Architecture, P.R.China
Mansfield University, USA
Email: [email protected]
Email: hiseri@mnsfld.edu
Editors Xueliang Li
Nankai University, P.R.China
Arindam Bhattacharyya Email: [email protected]
Jadavpur University, India Guodong Liu
Email: [email protected] Huizhou University
Said Broumi Email: [email protected]
Hassan II University Mohammedia W.B.Vasantha Kandasamy
Hay El Baraka Ben M’sik Casablanca Indian Institute of Technology, India
B.P.7951 Morocco Email: [email protected]
Junliang Cai Ion Patrascu
Beijing Normal University, P.R.China Fratii Buzesti National College
Email: [email protected] Craiova Romania
Yanxun Chang Han Ren
Beijing Jiaotong University, P.R.China East China Normal University, P.R.China
Email: [email protected] Email: [email protected]
Jingan Cui Ovidiu-Ilie Sandru
Beijing University of Civil Engineering and Politechnica University of Bucharest
Architecture, P.R.China Romania
Email: [email protected]
ii MathematicalCombinatorics(InternationalBookSeries)
Mingyao Xu
Peking University, P.R.China
Email: [email protected]
Guiying Yan
ChineseAcademyofMathematicsandSystem
Science, P.R.China
Email: [email protected]
Y. Zhang
Department of Computer Science
Georgia State University, Atlanta, USA
Famous Words:
We know nothing of what will happen in future, but by the analogy of past experience.
By Abraham Lincoln, an American president
Math.Combin.Book Ser. Vol.1(2020), 01-09
On k-Type Slant Helices Due to Bishop Frame in
Euclidean 4-Space E4
Yasin U¨nlu¨tu¨rk1, Hatice Tozak2 and Cumali Ekici3
1. DepartmentofMathematics,FacultyofArtandScience,KırklareliUniversity,39100,Kırklareli,Turkey
2. Du¨ndarUc¸arVocationalandTechnicalAnatolianHighSchool,34212,I˙stanbul,Turkey
3. Eski¸sehirOsmangaziUniversity,FacultyofScienceandLetters,DepartmentofMathematicsand
ComputerScience,TR-26040,Eskisehir,Turkey
E-mail: [email protected],[email protected],[email protected]
Abstract: In this work, we study k−type (k ∈ {0,1,2,3}) slant helices with non-zero
BishopcurvaturefunctionsduetoBishopframeinE4. Generalhelixisa0−typeslanthelix
within the notation of this study. We characterize all of slant helices in terms of Bishop
curvatures in E4.
Key Words: Bishop frame, regular curves, general helix, slant helix, Euclidean 4-space.
AMS(2010): 53A04.
§1. Introduction
Inthelocaldifferentialgeometryofspacecurves,itiswell-knownthatageneralhelixisacurve
whose tangent makes a constant angle with a non-zero constant vector field (the axis of the
helix). Moreover,thenecessaryandsufficientconditionforacurveisageneralhelixifandonly
iftheratioofthecurvatureandthetorsionofthatcurveisconstant. Aslanthelixisdefinedas
acurvewhoseprincipalnormalvectormakesaconstantanglewithafixeddirectionbyIzumiya
and Takeuchi in E3 [7]. Some characterizations of a slant helix are investigated in [9]. Ali and
TurguthavegeneralizedtheslanthelixtoEuclideann-spaceEn andhavegivensomeproperties
for a non-degenerate slant helix [2]. O¨ztu¨rk et.al. have considered the focal representation and
some properties of focal curves with their curvatures of k-slant helices in Em+1 [11]. Further,
some characterizations of slant helices in different spaces such as Minkowski and Galilean are
studied [12, 13, 14, 16].
Most of the study of curves are done by using Frenet-Serret frame in classical differential
geometryinEuclideanspace. In[4],BishopdefinedanalternativeoverFrenetframeforacurve
and called it Bishop frame. The advantage of Bishop frame is well-defined when the curve has
a vanishing second derivative in 3-dimensional Euclidean space E3 unlike Frenet frame. Also,
Bishop frame is used in many applications such as engineering, computer aided design, DNA
analysis etc. After defining this useful alternative frame, many studies have been done by
1ReceivedAugust23,2019,AcceptedMarch2,2020.
2 YasinU¨nlu¨tu¨rk,HaticeTozakandCumaliEkici
mathematicians using it [3, 8, 15]. O¨z¸celik et. al. have been introduced the parallel transport
frame of the curve in 4-dimensional Euclidean space E4 [10].
ThepresentstudyaimstodeterminethecharacterizeallofslanthelicesintermsofBishop
curvatures in E4 with the help of the literature.
§2. Preliminaries
Here, the basic definitions and theorems for the theory of curves in Euclidean 4-space E4 are
given for the next section (A more complete elementary treatment can be found in [5], [6]).
The standard flat metric in Euclidean 4-space E4 is given by
(cid:104) , (cid:105)=dx2+dx2+dx2+dx2,
1 2 3 4
where (x ,x ,x ,x ) is a rectangular coordinate system of Euclidean 4-space E4. The norm of
1 2 3 4
anarbitraryvectora∈E4 isgivenby(cid:107)a(cid:107)=(cid:112)(cid:104)a,a(cid:105). Thecurveαiscalledanunitspeedcurve
if a velocity vector v of α satisfies (cid:107)v(cid:107)=1. For vectors v,w ∈ E4, it is said to be orthogonal if
and only if (cid:104)v,w(cid:105)=0. Let α=α(s) be a regular curve in Euclidean 4-space E4. If the tangent
vector field of this curve forms a constant angle with a constant vector field U, then this curve
is called a general helix or an inclined curve.
Denoteby{T,N,B,E}themovingFrenet-Serretframealongthecurve α inthespace E4.
For an arbitrary curve α in Euclidean 4-space E4, the following Frenet-Serret formulae is given
with respect to the first curvature κ, the second curvature τ and the third curvature σ in [6]
    
T(cid:48) 0 κ 0 0 T
    
 N(cid:48)   −κ 0 τ 0  N 
 =  ,
    
 B(cid:48)   0 −τ 0 σ  B 
    
E(cid:48) 0 0 −σ 0 E
where T,N,B and E are called the tangent, the principal normal, the first and the second
binormal vectors of the curve α, respectively.
Theorem 2.1([14]) Let α = α(t) be an arbitrary curve in Euclidean 4-space E4 with above
Frenet-Serret equations. Frenet-Serret apparatus of α can be written as follows:
α(cid:48)
T = , (2.1)
(cid:107)α(cid:48)(cid:107)
(cid:107)α(cid:48)(cid:107)2α(cid:48)(cid:48)−(cid:104)α(cid:48),α(cid:48)(cid:48)(cid:105)α(cid:48)
N = (cid:13) (cid:13), (2.2)
(cid:13)(cid:107)α(cid:48)(cid:107)2α(cid:48)(cid:48)−(cid:104)α(cid:48),α(cid:48)(cid:48)(cid:105)α(cid:48)(cid:13)
(cid:13) (cid:13)
B =µN ∧T ∧B , (2.3)
2
Onk-TypeSlantHelicesDuetoBishopFrameinEuclidean4-SpaceE4 3
T ∧N ∧α(cid:48)(cid:48)(cid:48)
E =µ , (2.4)
(cid:107)T ∧N ∧α(cid:48)(cid:48)(cid:48)(cid:107)
(cid:13) (cid:13)
(cid:13)(cid:107)α(cid:48)(cid:107)2α(cid:48)(cid:48)−(cid:104)α(cid:48),α(cid:48)(cid:48)(cid:105)α(cid:48)(cid:13)
κ= (cid:13) (cid:13) (2.5)
(cid:107)α(cid:48)(cid:107)4
(cid:107)T ∧N ∧α(cid:48)(cid:48)(cid:48)(cid:107)(cid:107)α(cid:48)(cid:107)
τ = (cid:13) (cid:13) (2.6)
(cid:13)(cid:107)α(cid:48)(cid:107)2α(cid:48)(cid:48)−g(α(cid:48),α(cid:48)(cid:48))α(cid:48)(cid:13)
(cid:13) (cid:13)
and
(cid:10)α(IV),E(cid:11)
σ = , (2.7)
(cid:107)T ∧N ∧α(cid:48)(cid:48)(cid:48)(cid:107)(cid:107)α(cid:48)(cid:107)
where µ is taken −1 or +1 to make +1 the determinant of the matrix [T,N,B,E].
Bishop frame is also referred to as parallel transport that is an orthonormal frame formed
by transporting in parallel each component of the frame. The parallel transport is formed with
tangent vector and any convenient arbitrary basis for the remainder of the frame (for details,
see [4], [10]). Then, the relations between Frenet-Serret frame and parallel transport frame for
the curve α:I ⊂R→E4 are given as follows:
T(s)=T(s),
N(s)=cosθ(s)cosψ(s)M +(−cosφ(s)sinψ(s)+sinφ(s)sinθ(s)cosψ(s))M
1 2
+(sinφ(s)sinψ(s)+cosφ(s)sinθ(s)cosψ(s))M ,
3
B(s)=cosθ(s)sinψ(s)M +(cosφ(s)cosψ(s)+sinφ(s)sinθ(s)sinψ(s))M
1 2
+(−sinφ(s)cosψ(s)+cosφ(s)sinθ(s)sinψ(s))M ,
3
E(s)=−sinθ(s)M +sinφ(s)cosθ(s)M +cosφ(s)cosθ(s)M .
1 2 3
The parallel transport frame equations are expressed as [10]
    
T(cid:48) 0 k k k T
1 2 3
    
 M(cid:48)   −k 0 0 0  M 
 1 = 1  1 , (2.8)
    
 M2(cid:48)   −k2 0 0 0  M2 
M(cid:48) −k 0 0 0 M
3 3 3
where k ,k ,k are curvature functions according to parallel transport frame of the curve α,
1 2 3
and their expression as follows:
k =κcosθ(s)cosψ(s),
1
k =κ(−cosφ(s)sinψ(s)+sinφ(s)sinθ(s)cosψ(s)),
2
k =κ(sinφ(s)sinψ(s)+cosφ(s)sinθ(s)cosψ(s)),
3
4 YasinU¨nlu¨tu¨rk,HaticeTozakandCumaliEkici
where √ √
σ σ2−θ(cid:48)2 σ2−θ(cid:48)2
θ(cid:48) = √ , ψ(cid:48) =−τ −σ√ , φ(cid:48) =− ,
κ2+τ2 κ2+τ2 cosθ
and Frenet curvature functions are given as follows:
(cid:112) θ(cid:48)
κ(s)= k2+k2+k2, τ(s)=−ψ(cid:48)+φ(cid:48)sinθ, σ(s)= ,
1 2 3 sinψ
and
φ(cid:48)cosθ+θ(cid:48)cotψ =0,
in terms of the invariants of parallel transport frame.
§3. On k-Type Slant Helices Due to Bishop Frame in Euclidean 4-Space E4
Definition 3.1 Let α = α(s) be a curve parametrized by arc-length with {T,M ,M ,M } a
1 2 3
BishopframeinE4.Ifthereexistsanon-zeroconstantvectorfieldU inE4 suchthat(cid:104)M ,U(cid:105)=(cid:54) 0
k
is a constant for all s ∈ I, where M = T, then α is said to be k−type (k ∈ {0,1,2,3}) slant
0
helix, and U is called axis of α.
Theorem 3.2 Let α=α(s) be a unit speed curve with non zero Bishop curvatures k ,k , and
1 2
k due to Bishop frame in E4. There is no 0−type slant helix (general helix) due to Bishop
3
frame in E4.
Proof Let α=α(s) be 0−type slant helix in E4 and the axis of the curve be U. Then, we
have that
(cid:104)T,U(cid:105)=c (s)=constant (3.1)
1
alongthecurveα.Differentiating(3.1)withrespecttosandusingBishopframe,weknowthat
k (cid:104)M ,U(cid:105)+k (cid:104)M ,U(cid:105)+k (cid:104)M ,U(cid:105)=0,
1 1 2 2 3 3
which implies that the unit vector U lies on the subspace spanned by {T} and therefore, it can
be written as
U =c (s)T. (3.2)
1
Differentiation of (3.2) gives
c k M +c (s)k M +c (s)k M =0.
1 1 1 1 2 2 1 3 3
Since the vectors {M ,M ,M } are linearly independent, we have c =0 which yields
1 2 3 1
U =0. (3.3)
Sincetheresult(3.3)contradictswiththedefinitionofU,weclaimthatthereisno0−type
slant helix (general helix) due to Bishop frame in E4. (cid:3)
Onk-TypeSlantHelicesDuetoBishopFrameinEuclidean4-SpaceE4 5
Theorem 3.3 Let α=α(s) be a unit speed curve with non zero Bishop curvatures k ,k , and
1 2
k due to Bishop frame in E4. Then α is a 1−type slant helix if and only if the function
3
k k
−c 1 −c 2 (3.4)
0k 1k
3 3
is a constant, and c =const. and c =const.
0 1
Proof Letα=α(s)be1−typeslanthelixinE4 andU beafixednon-zerodirection. Then
we have
(cid:104)M ,U(cid:105)=c (s),c (s)∈R (3.5)
1 0 0
along the curve α. Using (3.5) and Bishop frame formulae, we have
−k (cid:104)T,U(cid:105)=0,
1
which implies that the unit vector U lies on the subspace spanned by {M ,M ,M } and it can
1 2 3
be written as
U =c (s)M +a(s)M +b(s)M . (3.6)
0 1 2 3
Differentiation of (3.6) gives
(−c k −ak −bk )T +a(cid:48)M +b(cid:48)M =0.
0 1 2 3 2 3
Since the vectors {T,M ,M } are linearly independent, we have
2 3
−c k −ak −bk =0,
0 1 2 3
a(cid:48) =0, (3.7)
b(cid:48) =0.
From (3.7), we obtain
k k
a=c , b=−c 1 −c 2,
1 0k 1k
3 3
where c is constant.
1
Conversely,if(3.4)holds,wecanfindafixednonzerovectorU satisfying(cid:104)M ,U(cid:105)=constant.
1
We consider the axis as
(cid:18) (cid:19)
k k
U =M +M − 1 + 2 M . (3.8)
1 2 k k 3
3 3
Differentiating U with the help of (3.4) gives U(cid:48) = 0. This means that U is a constant
vector. As a result, α is a 1−type slant helix in E4. (cid:3)
Using Theorem 3.3, we have the following result.
Corollary 3.4 Let α = α(s) be a 1−type slant helix with non zero Bishop curvatures k ,k ,
1 2