Low-dimensional Lie Algebras AndControl TheorybyOmar Mrani ZentarA THESIS SUBMITTED IN PARTIAL FULFILMENT OF THEREQUIREMENTS FOR THE DEGREE OFMASTER OF SCIENCEinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)The University of British Columbia(Vancouver)July 2019© Omar Mrani Zentar, 2019The following individuals certify that they have read, and recom-mend to the Faculty of Graduate and Postdoctoral Studies for ac-ceptance, a thesis entitled:Low-dimensional Lie algebras and control theorysubmitted by Omar Mrani Zentar in partial fulfilment of the re-quirements forthe degree of Master of Sciencein MathematicsExamining CommitteeGeorge BlumanSupervisorZinovy ReichsteinAdditional examineriiAbstractLie groups and Lie algebras are important mathematical constructsfirst developed by Sophus Lie in the late nineteenth century to unifyand extend known methods used to solve differential equations. Theproblem considered in this thesis emphasizes one way Lie groupsand Lie algebras can be used in control theory.Suppose an apparatus has mechanisms for moving in a limited num-ber of ways with other movements generated by compositions of al-lowed motions. The question is then how to get a targeted motionusing a minimal number of the allowed motions. Motions can oftenbe represented by Lie groups which have associated Lie algebras astheir building blocks. This research shows explicitly how one canobtain elements of Lie groups as compositions of products of otherelements based on the structure of the associated Lie algebras. Here,the structure of a Lie algebra refers to its commutators which are theresults that one gets by applying an operation known as the ”com-mutator” to each pair of elements of a Lie algebra.Two concrete examples of this problem, in control theory, are: (1)the restricted parallel parking problem where the commutator ofthe Lie algebra element representing translations in y and that rep-resenting rotations in the xy-plane yields translations in x. Here thecontrol problem involves a vehicle that can only perform a seriesinvolving translations in y and rotations with the aim of efficientlyobtaining a pure translation in x; (2) involves an apparatus that canonly perform rotations about two axes and the aim is to performa pure rotation about a third axis. Both examples involve three-dimensional Lie algebras.In this thesis, the composition problem is solved for the nine three-iiiand four-dimensional Lie algebras with non-trivial solutions. Threedifferent solution methods are presented. Two of these methodsdepend on operator and matrix representations of a Lie algebra.The other method is a differential equation method that dependssolely on the commutator properties of a Lie algebra. Remarkably,for these distinguished Lie algebras the solutions involve arbitraryfunctions and can be expressed in terms of elementary functions.ivLay SummaryLie groups can represent motion and have building blocks known asLie algebras. The objective is to explicitly show how one can com-bine certain ”allowed” motions (Lie group elements) to obtain a dif-ferent and desired type of motion (another Lie group element) in aminimal number of steps. An example is parallel parking where oneaims to perform a pure translation in x using a minimal combina-tion of translations in y and rotations. This problem was solved forthe nine relevant three- and four-dimensional Lie algebras.vPrefaceThe problem considered in this research is related to a problem thatarose in the 1967 thesis by George Bluman [4]. It consists of explic-itly showing how one can combine ”certain” allowed motions (Liegroup elements) to obtain a different and desired type of motion(another Lie group element) in a minimal number of steps. Blumandeveloped three methods to approach the problem which are the op-erator method, matrix method, and differential equation method. Usingthe differential equation method, Bluman worked with Deshin Fin-lay to solve the problem for all three-dimensional Lie algebras otherthan the Euler angles Lie algebra so(3,R). Additionally, they laidout some of the basic work needed to solve the problem using theDE method for the four-dimensional Lie algebra S4,7 . My work con-sists of solving the composition problem for all relevant three- andfour-dimensional Lie algebras using the DE method as well as twoother methods which depend on the representation of Lie algebrasusing operators or matrices. I also did research that enabled us tofind all matrix representations and some operator representations.Once my research was completed, George Bluman, Deshin Finlay,and I wrote a paper to publish the results: Composition of Lie groupelements from basis Lie algebra elements. JNMP.25, 528-557 (2018)[2]. During the revision process we were made aware of reference[12] which provided operator representations for low-dimensionalLie algebras. This was particularly useful as we were unaware ofany operator representations for certain Lie algebras and hence oneof the referees helped us make our research more complete. Chap-ters 3-6 of this thesis are significantly based on the paper.viTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . vPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . viiList of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . xList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . xi1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.1 Lie Groups of Transformations . . . . . . . . . . . . . . . 52.1.1 Group . . . . . . . . . . . . . . . . . . . . . . . . . 52.1.2 One-parameter Lie group of transformations . . . 72.1.3 Infinitesimal generator . . . . . . . . . . . . . . . . 132.2 Multiparameter Lie group of transformations . . . . . . 162.2.1 r-parameter Lie group of transformations . . . . . 172.3 Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . 223 Research problem . . . . . . . . . . . . . . . . . . . . . . . . 253.1 Operator method . . . . . . . . . . . . . . . . . . . . . . 283.1.1 Description of the operator method . . . . . . . . 283.1.2 Example sl (2,R) . . . . . . . . . . . . . . . . . . . 293.2 Matrix representation method . . . . . . . . . . . . . . . 313.2.1 Description of the procedure used for the matrixrepresentation method . . . . . . . . . . . . . . . . 32vii3.2.2 Example sl(2,R) . . . . . . . . . . . . . . . . . . . . 323.3 DE method . . . . . . . . . . . . . . . . . . . . . . . . . . 333.3.1 Description of the DE method . . . . . . . . . . . 343.3.2 Example sl(2,R) . . . . . . . . . . . . . . . . . . . 364 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395 Three-dimensional Lie algebras . . . . . . . . . . . . . . . . 435.1 Parallel parking problem (Lie algebra S3,3 with con-stant r = 0) . . . . . . . . . . . . . . . . . . . . . . . . . . 435.1.1 Model example . . . . . . . . . . . . . . . . . . . . 435.1.2 Solution using the operator method . . . . . . . . 445.1.3 Solution using the matrix representation method 465.1.4 Solution using the DE method . . . . . . . . . . . 475.2 Euler angles problem (Lie algebra so (3,R)) . . . . . . . 485.2.1 Solution using the operator method . . . . . . . . 485.2.2 Solution using the matrix representation method . 505.2.3 Solution using the DE method . . . . . . . . . . . 515.3 Lie algebra n3,1 . . . . . . . . . . . . . . . . . . . . . . . . 515.3.1 Solution using the operator method . . . . . . . . 515.3.2 Solution using the matrix representation method . 525.3.3 Solution using the DE method . . . . . . . . . . . 535.4 Lie algebra S3,1 . . . . . . . . . . . . . . . . . . . . . . . . 535.4.1 Solution using the operator method . . . . . . . . 535.4.2 Solution using the matrix representation method . 545.4.3 Solution using the DE method . . . . . . . . . . . 555.5 Lie algebra S3,2 . . . . . . . . . . . . . . . . . . . . . . . . 555.5.1 Solution using the operator method . . . . . . . . 555.5.2 Solution using the matrix representation method . 565.5.3 Solution using the DE method . . . . . . . . . . . 575.6 Lie algebra S3,3 . . . . . . . . . . . . . . . . . . . . . . . . 57viii5.6.1 Solution using the operator method . . . . . . . . 575.6.2 Solution using the matrix representation method . 585.6.3 Solution using the DE method . . . . . . . . . . . 596 Four-dimensional Lie algebras . . . . . . . . . . . . . . . . . 606.1 Lie algebra S4,2 . . . . . . . . . . . . . . . . . . . . . . . . 606.1.1 Solution using the operator method . . . . . . . . 606.1.2 Solution using the matrix representation method . 616.1.3 Solution using the DE method . . . . . . . . . . . 626.2 Lie algebra S4,7 . . . . . . . . . . . . . . . . . . . . . . . . 636.2.1 Solution using the operator method . . . . . . . . 636.2.2 Solution using the matrix representation method . 646.2.3 Solution using the DE method . . . . . . . . . . . 656.3 Lie algebra S4,9 . . . . . . . . . . . . . . . . . . . . . . . . 666.3.1 Solution using the operator method . . . . . . . . 666.3.2 Solution using the matrix representation method . 676.3.3 Solution using the DE method . . . . . . . . . . . 696.4 Lie algebra S4,10 . . . . . . . . . . . . . . . . . . . . . . . 696.4.1 Solution using the operator method . . . . . . . . 696.4.2 Solution using the matrix representation method . 716.4.3 Solution using the DE method . . . . . . . . . . . 727 Inverse Problem . . . . . . . . . . . . . . . . . . . . . . . . . 747.1 The inverse parking problem . . . . . . . . . . . . . . . . 778 Discussion and conclusions . . . . . . . . . . . . . . . . . . 79References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83ixList of Tables4.1 Results for three-dimensional Lie algebras . . . . . . . . 404.2 Results for four-dimensional Lie algebras . . . . . . . . 417.1 Results for the Inverse Problem . . . . . . . . . . . . . . 76xList of Figures5.1 Illustration of solution of parallel parking problem. (1)represents initial configuration of vehicle with succes-sive configurations represented by (2) to (5). . . . . . . . 457.1 Illustration of the solutions of the parallel parking prob-lem and its inverse. . . . . . . . . . . . . . . . . . . . . . 78xiChapter 1IntroductionLie groups and their representations play an important role in vari-ous applications. Lie groups of transformations describe rigid bodymotions (rotations and translations), scalings, as well as other trans-formations. This thesis is concerned with showing explicitly howone can obtain further elements of a low-dimensional Lie group ascompositions of other elements of a chosen basis. Problems of thiskind can arise naturally in control theory [5,10,11]. Here an appara-tus has mechanisms for moving in a limited number of ways and theaim is to generate efficiently other movements from compositions ofpossible motions. Two concrete examples are:(1) The restricted parallel parking problem where the commutatorof translations in y and rotations in the xy-plane yields translationsin x. Here the control problem involves a vehicle that only performstranslation in y and rotations in the xy-plane with the aim of effi-ciently obtaining a pure translation in x;(2) An apparatus that only performs rotations about two axes withthe aim of efficiently generating rotations about a third axis. Herethe commutator of rotations about two axes yields rotations about1the third axis.Both examples involve three-dimensional Lie algebras with the prop-erty that the commutator of two of its elements generates a third el-ement. In terms of the notation used in [13], examples (1) and (2)respectively include the Lie algebras S3,3 with its parameter set tozero and so(3,R).Three distinct methods are presented to solve the composition prob-lem. The first method (operator method) depends on realizing a Liegroup as a Lie group of transformations. Such realizations can befound in [8] for some and in [12] for all three- and four-dimensionalLie algebras.The second method (matrix representation method) involves matrixrepresentations of finite-dimensional Lie algebras which are knownto exist from Ado’s theorem [1]. This theorem states that there existsa faithful square matrix representation for every finite-dimensionalLie algebra. There are many existing algorithms that generate suchmatrices including one developed by Willem de Graaf [9]. Whilethe minimal dimension of a matrix representation is not known ingeneral, it is known for all three- and four-dimensional Lie algebras(See [6] and [7]). This method is applied to a control theory problemin [10].The third method (DE method) was initially presented in [4] forother purposes. This method only uses the commutator propertiesof a Lie algebra. In particular, it does not require the use of a rep-resentation of a Lie algebra. The DE method involves setting upand solving an initial value problem for a nonlinear system of firstorder ordinary differential equations. The DE method yields a nec-essary condition for solutions—it turns out that for all three- andfour-dimensional cases, the DE method yields all solutions.Remarkably, for all relevant n-dimensional Lie algebras, n = 3 or24, the considered composition has n + 1 Lie group elements andthe solution involves one arbitrary function and can be expressed interms of elementary functions.In Chapter 2, the necessary background is presented. We start bygiving historical motivations for the study of Lie groups and Lie al-gebras. First, we define one-parameter Lie groups of transforma-tions and present Lie’s fundamental theorems needed for solving theresearch problem with illustrative examples. Then we define mul-tiparameter Lie groups of transformations and associated Lie alge-bras. Here we present relevant fundamental theorems and illustratethem through examples.In Chapter 3, we give a precise mathematical statement of the re-search problem. Then we describe fully the three different methodsused to solve it. As an illustrative example, we focus on the Lie al-gebra sl(2,R). In Chapter 4, in two tables we summarize our resultsfor all relevant three- and four-dimensional Lie algebras. Follow-ing this, we show the essential details that yield these solutions forthree- and four-dimensional Lie algebras in Chapters 5 and 6 re-spectively. In Chapter 7 we consider the case where the compositionproblems are stated with the order of the Lie group elements re-versed. Finally in Chapter 8, we make further remarks and discussthe advantages and disadvantages of the three presented methods.3Chapter 2BackgroundThe problem considered in this thesis consists of explicitly deter-mining how one can obtain further elements of Lie groups as com-positions of other Lie group elements for all relevant three- andfour-dimensional Lie algebras. However, before introducing thespecifics of the research problem and the methods used to solve it,we begin by providing the reader with the necessary background onLie groups and Lie algebras. We provide definitions and examples ofLie groups and Lie algebras (taken from [3]) as well as fundamentaltheorems.In the late 19th century, Sophus Lie initially developed the conceptof Lie groups to unify the known techniques for solving ordinarydifferential equations. Lie’s work resulted in a systematic approachto solving ordinary differential equations and, in particular, unifiedthe topics of integrating factors, separable equation, homogeneousequation, variation of parameter, reduction of order, Fourier andLaplace transforms. Because of its systematic nature, Lie’s work andits advancements led to the development of symbolic manipulationsoftware that can be used to solve differential equations [3].The key element in Lie’s framework for solving differential equa-tions is the use of symmetries, i.e. transformations that map solutions4of a differential equation into other solutions. These symmetries areexpressed in terms of continuous groups of transformations knownas Lie groups of transformations which are defined in the next sub-section. It is worth noting that an essential part of Lie’s work is analgorithm to find all Lie groups of transformations admitted by adifferential equation.2.1 Lie Groups of TransformationsHere we introduce Lie groups of transformations acting on Rn afterdefining groups and groups of transformations.2.1.1 GroupDefinition.A group G is a set of elements with a law of composition φ betweenelements satisfying the following properties.(i) Closure property. For any elements a and b in G, φ(a,b) is an ele-ment of G.(ii) Associative property. For any elements a,b and c inG,φ(a,φ(b,c)) =φ(φ(a,b), c).(iii) Identity element. There exist an identity element e in G suchthat for every element a in G, φ(a,e) = φ(e,a) = a.(iv) Inverse element. For every element a in G there is an inverse5element a−1 such that φ(a,a−1) = (a−1, a) = e.Definition.A subset H of a group G that is a group under the same law of com-position φ is a subgroup of G.Examples of groups(1) The set of all positive real numbers R+, with the law of composi-tion φ given by multiplication φ(a,b) = ab and the identity elemente = 1. Here the inverse of each element a is a−1 = 1/a.(2) The set of all complex numbers C, with the law of compositionφ given by addition, φ(a,b) = a + b. Here the identity element ise = 0 and the inverse of each element a is a−1 = −a.[The set of all realnumbers R is a subgroup of C whereas the set of all integers Z is asubgroup of both R and C.](3) The set of all invertible n × n matrices, where the law of com-position φ is given by matrix multiplication, the identity element isthe identity matrix In, and the inverse of each element is the inverseof a matrix under multiplication.62.1.2 One-parameter Lie group of transformationsDefinition.Let x = (x1,x2, ...,xn) lie in a region D ⊂ Rn and let a parameter ε ∈S ⊂R. The set of transformationsx∗ = X(x;ε) (2.1.1)defined for each x ∈D and ε ∈ S with law of composition φ(ε,δ) de-fined for all ε and δ in S, is a one-parameter group of transformationsiff the following properties hold.(i) For each ε in S the transformations are one-to-one onto D.(ii) S with the law of composition φ is a group.(iii) For the identity element e = ε0 and each x ∈D: x∗ = x, i.e.x∗ = X(x;ε0) = x(iv) If x∗ = X(x, ε) and x∗∗ = X(x∗,δ), thenx∗∗ = X(x;φ(ε,δ)).Definition.A group of transformations is a one-parameter Lie group of transfor-mations if in addition to properties (i) to (iv) it also satisfies:(v) ε is a continuous parameter, i.e. S is an interval in R.7(vi) X is infinitely differentiable with respect to x in D and an an-alytic function of ε in S.(vii) φ(ε,δ) is an analytic function of ε and δ, for all ε and δ in S.Note that one can think of ε as a time parameter and x as a spatialvariable. Here x∗ = X(x, ε) represents the evolution over time of apoint in the region D.Examples of one-parameter Lie groups of transformations include:(1) Group of x-translations in the planeHere D = R2, S = R, and φ(ε,δ) = ε + δ and for any x = (x,y) ∈ Dand ε ∈ S,x∗ = X(x;ε) = (x+ ε,y).This group corresponds to translations parallel to the x-axis.(2) A group of scalings in the planeIn this example D = R2, S = (0,∞), φ(ε,δ) = εδ, the identity ele-ment is 1, and for any x = (x,y) ∈D and α ∈ S,x∗ = X˜(x,α) = (αx,α2y).In order to set the identity element to 0, one can reparameterize this8group by setting ε = α − 1. Hence S becomes (−1,∞) and the trans-formation becomesx∗ = X(x;ε) = X˜(x;1 + ε) =((1 + ε)x, (1 + ε)2y)with law of composition φ(ε,δ) = ε+ δ+ εδ.First Fundamental Theorem of Lie; infinitesimal generatorsDefinition. Consider a one-parameter Lie group of transformationsx∗ = X(x;ε) (2.1.2)with the identity ε = 0 and law of composition φ. From the Taylorexpansion of (2.1.2) about ε = 0, one getsx∗ = x+ ε(∂X(x;ε)∂ε)∣∣∣∣∣ε=0+O(ε2) (2.1.3)Letξ(x) =∂X(x;ε)∂ε∣∣∣ε=0. (2.1.4)The transformation x+εξ(x) is the infinitesimal transformation of theLie group of transformations (2.1.2). The components of ξ(x) are theinfinitesimals of (2.1.2).Theorem 2.1. First Fundamental Theorem of Lie. There exists aparameterization τ(ε) such that the Lie group of transformations (2.1.2)9is equivalent to the solution of an initial value problem for a system offirst-order ODEs given bydx∗dτ= ξ(x∗) (2.1.5)withx∗ = x when τ = 0. (2.1.6)In particular,τ(ε) =∫ ε0Γ (ε′)dε′, (2.1.7)whereΓ (ε) =∂φ(a,b)∂b∣∣∣(ε−1,ε) (2.1.8)andΓ (0) = 1. (2.1.9)Proof.See Ref [3].The First Fundamental Theorem of Lie shows that a one-parameterLie group of transformations can be determined from its infinitesi-mals. Next, we look at some examples.(1) Group of x-translations in the planeHere D = R2, S = R, φ(ε,δ) = ε + δ and for any x = (x,y) ∈ D andε ∈ S,x∗ = X(x;ε) = (x+ ε,y). (2.1.10)10Clearly, the infinitesimals areξ(x) = (1,0).Using equation (2.1.8), one can easily check thatΓ (ε) = 1.Hence equation (2.1.7) leads toτ(ε) = ε.Thus the system of equations (2.1.5), (2.1.6) becomes(dx∗dε,dy∗dε)= (1,0); (2.1.11)and when ε = 0(x∗, y∗) = (x,y). (2.1.12)One can easily check that the solution to the initial value problem(2.1.11), (2.1.12) generates the group of transformations (2.1.10).(2) A group of scalings in the planeIn this example D = R2, S = (−1,∞), φ(ε,δ) = ε + δ + εδ, and for11any x = (x,y) ∈D and ε ∈ S the transformation is given byx∗ = X(x;ε) =((1 + ε)x, (1 + ε)2y). (2.1.13)It is easy to check that for each ε ∈ S, its inverse is given byε−1 =11 + ε. (2.1.14)Clearly, the infinitesimals of this Lie group of transformations areξ(x) = (x,y).Using equations (2.1.8) and (2.1.14), one can show thatΓ (ε) =11 + ε.Hence equation (2.1.7) leads toτ(ε) = ln(1 + ε). (2.1.15)Thus the system of equations (2.1.5), (2.1.6) becomes(dx∗dτ,dy∗dτ)= (x∗,2y∗) (2.1.16)and when ε = 0,(x∗, y∗) = (x,y). (2.1.17)12One can easily check, using equation (2.1.15), that the solutionto the IVP (2.1.16), (2.1.17) generates the group of transformations(2.1.13).2.1.3 Infinitesimal generatorAs another consequence of the First Fundamental Theorem of Lie,one sees that it is always possible to reparameterize a one-parameterLie group of transformations so that, without loss of generality, onecan assume that its law of composition is given by φ(a,b) = a + bwith identity element 0, and the inverse for each ε ∈ S is given byε−1 = −ε. Hence the group of transformations in (2.1.2) becomes(τ = ε)dx∗dε= ξ(x∗) (2.1.18)withx∗ = x when ε = 0. (2.1.19)The infinitesimal generator of a one-parameter Lie group of transfor-mations is the operator given byX = X(x) = ξ(x).∇ =n∑i=1ξi(x)∂∂xi, (2.1.20)where ∇ is the gradient operator13∇ =(∂∂x1,∂∂x2, ...,∂∂xn). (2.1.21)For any differentiable function F(x) = F(x1,x2, ...,xn) one hasXF(x) = ξ(x).∇F(x) =n∑i=1ξi(x)∂F(x)∂xi.Note that Xxi = ξi(x) for i = 1, ...,n. It follows that the infinitesi-mal generator determines the infinitesimal transformation of a Liegroup and, from the First Fundamental Theorem of Lie, the infinites-imal generator can be used to determine the Lie group of transfor-mations. Next, we show another important and explicit method fordetermining a one-parameter Lie group of transformations from itsinfinitesimal generator.Theorem 2.1.1. The one-parameter Lie group of transformations (2.1.2)is equivalent tox∗ = eεXx = x+ εXx+12ε2X2x+ ... =∞∑0εkk!Xkx, (2.1.22)where the operator X = X(x) is defined by (2.1.20) and XkF(x) is thefunction obtained by applying the operator X k times with k = 1,2, ....Moreover, X0F(x) ≡ F(x).14Proof. See Ref [3].This theorem presents a computationally useful alternative to theFirst Fundamental Theorem of Lie that can be used to generate aone-parameter Lie group of transformations from its infinitesimaltransformation. The series shown in equation (2.1.22) is the Lie se-ries of a one-parameter Lie group of transformations.Theorem (2.1.1) leads to the following important corollary.Corollary 2.1.1.1. If F(x) is infinitely differentiable, then for a one-parameter Lie group of transformations (2.1.2) with infinitesimal gener-ator (2.1.20), one hasF(x∗) = F(eεXx) = eεXF(x) (2.1.23)Proof.See Ref [3].As an example consider the rotation groupx∗ = cosεx+ sinεy, (2.1.24)y∗ = −sinεx+ cosεy. (2.1.25)The infinitesimal for this group of transformations isξ(x) =(ξ1(x),ξ2(x))= (y,−x)15Hence the infinitesimal generator is given byX = X(x) = ξ(x).∇ = y ∂∂x− x ∂∂y(2.1.26)One can use equation (2.1.26) to obtain the Lie series of this groupof transformations as follows: Xx = y, X2x = −x. ThusXkx = (−1) k−12 x if k is odd(−1) k2y if k is evenHencex∗ = eεXx = x( ∞∑k=0(−1)kε2k(2k)!)+ y( ∞∑k=0(−1)kε2k+1(2k + 1)!)= cosεx+ sinεy.Similarly, one can show thaty∗ = −sinεx+ cosεy.2.2 Multiparameter Lie group of transformationsIn this section we present some key definitions and results related tomultiparameter Lie groups of transformations. We only consider Liegroups of transformations with a finite number of parameters. Eachparameter of an r-parameter Lie group of transformations leads toan infinitesimal generator which in turn leads to an r-dimensionalvector space that has an additional ”commutator” operation. The16vector space generated by the infinitesimals is a Lie algebra (r-dimensionalLie algebra).One can obtain a multiparameter Lie group of transformationsthrough the exponentiation of its associated Lie algebra elements,as in the case of one-parameter Lie groups of transformations.2.2.1 r-parameter Lie group of transformationsAn r-parameter Lie group of transformation is defined in a way that isanalogous to the definition of a one-parameter Lie group of trans-formations.DefinitionLet x = (x1,x2, ...,xn) ∈D ⊂Rn and ε = (ε1, ε2, ..., εr), δ = (δ1,δ2, ...,δr) ∈S ⊆Rr . The transformationx∗ = X(x;ε) (2.2.1)with law composition φ(ε,δ) is an r-parameter Lie group of transfor-mation if it is a one-to-one transformation onto D such that(i) S with law of composition φ(ε,δ) is a group with identity ele-ment ε = 0 which corresponds to ε1 = ε2 = ... = εr = 0.(ii) φ(ε,δ) is analytic in ε and δ.(iii) For each x ∈D,x∗ = X(x;0) = x.17(iv) If x∗ = X(x;ε) and x∗∗ = X(x∗;δ), thenx∗∗ = X(x;φ(ε,δ)).Next, we present the First Fundamental Theorem of Lie for r-parameterLie groups of transformations after defining the infinitesimal matrixof the r-parameter Lie group of transformations (2.2.1).Definition.The infinitesimal matrix is the r ×n matrix Ξ(x) with entriesξij(x) =∂x∗j∂εi∣∣∣∣∣ε=0=∂Xj(x, ε)∂εi∣∣∣∣∣ε=0, i = 1,2, ..., r, j = 1,2, ...,n. (2.2.2)Theorem 2.3. The First Fundamental Theorem of Lie. Let Θ(ε) bethe r × r matrix with elementsΘkl(ε) =∂φl∂δk∣∣∣∣∣δ=0, k, l = 1,2, ...r. (2.2.3)Let its inverse matrix be given byΨ (ε) =Θ−1(ε). (2.2.4)In some neighborhood of ε = 0, the Lie group of transformations (2.2.1)is equivalent to the solution of the initial value problem for the system of18nr first-order PDEs given by∂x∗1∂ε1∂x∗2∂ε1. . . ∂x∗n∂ε1∂x∗1∂ε2∂x∗2∂ε2. . . ∂x∗n∂ε2...... . . ....∂x∗1∂εr∂x∗2∂εr. . . ∂x∗n∂εr= Ψ (ε)Ξ(x∗), (2.2.5)withx∗ = x at ε = 0. (2.2.6)Proof.See Ref [3].Definition.The infinitesimal generator Xα corresponding to the parameter εα ofthe r-parameter Lie group of transformations (2.2.1), is given byXα =n∑j=1ξαj∂∂xj, α = 1, ..., r. (2.2.7)One can show that the Lie group of transformations (2.2.1) is equiv-alent tox∗ =r∏α=1eµαXαx = eµ1X1eµ2X2 . . . eµrXrx, (2.2.8)19where, µ1,µ2, . . . ,µr are arbitrary real constants [3]. Note that the or-der of the operations in (2.2.8) can be rearranged by renumberingthe infinitesimal generators. This corresponds to a different param-eterization, i.e., Ψ (ε) changes.As an example, consider the two-parameter Lie group of transfor-mationsx∗ = eε1x+ ε2,y∗ = e2ε1y.(2.2.9)Here D = S =R2 and the law of composition isφ(ε,δ) = φ((ε1, ε2), (δ1,δ2))=(ε1 + ε2, ε2eδ1 + δ2); ε,δ ∈ S. (2.2.10)It is easy to check that the infinitesimal matrix isΞ(x,y) =(x y1 0).Additionally,Θ(ε1, ε2) =(1 ε20 1),with its inverse given byΨ (ε1, ε2) =(1 −ε20 1).Thus the system of PDEs (2.2.5) with initial conditions (2.2.6) be-20comes ∂x∗∂ε1 ∂y∗∂ε1∂x∗∂ε2∂y∗∂ε2 = Ψ (ε1, ε2)Ξ(x∗, y∗) = (x∗ − ε2 2y∗1 0), (2.2.11)with(x∗, y∗) = (x,y) when (ε1, ε2) = (0,0). (2.2.12)One can easily check that the solution to the system (2.2.11), (2.2.12)is given by (2.2.9). Next, we check that (2.2.9) can also be generatedusing equation (2.2.8).The infinitesimal generators for the two-parameter Lie group of trans-formations (2.2.9) are given byX1 = x∂∂x+ 2y∂∂y,X2 =∂∂x.It is useful to note that by analogy with equation (2.1.23), one hasfor every differentiable function F(x,y)eεX1F(x,y) = F(eεX1x,eεX1y)= F(eεx,e2εy),eεX2F(x,y) = F(eεX2x,eεX2y)= F(x+ ε,y).Hence equation (2.2.8) becomes21eµ1X1eµ2X2(x,y) = eµ1X1(x+µ2, y) = (eµ1x+µ2, e2µ1y) = (x∗, y∗). (2.2.13)2.3 Lie AlgebrasDefinition.Consider an r-parameter Lie group of transformations (2.2.1) withinfinitesimal generators Xα, α = 1, ..., r defined by (2.2.7). The com-mutator (Lie bracket) of Xα and Xβ is a first-order operator given by[Xα,Xβ] = XαXβ −XβXα=n∑i,j=1(ξαi(x) ∂∂xi)(ξβj(x)∂∂xj)−(ξβi(x)∂∂xi)(ξαj(x)∂∂xj)=n∑j=1ηj(x)∂∂xj,(2.3.1)whereηj =n∑i=1(ξαi(x)∂ξβj∂xi− ξβi(x)∂ξαj∂xi). (2.3.2)From equation (2.3.2), it is clear that the commutator operation isanti-symmetric, i.e,[Xα,Xβ] = −[Xβ,Xα]. (2.3.3)22By direct computation, one can show that for any three infinitesimalgenerators Xα, Xβ, Xγ , one has the Jacobi identity[Xα[Xβ,Xγ]] + [Xβ[Xγ ,Xα]] + [Xγ[Xα,Xβ]] = 0. (2.3.4)Theorem 2.4. The Second Fundamental Theorem of Lie. The com-mutator of any two infinitesimal generators of an r-parameter Lie groupof transformations is also an infinitesimal generator. In particular,[Xα,Xβ] =r∑γ=1CγαβXγ , (2.3.5)where the coefficients Cγαβ are real constants called the structure con-stants, α, β, γ = 1, ..., r.Proof.See ref [3].Equations (2.3.5) are the commutation relations of the r-parameterLie group of transformations (2.2.1).Theorem 2.5. The anti-symmetric property (2.3.3) of the infinitesimalgenerators in (2.2.7) and Jacobi’s identity (2.3.4) lead to the followingrelations satisfied by the structure constants.Cγαβ = Cγβα, (2.3.6)23r∑ρ=1[CραβCδργ +CρβγCδρα +CργαCδρβ]= 0. (2.3.7)Definition.A Lie algebra L is a vector space with a bilinear bracket operation(the commutator) satisfying the properties (2.3.3), (2.3.4) and (2.3.5).In particular, the set of infinitesimal generators {Xα}, α = 1,2, . . . , r,of an r-parameter Lie group of transformations (2.2.1) forms an r-dimensional Lie algebra over R.It is important to note that in this thesis, we are interested in amore abstract definition of a Lie algebra. In particular, we definea Lie algebra solely based on its properties (2.3.3)-(2.3.5). Startingfrom this abstract definition, we then seek natural representations ofa Lie algebra, e.g. an operator and matrix representations, in orderto solve the research problem. Most importantly, one can also obtaina necessary condition to solving the research problem, without anyrepresentation i.e., just using the algebraic properties (2.3.3)-(2.3.5)(differential equation method).24Chapter 3Research problemConsider a three-dimensional Lie algebra L with basis elements B1,B2, and B3 such that the commutator of B1 and B2 is a linear combi-nation of the basis elements of L, i.e.,[B1,B2] = B1B2 −B2B1 =3∑k=1Ck12Bk, (3.0.1)in terms of real structure constants C112, C212, and C312, with C312 , 0.In other words, the Lie algebra element B3 is generated by the othertwo elements. All six three-dimensional Lie algebras presented in[13] have at least one commutator that satisfies this non-zero prop-erty. In particular, for the problem under consideration, it does notmatter what the other commutators of L are. The question of interestis whether the Lie group element generated by B3 can be obtainedfrom the Lie group elements generated by B1 and B2 as illustratedby the two examples mentioned in the Introduction. Motivated bythe commutator property (3.0.1) with C312 , 0, the aim is to find con-tinuous functions a (ε) ,b (ε) , c (ε), and d (ε) so that the equationea(ε)B1eb(ε)B2ec(ε)B1ed(ε)B2 = eεB325witha(0) = b(0) = c(0) = d(0) = 0 (3.0.2)holds for an arbitrary value of ε. Next, to clarify how the researchproblem given by (3.0.2) follows from (3.0.1) with C312 , 0, a con-crete example is presented.The parallel parking problem has commutators given by [R,Y ] =X, [R,X] = −Y , [X,Y ] = 0, whereX = − ∂∂x, Y =∂∂y, R = y∂∂x− x ∂∂y.The first commutator indicates thatX can be generated from R andY . This leads one to consider either equationea(ε)Reb(ε)Y ec(ε)Red(ε)Y = eεX ; (3.0.3)orea(ε)Y eb(ε)Rec(ε)Y ed(ε)R = eεX . (3.0.4)In equation (3.0.3) R,Y , and X correspond to B1,B2, and B3, respec-tively. The solution to equation (3.3) will be presented in Chapter4. Note that when the commutator of a Lie algebra used in equation(3.0.2) does not satisfy the assumptions stated above, one would ex-pect only trivial solutions. For example, since in the parking prob-lem the commutators of X and Y do not generate R, the equationea(ε)Xeb(ε)Y ec(ε)Xed(ε)Y = eεRonly has the trivial solution.When the DE method is used, an additional assumption aboutthese functions is needed. In particular, here a (ε) ,b (ε) , c (ε), andd (ε) are differentiable everywhere except at ε = 0. This assumption26is needed since the DE method relies on finding a system of differ-ential equations that the four functions must satisfy.In general, it turns out that the problem as stated always has a de-gree of freedom in its solution. Moreover, a minimum number offour terms are needed on the left hand side of equation (3.0.2). Thisfollows from the origin of the commutator equation (3.0.1). In par-ticular, from the form of equation (3.0.2), one would expect, as willbe seen later in this thesis, that there are solutions for which a (ε) ,b (ε) , c (ε), and d (ε) are of order√ε as ε→ 0 to generate a commuta-tor element of order ε on the right-hand side of equation (3.0.2).Now consider a four-dimensional Lie algebra L with basis elementsB1, B2, B3, and B4 such that the elements B1, B2, and B3 do not forma subalgebra. For the research problem we require that the commu-tator of B1 and B2 satisfies[B1,B2] = B1B2 −B2B1 =4∑k=1Ck12Bk, (3.0.5)with real structure constants C112, ...,C412 where C312 , 0 and C412 = 0.Here the Lie algebra element B3 can be generated from the ele-ments B1 and B2. The problem of interest is whether the Lie groupelement generated by B3 can be obtained from the Lie group ele-ments generated by B1 and B2. In particular, in view of the com-mutator property (3.0.5), we are interested in finding continuousfunctions a (ε) ,b (ε) , c (ε) ,d (ε) , f (ε), and g(ε) so that the equationea(ε)B1eb(ε)B2ec(ε)B1ed(ε)B2ef (ε)B1eg(ε)B2 = eεB3, (3.0.6)with a(0) = b(0) = c(0) = d(0) = f (0)) = g(0) = 0 holds for an arbitraryvalue of ε.It is important to note that it is essential that the left hand sideof (3.0.6) is composed of the product of six Lie group elements: in27all considered cases there exist solutions where one of a (ε) or g(ε) iszero, but this is not obvious a priori.One should also note that one can also state the problems in (3.0.2)and (3.0.6) with the roles of B1 and B2 interchanged when the num-ber of terms to the left of these equations is even. This does notchange the nature of the problem and, in fact, it leads to isomorphicsolutions as will be shown in Chapter 7.In this thesis, for all relevant three- and four-dimensional Lie alge-bras, we present three different methods that can yield the generalsolution for their respective equations (3.0.2) and (3.0.6). In whatfollows, we will describe the different methods used to solve (3.0.2)and (3.0.6). As a simple example, in this Chapter we solve the com-position problem for the three-dimensional Lie algebra sl(2,R) to il-lustrate how these different methods work. The solution of the com-position problem for the other relevant three- and four-dimensionalLie algebras will be presented in Chapter 4.We first note that sl (2,R) has the commutators[X,Y ] = Z, [Z,X] = 2X, [Y ,Z] = 2Y . (3.0.7)3.1 Operator methodThe operator method requires a representation of a Lie algebra interms of differential operators. The operator representation is notnecessarily unique. This lack of uniqueness is illustrated by the ex-ample of sl (2,R) .3.1.1 Description of the operator methodLet {∆1, . . . ,∆r} be a differential operator representation of a Lie al-gebra L which respectively has basis elements {B1, . . . ,Br}, where28∆i = γij (x)∂∂xj, i = 1, . . . , r, and x = (x1, . . . ,xm) .Then equations (3.0.2) and (3.0.6) become respectively the equa-tionsea(ε)∆1eb(ε)∆2ec(ε)∆1ed(ε)∆2x = eε∆3x, (3.1.1)andea(ε)∆1eb(ε)∆2ec(ε)∆1ed(ε)∆2ef (ε)∆1eg(ε)∆2x = eε∆3x, (3.1.2)where eε∆ix =∞∑n=0εnn!∆ni x = (xi)∗ for i = 1, . . . , r; This corresponds togenerating the Lie algebra elements using Theorem 2.1.2.(xi)∗=((x1i)∗, . . . , (xmi)∗)is the image of x with respect to the ithbasis element of the Lie group of transformations connected withthe differential operator ∆i, with i = 1, . . . , r.From the First Fundamental Theorem of Lie 2.1, one can also ob-tain (xi)∗by solving the system of differential equationsd(xji)∗dε = γji (x), j = 1, . . . ,m, with initial condition (xj i)∗ (0) = xj .3.1.2 Example sl (2,R)Operator representations for sl (2,R) include∆1 = X = −x2 ∂∂x, ∆2 = Y =∂∂x, ∆3 = Z = 2x∂∂x, in R; (3.1.3)∆1 = X = −y ∂∂x, ∆2 = Y = −x∂∂y, ∆3 = Z = y∂∂y−x ∂∂x, in R2. (3.1.4)For the operator representation (3.1.3) for sl (2,R) , one obtainseεX (x) =x1 + εx, eεY (x) = x+ ε, eεZ(x) = e2εx. (3.1.5)29There are two well-known methods to obtain (3.1.5).Method I. From the First Fundamental Theorem of Lie (Theorem2.1), the operator representation (3.1.3) leads to solving separatelythe three IVPsdx∗dε= −x∗2,x∗ (0) = x; (3.1.6)dx∗dε= 1, x∗ (0) = x; (3.1.7)dx∗dε= 2x∗,x∗ (0) = x. (3.1.8)It is easy to show that the three one-parameter Lie groups of trans-formations (3.1.5) respectively solve the IVPs (3.1.6)-(3.1.8).Method II. Using induction, it is easy to show thatXnx = (−1)nn!xn+1, n ≥ 0.Hence eεXx =∞∑n=0εnn!Xnx =∞∑n=0εn(−1)nxn+1 = x1+εx . Since Y x = 1,Y nx =0, n ≥ 2, it follows that eεYx = x + ε. It is easy to show that Znx =2nx, n ≥ 0. Hence eεZx = e2εx.To proceed, it is convenient to rewrite expression (3.1.1) in theformeb(ε)Y ec(ε)Xed(ε)Yx = e−a(ε)XeεZx. (3.1.9)Then from (3.1.5), one obtainseb(ε)Y ec(ε)Xed(ε)Yx = eb(ε)Y ec(ε)X (x+ d) = eb(ε)Y(d +x1 + cx)=30d +x+ b1 + c(x+ b), e−a(ε)XeεZ (x) = e−a(ε)X(e2εx)= e2εx1− ax.Then expression (3.1.9) becomes d+ x+b1+c(x+b) = e2ε x1−ax . Hence for allx, one has−(a (dc+ 1) + ce2ε)x2 +(−a (d + b+ bdc) + dc+ 1− e2ε (1 + bc))x+ b+ d+bdc = 0.This yields the set of equationsb+ d + bdc = 0, dc+ 1 = e2ε(1 + bc), −a (dc+ 1) = ce2ε. (3.1.10)The solution to the system of equations (3.1.10) is given bya (ε) =eε − e2εd(ε), b (ε) = −e−εd(ε), c (ε) = eε − 1d (ε), (3.1.11)where d (ε) is any continuous function chosen so that a(ε) and c(ε)are continuous functions, and satisfying d (ε) , 0 for any ε , 0 witha(0) = b(0) = c(0) = d(0) = 0.3.2 Matrix representation methodThe matrix representation method involves a matrix representation ofa Lie algebra L. In particular, in this thesis we seek an appropri-ate matrix for each basis element of L using the Lie algebra pack-age of the computer software GAP (Group, Algorithms, Program-ming) [9]. A difficulty arose in the case of the four-dimensional Liealgebra S4,7 (in terms of the nomenclature used in the classification31of Lie algebras in [11]). Here the matrix representation obtainedfrom the software package [9] could not be used since the obtainedrepresentation is not isomorphic to S4,7. For this Lie algebra, weused a matrix representation given in [6].3.2.1 Description of the procedure used for the matrix represen-tation methodLet L be a k-dimensional Lie algebra with basis elements Bi with Midenoting a matrix representation of Bi , i = 1, . . . , k.Step 1. Find matrices {Mi} that represent L using computer soft-ware or relevant literature ([6], [9]).Step 2. Attempt to find a closed form representation for each ele-ment eεBi of the Lie group associated with L from the Taylorexpansion eεMi =∞∑n=0εnn!Mni where M i0 = I for i = 1, . . . , k. (In allthree- and four-dimensional cases, this step was successful in lead-ing to such a closed form representation.)Step 3. Compute∏2k−2i=1 eai(ε)Bj where{j = 1 if i is oddj = 2 if i is even.Step 4. Solve equations (3.0.2) and (3.0.6).3.2.2 Example sl(2,R)Although sl(2,R) is a three-dimensional Lie algebra, a matrix repre-sentation is given by the 2× 2 matrices X =(0 10 0), Y =(0 01 0),32Z =(1 00 −1). Then X2 =(0 00 0). Hence Xn =(0 00 0), n ≥ 2. ThuseεX = I + εX =(1 ε0 1). Similarly, one can show that eεY =(1 0ε 1).One can easily show that Zn=(1 00 (−1)n). Hence eεZ =(eε 00 e−ε).Consequently,M = ea()Xeb()Y ec()Xed()Y =(abcd + ab+ ad + cd + 1 a+ c+ abcb+ d + bcd bc+ 1).After setting M = eεZ , one obtains the equationsbc+ 1 = e−ε, b+d +bcd = 0, a+ c+abc = 0, abcd +ab+ad + cd + 1 = eε,whose solution is given by (3.1.11).3.3 DE methodThe DE method requires differentiation of the unknown functions inequations (3.0.2) and (3.0.6). It involves setting up a nonlinear sys-tem of first order ordinary differential equations that must be sat-isfied by all differentiable solutions of equations (3.0.2) and (3.0.6).Here the solutions respectively satisfy initial conditionsa (0) = b (0) = c (0) = d (0) = 0, (3.3.1)a (0) = b (0) = c (0) = d (0) = f (0) = g (0) = 0 , (3.3.2)in the three- and four-dimensional cases. Note that when ε = 0, theunknown functions will be continuous but not differentiable. TheDE method yields a necessary condition for solutions. Sufficiency isshown from the solutions obtained by the other two methods.333.3.1 Description of the DE methodAfter differentiating equations (3.0.2) and (3.0.6) with respect to ε,one obtains respectively,a′1 (ε)B14∏i=1eaiBi + a′2 (ε)ea1B1B24∏i=2eaiBi + a′3 (ε)2∏i=1eaiBiB14∏i=3eaiBi+ a′4 (ε)3∏i=1eaiBiB2ea4B2 = B3eεB3,(3.3.3)a′1 (ε)B16∏i=1eaiBi + a′2 (ε)ea1B1B26∏i=2eaiBi + · · ·+ a′6 (ε)5∏i=1eaiBiB2ea6B2= B3eεB3,(3.3.4)where Bi ={B1 if i is oddB2 if i is evenand a1 = a, a2 = b, a3 = c, a4 = d, a5 =f , a6 = g.From equations (3.3.3) and (3.3.4), one sees that a formula is neededfor pulling the products of the exponentials appropriately to theright of each Bp in order to get back∏ni=1 eai(ε)Bi = eεB3, n = 4, 6,respectively.In general, one proceeds as follows.Step 1. Find {fj} so thateεBiBp =k∑j=1fjBjeεBi (3.3.5)34where i = 1, 2, p = 1, . . . , k, and k is the dimension of the Lie algebraL. Since Ado’s theorem [?] guarantees the existence of a matrix rep-resentation for every finite-dimensional Lie algebra, one can treatthe operations in L as matrix elements so that, without loss of gen-erality, L is associative.Step 2. Differentiate equations (3.0.2) and (3.0.6) with respect toε to obtain equations (3.3.3) and (3.3.4), respectively. Then appro-priately and recursively substitute equation (3.3.5) into equations(3.3.3) and (3.3.4). Thus in each case one obtains an equation of theformk∑j=1αj (ε)Bjk∏i=1eaiBi = B3eεB3, (3.3.6)for specific functions αj (ε).Step 3. Assume that expressions (3.0.2) and (3.0.6) hold. Con-sequently, this yields necessary conditions that {αj(ε)} must satisfy,namely the nonlinear system of first order ODEsα3(ε) = 1,αi(ε) = 0, i , 3,(3.3.7)with initial conditions (3.3.1) and (3.3.2), respectively.Step 4. Check that the solution of the ODE system (3.3.7) solvesrespectively expressions (3.0.2) or (3.0.6) using either the matrix oroperator method.353.3.2 Example sl(2,R)Theorem 3.3.1. For sl(2,R) the following identities hold for any ε.eεXY ≡ (Y + εZ − ε2X)eεX , eεXZ ≡ (Z − 2εX)eεX ,eεYX ≡ (X − εZ − ε2Y )eεY , eεYZ = (Z + 2εY )eεY . (3.3.8)Proof. From the commutator relations (3.0.7), one directly obtainsXY = YX +Z,XZ = ZX − 2X,YZ = ZY + 2Y .Hence X2Y = YX2 + 2ZX − 2X. Then it is easy to show thatXnY = YXn +nZXn−1 −n(n− 1)Xn−1, n ∈N.Similarly, one can show that the following relations hold.XnZ = ZXn − 2nXn,Y nX = XY n −nZY n−1 −n(n− 1)Y n−1,Y nZ = ZY n + 2nY n.Consequently,eεXY = Y eεX + εZ∞∑n=1εn−1(n− 1)!Xn−1 − ε2X∞∑n=2εn−2(n− 2)!Xn−2 =(Y + εZ − ε2X)eεX .Similarly, one obtains the remaining relations in (3.3.8), complet-ing the proof. 36Now to proceed further, we differentiate equation (3.0.6) with re-spect to ε. This yieldsa′(ε)XeεZ + b′(ε)ea(ε)XY eb(ε)Y ec(ε)Xed(ε)Y + c′(ε)ea(ε)Xeb(ε)YXec(ε)Xed(ε)Y+ d ′(ε)ea(ε)Xeb(ε)Y ec(ε)XY ed(ε)Y = ZeεZ .(3.3.9)Using the relations in Theorem (3.3.1), one finds that equation(3.3.9) becomes[α1 (ε)X +α2 (ε)Y +α3 (ε)Z]eεZ = ZeεZ withα1 (ε) = a′ − a2(b′ + d ′ + b(−bc′ + bc2d ′ + 2cd ′))+ c′+ 2abc′ − 2acd ′ − c2d ′ − 2abc2d ′ = 0, (3.3.10)α2 (ε) = b′ + d ′ + b(−bc′ + bc2d ′ + 2cd ′)= 0, (3.3.11)α3 (ε) = a(b′ + d ′ + b(−bc′ + bc2d ′ + 2cd ′))− bc′ + cd ′ + bc2d ′ = 1.(3.3.12)After substituting equation (3.3.11) into equations (3.3.10) and(3.3.12), one gets respectivelya′ + c′ − c2d ′ − 2a(−bc′ + cd ′ + bc2d ′ ) = 0, (3.3.13)− bc′ + cd ′ + bc2d ′ = 1. (3.3.14)After substituting equation (3.3.14) into each of equations (3.3.13)and (3.3.11), one obtainsa′ + c′ − c2d ′ − 2a = 0, (3.3.15)b′ + d ′ + bcd ′ + b = 0. (3.3.16)37After subtracting c times equation (3.3.16) from equation (3.3.14),one finds that(bc)′ + bc = −1. Henceb (ε) =e−ε − 1c(ε). (3.3.17)Substitution of equation (3.3.17) into equation (3.3.16) leads to d ′ =ceε−c′ (eε−1)c2.Consequently,d(ε) =eε − 1c(ε). (3.3.18)After substituting equation (3.3.18) into equation (3.3.15), one gets2(a+ eεc) = (a+ eεc)′ . Hence a (ε) = −c(ε)eε.Thus the solution to the system of differential equations (3.28)-(3.30) with initial condition (3.19) is given by (3.1.11).38Chapter 4ResultsUsing the procedures described in chapter three, the results for allrelevant three- and four-dimensional Lie algebras are presented inTables 1 and 2, respectively.39Lie algebra; commutators Composition equation Solutionsl (2,R)[X,Y ] = Z[X,Z] = −2X[Y ,Z] = 2Yea(ε)Xeb(ε)Y ec(ε)Xed(ε)Y = eεZ d (ε) is an arbitrary functionsatisfying d (ε) , 0 whenε , 0a(ε) = eε−e2εd(ε)b (ε) = e−ε−1eε−1 d(ε)c (ε) = eε−1d(ε)Parallel parking problem,S3,3 with constant r = 0[R,Y ] = X[R,X] = −Y[X,Y ] = 0ea(ε)Reb(ε)Y ec(ε)Red(ε)Y = eεX d (ε) is an arbitrary functionsatisfying d(ε) , kpi for everyk ∈Z when ε , 0a (ε) = −arctan(εd(ε))|b (ε) | =√d2 + ε2c(ε) = arctan(εd(ε))Euler angles problem,so(3,R)[X,Y ] = Z[X,Z] = −Y[Y ,Z] = Xea(ε)Xeb(ε)Y ec(ε)Xed(ε)Y = eεZ Any c (ε) satisfying c (ε) , kpifor every k ∈ Z when ε , 0and∣∣∣∣ sin εsin c(ε) ∣∣∣∣ ≤ 1 witha (ε) = −arccos(cos c(ε)cos ε)b (ε) = −arcsin(sin εsinc(ε))d (ε) = arccos(− sin a (ε)sin c(ε));a (ε) = arccos(cos c(ε)cos ε)b (ε) = pi+ arcsin(sin εsinc(ε))d (ε) = arccos(− sin a(ε)sin c(ε))n3,1[X,Y ] = Z[X,Z] = 0[Z,Y ] = 0ea(ε)Xeb(ε)Y ec(ε)Xed(ε)Y = eεZ d (ε) is an arbitrary functionsatisfying d (ε) , 0 when ε , 0a(ε) = − εd(ε)b (ε) = −d (ε)c (ε) = εd(ε)S3,1[Y ,Z] = −Y[Y ,X] = 0[Z,X] = rX +Ywhere r is a constant satis-fying |r | ≤ 1ea(ε)Xeb(ε)Zec(ε)Xed(ε)Z = eεY d (ε) is an arbitrary functionsatisfying d (ε) , 0 when ε , 0b (ε) = −d (ε)a (ε) = ε(1−r)1−e(r−1)d(ε)c (ε) = −a erd(ε)S3,2[Z,X] = X[Z,Y ] = X +Y[X,Y ] = 0ea(ε)Y eb(ε)Zec(ε)Y ed(ε)Z = eεX d (ε) is an arbitrary functionsatisfying d (ε) , 0 when ε , 0b (ε) = −d (ε)c (ε) = − εed(ε)d(ε)a (ε) = εd(ε)S3,3 general case[R,X] = rX −Y[R,Y ] = X + rY[X,Y ] = 0where r is a non-negativeconstant.ea(ε)Reb(ε)Y ec(ε)Red(ε)Y = eεX d (ε) is an arbitrary functionsatisfying d(ε) , kpi for everyk ∈Z when ε , 0a (ε) = −arctan(εd(ε))|b (ε) | =√d2 + ε2er arctan( εd(ε) )c(ε) = arctan(εd(ε))Table 4.1: Results for three-dimensional Lie algebras40Lie algebra;commutatorsComposition equation SolutionS4,2[W,X] = X[W,Y ] = X +Y[W,Z] = Y +Z[X,Y ] = 0[X,Z] = 0[Z,Y ] = 0ea(ε)W eb(ε)Zec(ε)W ed(ε)Zef (ε)W = eεY f (ε) is an arbitrary function sat-isfying f (ε) , 0 when ε , 0a (ε) = f (ε)b (ε) = ε2f (ε) e−f (ε)c(ε) = −2f (ε)d (ε) = − ε2f (ε) ef (ε)S4,7[Y ,Z] = X[W,Y ] = −Z[W,Z] = Y[W,X] = 0[X,Y ] = 0[X,Z] = 0ea(ε)Zeb(ε)W ec(ε)Zed(ε)W ef (ε)Z= eεY f (ε) is an arbitrary functionsatisfying f (ε) , kpi for every k∈Z when ε , 0a (ε) = f (ε)b(ε) = −arctan( ε2f (ε) )c(ε) = ε√ε2+4f (ε)22f (ε)d (ε) = arctan( ε2f (ε) )S4,9[Y ,Z] = X[W,Y ] = rY −Z[W,Z] = Y + rZ[W,X] = 2rX[X,Y ] = 0[X,Z] = 0ea(ε)W eb(ε)Zec(ε)W ed(ε)Zef (ε)W = eεY f (ε) is an arbitrary function sat-isfying f (ε) , kpi for every k ∈ Zwhen ε , 0a (ε) = f (ε)b(ε) = −arctan( ε2f (ε) )c(ε) = ε√ε2+4f (ε)22f (ε) e−rbd (ε) = arctan( ε2f (ε) )S4,10[Y ,Z] = X[W,Y ] = Y[W,Z] = Y +Z[W,X] = 2X[X,Y ] = 0[X,Z] = 0ea(ε)W eb(ε)Zec(ε)W ed(ε)Zef (ε)W eg(ε)Z = eεY a (ε) and c (ε) are arbi-trary functions satisfyinga (ε)c (ε) , 0, a (ε) + c (ε) , 0,and c (ε)2 + a (ε)c (ε) ≥ 0f (ε) = −(a (ε) + c (ε))b (ε) =ε c(ε)±√c(ε)2+a(ε)c(ε)a(ε)c(ε)e−a(ε)g (ε) = − ε+b(ε)c(ε)ea(ε)a(ε)+c(ε)d (ε) = −g (ε)e−(a(ε)+c(ε)) −b (ε)e−c(ε)and in the limiting case whena (ε) = 0,f (ε) = −c(ε)d (ε) = εe−c(ε)c(ε)b (ε) = g (ε) = − ε2c(ε)Table 4.2: Results for four-dimensional Lie algebras41The sketch of the proofs of the results, exhibited in Tables 1 and 2,follow in Chapters 5 and 6, respectively.42Chapter 5Three-dimensional Lie algebrasIn this chapter, we present the proofs of the results presented inTable 1.5.1 Parallel parking problem (Lie algebra S3,3 withconstant r = 0)5.1.1 Model exampleTo illustrate parallel parking, as an example, consider a unicyclethat performs forward and backward translations as well as rota-tions. The unicycle is represented by a straight line with centre lo-cated at (x,y) and initial orientation parallel to the y-axis with itscentre located at (0,0) in Figure 5.1. The aim is to move the unicycleso that its centre finishes at (ε,0) with the vehicle parallel to its ini-tial orientation by a succession of rotations and translations in thesame direction as the straight line. As will be illustrated in Figure5.1, the minimum number of steps that start with a non-zero trans-lation is four. Let d(ε) and b(ε) be the translations in the first andthird steps, respectively and let c(ε) and a(ε) be the angles of rota-tion in the second and last steps, respectively. Since the direction the43vehicle is facing at the end must be the same as at the start, one musthave a(ε) = −c(ε). From Figure 5.1, one sees that d(ε)+b(ε)cosc(ε) = 0and b(ε)sinc(ε) = ε. Hence the solution to this problem is given bya(ε) = −c(ε), b(ε) = εsin c(ε), d(ε) = −εcotc(ε), (5.1.1)where c (ε) is any continuous function chosen so that b(ε) and d(ε)are continuous functions, and satisfying c(ε) , kpi for every k ∈ Zand ε , 0 with a(0) = b(0) = c(0) = d(0) = 0.Alternatively, one could treat d(ε) as an arbitrary function. Herethe solution as reflected by Figure 5.1 is given bya(ε) = −c(ε), c(ε) = −arctan(εd), |b(ε)| =√d2 + ε2,where d (ε) is any continuous function chosen so that c(ε) is a contin-uous function, and satisfying d(ε) , 0 and − 2piε < d(ε) < 2piε for everyε , 0 with a(0) = b(0) = c(0) = d(0) = 0.Note that, as a special case, one obtains the trivial solution d = 0, c =pi2, b = ε, a = −pi2.Next, the solution to the parallel parking problem is presented us-ing the methods described in Chapter 3 which can be applied to anyLie algebra.5.1.2 Solution using the operator methodAn operator representation for this Lie algebra is given byX = − ∂∂x, Y =∂∂y, R = y∂∂x− x ∂∂y.44Figure 5.1: Illustration of solution of parallel parking problem. (1) represents initialconfiguration of vehicle with successive configurations represented by (2) to (5).Consequently,M = ea(ε)Reb(ε)Y ec(ε)Red(ε)Y (x,y) = ea(ε)Reb(ε)Y ec(ε)R (x,y + d)= ea(ε)Reb(ε)Y (xcos c + y sin c ,d + y cos c − x sin c )= ea(ε)R (b sin c + x cos c + y sin c ,d + bcosc+ y cos c − x sin c )= (b sin c + x cos (a+ c) + y sin (a+ c) ,d + b cos c − x sin (a+ c)+ y cos (a+ c)).Then M = eεX (x,y) = (x − ε,y) iffsin (a+ c) = 0, cos (a+ c) = 1,d + b cos c = 0,−b sin c = ε. (5.1.2)45The solution to the system of equations (5.1.2) is given by (5.1.1).5.1.3 Solution using the matrix representation methodA matrix representation of S3,3 with constant r = 0 is given byX =0 0 10 0 00 0 0 , Y =0 0 00 0 −10 0 0 , R =0 −1 01 0 00 0 0 .Hence Xn = Y n = 0 , n ≥ 2.Thus eεY = I + εY =1 0 00 1 −ε0 0 1 , and eεX =1 0 ε0 1 00 0 1.For all k > 0, one has0 −1 01 0 00 0 0k=0 −1 01 0 00 0 0 if k = 1 (mod 4)−1 0 00 −1 00 0 0 if k = 2 (mod 4)0 1 0−1 0 00 0 0 if k = 3 (mod 4)1 0 00 1 00 0 0 if k = 4 (mod 4).Consequently,eεR =cos ε − sin ε 0sin ε cos ε 00 0 1 .46HenceM = ea(ε)Reb(ε)Y ec(ε)Red(ε)Y =cos (a+ c) − sin (a+ c) d sin (a+ c) + b sin asin (a+ c) cos (a+ c) −d cos (a+ c) − b cos a0 0 1 .Then M = eεX is satisfied iffcos (a+ c) = 1, sin (a+ c) = 0, b sin a = ε, d+b cos a = 0. (5.1.3)The solution to the system of equations (5.1.3) is given by (5.1.1).5.1.4 Solution using the DE methodFor all n ≥ 0, one can show thatXnY = YXn, XnR = RXn −nYXn−1,RnX = Xn∑i = 0i even(−1) i2(ni)Rn−i −Yn∑i = 1i odd(−1) i+12(ni)Rn−i ,RnY = Yn∑i = 0i even(−1) i2(ni)Rn−i +Xn∑i = 1i odd(−1) i+12(ni)Rn−i .(5.1.4)Using these results, one can easily obtain the identities47eεYX ≡ XeεY , eεYR ≡ (R− εX)eεY , eεRX ≡ ( cos ε X − sin ε Y )eεR,eεRY ≡ ( cos ε Y + sin ε X)eεR.(5.1.5)Now to proceed, we differentiate with respect to ε the equationea(ε)Reb(ε)Y ec(ε)Red(ε)Y = eεX . (5.1.6)Thusa′Rea(ε)Reb(ε)Y ec(ε)Red(ε)Y + b′ea(ε)RY eb(ε)Y ec(ε)Red(ε)Y+c′ea(ε)Reb(ε)YRec(ε)Red(ε)Y + d ′ea(ε)Reb(ε)Y ec(ε)RY ed(ε)Y = XeεX .(5.1.7)Using the identities in (5.1.5), one can show that equation (5.1.7)leads to the ODE systema′ + c′ = 0, cos a b′ + b sin a c′ + cos (a+ c) d ′ = 0,sin a b′ − b cos a c′ + sin (a+ c)d ′ = 1. (5.1.8)It is easy to show that the solution to the ODE system (5.1.8) isgiven by (5.1.1).5.2 Euler angles problem (Lie algebra so (3,R))5.2.1 Solution using the operator methodAn operator representation for this Lie algebra is given byX = x∂∂y− y ∂∂x, Y = y∂∂z− z ∂∂y, Z = x∂∂z− z ∂∂x.48Here setting ea(ε)Xeb(ε)Y ec(ε)Xed(ε)Y = eεZ leads to the system of nineequationssin c sin b = − sin ε , cos c − cos a cos ε = 0, sin c cos b +sin a cos ε = 0,cos d sin c + sin a = 0, cos b cos c cosd − sin b sin d = cos a ,sin b cos c cos d + cos b sin d = 0, sin b cos d +cos b cos c sin d = sin a sin ε ,sin c sin d − cos a sin ε = 0, cos b cos d− sin b cos c sin d = cos ε ,(5.2.1)whose solutions are given bya(ε) = −arccos(cos c (ε)cos ε), b(ε) = −arcsin(sin εsin c (ε)),d(ε) = arccos(− sin a(ε)sin c(ε));(5.2.2)a(ε) = arccos(cos c (ε)cos ε), b(ε) = pi+ arcsin(sin εsin c (ε)),d(ε) = arccos(− sin a(ε)sin c(ε)).(5.2.3)In both solutions, for any ε , 0, c(ε) is any continuous functionchosen so that a(ε) and d(ε) are continuous, and satisfying∣∣∣∣ sinεsin c(ε) ∣∣∣∣ ≤1 with c (ε) , kpi for every k inZ and such that a(0) = b(0) = c(0) =d(0) = 0.495.2.2 Solution using the matrix representation methodA matrix representation of so(3,R) is given byX =0 0 00 0 −10 1 0 , Y =0 0 10 0 0−1 0 0 , Z =0 −1 01 0 00 0 0 .Analogous to the way of obtaining the rotation matrix ecR in theparking problem, one finds thateεX =1 0 00 cos ε − sin ε0 sin ε cos ε , eεY =cos ε 0 sin ε0 1 0− sin ε 0 cos ε ,eεZ =cos ε − sin ε 0sin ε cos ε 00 0 1 .Consequently, one can show that the entries{aij}of the matrixM =ea(ε)Xeb(ε)Y ec(ε)Xed(ε)Y are given bya11 = cos b cos d − sin b cos c sin d , a12 = sin b sin c ,a13 = cos b sin d + sin b cos c cos d ,a21 = sin a sin b cos d + cos a sin c sin d +sin a cos b cos c sin d ,a22 = cos a cos c − sin a cos b sin c ,a23 = sin a sin b sin d − cos a sin c cos d −sin a cos b cos c cos d ,a31 = − cos a sin b cos d + sin a sin c sin d −cos a cos b cos c sin a ,a32 = sin a cos c + sin c cos a cos b ,a33 = − sin d cos a sin b − sin a sin c cos d +cos a cos b cos c cos d .50From the matrix equation M = eεZ , one obtains a system of equa-tions that can be simplified to (5.2.1). Hence the solutions are givenby (5.2.2) and (5.2.3).5.2.3 Solution using the DE methodOne can show the following identities hold for all ε.eεXY ≡ ( cos ε Y + sin ε Z)eεX , eεXZ ≡ ( cos ε Z − sin ε Y )eεX ,eεYX ≡ ( cos ε X − sin ε Z)eεY , eεYZ ≡ ( cos ε Z + sin ε X)eεY .(5.2.4)After differentiating with respect to ε the equationea(ε)Xeb(ε)Y ec(ε)Xed(ε)Y = eεZ and using the identities in (5.2.4), one ob-tains the simplified ODE systemcos b a′ + c′ = − sin b cos a , sin b a′ + sin c d ′= cos a cos b, b′ + cos c d ′= sin a .(5.2.5)The ODE system (5.2.5) admits (5.2.2) and (5.2.3) as solutions.5.3 Lie algebra n3,15.3.1 Solution using the operator methodFrom [12], an operator representation for n3,1is given byX = ∂∂x , Y = x∂∂z , Z =∂∂z .Consequently, equation ea(ε)Xeb(ε)Y ec(ε)Xed(ε)Y (x,z) = eεZ (x,z) leadsto the equation(x+ a+ c, (b+ d) (x+ a) + dc+ z) = (x,z+ ε) . (5.3.1)51It is easy to see that the solution to equation (5.3.1) is given bya (ε) = − εd (ε), b (ε) = −d (ε) , c (ε) = εd (ε), (5.3.2)where d (ε) is any continuous function chosen so that a(ε) and c(ε)are continuous functions, and satisfying d (ε) , 0 for any ε , 0 witha(0) = b(0) = c(0) = d(0) = 0.5.3.2 Solution using the matrix representation methodFrom [9], a matrix representation of n3,1is given byX =0 1 00 0 00 0 0 , Y =0 0 00 0 10 0 0 , Z =0 0 10 0 00 0 0 .Hence X2 = Y 2 = Z2 = 0. Then one can show thateεX =1 ε 00 1 00 0 1 , eεY =1 0 00 1 ε0 0 1 , eεZ =1 0 ε0 1 00 0 1 .Accordingly, one can show thatM = ea(ε)Xeb(ε)Y ec(ε)Xed(ε)Y =1 a+ c cd + ad + ab0 1 b+ d0 0 1 .Consequently M = eεZ yields the system of equationsa+ c = 0, b+ d = 0, cd + ad + ab = ε,whose solution is given by (5.3.2).525.3.3 Solution using the DE methodOne can readily obtain the following identities which hold for all ε.eεXY ≡ (X + εZ)eεX , eεXZ ≡ ZeεX , eεYX ≡ (Y − εZ)eεY , eεYZ ≡ ZeεY .(5.3.3)After differentiating with respect to ε the equationea(ε)Xeb(ε)Y ec(ε)Xed(ε)Z = eεZ and using the identities (5.3.3), one ob-tains the ODE system a′ + c′ = 0, b′ + d ′ = 0, ab′ − bc′ + (a+c)d ′ = 1,whose solution is given by (5.3.2).5.4 Lie algebra S3,15.4.1 Solution using the operator methodAn operator representation [12] for S3,1 is given byX =∂∂x+∂∂y, Y = (1− r) ∂∂x, Z = −x ∂∂x− ry ∂∂y.Consequently, the equation ea(ε)Xeb(ε)Zec(ε)Xed(ε)Z (x,y) = eεY (x,y)leads to the equation(e−(b+d)(x+ a+ ceb), e−r(b+d)(y + a+ cerb))= ((1− r)ε+ x,y). (5.4.1)It is easy to see that equation (5.4.1) is satisfied iffb = −d, a+ ce−d = (1− r)ε, a+ ce−rd = 0. (5.4.2)The solution to the system of equations (5.4.2) is given bya (ε) =ε(1− r)1− e(r−1)d(ε) , b (ε) = −d (ε) , c (ε) = −a erd(ε), (5.4.3)where d (ε) is any continuous function chosen so that a(ε) is a con-tinuous function, and satisfying d (ε) , 0 for any ε , 0 with a(0) =b(0) = c(0) = d(0) = 0.53Note that in the limiting case where r → 1, equation (5.4.3) be-comes a (ε) = − εd(ε) , b (ε) = −d (ε), and c (ε) = −a ed(ε).5.4.2 Solution using the matrix representation methodA matrix representation of S3,1 is given byX =0 −r 00 0 00 −1 0 , Y =0 0 00 0 00 −1 0 , Z =r 0 00 0 01 0 1 .Hence for all n ≥ 2 one has Xn = Y n = 0 and Zn =rn 0 00 0 0rn−1 + rn−2 + ...+ r + 1 0 1.Then one can easily show thateεX =1 −rε 00 1 00 −ε 1 , eεY =1 0 00 1 00 −ε 1 , eεZ =eεr 0 00 1 0eε−eεr1−r 0 eε .HenceM = ea(ε)Xeb(ε)Zec(ε)Xed(ε)Z =er(b+d) −ra− crebr 00 1 0eb+d−er(b+d)1−rcr(eb−ebr )r−1 − a− ceb eb+d .Then equation M = eεY leads to the system of equationsa+ cebr = 0, b+ d = 0,cr(eb − ebr)1− r + a+ ceb = ε, (5.4.4)with solution given by (5.4.3).545.4.3 Solution using the DE methodOne can show that the following identities hold for all ε.eεXY ≡ Y eεX , eεXZ ≡ (Z − rεX − εY )eεX ,eεZX ≡ (eεrX + eε − eεr1− r Y )eεZ , eεZY ≡ (Y + εY )eεZ . (5.4.5)After differentiating with respect to ε the equationea(ε)Xeb(ε)Zec(ε)Xed(ε)Z = eεY and using the identities (5.4.5), one ob-tains the ODE systema′−rab′+erbc′−r(a+cerb)d ′ = 0, b′+d ′ = 0, eb − erb1− r (c′ − rcd ′)−cebd ′ = 1,(5.4.6)whose solution is given by (5.4.3).5.5 Lie algebra S3,25.5.1 Solution using the operator methodFrom [12], an operator representation for S3,2 is given byX =∂∂x , Y =y ∂∂x , Z = −x ∂∂x + ∂∂y .Consequently, equation ea(ε)Y eb(ε)Zec(ε)Y ed(ε)Z (x,y) = eεX (x,y) leadsto equation(e−(b+d)x+ e−(b+d)(a+ ceb)y + bce−d , y + b+ d)= (x+ ε,y). (5.5.1)It is easy to see that equation (5.5.1) is satisfied iffb = −d, a+ ce−d = 0, bce−d = ε. (5.5.2)The solution to the system of equations (5.5.2) is given bya (ε) =εd(ε), b (ε) = −d (ε) , c (ε) = −εed(ε)d(ε), (5.5.3)55where d (ε) is any continuous function chosen so that a(ε) and c(ε)are continuous functions, and satisfying d (ε) , 0 for any ε , 0 witha(0) = b(0) = c(0) = d(0) = 0.5.5.2 Solution using the matrix representation methodA matrix representation for S3,2 is given byX =0 0 −10 0 00 0 0 , Y =0 0 −10 0 −10 0 0 , Z =1 1 00 1 00 0 0 .Hence for all n ≥ 2 one has Xn = Y n = 0 and Zn =1 n 00 1 00 0 0.Then one can easily show thateεX =1 0 −ε0 1 00 0 1 , eεY =1 0 −ε0 1 −ε0 0 1 , eεZ =eε εeε 00 eε 00 0 1 .HenceM = ea(ε)Y eb(ε)Zec(ε)Y ed(ε)Z =eb+d eb+d(b+ d) −ceb − a− bceb0 eb+d −ceb − a0 0 1 .Consequently, the equation M = eεX leads to the system of equa-tionsceb + a = 0, b+ d = 0, ceb + a+ bceb = ε,whose solution is given by (5.5.3).565.5.3 Solution using the DE methodOne can show that the following identities hold for all ε.eεYX ≡ XeεY , eεYZ ≡ (Z − εX − εY )eεY , eεZX ≡ (eεX)eεZ ,eεZY ≡ (eεY + εeεX)eεZ . (5.5.4)The differentiation with respect to ε of the equationea(ε)Y eb(ε)Zec(ε)Y ed(ε)Z = eεX and the repeated use of the identities(5.5.4) leads to the ODE systema′−ab′+ebc′− (a+ceb)d ′ = 0, b′+d ′ = 0, −ab′+bebc′− (a+cbeb)d ′ = 1,whose solution is given by (5.5.3).5.6 Lie algebra S3,35.6.1 Solution using the operator methodFrom [8] and [12], an operator representation for S3,3 is given byX = − ∂∂x, Y = − ∂∂y, R = (rx+ y)∂∂x+ (ry − x) ∂∂y.Consequently, equation ea(ε)Reb(ε)Y ec(ε)Red(ε)Y (x,y) = eεX (x,y) leadsto equation( cos (a+ c) e−r(a+c)x − sin (a+ c) e−r(a+c)y + b sin c e−rc,cos (a+ c) e−r(a+c) y + sin (a+ c) e−r(a+c)x − d − b cos c e−rc ) =(x − ε,y).(5.6.1)It is easy to see that equation (5.6.1) is satisfied iffsin (a+ c)e−r(a+c) = 0, cos (a+ c)e−r(a+c) = 1, d + b cos c e−rc = 0,b sin c e−rc = −ε.(5.6.2)57The solution to the system of equations (5.6.2) is given bya = −c(ε), b = − εerc(ε)sin c (ε), d =εtan c (ε), (5.6.3)where c (ε) is any continuous function chosen so that b(ε) and d(ε)are continuous functions, and satisfying c (ε) , kpi for every k ∈ Zand for any ε , 0 with a(0) = b(0) = c(0) = d(0) = 0.5.6.2 Solution using the matrix representation methodA matrix representation of S3,3 is given byX =0 0 10 0 −r0 0 0 , Y =0 0 −r0 0 −10 0 0 , R =r −1 01 r 00 0 0 .Hence one can show thateεX =1 0 ε0 1 −rε0 0 1 , eεY =1 0 −rε0 1 −ε0 0 1 ,eεR =cos ε erε − sin ε erε 0sin ε erε cos ε erε 00 0 1 .Consequently, the entries{aij}of the matrixM = ea(ε)Reb(ε)Y ec(ε)Red(ε)Yare given bya11= a22 = cos (a+ c)er(a+c) , a12 = − sin (a+ c)er(a+c) ,a21 = sin (a+ c)er(a+c) ,a13 = d(−r cos (a+ c) + sin (a+ c))er(a+c) + b(−r cos a + sin a)era ,a23 = −d( cos (a+ c) + r sin (a+ c))er(a+c) − b( cos a + r sin a)era ,a31 = a32 = 0, a33 = 1.58Thus the relation M = eεX yields the equationscos (a+ c)er(a+c) = 1, sin (a+ c) er(a+c) = 0,d(−r cos (a+ c) + sin (a+ c))er(a+c) + b(−r cos a + sin a)era = ε,d( cos (a+ c) + r sin (a+ c))er(a+c) + b( cos a + r sin a)era = rε,whose solution is given by (5.6.3).5.6.3 Solution using the DE methodOne can show that the following identities hold for all ε.eεYX ≡ XeεY , eεYR ≡ (R− εX − rεY )eεY ,eεRX ≡ erε( cos ε X − sin ε Y )eεR,eεRY ≡ erε( cos ε Y + sin ε X)eεR.(5.6.4)To proceed, one differentiates with respect to ε the equationea(ε)Reb(ε)Y ec(ε)Red(ε)Y = eεX and then uses the identities (5.6.4) recur-sively. This yields the ODE systema′ + c′ = 0,era(cos a b′ + (b sin a − rb cos a )c′ + cos (a+ c)ercd ′)= 0,era ( sin a b′ − (rb sin a + b cos a )c′ + sin (a+ c)ercd ′ ) = 1.(5.6.5)It is easy to show that the solution to the ODE system (5.6.5) isgiven by (5.6.3).59Chapter 6Four-dimensional Lie algebrasIn this chapter, we present the proofs of the results presented inTable 2.6.1 Lie algebra S4,26.1.1 Solution using the operator methodFrom [12], an operator representation for S4,2 is given byW = −x ∂∂x+∂∂y+ y∂∂z, X =∂∂x, Y = y∂∂x, Z = z∂∂x.Consequently, equationea(ε)W eb(ε)Zec(ε)W ed(ε)Zef (ε)W (x,y,z) = eεY (x,y,z) leads to((cd + ad + abe−c)e−f y +((d + be−c)e−f)z+ e−(a+f +c)x+ (acd + 12c2d)e−f ,y + a+ c+ f , (a+ f + c)y + z+ 12(a+ f + c)2) = (x+ εy,y,z).(6.1.1)It is easy to see that equation (6.1.1) is satisfied iffa+ f + c = 0, d + be−c = 0, cde−f = ε, a+ c = 0. (6.1.2)60The solution to the system of equations (6.1.2) is given bya (ε) = f (ε) = −12c(ε), b (ε) = −εc (ε)e12c(ε), d (ε) =εc (ε)e−12c(ε), (6.1.3)where c (ε) is any continuous function chosen so that b(ε) and d(ε)are continuous functions, and satisfying c (ε) , 0 for any ε , 0 witha(0) = b(0) = c(0) = d(0) = f (0) = 0.6.1.2 Solution using the matrix representation methodA matrix representation for S4,2 is given byX =0 0 0 −10 0 0 00 0 0 00 0 0 0 , Y =0 0 0 −10 0 0 −10 0 0 00 0 0 0 , Z =0 0 0 00 0 0 −10 0 0 −10 0 0 0 ,W =1 1 0 00 1 1 00 0 1 00 0 0 0 .Hence for all n ≥ 2 one has Y n = Zn = 0 and W n =1 n n(n−1)2 00 1 n 00 0 0 00 0 0 0.Thus one can show thateεY =1 0 0 −ε0 1 0 −ε0 0 1 00 0 0 1 , eεZ =1 0 0 00 1 0 −ε0 0 1 −ε0 0 0 1 ,61eεW =eε εeε 12ε2eε 00 eε εeε 00 0 eε 00 0 0 1 .Consequently, the entries{aij}of the matrixM = ea()W eb()Zec()W ed()Zef (ε)W are given bya11 = a22 = a33 = ea+c+f , a12 = a23 = (a+ c+ f )ea+c+f ,a13 =(af + f c+ ac+ 12(a2 + f 2 + c2))ea+c+f ,a14 = −d(a+ c+ ac+ 12(a2 + c2))ea+c − ba(1 + 12a)ea,a24 = −d (1 + c+ a)ea+c − b (1 + a)ea,a34 = −dea+c − bea, a44 = 1, aij = 0 for j < i.After simplification, the relation M = eεY leads to the system ofequationsa+ c+ f = 0, d(a+ c+ ac+ 12(a2 + c2))ea+c + ba(1 + 12a)ea = ε,d (1 + a+ c)ea+c + b (1 + a)ea = ε, dea+c + bea = 0, whose solution isgiven by (6.1.3).6.1.3 Solution using the DE methodOne can show that the following identities hold for all ε.eεWX ≡ eεXeεW , eεWY ≡ (eεY + εeεX)eεW ,eεWZ ≡(eεZ + εeεY + 12ε2eεX)eεW ,eεZX ≡ XeεZ , eεZY ≡ Y eεZ , eεZW ≡ (W − εZ − εY )eεZ .(6.1.4)62After differentiating with respect to ε the equationea(ε)W eb(ε)Zec(ε)W ed(ε)Zef (ε)W = eεY and using the identities (6.1.4), oneobtains the system of differential equationsa′ + c′ + f ′ = 0,a2eab′ − ba(2 + a)eac′ + (a+ c)2ea+cd ′ − (2ab+ 2(a+ c)dec + ba2+ d(a+ c)2ec)eaf ′ = 0,eab′ − beac′ + ea+cd ′ − (bea + dea+c)f ′ = 0,aeab′ − b(1 + a)eac′ + (a+ c)ea+cd ′ − (b(1 + a) + d(1 + a+ c)ec)eaf ′ = 1.(6.1.5)The solution to the ODE system (6.1.5) is given by (6.1.3).6.2 Lie algebra S4,76.2.1 Solution using the operator methodFrom [12], an operator representation for S4,7 is given byW =12(y2 − z2) ∂∂x− z ∂∂y+ y∂∂z, X =∂∂x, Y =∂∂y, Z = y∂∂x+∂∂z.One can show that for all ε the following relations hold.eεY (x,y,z) = (x,y + ε,z), eεZ (x,y,z) = (x+ εy,y,z+ ε),eεW (x,y,z) = (14( sin (2ε)(y2 − z2) + 2 cos (2ε) zy) + x − 12zy,cos ε y − sin ε z, cos ε z + sin ε y).After much calculation and the use of the relations above, one canshow that the equation63ea(ε)Zeb(ε)W ec(ε)Zed(ε)W ef (ε)Z (x,y,z) = eεY (x,y,z) leads to the simplifiedsystem of equationsf +c = 0, d sin f + ε = 0, b+g+d cos f = 0, g sin f − 12ε cos f = 0.(6.2.1)The solution to the system of equations (6.2.1) is given bya(ε) = f (ε) = − ε2tanb(ε), c(ε) =εsinb(ε), d(ε) = −b(ε) , (6.2.2)where b(ε) is any continuous function chosen so that a(ε), f (ε) andc(ε) are continuous functions, and satisfying b(ε) , 0 for every ε , 0with a(0) = b(0) = c(0) = d(0) = f (0) = 0.6.2.2 Solution using the matrix representation methodThe following matrix representation for S4,7 was found after correc-tion of the matrix representation of S4,7 given in [8].eεY =1 0 −ε 00 1 0 −ε0 0 1 00 0 0 1 , eεZ =1 ε 0 00 1 0 00 0 1 −ε0 0 0 1 ,eεW =1 0 0 00 cos ε sin ε 00 − sin ε cos ε 00 0 0 1 .Consequently the entries{aij}of the matrixM = ea(ε)Zeb(ε)W ec(ε)Zed(ε)W ef (ε)Z are found to be64a11 = a44 = 1, a12 = f + c cos d + a cos (b+ d) ,a13 = c sin d + a sin (b+ d) ,a14 = −f c sin d − af sin (b+ d) − ac sin b , a22 = a33 = cos (b+ d),a23 = −a32 = sin (b+ d), a24 = −f sin (b+ d) − c sin b ,a34 = −f cos (b+ d) − c cos b − b, a21 = a31 = a41 = a42 = a43 = 0.Consequently, the relation M = eεY yields the system of equationsb+ d = 0, f c sin d + ac sin b = 0, c sin d = −ε,f + c cos d + a = 0.whose solution is given by (6.2.1).6.2.3 Solution using the DE methodThe following identities hold for all ε.eεWY ≡ ( cos ε Y − sin ε Z)eεW , eεWZ ≡ ( cos ε Z + sin ε Y )eεW ,eεZY ≡ (Y − εX)eεZ , eεZW ≡ (W − εY + 12ε2X)eεZ .(6.2.3)After differentiating with respect to ε the equationea(ε)Zeb(ε)W ec(ε)Zed(ε)W ef (ε)Z = eεY and repeatedly using the identities(6.2.3), one obtains the ODE systemb′ + d ′ = 0, a′ + cos b c′ + c sin b d ′ + cos (b+ d) f ′ = 0,− ab′ + sin b c′ − (a+ c cos b) d ′ + sin (b+ d) f ′ = 1,12a2b′ − a sin b c′ +(ac cos b+ 12(a2 + c2))d ′− (c sin d + a sin (b+ d)) f ′ = 0.(6.2.4)The solution to the ODE system (6.2.4) is given by (6.2.1).656.3 Lie algebra S4,96.3.1 Solution using the operator methodFrom [12], an operator representation for S4,9 is given byW = 12(y2 − z2 − 4rx) ∂∂x− (ry + z) ∂∂y+ (y − rz) ∂∂z, X =∂∂x, Y =∂∂y,Z = y∂∂x+∂∂z.One can show that for all ε the following relations hold.eεY (x,y,z) = (x,y + ε,z), eεZ (x,y,z) = (x+ εy,y,z+ ε),eεW (x,y,z) = (14( sin (2ε)(y2 − z2) + 2 cos (2ε) zy) + x − 12zy)e−2εr ,e−εr( cos ε y − sin ε z), e−εr( cos ε z+ sin ε y)).These relations allow one to show that the equationea(ε)Zeb(ε)W ec(ε)Zed(ε)W ef (ε)Z (x,y,z) = eεY (x,y,z) leads to the equationsf + c = 0, d sin f + εef r = 0, b+ g + d cos f e−f r = 0,2g sin f − ε cos f = 0. (6.3.1)The solution to the system of equations (6.3.1) is given bya(ε) = f (ε) = − ε2tanb(ε), c(ε) =εe−rbsinb(ε), d(ε) = −b(ε) , (6.3.2)where b(ε) is any continuous function chosen so that a(ε), f (ε) andc(ε) are continuous functions, and satisfying b(ε) , 0 for any ε , 0with a(0) = b(0) = c(0) = d(0) = f (0) = 0.666.3.2 Solution using the matrix representation methodA matrix representation for S4,9 is given byX =0 0 0 −2r0 0 0 00 0 0 00 0 0 0 , Y =0 0 1 00 0 0 −r0 0 0 10 0 0 0 , Z =0 −1 0 00 0 0 −10 0 0 −r0 0 0 0 ,W =2r 0 0 00 r 1 00 −1 r 00 0 0 0 .Hence for all n ≥ 3 one has Y n = Zn = 0 and Y 2 = Z2 =0 0 0 10 0 0 00 0 0 00 0 0 0.Thus the corresponding matrix representation of the associated Liegroup is given byeεY =1 0 ε 12ε20 1 0 −rε0 0 1 ε0 0 0 1 , eεZ =1 −ε 0 12ε20 1 0 −ε0 0 1 −rε0 0 0 1 ,Analogous to the way of obtaining the rotation matrix ecR in theparking problem, one finds thateεW =e2rε 0 0 00 cos ε erε sin ε erε 00 − sin ε erε cos ε erε 00 0 0 1 .67Consequently the entries{aij}of the matrixM = ea(ε)Zeb(ε)W ec(ε)Zed(ε)W ef (ε)Z are given bya11 = e2r(b+d),a12 = −f e2r(b+d) − c cos d er(2b+d) − a cos (b+ d)er(b+d) ,a13 = −c sin der(2b+d) − a sin (b+ d) er(b+d),a14 =12f2e2r(b+d)+ cf cos d er(2b+d) + rcf sin d er(2b+d)+af cos (b+ d)er(b+d) + raf sin (b+ d)er(b+d) + 12c2e2rb + ac cos b erb+rac sin b erb + 12a2,a22 = a33 = cos (b+ d)er(b+d), a23 = −a32 = sin (b+ d)er(b+d),a24 = −f cos (b+ d)er(b+d) − rf sin (b+ d)er(b+d) − c cos b erb− rc sin b erb − a,a34 = f sin (b+ d)er(b+d) − rf cos (b+ d)er(b+d) + c sin b erb− rc cos b erb − rb,a44 = 1, a21 = a31 = a41 = a42 = a43 = 0.The relation M = eεY yields, after simplification, the system ofequationsb+ d = 0, c sin b erb = ε, f + c cos d erb + a = 0,12f2 + (a+ f )c cos b erb + r(a− f )c sin b erb + af + 12c2e2rb + 12a2 = 12ε2.(6.3.3)The solution to the system of equations (6.3.3) is given by (6.3.1).686.3.3 Solution using the DE methodThe following identities hold for all ε.eεWY ≡ (cosεerε Y − sinε erεZ)eεW , eεWZ ≡ (cosε erεZ + sinεerε Y )eεW ,eεWX ≡ e2rεXeεW , eεZY ≡ (Y − εX)eεZ ,eεZW ≡ (W − rεZ − εY + 12ε2X)eεZ , eεZX ≡ XeεZ .(6.3.4)After differentiating with respect to ε the equationea(ε)Zeb(ε)W ec(ε)Zed(ε)W ef (ε)Z = eεY and using the identities (6.3.4) oneobtains the ODE systemb′ + d ′ = 0,a′ + cos b erbc′ + (−rc cos b erb + c sin b erb) d ′ + er(b+d)f ′ = 0,−ab′ + sin b erbc′ + (−rc sin b erb − c cos b erb) d ′ = 1,− a sin b erbc′ +(rac sin b erb + ac cos b erb + 12c2e2rb)d ′− c sin d erb f ′ = 0.(6.3.5)The solution to the ODE system (6.3.4) is given by (6.3.1).6.4 Lie algebra S4,106.4.1 Solution using the operator methodFrom [12], an operator representation for S4,10 is given byW = −2x ∂∂x− y ∂∂y+∂∂z, X =∂∂x, Y =∂∂y, Z = y∂∂x+ z∂∂y.69Consequently, the equationea(ε)W eb(ε)Zec(ε)W ed(ε)Zef (ε)W eg(ε)W (x,y,z) = eεY (x,y,z) leads to the sim-plified system of equationsa+ c+ f = 0, g + dea+c + bea = 0, cde−f + cg + f g = ε,d2ce−2f+ 2gdce−f − ag2 = 0, (6.4.1)whose solutions are given byf (ε) = − (a (ε) + c (ε)) , b(ε) = εe−ac (ε)±√c (ε)2 + a (ε)c (ε)a (ε)c (ε) ,g (ε) = −ε+ b (ε)c (ε)ea(ε)a (ε) + c (ε), d (ε) = −g (ε)e−(a(ε)+c(ε)) − b (ε)e−c(ε),(6.4.2)where a (ε) and c (ε) are any continuous function chosen so thatb(ε),d(ε) and g(ε) are continuous functions, and satisfying a (ε)c (ε) ,0, a (ε) + c (ε) , 0, and c (ε)2 + a (ε)c (ε) ≥ 0 for any ε , 0 with a(0) =b(0) = c(0) = d(0) = f (0) = g(0) = 0.In the limiting case when a (ε) = 0,f (ε) = −c (ε) , d (ε) = εe−c(ε)c(ε) , b (ε) = g (ε) = − ε2c(ε) , with c (ε) any con-tinuous function chosen so that b(ε), g(ε) and d(ε) are continuousfunctions, and c (ε) , 0 when ε , 0.706.4.2 Solution using the matrix representation methodA matrix representation of S4,10 is given byX =0 0 0 −20 0 0 00 0 0 00 0 0 0 , Y =0 0 1 00 0 0 −10 0 0 00 0 0 0 , Z =0 −1 0 00 0 0 −10 0 0 −10 0 0 0 ,W =2 0 0 00 1 1 00 0 1 00 0 0 0 .Hence for all n ≥ 2 one has Xn = Y n = 0 and Z2 =0 0 0 10 0 0 00 0 0 00 0 0 0.For all n ≥ 3 Z3 = 0 and W n =2n 0 0 10 1 n 00 0 1 00 0 0 0.One can show thateεX =1 0 0 −2ε0 1 0 00 0 1 00 0 0 1 , eεY =1 0 ε 00 1 0 −ε0 0 1 00 0 0 1 , eεZ =1 −ε 0 12ε20 1 0 −ε0 0 1 −ε0 0 0 1 ,eεW =e2ε 0 0 00 eε εeε 00 0 eε 00 0 0 1 .71Then the matrix M = ea()W eb()Zec()W ed()Zef (ε)W eg(ε)Z has entries{aij}given bya11 = e2(a+c+f ), a12 = −(gec+f + dec + b)e2a+c+f ,a13 = − (f dec + f b+ bc)e2a+c+f ,a14 =(12g2ec+f+ g (1 + f )dec + gb (1 + f ) + bcg)e2a+c+f +(12d2e2c+ bd (1 + c)ec + 12b2)e2a,a22 = a33 = ea+c+f , a23 = (a+ c+ f )ea+c+f ,a24 = −g (1 + a+ c+ f )ea+c+f − (d (1 + a+ c)ec + b (1 + a))ea,a34 = −(gea+c+f + dea+c + bea), a44 = 1, aij = 0 for j < i.Consequently, after simplification, the relation M = eεY leads tothe system of equations (6.4.1) whose solutions are given by (6.4.2).6.4.3 Solution using the DE methodOne can show that the following identities hold for all ε.eεWX ≡ (e2εX)eεW , eεWY ≡ (eεY )eεW , eεWZ ≡ (eεZ + εeεY )eεW ,eεZX ≡ XeεZ , eεZY ≡ (Y − εX)eεZ , eεZW ≡ (W − εZ − εY + 12ε2X)eεZ .(6.4.3)After differentiating with respect to ε the equationea(ε)W eb(ε)Zec(ε)W ed(ε)Zef (ε)W eg(ε)Z = eεY and using the identities in72(6.4.3), one obtains the ODE systema′ + c′ + f ′ = 0,ea(ab′ − b (1 + a)c′ + (a+ c)ecd ′ − (ab+ adec + b+ d (1 + c)ec)f ′) = 1,eab′ − beac′ + ea+cd ′ − (b+ ecd )eaf ′ + g ′ = 0,12b2e2ac′ − bce2a+cd ′ +(12(b2 + d2e2c) + bd(1 + c)ec)e2af ′−((bf + bc)ef +c+2a + df ef +2c+2a)g ′ = 0.(6.4.4)The solution to ODE system (6.4.3) is given by (6.4.2).73Chapter 7Inverse ProblemHere we show that reversing the order of elements in the LHS ofboth equation (3.2) and equation (3.6) leads to an isomorphic solu-tion if the number of elements that are not identically zero is even.Additionally, we present the isomorphism relating the inverse prob-lem to the original problem. Then we present a table that shows theresults for the inverse problems for all three-dimensional Lie alge-bras. Finally, we will examine the relation between the solution tothe parking problem and the solution to its inverse problem.The original problem is given byea(ε)B1eb(ε)B2ec(ε)B1ed(ε)B2ef (ε)B1eg(ε)B2 = eεB3. (7.0.1)To obtain the solution of the inverse problem from equation ( 7.0.1),we first multiply equation ( 7.0.1) bye−g(ε)B2e−f (ε)B1e−d(ε)B2e−c(ε)B1e−b(ε)B2e−a(ε)B1. This yields1 = e−g(ε)B2e−f (ε)B1e−d(ε)B2e−c(ε)B1e−b(ε)B2e−a(ε)B1eεB3.74Next, we multiply both sides by e−εB3. This leads toe−g(ε)B2e−f (ε)B1e−d(ε)B2e−c(ε)B1e−b(ε)B2e−a(ε)B1 = e−εB3. (7.0.2)Thus the inverse problem given byea˜(ε)B2eb˜(ε)B1ec˜(ε)B2ed˜(ε)B1ef˜ (ε)B2eg˜(ε)B1 = eεB3. (7.0.3)has a solution isomorphic to that of equation (7.0.1) and the isomor-phism is given bya˜(ε) = −g(−ε),b˜(ε) = −f (−ε),c˜(ε) = −d(−ε),d˜(ε) = −c(−ε),f˜ (ε) = −b(−ε),g˜(ε) = −a(−ε).(7.0.4)The solution to the inverse problem is presented in the followingtable for all relevant Lie algebras75Lie algebra; com-mutatorsComposition equation Solutionsl (2,R)[X,Y ] = Z[X,Z] = −2X[Y ,Z] = 2Yea(ε)Y eb(ε)Xec(ε)Y ed(ε)X = eεZ d (ε) is an arbitrary functionsatisfying d (ε) , 0 when ε , 0a(ε) = −d (ε)b (ε) = − e−ε−1d(ε)c (ε) = − eε−1e−ε−1d(ε)Parallel parkingproblem,S3,3 withconstant r = 0[R,Y ] = X[R,X] = −Y[X,Y ] = 0ea(ε)Y eb(ε)Rec(ε)Y ed(ε)R = eεX d (ε) is an arbitrary functionsatisfying d(ε) , kpi for everyk ∈Z when ε , 0a (ε) = εcotd(ε)b (ε) = −d(ε)c (ε) = −εcscd(ε)Euler anglesproblem,so(3,R)[X,Y ] = Z[X,Z] = −Y[Y ,Z] = Xea(ε)Y eb(ε)Xec(ε)Y ed(ε)X = eεZ Any b (ε) satisfying b (ε) , kpifor every k ∈ Z when ε , 0and∣∣∣∣ sin εsin b(ε) ∣∣∣∣ ≤ 1 witha (ε) = −arccos(sin d(ε)sin b(ε))c (ε) = arcsin(sin εsin b(ε))d (ε) = arccos(cos b(ε)cos ε);a (ε) = −arccos(sin d(ε)sin b(ε))c (ε) = pi − arcsin(sin εsinb(ε))d (ε) = −arccos(cos b(ε)cos ε)n3,1[X,Y ] = Z[X,Z] = 0[Z,Y ] = 0ea(ε)Y eb(ε)Xec(ε)Y ed(ε)X = eεZ d (ε) is an arbitrary functionsatisfying d (ε) , 0 when ε , 0a(ε) = εd(ε)b (ε) = −d (ε)c (ε) = − εd(ε)S3,1[Y ,Z] = −Y[Y ,X] = 0[Z,X] = rX +Ywhere r is a con-stant satisfying|r | ≤ 1ea(ε)Zeb(ε)Xec(ε)Zed(ε)X = eεY a (ε) is an arbitrary functionsatisfying a (ε) , 0 when ε , 0d (ε) = ε(1−r)1−e(1−r)a(ε)b (ε) = d(ε) e−ra(ε)c (ε) = −a (ε)S3,2[Z,X] = X[Z,Y ] = X +Y[X,Y ] = 0ea(ε)Zeb(ε)Y ec(ε)Zed(ε)Y = eεX d (ε) is an arbitrary functionsatisfying d (ε) , 0 when ε , 0a (ε) = − εd(ε)b (ε) = −d(ε)e−a(ε)c (ε) = εd(ε)S3,3 general case[R,X] = rX −Y[R,Y ] = X + rY[X,Y ] = 0where r is a non-negative constant.ea(ε)Y eb(ε)Rec(ε)Y ed(ε)R = eεX d (ε) is an arbitrary functionsatisfying d(ε) , kpi for everyk ∈Z when ε , 0a (ε) = εcotd(ε)b (ε) = −d(ε)c (ε) = −εcscd(ε)erd(ε)Table 7.1: Results for the Inverse Problem767.1 The inverse parking problemAs mentioned in Chapter 5 the parking problem models a vehiclethat aims to perform a parallel translation in x by ε through trans-lations in y and rotations starting with a translation. This was illus-trated in Figure (5.1) and the solution was given by equation (5.1.1).By applying the isomorphism given in equation (7.1.1), one can seethat the solution to the inverse problem, i.e, parking which startswith a rotation, is given bya1 (ε) = −εcotb1(ε), c1 (ε) = εcscb1(ε), d1 (ε) = −b1(ε), (7.1.1)where b1 (ε) is an arbitrary function satisfying b1 (ε) , kpi for every k∈Z when ε , 0.In the diagram below we represent both solutions to the parkingproblem. In red is the solution that starts with an arbitrary trans-lation while in blue is the inverse solution that begins with an arbi-trary rotation. As one can see, such solutions are reflections of eachother with respect to the line x =ε2. In practice, one solution mightbe more desirable than the other depending on other constraintsthat one might have. The solution in red would be most useful ina situation where one has to park between two vehicles with lim-ited space between the vehicles. The inverse solution in blue, on theother hand, is practical when there is a lot of parking space avail-able. The red solution requires a more skilled driver since it requiresa precise rotation after the arbitrary translation.77Figure 7.1: Illustration of the solutions of the parallel parking problem and its inverse.78Chapter 8Discussion and conclusionsIn this thesis, for all relevant three- and four-dimensional Lie alge-bras, we have shown explicitly how one can obtain elements of theassociated Lie groups as compositions of products of other elementsfrom the commutator properties of their Lie algebras. Three meth-ods have been presented to accomplish this: an operator method, amatrix representation method, and a DE method. It turns out thatin all cases solutions contain an arbitrary function of a parameter ε.In the parallel parking problem, the parameter ε is a translation in xarising from translations in y and rotations in the xy-plane and thearbitrary continuous function can be the angle of rotation or the ini-tial translation. Interestingly, in all cases solutions can be expressedin terms of elementary functions involving an arbitrary continuousfunction. In practical applications, other constraints could be satis-fied by appropriately restricting associated arbitrary functions.There is an “initial condition” that constrains the arbitrary func-tion. In particular as ε → 0, if the arbitrary function is O(εp) thenit is easy to check that 0 < p < 1 and that all other functions in thecompositions are either O(εp) or O(ε1−p).As noted earlier in the thesis, one can also state the problems in(3.0.2) and (3.0.6) with the roles of B1 and B2 interchanged. It was79shown that doing so, when the number of terms to the left of equa-tions (3.0.2) and (3.0.6) is even, leads to isomorphic solutions. How-ever, if the number of terms to the left of (3.0.2) and (3.0.6) is odd,then interchanging the roles of B1 and B2 may lead to a problemwith no solutions. For instance, considering the S4,7 problem withthe alternative order of Z and W leads to no solutions.Each of the three methods, used to solve equations (3.0.2) and(3.0.6), have different strengths and challenges. When a useful op-erator representation exists, the operator method offers a computa-tionally very simple and complete approach to solving (3.0.2) and(3.0.6). However, an appropriate operator representation of a Liealgebra is only known for three- and four-dimensional Lie algebras[12]. But one would expect an operator representation to exist forLie algebras that arise in practical problems.The matrix representation method requires a matrix representa-tion of a Lie algebra. Such a representation may not always be read-ily available. In the case of S4,7, for example, the matrix represen-tation found using the software [9] was not isomorphic to S4,7, andthus could not be used. Instead, our correction of the adjoint ma-trix representation found in [6] was used. Another issue with thematrix representation method is that the software GAP [9] cannothandle Lie algebras with algebraic values or non-integers in theirstructure constants. Accordingly, we had to make adjustments forthe matrix representations for the Lie algebras S3,1, S3,3, and, S4,9.Moreover, without carrying out all calculations, the number of in-dependent equations one obtains from the matrix representationmethod and whether a solution exists cannot be determined a pri-ori. The main strength of the matrix representation method is thatin all cases it resulted in algebraic systems of equations that we wereable to solve. Most importantly, the matrix representation methodis complete since it leads to necessary and sufficient conditions for80solutions.Unlike the matrix and operator representation methods, the dif-ferential equation method (DE method) requires no Lie algebra rep-resentation. Moreover, it can handle all forms of structure con-stants. Furthermore, in the DE method, unlike the other two meth-ods, one can see that the solution should depend on an arbitraryfunction before calculations are performed since the resulting sys-tem of ODEs has more unknowns than the number of ODEs in thesystem. However, the resulting first order system of nonlinear ODEsoften presents a more significant challenge to solve than the systemof equations obtained through the other two methods. For instance,we were unable to solve directly the ODE system associated withthe Euler angles problem but obtained its solution using the opera-tor and matrix representation methods. The most crucial issue withthe DE method remains that it only yields a necessary condition.But for all cases considered, it turns out that the obtained solutionssatisfied both necessary and sufficient conditions. Related to this, itis an open problem to prove the existence and uniqueness theoremfor the nonlinear systems of first order ODEs that result from theDE method for any relevant n-dimensional Lie algebra without useof solutions arising from matrix or operator representations.One should note that it is possible to extend the solutions pre-sented in this thesis by not requiring the initial conditions (3.3.1)or (3.3.2) to be satisfied. For example, the parallel parking problemalso has the solutiond(ε) = ε, c(ε) = −a(ε) = pi4, b(ε) =√2ε.It is of interest to note that the operator and matrix representationmethods are algebraic ways of solving nonlinear ODE systems aris-ing from the DE method!81References[1] I. D. Ado, The representation of Lie algebras by matrices. Amer.Math. Soc. Translations (2), 21 pp. (1949).[2] G. Bluman, D. Finlay, and O. Mrani-Zentar, Composition of Liegroup elements from basis Lie algebra elements. JNMP. 25, 528-557(2018).[3] G. Bluman and S. Kumei, Symmetries and Differential Equations,(Springer, Applied Mathematical Sciences, 1989).[4] G. Bluman, Construction of Solutions to Partial Differential Equa-tions by the Use of Transformation Groups, Ph.D. Thesis, Caltech(1967).[5] R.W. Brockett, Lie algebras and Lie groups in control theory, Ge-ometric Methods in System Theory: Proceedings of the NATO AdvancedStudy Institute,(Dordrecht: D. Reidel Pub. Co., D.Q. Mayne andR.W. Brockett eds., 1973), pp. 43-82 (1973).[6] R. Ghanam and G.Thompson, Minimal matrix representation offour-dimensional Lie algebras. Bull. Malays. Math. Sci. Soc. 36,343-349 (2013).82[7] R. Ghanam and G. Thompson, Minimal matrix representation offour-dimensional Lie algebras. Extracta Math.30, 95-133 (2015).[8] A. Gonza´lez-Lo´pez, N. Kamran, and P. Olver, Lie algebras of dif-ferential operators in two complex variables. Amer. J. Math. 114,1163-1185 (1992).[9] W.A. de Graaf, Constructing faithful matrix representations ofLie algebras, Proceedings of the 1997 International Symposium onSymbolic and Algebraic Computation, pp.54-59 (1997).[10] E.Kirillova and K. Spindler, Optimal control of a vehicle duringparking. IFAC Proceedings Volumes, 37(13), 925-930 (2004).[11] E. Nelson, Tensor Analysis, (Princeton University Press, Prince-ton, NJ, 1967).[12] R.O. Popovych, V.M. Boyko, M.O.Nesterenko, and M. W. Lut-fullin, Realizations of real low-dimensional Lie algebras. J. Phys. A36, 7337- 7360 (2003).[13] L. Snobl and P. Winternitz, Classification and Identification of LieAlgebras, (American Mathematical Society, Providence, RI, 2014).83
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Title | Low-dimensional Lie algebras and control theory |
Creator |
Mrani Zentar, Omar |
Publisher | University of British Columbia |
Date Issued | 2019 |
Description | Lie groups and Lie algebras are important mathematical constructs first developed by Sophus Lie in the late nineteenth century to unify and extend known methods used to solve differential equations. The problem considered in this thesis emphasizes one way Lie groups and Lie algebras can be used in control theory. Suppose an apparatus has mechanisms for moving in a limited number of ways with other movements generated by compositions of allowed motions. The question is then how to get a targeted motion using a minimal number of the allowed motions. Motions can often be represented by Lie groups which have associated Lie algebras as their building blocks. This research shows explicitly how one can obtain elements of Lie groups as compositions of products of other elements based on the structure of the associated Lie algebras. Here, the structure of a Lie algebra refers to its commutators which are the results that one gets by applying an operation known as the ”commutator” to each pair of elements of a Lie algebra. Two concrete examples of this problem, in control theory, are: (1) the restricted parallel parking problem where the commutator of the Lie algebra element representing translations in y and that representing rotations in the xy-plane yields translations in x. Here the control problem involves a vehicle that can only perform a series involving translations in y and rotations with the aim of efficiently obtaining a pure translation in x; (2) involves an apparatus that can only perform rotations about two axes and the aim is to perform a pure rotation about a third axis. Both examples involve threedimensional Lie algebras. In this thesis, the composition problem is solved for the nine three and four-dimensional Lie algebras with non-trivial solutions. Three different solution methods are presented. Two of these methods depend on operator and matrix representations of a Lie algebra. The other method is a differential equation method that depends solely on the commutator properties of a Lie algebra. Remarkably, for these distinguished Lie algebras the solutions involve arbitrary functions and can be expressed in terms of elementary functions. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2019-07-24 |
Provider | Vancouver : University of British Columbia Library |
Rights | Attribution-NonCommercial-NoDerivatives 4.0 International |
DOI | 10.14288/1.0380064 |
URI | http://hdl.handle.net/2429/71090 |
Degree |
Master of Science - MSc |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
GraduationDate | 2019-09 |
Campus |
UBCV |
Scholarly Level | Graduate |
Rights URI | http://creativecommons.org/licenses/by-nc-nd/4.0/ |
AggregatedSourceRepository | DSpace |
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