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A Maximum Principle in the Engel Group James Klinedinst University of South Florida, [email protected]

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A Maximum Principle in the Engel Group James Klinedinst University of South Florida, Klinedinstjames@Gmail.Com University of South Florida Scholar Commons Graduate Theses and Dissertations Graduate School 4-4-2014 A Maximum Principle in the Engel Group James Klinedinst University of South Florida, [email protected] Follow this and additional works at: https://scholarcommons.usf.edu/etd Part of the Mathematics Commons Scholar Commons Citation Klinedinst, James, "A Maximum Principle in the Engel Group" (2014). Graduate Theses and Dissertations. https://scholarcommons.usf.edu/etd/5248 This Thesis is brought to you for free and open access by the Graduate School at Scholar Commons. It has been accepted for inclusion in Graduate Theses and Dissertations by an authorized administrator of Scholar Commons. For more information, please contact [email protected]. A Maximum Principle in the Engel Group by James Klinedinst A thesis submitted in partial fulfillment of the requirements for the degree of Master of Arts Department of Mathematics & Statistics College of Arts and Sciences University of South Florida Major Professor: Thomas Bieske, Ph.D. Razvan Teodorescu, Ph.D. Seung-Yeop Lee, Ph.D. Date of Approval: April 04, 2014 Keywords: Viscosity Solutions, Partial Differential Equations, Carnot Groups, Heisenberg Group Copyright c 2014, James Klinedinst Dedication This work is dedicated to my son, Blaise. May his discov- eries be even better. Table of Contents Abstract . ii Chapter 1 Background on the Engel Environment . 1 Chapter 2 Carnot Jets and Viscosity Solutions . 10 Chapter 3 Sub-Riemannian Maximum Principle . 14 References . 26 i Abstract In this thesis, we will examine the properties of subelliptic jets in the Engel group of step 3. Step-2 groups, such as the Heisenberg group, do not provide insight into the general abstract calculations. This thesis then, is the first explicit non-trivial computation of the abstract results. ii Chapter 1 Background on the Engel Environment One of the properties of Riemannian manifolds is that at each point, the dimension of the tangent n space is equal to the topological dimension of the manifold. For example, if the manifold is R , n 3 the tangent space is also R . If the manifold is a surface in R , then the tangent space at a point is 2 R . Thus, there are no restricted directions in the tangent space. However, many real-world models require restricted directions, because movement is limited. An example is driving a car [1]. This is because a car cannot move laterally, restricting the direction of motion. One other model is how the human brain processes visual images [5]. We then use sub-Riemannian spaces, which are spaces having points where the dimension of the tangent space is strictly less than the topological dimension of the manifold. We will consider sub- Riemannian manifolds with an algebraic group law, called Carnot groups. In Bieske’s paper [3], a sub-Riemannian maximum principle is proved for Heisenberg groups. Later, in [2], this maximum principle is proved for general Carnot groups. In the time between these two papers, it was discovered that the Heisenberg group and so-called “step-2 groups” do not have a sufficiently mellifluous geometry, resulting in these groups being inadequate concrete examples. In this thesis, we explore the step-3 Engel group, which allows us a more concrete understanding of the abstraction found in [2]. 4 We start in R with coordinates (x1; x2; x3; x4) and for α 2 R, let 1 1 u = u(x ; x ; x ; α) = x + x (x + αx ) 1 2 3 2 3 12 2 1 2 1 1 m = m(x ; x ; x ; α) = − αx + x (x + αx ) 1 2 3 2 3 12 1 1 2 1 1 n = n(x ; x ; α) = x + αx : 1 2 2 1 2 2 1 Consider the linearly independent vector fields @ x2 @ @ X1 = − − u @x1 2 @x3 @x4 @ x1 @ @ X2 = + + m @x2 2 @x3 @x4 @ @ X3 = + n @x3 @x4 @ and X4 = : @x4 These vector fields obey the relations [X1;X2] = X3; [X1;X3] = X4; [X2;X3] = αX4, and [Xi;X4] = 0 for i = 1; 2; 3: The vectors and these brackets form a Lie algebra, represented by g, that has a corresponding de- composition given by g = V1 ⊕ V2 ⊕ V3 where the vector space Vi is given by V1 = span fX1;X2g V2 = span fX3g and V3 = span fX4g: Note that [V1;V1] = V2; [V1;V2] = V3; and [V1;V3] = 0. This Lie algebra, has an inner product that orthonomalizes the basis fX1;X2;X3;X4g denoted by h·; ·i. It is given by the symmetric matrix 2 3 k11 k12 k13 k14 6 7 6 7 6k12 k22 k23 k247 6 7 6 7 6k13 k23 k33 k347 4 5 k14 k24 k34 k44 2 where x2(1 + n2) k = 1 + u(u − nx ) + 2 11 2 4 (1 + n2)x x 1 k = −mu − 1 2 + n(ux + mx ) 12 4 2 1 2 (1 + n2)x k = −nu + 2 13 2 nx k = u − 2 14 2 (1 + n2)x2 k = 1 + m2 − mnx + 1 22 1 4 (1 + n2)x k = mn − 1 23 2 nx k = −m + 1 ; k = 1 + n2; k = −n 24 2 33 34 k44 = 1: The exponential map, which allows us to relate this Lie algebra to a Lie group called the step-3 Engel group, takes a vector from the Lie algebra at point p, say Xp and relates it to a unique integral 0 curve given by γ(t). The relationship is defined as exp (Xp) = γ(1) where γ (0) = Xp and γ(0) = p. THEOREM 1.1 The exponential map exp: g ! G is the identity map. Proof. Let γ(t) = (γ1(t); γ2(t); γ3(t); γ4(t)) be a curve. Suppressing the parameter t, we compute: 4 X @ γ2 @ @ a X = a − − u(γ ; γ ; γ ; α) i i 1 @x 2 @x 1 2 3 @x i=1 1 3 4 @ γ1 @ @ + a2 + + m(γ1; γ2; γ3; α) @x2 2 @x3 @x4 @ @ @ + a3 + n(γ1; γ2; α) + a4 @x3 @x4 @x4 @ @ −a1γ2 γ1a2 @ = a1 + a2 + + + a3 @x1 @x2 2 2 @x3 a γ a γ γ αa γ2 αa γ a γ2 + − 1 3 − 1 1 2 − 1 2 − 2 3 + 2 1 2 12 12 2 12 αa2γ1γ2 a3γ1 αa3γ2 @ + + + + a4 : 12 2 2 @x4 3 Thus the initial value problem 8 4 > 0 P <>γ (t) = aiXi(γ(t)) i=1 > :>γ(0) = 0 is equivalent to the initial value problem 8 >γ0 (t) = a > 1 1 > > >γ0 (t) = a > 2 2 <> 0 −a1γ2 γ1a2 γ3(t) = + + a3 > 2 2 > 2 2 > 0 a1γ3 a1γ1γ2 αa1γ2 αa2γ3 a2γ1 αa2γ1γ2 a3γ1 αa3γ2 >γ (t) = − − − − + + + + + x4 > 4 2 12 12 2 12 12 2 2 > > :γi(0) = 0 for i = 1; 2; 3; 4: Through integration, γ1(t) = a1t and γ2(t) = a2t: 0 Substituting, we get γ3(t) = a3 and so γ3(t) = a3t 0 and γ4(t) = a4, giving γ4(t) = a4t: Thus, γ(t) = (a1t; a2t; a3t; a4t) and so γ(1) = (a1; a2; a3; a4). So exp(a1X1 + a2X2 + a3X3 + a4X4) = (a1; a2; a3; a4): The non-abelian algebraic group law is supplied by the Baker-Campbell-Hausdorff formula [10], which is given by 1 1 1 expX? expY = exp(X + Y + [X; Y ] + [X; [X; Y ]] − [Y; [X; Y ]]): 2 12 12 The higher order brackets are zero because they will all involve bracketing with X4. PROPOSITION 1 For p = (x1; x2; x3; x4); q = (y1; y2; y3; y4) 2 G, the group law is 1 1 1 p ? q = x + y ; x + y ; x + y + µ, x + y + ν + µ(x + αx − y − αy ) 1 1 2 2 3 3 2 4 4 2 12 1 2 1 2 4 where µ = (x1y2 − x2y1) and ν = x1y3 − x3y1 + α(x2y3 − x3y2) 4 in the embedding space R . Proof. p ? q = (x1; x2; x3; x4) ? (y1; y2; y3; y4) = exp x1X1 + x2X2 + x3X3 + x4X4 exp y1X1 + y2X2 + y3X3 + y4X4 = exp (x1X1 + x2X2 + x3X3 + x4X4) + (y1X1 + y2X2 + y3X3 + y4X4) 1 + [x X + x X + x X + x X ; y X + y X + y X + y X ] 2 1 1 2 2 3 3 4 4 1 1 2 2 3 3 4 4 1 + [x X + x X + x X + x X ; 12 1 1 2 2 3 3 4 4 [x1X1 + x2X2 + x3X3 + x4X4; y1X1 + y2X2 + y3X3 + y4X4]] 1 − [y X + y X + y X + y X ; 12 1 1 2 2 3 3 4 4 [x1X1 + x2X2 + x3X3 + x4X4; y1X1 + y2X2 + y3X3 + y4X4]] : The only non-zero Lie brackets for [x1X1 + x2X2 + x3X3 + x4X4; y1X1 + y2X2 + y3X3 + y4X4] are given by x1y2[X1;X2] + x1y3[X1;X3] + x2y1[X2;X1] + x2y3[X2;X3] + x3y1[X3;X1] + x3y2[X3;X2] = x1y2X3 + x1y3X4 − x2y1X3 + αx2y3X4 − x3y1X4 − αx3y2X4 = x1y2 − x2y1 X3 + x1y3 − x3y1 + α(x2y3 − x3y2) X4 = µX3 + νX4: The only non-zero Lie brackets for [x1X1 + x2X2 + x3X3 + x4X4; [x1X1 + x2X2 + x3X3 + x4X4; y1X1 + y2X2 + y3X3 + y4X4]] are x1µ[X1;X3] + x2µ[X2;X3] = x1µ + αx2µ X4: Finally, the non-zero brackets for [y1X1 + y2X2 + y3X3 + y4X4; [x1X1 + x2X2 + x3X3 + x4X4; y1X1 + y2X2 + y3X3 + y4X4]] are y1µ[X1;X3] + y2µ[X2;X3] = y1µ + αy2µ X4: 5 Plugging these values into the formula, we get exp (x1X1 + x2X2 + x3X3 + x4X4) + (y1X1 + 1 1 1 y2X2 + y3X3 + y4X4) + 2 (µX3 + νX4) + 12 x1µ + αx2µ X4 − 12 y1µ + αy2µ X4 . Combining like terms, we get 1 1 1 exp (x1 + y1)X1 + (x2 + y2)X2 + x3 + y3 + 2 µ X3 + x4 + y4 + 2 ν + 12 µ(x1 + αx2) − 1 12 µ(y1 + αy2) X4 . The proposition then follows since the exponential is the identity. COROLLARY 1.0.1 The identity element under group multiplication is (0; 0; 0; 0) and the inverse element is (−x1; −x2; −x3; −x4).
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