A Beckman-Quarles type theorem for linear fractional transformations of the extended double plane

Rose- Hulman Undergraduate Mathematics Journal A Beckman-Quarles tpe theorem for linear fractional transformations of the extended double plane Andrew Jullian Mis a Josh Keilman b Volume 12, No. 2, Fall 2011 Sponsored b Rose-Hulman Institute of Technolog Department of Mathematics Terre Haute, IN 47803 Email: mathjournal@rose-hulman.edu http://www.rose-hulman.edu/mathjournal a Calvin College, andrew.jullian.mis@alumni.calvin.edu b Calvin College, jkkeilman@gmail.com

Rose-Hulman Undergraduate Mathematics Journal Volume 12, No. 2, Fall 2011 A Beckman-Quarles tpe theorem for linear fractional transformations of the extended double plane Andrew Jullian Mis Josh Keilman Abstract. In this presentation, we consider the problem of characterizing maps that preserve pairs of right hperbolas or lines in the extended double plane whose hperbolic angle of intersection is zero. We consider two disjoint spaces of right hperbolas and lines in the extended double plane H + and H and prove that bijective mappings on the respective spaces that preserve tangenc between pairs of hperbolas or lines must be induced b a linear fractional transformation. Acknowledgements: The research presented in this article was supported b the National Science Foundation under Grant No. DMS-1002453. We thank our Facult Advisor Michael Bolt for his support and encouragement during the time this research took place as well as beond it. We also give a special thanks to Tim Ferdinands and Landon Kavlie for their guidance and support over the ears.

Page 116 RHIT Undergrad. Math. J., Vol. 12, No. 2 1 Introduction Beckman and Quarles proved that an mapping from R n to itself, n 2, that preserves a fixed distance between two points is necessaril a rigid motion [1]. This theorem was among the first results of which are now called characterizations of geometrical mappings under mild hpotheses [4]. Theorems of this sort require no regularit conditions, such as differentiabilit or continuit. Toda there are man theorems that belong to this area. Another example is Lester s characterization of Möbius transformations of the extended complex plane Ĉ = C { }. It is well known in complex analsis that Möbius transformations preserve the space of circles and lines C as well as the angle of intersection θ AB for an pair A, B C. She showed that the converse is also true. Lester proved the following Beckman-Quarles tpe theorem. Theorem 1 (Lester [7]). For a fixed real 0 α < π, let X X be a bijective mapping from C to itself such that, for all A, B in C, θ AB = α if and onl if θ Ā B = α. Then the mapping is induced on C b a Möbius transformation of Ĉ. z = x + i w = u + vi Figure 1: Möbius transformations preserve the space of circles and lines and angles of intersection. We consider the analogous problem for the extended double plane ˆP = P H, where P = {x + k : x, R, k 2 = 1} and H = {(α ± αk) 1 : α R { }} (see Section 2.2). As in complex analsis, we found that in the extended double plane, linear fractional transformations preserve both the space of vertical right hperbolas and lines with slopes greater than -1 and less than 1, denoted H +, and the space of horizontal right hperbolas and lines with slopes less than -1 or greater than 1, denoted H. Furthermore, the hperbolic angle of intersection ϕ h1 h 2 for an pair right hperbolas or lines h 1, h 2 in H + or H is preserved b linear fractional transformations. Following Lester s model, we prove the following Beckman-Quarles tpe theorem.

RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 117 Theorem 2. Let T be a bijective mapping from H + (resp. H ) to itself such that, for all h 1, h 2 H + (resp. H ), ϕ h1 h 2 = 0 if and onl if ϕ T (h1 )T (h 2 ) = 0. Then T is induced on H + (resp. H ) b a linear fractional transformation of ˆP. z = x + k w = u + vk Figure 2: Linear fractional transformations of the extended double plane preserve vertical and horizontal right hperbolas and lines and hperbolic angles of intersection. 2 Geometr in the Extended Double Plane ˆP 2.1 The Double Plane P Definition 1. A double number 1 is a formal expression x + k where x, R and k 2 = 1, but k / R (k is known as a unipotent). Ever double number z = x + k has a real component Re(z) = x, a double component Im(z) =, and conjugate z = x k. The double plane P is the set of all double numbers, which we call points. This ma be thought of as a two-dimensional vector space over R, where each double number x + k corresponds to the vector (x, ) in R 2 with Addition: (x 1, 1 ) + (x 2, 2 ) = (x 1 + x 2, 1 + 2 ); Scalar Multiplication: c(x 1, 1 ) = (cx 1, c 1 ); and Multiplication: (x 1, 1 ) (x 2, 2 ) = (x 1 x 2 + 1 2, x 1 2 + x 2 1 ). 1 The name double number was used b Yaglom. Others have called them perplex, split-complex, spacetime, or hperbolic numbers.

Page 118 RHIT Undergrad. Math. J., Vol. 12, No. 2 This wa we can see that double numbers form a commutative algebra over R. It is straightforward to verif that for z = x + k, 1 def z = 1 z = z z z = x k x 2 2 is the multiplicative inverse of z whenever x ±. Therefore, while P is a commutative ring, it is neither a field nor even an integral domain, because ever nonzero number with form α ± αk, α R, is a zero-divisor. The double modulus of z = x + k is z P def = z z = x 2 2. This is considered the double distance of the point z from the origin. Numbers with form α ± αk are in a sense isotropic, since α ± αk P = 0, even when α 0. So there are points z 1, z 2 P, where z 1 z 2, such that z 1 z 2 P = 0. Therefore, strictl speaking, the double distance given b the modulus is not a metric. But it gives a geometr on R 2 that is quite different from the Euclidean geometr of the complex plane, where z C = x 2 + 2 = 0 if and onl if z = 0. In the complex plane, the set of all points z = x + i C satisfing z C = r > 0 forms a circle with radius r. Similarl, we have that the set of all points z P with z P = ρ > 0 forms a four-branched right hperbola with semi-axes of length ρ, whose asmptotes are the isotropic lines = ±x. As the parameter ϕ increases, < ϕ <, each branch is drawn exactl once. See Figure 3. 0 + ρk ϕ ϕ θ r + 0i x ρ + 0k ϕ ϕ ϕ ϕ ρ + 0k x ϕ ϕ 0 ρk z C = r > 0 0 θ < 2π z P = ρ > 0 0 ϕ < Figure 3: The r-circle and the ρ-right hperbola

RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 119 The double argument of z = x + k is arg(z) def = ϕ where tanh 1 : < x ϕ = tanh 1 x : > x. undefined : = x It is interesting to note that r > 0 and fixed 0 θ < π determine a unique point in C, whereas ρ > 0 and ϕ R determine four points in P. Therefore, the double number version of polar form does not allow for unique representation of points. For more on double numbers, see [9, 8, 10]. 2.2 Hperbolas in the Extended Double Plane ˆP. The extended double plane ˆP is the union P H, where H = {(α ± αk) 1 : α R { }}. We sometimes refer to the points in H as the points at infinit. This set ma be thought of as two lines at infinit that intersect at (0 + 0k) 1. A hperbola h in H + or H is the subset of ˆP that includes ever point z = x + k P satisfing Az z + Re[(B + Ck)z] + D = 0, (1) where A, B, C, D R, 4AD + C 2 B 2 0. For convenience, we denote h b [A : B : C : D]. Notice that for an nonzero λ R, [λa : λb : λc : λd] defines the same hperbola. This means that there are man choices A, B, C, D R, which give us the same hperbola. For the remainder of this article unless otherwise stated the reader ma assume that A, B, C, D R are chosen such that 4AD + C 2 B 2 = ±1. This normalization makes calculations later much easier and neither affects H + nor H. Proposition 1. Let h = [A : B : C : D], where A, B, C, D R. Then (i) h H + if and onl if 4AD + C 2 B 2 = 1; (ii) h H if and onl if 4AD + C 2 B 2 = 1. A hperbola h = [A : B : C : D] also includes point(s) at infinit. Using linear fractional transformations, we can verif the point(s) in H which h intersects. (i) (α 1 + α 1 k) 1 where α 1 = (ii) (α 2 α 2 k) 1 where α 2 = { A B C { A B+C : B C : B = C : B C : B = C (Notice that (0 + 0k) 1 = (0 0k) 1, but ( + k) 1 ( k) 1.) If A 0, then h is a vertical or horizontal right hperbola and includes exactl two points at infinit. But if A = 0, then h is a line and onl includes one point at infinit. Moreover, h is a line if and onl if h (0 + 0k) 1.

Page 120 RHIT Undergrad. Math. J., Vol. 12, No. 2 Using stereographic projection, the extended double plane can be viewed as an infinite hperboloid. Consider the hperboloid x 2 2 + (z 1) 2 = 1, where the x-plane is the double plane, and take the point (0, 0, 2) as the projection point. An line drawn through a point on the extended double plane and the projection point intersects the hperboloid at a second point. Hperbolas in H + correspond with planar cross sections of the hperboloid that take the form of ellipses and hperbolas. A second hperboloid x 2 + 2 + (z 1) 2 = 1 corresponds with hperbolas in H. Figure 4: Projection of a hperbola onto the x-plane. 2.3 Linear Fractional Transformations of ˆP Definition 2. Direct and indirect linear fractional transformations of the extended double plane are mappings µ : ˆP ˆP with forms µ(z) = az + b cz + d and µ(z) = a z + b c z + d, respectivel, where a, b, c, d P and ad bc α ± αk for α R. The restriction ad bc α ± αk guarantees that we avoid transformations which take the entire double plane to a single point or to lines with slopes ±1. The set of direct and indirect linear fractional transformations of the extended double number plane form a group under composition. We denote this group b LFT (ˆP). An direct linear fractional transformation is composed of at most four of the following simple transformations. Translation: µ(z) = z + b, for b P Rotation and Dilation: µ(z) = az, for a P, a α ± αk, where α R Inversion: µ(z) = 1 z

RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 121 The full group is obtained b including conjugation z z. B further restricting ad bc = ±1, we find that LFT (ˆP) is homomorphic to the group SL(2, P) with the two-to-one correspondence az+b, a z+b ( ) a b cz+d c z+d c d. This restriction is for normalization purposes onl and does not affect the group in an wa. Proposition 2. Let [A : B : C : D] be a hperbola in H +, and let µ be a simple linear fractional transformation. (i) (Translation) If µ(z) = z + b, where b = x 0 + 0 k for x 0, 0 R, then [A : B : C : D] µ [A : B 2Ax 0 : C + 2A 0 : A(x 2 0 2 0) Bx 0 C 0 + D]. (ii) (Rotation or Dilation) If µ(z) = az, where a = x 0 + 0 k for x 0, 0 R but 0 ±x 0, then µ [A : B : C : D] [A : Bx 0 C 0 : Cx 0 B 0 : D(x 2 0 0)]. 2 (iii) (Inversion) If µ(z) = 1 z, then (iv) (Conjugation) If µ(z) = z, then [A : B : C : D] [A : B : C : D] µ [D : B : C : A]. µ [A : B : C : D]. As mentioned prior, ever µ LFT (ˆP) maps hperbolas in H + or H onto hperbolas in H + or H. One ma verif this b showing that each of the four simple transformations performs this. A pair of hperbolas in H + (resp. H ) are said to be disjoint, tangent or intersecting if the share, respectivel, 0, 1 or 2 point(s) in ˆP (note that intersecting excludes tangent). Because linear fractional transformations are bijections, the preserve the number of intersection points. 2.4 The Hperbolic Angle of Intersection in ˆP Definition 3. Let h 1 = [A : B : C : D] and h 2 = [E : F : G : H] be distinct hperbolas in H + (resp. H ). Then the hperbolic angle of intersection ϕ h1 h 2 is defined b cosh 2 ϕ h1 h 2 = (2AH + 2DE + CG BF ) 2 (4AD + C 2 B 2 )(4EH + G 2 F 2 ). (2) It is straightforward to verif that for an µ LFT (ˆP), ϕ h1 h 2 = ϕ µ(h1 )µ(h 2 ). In particular, one need onl check that this holds for the four simple transformations described in Proposition 2. Since µ is composed of finitel man simple transformations, it follows the hperbolic angle of intersection for a pair of hperbolas is preserved b µ. We find that the value of ϕ h1 h 2 is particularl helpful for classifing the relationship between two hperbolas h 1 and h 2.

Page 122 RHIT Undergrad. Math. J., Vol. 12, No. 2 Proposition 3. If h 1, h 2 H + (resp. H ), then (i) h 1 and h 2 are disjoint if and onl if ϕ h1 h 2 is undefined; (ii) h 1 and h 2 are tangent if and onl if ϕ h1 h 2 = 0; and (iii) h 1 and h 2 are intersecting if and onl if ϕ h1 h 2 > 0. 2.5 Canonical Forms In our argument, we use canonical representation of hperbola pairs. The conception is that after a suitable linear fractional transformation, a pair of hperbolas ma be assumed to have a simplified form. See Figure 5. To demonstrate this, we let h 1, h 2 be distinct hperbolas in H +. Next, we select three distinct points p 0, p 1, p h 1 and define µ 0 b µ 0 (z) = (z p 0)(p 1 p ) (z p )(p 1 p 0 ). This takes h 1 to the line [0 : 0 : 1 : 0]. We wish to use a subsequent linear fractional transformation to simplif µ 0 (h 2 ) = [E : F : G : H]. Note that the onl µ LFT (ˆP) which preserve µ 0 (h 1 ) = [0 : 0 : 1 : 0] are µ(z) = az+b, cz+d where a, b, c, d R and ad bc = ±1. (It is straightforward to verif that an such µ takes the line = 0 to itself; moreover, a linear fractional transformation that preserves [0 : 0 : 1 : 0] must have this form.) We assume next that G 0 if necessar, multipl E, F, G, and H b -1, as this has no effect on the hperbola. At this point, our simplification is divided into three cases. Case 1. If G < 1, then 4EH + G 2 F 2 = 1 implies that F 2 4EH < 0, thus E, H 0 lest F 2 < 0. Define µ 1 and µ 2 b µ 1 (z) = z + F 2E and µ 2(z) = 2E 4EH F 2 z. Then µ 2 µ 1 µ 0 (h 1 ) = [0 : 0 : 1 : 0] and µ 2 µ 1 µ 0 (h 2 ) = [1 : 0 : λ : 1], where 2G λ = 4EH F 2 0. We call these a canonical pair of disjoint hperbolas in H +. Case 2. If G = 1, then 4EH + G 2 F 2 = 1 implies that F 2 4EH = 0. If E = 0, then F = 0, so µ 0 (h 2 ) = [0 : 0 : 1 : H], H 0. Let µ 1 (z) = 1 z, then µ H 1 µ 0 (h 1 ) = [0 : 0 : 1 : 0] and µ 1 µ 0 (h 2 ) = [0 : 0 : 1 : 1]. If E 0, then solving the equations = 0 and E(x 2 2 ) + F x + + H = 0 for x ields x = F. Let 2E µ 1 (z) = 1 E, z + F 2E then µ 1 µ 0 (h 1 ) = [0 : 0 : 1 : 0] and µ 1 µ 0 (h 2 ) = [0 : 0 : 1 : 1].

RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 123 Follow these with µ 2 (z) = 2z+k to get µ 2 µ 1 µ 0 (h 1 ) = [0 : 0 : 1 : 1] and µ 2 µ 1 µ 0 (h 2 ) = [0 : 0 : 1 : 1]. We call these a canonical pair of tangent hperbolas in H +. Case 3. If G > 1, then 4EH + G 2 F 2 = 1 implies that F 2 4EH > 0. If E = 0, then F 0, so µ 0 (h 2 ) = [0 : F : G : H], F, G 0. Let µ 1 (z) = z + H, then µ F 1 µ 0 (h 1 ) = [0 : 0 : 1 : 0] and µ 1 µ 0 (h 2 ) = [0 : λ : 1 : 0], where λ = F 0. G If E 0, then µ 0 (h 1 ) and µ 0 (h 2 ) intersect at z ± = F ± F 2 4EH. Let 2E µ 1 (z) = z z+ z z, then µ 1 µ 0 (h 1 ) = [0 : 0 : 1 : 0] and µ 1 µ 0 (h 2 ) = [0 : λ : 1 : 0], where λ = F 2 4EH. We call G these a canonical pair of intersecting hperbolas in H +. (i) (ii) (iii) Figure 5: Canonical forms for (i) disjoint; (ii) tangent, and; (iii) intersecting hperbolas in H +. 3 Adaptation of the Has-Mitchell Theorem In 2009, Has and Mitchell extended research on geometrical mappings on the extended double plane. The showed that injective mappings that are restricted to closed middle regions and send hperbolas in H + or H to other hperbolas in H + or H are linear fractional transformations. In their article, the proved the following theorem. Theorem 3 (Has-Mitchell [5]). Ever injective mapping from a closed middle region bounded b a horizontal or vertical right hperbola that sends hperbolas in H + H to hperbolas in H + H is a linear fractional transformation. In order to make a precise understanding of the term closed middle region used b Has and Mitchell, we give the following definition. Definition 4. Let h = [A : B : C : D] be a vertical or horizontal right hperbola and let P denote the proper subset of points in P {x + k : A(x 2 2 ) + Bx + C + D 0}. We call P the closed middle region bounded b h.

Page 124 RHIT Undergrad. Math. J., Vol. 12, No. 2 During our research, we anticipated that Theorem 3 would be a powerful tool for our proof of Theorem 2 and aspired to appl the work of Has and Mitchell to our own however, we found that we could not directl appl their result to our statement in Theorem 2. At the pith of our proof for Theorem 2, we will define a certain injection on ˆP that consequentl maps hperbolas in H + to hperbolas H + (see section 4.2). But the hpothesis in Theorem 3 requires that hperbolas in H + H are sent to hperbolas in H + H, and the injective mapping in our argument has no control over hperbolas in H. Therefore, we give a following slightl modified statement of Theorem 3, which we will later invoke in section 4. Lemma 1 (The Modified Has-Mitchell Theorem). If f : ˆP ˆP is an injective mapping that sends hperbolas in H + to hperbolas in H +, and P is a closed middle region bounded b a vertical right hperbola, then the restriction f P is a linear fractional transformation. To prove Lemma 1, we will adopt the majorit of the proof for Theorem 3 given in [5] and make changes in areas which are not true for our injective mapping f. This ma seem a bit confusing on the surface, but as the argument proceeds reasoning will become more clear. The proof in [5] is constructive, and in one of the steps a hperbola in H is used, to which we do not have access; however, we have an advantage, since f is injective on all ˆP and not just on a closed middle region, so we ma use hperbolas which extended beond the boundar of P, whereas Has and Mitchell ma onl use hperbolas which P includes. In our situation, two modifications to the proof in [5] are needed. First, where Has and Mitchell use the fact that [1 : 0 : 0 : 4] is preserved in order to arrive at a special preserved point of [1 : 0 : 0 : 1] ([5, 3.7]), we use the preservation of [1 : 2 : 0 : 2] to arrive at a different preserved point that plas the same role. Second, where Has and Mitchell again use a horizontal hperbola to argue for the preservation of [ 1 : 0 : 0 : 4 9] ([5, 3.7]), we give a new argument for the preservation of this hperbola. With these modifications, the remainder of the Has-Mitchell arguments applies to our situation, and we ma conclude that f is a linear fractional transformation when restricted to P. Proof of Lemma 1. An immediate result of the hpothesis is that f preserves the number of intersection points shared b a pair of hperbolas h 1, h 2, for injectivit implies that a point z h 1 h 2 if and onl if f(z) f(h 1 ) f(h 2 ). Now, consider h 0 = [1 : 0 : 0 : 1] and the closed middle region bounded b it, P = { x + k P : x 2 2 + 1 0 }. To further clarif the two modifications, we revisit the proof of Theorem 3 at the stage of [5, 3.7], where the following ma be assumed of f after having been pre- and post-composed with the suitable linear fractional transformations. 1. f preserves h 0 = [1 : 0 : 0 : 1] and the points k and (0 + 0k) 1 ([5, 3.2]). 2. f maps parallel lines to parallel lines ([5, 3.3]).

RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 125 3. f preserves l 1 = [0 : 0 : 1 : 1], l 2 = [0 : 0 : 1 : 1] and l 3 = [0 : 0 : 1 : 0] ([5, 3.4]). 4. f preserves the points 1 + k, 1 k, 1 + k, and 1 k ([5, 3.5-6]). The hperbola h 1 = [1 : 2 : 0 : 2] is tangent with l 1 at 1 + k and tangent with l 2 at 1 k. It follows that f maps h 1 to itself, and consequentl, { 1 5 f(h 0 ) f(h 1 ) = h 0 h 1 = 2 + 2 k, 1 } 5 2 2 k. We will show that f preserves not onl this set, but each point in it. Construct the line which passes through the points 1 + 5k and 1 + k, 2 2 [ l 4 = 0 : 2 5 : 1 : ] 5 1 H +. Since f(l 4 ) must be a line with slope in ( 1, 1), we conclude that f( 1 + 5 k) = 1 + 5k. See 2 2 2 2 Figure 6. (Moreover, because f is injective, we also conclude that f( 1 5 k) = 1 5 k.) 2 2 2 2 Therefore, we have found a point on h 0 which is preserved b f, thus fulfilling our first goal. l 4 h 0 h 1 l 1 p 1 p 2 x l 3 l 2 Figure 6: The line l 4, which passes through the points p 1 = 1 + 5 k and p 2 2 2 = 1 + k. Next, we will show that [ 1 : 0 : 0 : 9] 4 is preserved b f. B a similar argument as before, we claim that 1 + 5 k and 1 5k are preserved 2 2 2 2 b f. (These are the points of intersection of h 0 and [1 : 2 : 0 : 2]). Proceed to construct the line tangent to h 0 at 1 + 5 k and the line tangent to h 2 2 0 at 1 + 5k. The are 2 2 [ ] [ 1 2 l 5 = 0 : : 1 : and l 6 = 0 : 1 ] 2 : 1 :, 5 5 5 5 respectfull. Because f preserves tangenc, maps lines to lines, and preserves the points 1 + 5 k and 1 + 5 k, it follows that f(l 2 2 2 2 5) = l 5 and f(l 6 ) = l 6. Since l 3 l 5 2 + 0k and l 3 l 6 2 + 0k, it follows that f ( 2 + 0k) = 2 + 0k and f (2 + 0k) = 2 + 0k.

Page 126 RHIT Undergrad. Math. J., Vol. 12, No. 2 Next, construct two lines: one through the points 2+0k and 1+k and the other through the points 2+0k and 1+k. These lines intersect at 0+ 2 k. Each of these lines is preserved, 3 which implies that f ( 0 + 2k) = 0 + 2 k. Similarl, b constructing the line through 2 + 0k 3 3 and 1 k and the line through 2+0k and 1 k, we find that their intersection point 0 2k 3 is preserved. See Figure 7. Consequentl, the horizontal lines [0 : 0 : 1 : 2] and [0 : 0 : 1 : 2 ] are preserved b 3 3 f. These are tangent to [1 : 0 : 0 : 4] at 0 + 2k and 0 2 k, respectfull. It follows that 9 3 3 f ([ [ 1 : 0 : 0 : 9]) 4 = 1 : 0 : 0 : 4 9], thus fulfilling our second goal. As mentioned prior, the remaining portions of the Has-Mitchell argument follow. Therefore, we ma conclude that f is a direct or indirect linear fractional transformation when restricted to a closed middle region bounded b a vertical right hperbola. 1 2 + 5 2 k 1 + 5 k 2 2 1 + k 1 + k 0 + 2 3 k 2 + 0k 2 + 0k 0 2 3 k 1 k 1 k 1 2 5 2 k 1 5 k 2 2 Figure 7: Lines and points preserved b f.

RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 127 4 Proof of Theorem 2 4.1 T preserves families of tangent hperbolas The hpothesis of Theorem 2 sas that T is a bijection on the space of hperbolas H + that preserves hperbolic angle of intersection zero. 2 This is equivalent to saing that a pair of hperbolas in H + are tangent if and onl if their images under T are tangent. This means that neither T nor T 1 can map a pair of tangent hperbolas to a pair of disjoint or intersecting hperbolas. This further implies that given an collection of mutuall tangent hperbolas, their images will remain mutuall tangent. Lemma 2. An four pairwise tangent hperbolas in H + share exactl one, mutual point. Proof. Let h 1, h 2, h 3, h 4 H + be pairwise tangent. In canonical form, we take h 1 = [0 : 0 : 1 : 1] and h 2 = [0 : 0 : 1 : 1], and characterize ever hperbola [A : B : C : D] in H + that is tangent to both h 1 and h 2. From (2) in section 2.4 and the normalization on [A : B : C : D], we get the following sstem equations. ( 2A + C) 2 = 1 (3) (2A + C) 2 = 1 (4) 4AD + C 2 B 2 = 1 (5) Equations (3) and (4) tell us that either A = 0 or C = 0. If A = 0, then we get a oneparameter famil of horizontal lines [[0 : 0 : 1 : D], where ] D ±1. If C = 0, then we get a one-parameter famil of hperbolas 1 : λ : 0 : 1 + λ2, where λ is an real number. At this 4 point our argument is divided into three cases. Case 1. Recall that both h 3 and h 4 are each tangent to h 1 and h 2 and that the are also tangent to each other. If h 3 and h 4 belong to the first famil, then h 1, h 2, h 3, are h 4 are all horizontal lines, and thus 4 i=1 h i = {(0 + 0k) 1 }, which contains exactl one member. [ Case 2. Now] suppose that [ h 3 and h 4 both ] belong to the latter famil. We let h 3 = 1 : λ : 0 : 1 + λ2 and h 4 4 = 1 : η : 0 : 1 + η2, where λ, η R, and find the intersection 4 point(s) using the following sstem of equations. x 2 2 + λx + 1 + λ2 4 = 0 (6) x 2 2 + ηx + 1 + η2 = 0 4 { } (7) Solving for (6) and (7) for x and tells us that h 3 h 4 = λ+η 16+(λ η) ± k 4 4. This contradicts that h 3 and h 4 are tangent. 2 The reader should take note that we will onl prove the H + version of Theorem 2. However, the argument for the H version is completel analogous. Therefore, one should be able to draw further analogous statements and definitions for the H version. The reader should also take note that when we use the term hperbola it is understood that object under discussion is either a vertical right hperbola or a line with slope greater than -1 and less than 1 that is, a member of H +.

Page 128 RHIT Undergrad. Math. J., Vol. 12, No. 2 h 3 h 3 h 4 h 1 x h 1 x h 2 h 4 h 2 Figure 8: (Case 1.) Hperbolas h 1, h 2, h 3, and h 4 are pairwise tangent at exactl one mutual point (left); (Case 2.) Hperbolas h 3 and h 4 are intersecting (right). Case 3. Finall, [ suppose that h 3 ] and h 4 each belong to different familes. We let h 3 = [0 : 0 : 1 : D] and h 4 = 1 : λ : 0 : 1 + λ2, and find the intersection point(s) using the following 4 sstem of equations. x 2 2 + λx + 1 + λ2 4 + D = 0 (8) = 0 (9) Solving for (8) and (9) we find that if D < 1, then h 3 h 4 =, whereas if D > 1, then h 3 h 4 = { ( λ 2 ± D 2 1) + Dk }. These both contradict that h 3 and h 4 are tangent. h 3 h 3 h 1 x h 1 h 4 x h 2 h 2 Figure 9: (Case 3.) Hperbolas h 3 and h 4 are either disjoint, or intersecting. h 4 So Case 1 is the onl valid generalization of h 1, h 2, h 3 and h 4. Therefore, we ma conclude that four distinct pairwise tangent hperbolas must all meet at exactl one mutual point. It immediatel follows from Lemma 2 that an arbitrar collection of four or more pairwise tangent hperbolas all share exactl one, mutual point. We extend this concept of

RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 129 a collection of pairwise tangent hperbolas and classif special families of infinitel man pairwise tangent hperbolas. If we pick an point p in ˆP and a real number m with m < 1, then we can construct a famil of tangent hperbolas in H + that we denote T p,m. We divide families into four tpes. First, whenever p is finite (i.e. p P), we call T p,m the famil of hperbolas whose slope is m at point p. 3 In particular, if p = x 0 + 0 k, then T p,m def = {[A : m 2Ax 0 : 1 + 2A 0 : A(x 0 2 0 2 ) + mx 0 0 ] : A R}. Next, if p H \{( ± k) 1 }, then p 1 is finite. Whenever this is the case we define T p,m as the famil of hperbolas obtained as images of T p 1,m under the inversion mapping z 1. In this case, m does not represent a slope, but it does allow us to identif a specific z set of hperbolas at p. In particular, if p = (x 0 ± x 0 k) 1, with x 0, then T p,m def = {[mx 0 x 0 : m 2Ax 0 : 1 2Ax 0 : A] : A R}. We have left to construct a famil of hperbolas for p = ( ± k) 1. To accomplish this, we begin with the point (1 + k) 1 and construct the set T (1+k) 1,m, and then use a subsequent µ LFT (ˆP) so that µ((1 + k) 1 ) = ( + k) 1. The approach is somewhat indirect; we claim that since a point in H corresponds to the asmptote of the hperbolas containing it, if we know where the asmptote on P goes, then we can identif where points at infinit go. The µ we choose to use is µ(z) = z 1 2. So when p = ( + k) 1, then T p,m def = µ(t (1+k) 1,m) {[ = m 1 : 1 2A : 1 2A : m + 1 ] 4 } : A R. Similarl, when p = ( k) 1, we begin with (1 k) 1 and use the same µ to get T p,m def = µ(t (1 k) 1,m) {[ = m + 1 : 1 2A : 1 + 2A : m 1 ] 4 } : A R. It is straightforward to verif that for an of these tpes, hperbolas in the famil T p,m are pairwise tangent namel, at the point p. Lemma 3. Let p 1, p 2 ˆP and let m 1, m 2 R such that m j < 1. Then T p1,m 1 = T p2,m 2 if and onl if p 1 = p 2 and m 1 = m 2. Proof. If p 1 = p 2 and m 1 = m 2, then obviousl T p1,m 1 = T p2,m 2. 3 In the argument for the H version, we permit m = in order to account for vertical lines and the hperbolas tangent to them, in which case the slope is undefined.

Page 130 RHIT Undergrad. Math. J., Vol. 12, No. 2 Conversel, suppose we begin with T p1,m 1 = T p2,m 2. Then h 1, h 2 T p1,m 1 implies that h 1, h 2 T p2,m 2. If p 1 p 2, then h 1 and h 2 share two points namel, p 1 and p 2. This is a set contradiction since h 1 and h 2 are tangent. Thus p 1 = p 2 = p. B pre-composing with an appropriate linear fractional transformation, we ma assume that p is finite (i.e. p P). Since all hperbolas in T p,m1 = T p,m2 are tangent at one mutual point, the must share the same slope at p. Hence m 1 = m 2. Therefore, we conclude that T p1,m 1 = T p2,m 2 if and onl if p 1 = p 2 and m 1 = m 2. p x Figure 10: The famil of hperbolas T p,m. Lemma 4. If p ˆP and m R such that m < 1, then there exist p ˆP and m R, m < 1 so that T (T p,m ) = T p,m Proof. If h 1, h 2, h 3, h 4 T p,m, + then T (h 1 ), T (h 2 ), T (h 3 ), and T (h 4 ) are pairwise tangent. Moreover, b Lemma 2, the are tangent a mutual point p ˆP. Next, we post-compose T with a suitable linear fractional transformation in order that we ma assume that p is finite, and let m R, with m < 1, be the slope of T (h j ) at p, 1 j 4. Then T (h j ) T + p,m. We show that T (T p,m) + = T + p,m. Suppose that h 5 T p,m, but h 5 / {h 1, h 2, h 3, h 4 }. Then T (h 5 ) T (T p,m ) and, moreover, T (h 5 ) is tangent to T (h j ), 1 j 4. It follows that T (h 5 ) T p,m, thus T (T p,m) T p,m. Now let h 5 T p,m, but h 5 / {T (h 1 ), T (h 2 ), T (h 3 ), T (h 4 )}. Because T is surjective, there is an h 5 H + such that h 5 = T (h 5 ); furthermore, because T is injective, h 5 / {h 1, h 2, h 3, h 4 }. So T (h 5 ) is tangent to T (h j ), 1 j 4, thus T 1 (T (h 5 )) = h 5 is tangent to T 1 (T (h j )) = h j T p,m. Then h 5 T p,m and therefore, h 5 = T (h 5 ) T (T p,m ). Hence T p,m T (T p,m).

RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 131 4.2 T : H + H + is induced via p p We have just shown that a famil of mutuall tangent hperbolas T p,m maps onto a famil of mutuall tangent of hperbolas T p,m. This suggests the existence of a mapping p p. We show that this mapping is a well-defined that is, p is alwas mapped to p, regardless the choice of m. Lemma 5. The mapping p p is well-defined on ˆP Proof. We begin b selecting two distinct real numbers m 1 and m 2 with m j < 1, and construct the two families T p,m1 and T p,m2. B Lemma 4, there exist p 1, p 2 ˆP and m 1, m 2 with m j < 1 such that T (T p,m1 ) = T p 1,m 1 and T (T p,m 2 ) = T p 2,m 2. We must show that p 1 = p 2. Our reasoning is as follows. If p 1 p 2 we will construct hperbolas h 1 T p 1,m and h 1 2 T p 2,m that are tangent to 2 one another. Since T is a bijection that preserves tangenc, this would mean that hperbolas T 1 (h 1 ) T p,m1 and T 1 (h 2 ) T p,m2 exist and are tangent to each other where m 1 m 2, which is clearl false. To simplif matters, we post-compose T with a suitable linear fractional transformation µ in order that we ma assume that p 1 = 0, m 1 = 0 and p 2 is finite. This µ can be written as a composition µ 1 µ 2 where µ 2 sends p 1 to 0 and p 2 to some finite point; and µ 1 is a suitable rotation µ 2 (z) = az, where a is some real number. The hperbola h 1 T p 1,m = T 1 0,0 can then be written [λ : 0 : 1 : 0] for some λ R. At this point our argument is divided into two cases. Case 1. We assume that p 2 = x 0 + 0 k where 0 0. To simplif notation we set m 2 = m. We choose λ = 0 so that h 1 = [0 : 0 : 1 : 0], and will look for numbers E, F, G, H R so that h 2 = [E : F : G : H] is tangent to h 1 and belongs to T p 2,m = T 2 x 0 + 0 k,m. The normalization on h 2 and the tangenc with h 1 then require 4EH + G 2 F 2 = 1 (10) G 2 = 1 (11) In addition, as h 2 can be expressed in coordinates b E(x 2 2 ) + F x + G + H = 0, we have that h 2 T x0 + 0 k,m requires E(x 2 0 2 0) + F x 0 + G 0 + H = 0 (12) E(2x 0 2 0 m) + F + Gm = 0. (13) (The latter equation is obtained b implicit differentiation. Recall that m is the slope of h 2 at x 0 + 0 k.) From (11), we choose G = 1 and substitute into the remaining equations. (The choice of +1 or 1 makes no difference since multipling all components of h 2 b 1 has no effect

Page 132 RHIT Undergrad. Math. J., Vol. 12, No. 2 on h 2.) From (13), we find F = 2E(x 0 0 m) m and substitute into the remaining equations. Then from (12), we also find H = E(x 2 0 + 2 0 2x 0 0 m) + x 0 m 0 and substitute into the remaining equations. Thus from (10) we get the following equation. E 2 1 0 E m 2 4 02 (1 m 2 ) = 0 This is quadratic in E and has real solutions. So h 2 exists. Case 2. We assume that p 2 = x 0 + 0k with x 0 0 and we again let m 2 = m. B post-composing T with a further dilation we ma assume that x 0 = 1. We choose λ = 1 and will look for numbers E, F, G, H R so that h 2 = [E : F : G : H] is tangent to h 1 and belongs to T p 2,m 2 = T 1,m. The normalization on h 2 and the tangenc with h 1 then require that 4EH + G 2 F 2 = 1 (14) (2H + G) 2 = 1. (15) In addition, as h 2 can be expressed in coordinates b E(x 2 2 ) + F x + G + H = 0, we have that h 2 T 1,m requires that E + F + H = 0 (16) 2E + F + Gm = 0. (17) (The latter equation is obtained b implicit differentiation.) From (15), we choose 2H + G = 1 and substitute G = 1 2H into the remaining equations. (The choice of +1 or 1 makes no difference since multipling all components of h 2 b 1 has no effect on h 2.) From (16), we also find F = (E + H) and substitute into the remaining equations. We then have two equations 4H 2 4H (E H) 2 = 0 (18) (E H) + (1 2H)m = 0. (19) B solving (19) for E H and substituting into (18), and simplifing, we obtain the following equation for H. 4(1 m 2 )H 2 4(1 m 2 )H m 2 = 0. This is quadratic in H and has real solutions. So h 2 exists. Therefore, regardless of the choice of slope m for the famil T p,m, the point p will alwas be mapped the same image point p that is, p p is well-defined. Definition 5. Following the well-defined mapping p p in Lemma 5, we define ˆT : ˆP ˆP b ˆT (p) = p. We call this the mapping inducing T : H + H +.

RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 133 This next lemma will show that the pointwise mapping ˆT actuall determines the hperbola mapping T. Lemma 6. If h H +, then T (h) = { ˆT (p) : p h}. Proof. Let p h and let m R, such that m < 1, and construct T p,m. Then h T p,m. B Lemma 4, there is a p ˆP and an m R, with m < 1, so that T (T p,m ) = T p,m. Then T (h) T p,m = T ˆT (p),m and ˆT (p) T (h). Thus { ˆT (p) : p h} T (h). Now let p T (h) and suppose that T (h) T p,m, for some m. Then for T 1, b Lemma 4, there is p ˆP and m R, m < 1, so that T (T p,m ) = T p,m. 1 Furthermore, because T is bijective, there is a unique h H + such that h = T 1 (T (h)) T 1 (T p,m ) = T p,m, and thus p h. B definition of ˆT, we also have that ˆT (p) = p, which means p { ˆT (p) : p h}. Hence T (h) { ˆT (p) : p h}. The second part of the proof of Lemma 6 gives us that ˆT is surjective, since ever p T (h) has a preimage p h for an h H +. Surjectivit tells us that intersection points cannot be created b ˆT ; in order to show that intersection points cannot be destroed, we must show that ˆT is also injective. Lemma 7. ˆT is injective. Proof. Let p 1, p 2 ˆP and suppose that ˆT (p 1 ) = ˆT (p 2 ). Let m R, m < 1, and construct the families T ˆT (p1 ),m and T ˆT (p2 ),m. B Lemma 3, T ˆT (p1 ),m = T ˆT (p2 ),m. Then appling T 1, there is an m R with m < 1, such that T p1,m = T 1 (T ˆT (p1 ),m ) = T 1 (T ˆT (p2 ),m ) = T p2,m. B Lemma 3, p 1 = p 2. Hence ˆT is injective. We now have that ˆT is an injective mapping on ˆP which sends hperbolas in H + to hperbolas in H +. Therefore, b Lemma 1 from section 3, we know that ˆT is a linear fractional transformation when restricted to a closed middle region. We have left to show that ˆT is linear fractional on the entire extended double number plane and not onl on some closed middle region. 4.3 ˆT is a linear fractional transformation of ˆP Let P be a closed middle region bounded b some vertical right hperbola. Define µ LFT (ˆP) such that µ = ˆT P and let P = ˆT (P ). Then µ 1 (P ) = P and, furthermore, µ 1 ˆT fixes ever point in the region P. See Figure 11. We will show that µ 1 ˆT fixes ever point outside of P as well. Completion of the Proof of Theorem 2. Suppose that p is a finite number. Then we can construct two distinct lines which intersect at p. We claim that ever line in H + must intersect the closed middle region P at least twice (in fact, infinitel man times). Then

Page 134 RHIT Undergrad. Math. J., Vol. 12, No. 2 P x µ = ˆT P P = ˆT (P ) x µ 1 µ 1 ˆT P P x Figure 11: The mapping µ 1 ˆT P is the identit on the region P. ever line is preserved b µ 1 ˆT, and thus intersection points of an pair of lines are preserved. Hence µ 1 ˆT (p) = p. Now suppose p H. Since a hperbola is uniquel determined b its points in P, ever hperbola in H + is preserved b µ 1 ˆT. Then it follows that T p,m = µ 1 ˆT (T p,m ) = T µ 1 ˆT (p),m. Therefore, b Lemma 3, µ 1 ˆT (p) = p. Since µ 1 ˆT fixes ever point on the extended double plane, it follows that it is the identit mapping, and hence a linear fractional transformation. Thus, b the group structure of LFT (ˆP), ˆT = µ µ 1 ˆT LFT (ˆP). Therefore, bijective mappings that sends tangent hperbolas in H + to tangent hperbolas in H + are induced b a linear fractional transformation of the extended double plane.

RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 135 5 Conclusion This result furthers a connection between the complex numbers, dual numbers 4 and double numbers. Ferdinands and Kavlie showed in [3] that a bijection on the space of parabolas that preserves a fixed distance 1 between intersecting parabolas is induced b a linear fractional transformation of the dual plane D. This result along with Lester s and our own show how intrinsic the linear fractional transformations are with the geometrical spaces in which the act. We further wonder if there is a more unified wa to set up Theorem 2 that is, what are the necessar and sufficient conditions for T if we take into consideration the entire space of right hperbolas and lines H = H + H? This question came into mind while determining whether or not ˆT is well-defined when onl assuming that T : H H is a bijection. It also stands to show whether or not a stronger version of Theorem 2 is true b assuming a fixed angle > 0 is preserved. References [1] F. S. Beckman, D. A. Quarles, Jr. On isometries of Euclidean spaces, Proc. Amer. Math. Soc., 4:810-815, 1953. [2] H. S. M. Coxeter. Inversive distance, Annali Matematica, 4:73-83, 1966. [3] T. Ferdinands, L. Kavlie. A Beckman-Quarles tpe theorem for Laguerre transformations in the dual plane. Rose-Hulman Undergraduate Math Journal, 10(1), 2009. [4] W.-l. Huang. Characterizations of geometrical mappings under mild hpotheses, http://www.math.uni-hamburg.de/home/huang/huang files/amss.pdf [5] J. Has, T. Mitchell. The most general planar transformations that map right hperbolas into right hperbolas. Rose-Hulman Undergraduate Math Journal, 10(2), 2009. [6] V. V. Kisil. Starting with the group SL 2 (R). Notices Amer. Math. Soc., 54(11):1458-1465, 2007. [7] J. A. Lester. A Beckman-Quarles tpe theorem for Coxeter s inversive distance. Canad. Math. Bull., 34(4):492-498, 1991. [8] C. T. Miller. Geometr in the Hperbolic Number Plane, http://people.unt.edu/ctm0055/paper1.pdf [9] G. Sobczk. The Hperbolic Number Plane. The College Mathematics Journal, 4:268-280, 1995. [10] I. M. Yaglom. Complex numbers in geometr. Translated from the Russian b Eric J. F. Primrose. Academic Press, New York, 1968. 4 The dual numbers are of form x + j where j 2 = 0.