ANDREA HEYMAN, HAOYUAN LI, CORINA RAY

Transcription

1 POLYGONL OUTER ILLIRDS IN THE HYPEROLI PLNE NDRE HEYMN, HOYUN LI, ORIN RY bstract. This project was conducted at the summer REU at Penn State University in 009 under the supervision of ndrey Gogolev and Misha Guysinski. Our task was to investigate polygonal outer billiards in the hyperbolic plane, looking for periodic orbits, with an emphasis on triangles. Our main accomplishment was to demonstrate the existence of a three and five-periodic orbit for any non-large triangle... Outer illiards.. Introductory Material Definition. We define outer billiards as follows: Take a convex plane domain P and x a point outside P. There are two support lines to P through x. hoose, say, the left one from x s viewpoint, and define F (x) to be the reflection of x in the support point. F(x) ẋ This definition is not given if the support point is not unique... Hyperbolic Geometry. Hyperbolic geometry is different from Euclidean geometry, in the sense that Euclid s Fifth Postulate does not hold. That is, given a line l and a point P not on the line, we can draw more than one parallel line to l through P. In our analysis we used the Poincaré disk model. In this model, we can visualize lines as any straignt line passing through the origin O or lines are defined as arcs of circles that are perpendicular to the boundary Γ. lso, the boundary is defined to be. Date: ugust 3, 009.

2 NDRE HEYMN, HOYUN LI, ORIN RY In this model, there is a metric used to define the distance between two points, µ = (, P Q), the inverse of the crossratio. Given four points z 0, z, z, z 3, (z 0, z ; z, z 3 ) = z 0 z z z z 0 z 3 z z 3 The distance is multiplicative, that is given three colinear points,, such that is between and, then µ() = µ() µ(). s in Euclidean geometry, a triangle is defined by the intersection of three geodesics. nother important fact is that the sum of angles in a triangle is always less than π or Möbius Transformations. Definition. Möbius transformation is defined on the extended complex plane as a fractional linear function of the form f(z) = az + b (ad bc 0) cz + d We can view the transformation as a x matrix. The trace of the matrix is invariant under conjugation. The trace of our transformation is an important tool for determining the nature of its fixed points, or points such that f(z) = z Nature of Fixed Points Trace Elliptic: One fixed point in the interior of the disc 0, < 4 Parabolic: One fixed point on boundary =4 Hyperbolic: Two fixed points on boundary 4.4. Rotation Number. Definition 3. Suppose that f : S S is a homeomorphism of the circle. Then f may be lifted to a continuous map F : R R of the real line, where F (x) mod = f(x mod ). The rotation number of f is defined in terms of the iterates of F : ρ(f) = lim n F n (x) x. n The rotation number is well-defined and independent of both the point x and the lift F. Two important facts about the rotation number is that: f f : S S is a homeomorphism, then ρ(f) is rational if and only if f has a periodic point The rotation number of the outer billiard orbit is a continuous function of the polygon in the compact open topology.

3 POLYGONL OUTER ILLIRDS IN THE HYPEROLI PLNE 3.5. Large Polygons. Dogru and Tabachnikov analyzed large polygons in the hyperbolic plane. They proved that for large polygons, all orbits of the dual billiard map escape to infinity, but there exists a periodic orbit on the boundary. In this case, they referred to large polygons as convex n-gons such that ρ(p ) = n. In a more specific way, a triangle is large if and only if H >, where for all i =,, 3 where: H = sinh(h i ) sinh(a i ) = sin(α i ) sinh(a i+ ) sinh(a i+ ) = sinh(s) tanh(r) = sinh(s) sinh(s a ) sinh(s a ) sinh(s a 3 ) = 4 sin(k/) cosh(a/) cosh(b/) cosh(c/) a i = sides of triangle s = a + a + a 3 α i = the angle opposing side a i h i = altitude dropped on the i th side K = area r = radius of incircle n equivalent condition in terms of angles is: cos (α ) + cos (α ) + cos (α 3 ) + cos(α ) cos(α ) cos(α 3 ) > + sin(α ) sin(α ) sin(α 3 ). Since the case of large polygons has been answered, we consider only polygons which are not large.. Main Result The main result we have achieved is to demonstrate the existence of a three-periodic orbit for any triangle. We considered first the simplest case of the equilateral triangle, followed by the isosceles triangle, and finally considered an arbitrary triangle... Equilateral Triangle. Our proof is by construction and similar to that in the Euclidean case.

4 4 NDRE HEYMN, HOYUN LI, ORIN RY Given an equilateral triangle as shown in red, we wish to obtain a three-periodic outer orbit, shown in blue. We obtain this orbit by first drawing perpendicular bisectors through the triangle at each of the three vertices. We then draw geodesics at each of the three vertices, perpendicular to the bisectors. The three intersection points of these geodesics form the points of the desired orbit... Isosceles Triangle. Given an isosceles triangle, we relocate it by isometry so that vertex is at the origin of the disk. The dotted lines represent diameters of the disk. We claim that we can form the desired outer orbit starting at by placing appropriately along the horizontal diameter so that when we connect and 3 by a geodesic, the resulting point is such that 3 = 3. 3 The two pictures below illustrate the argument. In the picture on the left, we see that by locating close to the boundary at infinity, the segment 3 is much longer than the segment 3. The picture on the right shows that by locating closer to the origin, 3 is longer than 3. Since this is a continuous motion, by the Intermediate Value Theorem we can conclude that there must be a point at which 3 = rbitrary Triangle. For the case of an arbitrary triangle, we had to switch to a more analytic method. We developed a transformation to describe the motion of a point z when given a triangle with vertices v, v, v 3. To reflect the point z through vertex v, we first developed a function F v (z) that maps v to the origin. Then we found the desired transformation: F v (z) = z v vz + T v (z) = F v (z) id F v (z) = (v v + )z v v z v v

5 POLYGONL OUTER ILLIRDS IN THE HYPEROLI PLNE 5 We then composed T v T v T v3 (z) to look for a three-periodic orbit. We calculated the trace of this composition: 4(v (v v 3 ) + v 3 (v v v v v + v v v 3 ) + v (v 3 v v v 3 + v ( + v 3 v + v v 3 v 3 v 3 ))) Notice that for a small triangle, in which the vertices are close to the origin, this expression will be less than four. Therefore we have a fixed point, and thus a periodic orbit. However, we must check that our composition corresponds to a true outer orbit of the triangle and does not intersect any of the sides. In the three-periodic case this is easy, since traveling around the three vertices in order guarantees that we stay outside the triangle. 3. Recently solved onjectures 3.. The Five-Periodic Orbit. In the Euclidean plane, no triangle admits a five-periodic orbit, except the degenerate case, where we allow the billiard map to travel along the sides of the triangle. In hyperbolic geometry, it may be possible to admit an orbit of period five, by hitting the vertices v, v 3, v, v, v 3. 3 z0 y computing the trace of this transformation, one can observe that this may indicate an elliptic transformation. 8( + v 3 ( v + v 3 ) + v (v v 3 + v 3 v v 3 ))(v (v v 3 )+ v 3 (v v v v v + v v v 3 ) + v (v 3 v v v 3 + v ( + v 3 v + v v 3 v 3 v 3 ) However, we still have to check for the bad cases, where the orbit would follow the same combinatorics, but cross the triangle. We identified three cases, and focused on an isosceles triangle. 3.. ad ases and. z3 z0 z0 z z3 z z z z4 z4

6 6 NDRE HEYMN, HOYUN LI, ORIN RY These two cases seem different from one another, but they are actually symmetric with respect to the angle bisector. Since we cannot have both of them, but they can be obtained from one another, we concluded that these cases do not occur ad ase 3. The third bad case is similar to the Euclidean ase, i.e., we reflect with respect to vertices and, we obtained points P and Q on geodesics O and O, respectively. When we reflect point Q on, we obtain N and similarly, M by reflecting point P on. The main problem is to show that N,O,M do not define a geodesic. MO = P Q MO = P Q NO = QP NOP + O + MO = P Q + QP + O and < R Therefore NOM < R so the points N, O, M do not define a geodesic, so we can eliminate the bad case. N O M P Q

7 POLYGONL OUTER ILLIRDS IN THE HYPEROLI PLNE 7 4. lternative approach nalyze the behavior of the ɛ ball that starts in a particular white area of the graph and we are hoping that, under the transformation, it will follow the pattern of moving from one white-area to another one of the same surface. We interpret the different circles obtained as patterns for different orbits: white circles with the same area must be part of the same orbit, even if we are not able to come up with all the orbits. The dark surfaces are the singularities; if under the transformation, the end-point lands on any dark surface, then we say that the orbit escapes to infinity.