monograph guide

This demo explains the concept of the arithmetic graph. To read this demo, you should know the basic definitions. Also, after reading the basic definitions, you should play a bit with the outer billiards demo. The applet at right goes along with this demo. Please initiate the applet by pressing thenew kitebutton. You should see see a red polygonal shape appear in the bottom half of the applet.## selecting new kites

The top half of the applet does some of the outer billiards dynamics on a kite. You select a new kite by changing the numbers (initially 7/11). You change these numbers using mouse clicks and the keyboard. You want to choose a rational number in the interval (0,1), such that the denominator is less than 30. The number you select is the kite parameter A.## The parallel lines

The top half of the applet shows a family of blue parallel lines. These lines are the ones of the form y=2k+1, for k an integer. The o.m.b. preserves these lines, and hence so does the square o.b.m. We let o2 denote the square o.b.m.## The return map

Let S be the union of the two light blue rays in the applet. If we start with some X in S, then there is some first power Y=o2^k(X) which again belongs to S. We write R(X)=Y. So, R is the first return map to S. Push thereturn mapbutton on the applet to see the o2-orbit of X between X and R(X). Here X is marked by a yellow square and R(X) is marked by a light blue one. It turns out that the vector R(X)-X always has the form 2(A e1 + e2, e3); Here e1,e2,e3 are all numbers in the set {-1,0,1}. Moreover, e1+e2+e3 is even. Thus, really the pair (e1,e2) carries all the information about the return map. The numbers (e1,e2) depend on X.## Main construction

Let's start with the point X0=(1/q,1). Let the R-orbit be given by X1,X2,X3... Then X0=X0+2(Am0+n0,--) (m0,n0)=(0,0) X1=X0+2(Am1+n1,--) X2=X0+2(Am2+n2,--) X3=X0+2(Am3+n3,--), etc. The second coordinate are in the set {-1,1} and we ignore them. We get the arithmetic graph by connecting the vertices (m0,n0) to (m1,n1) to (m2,n2) etc., We call this graph G(A). The first coord of X_k is given M(m_k,n_k), as shown on the applet. If you click the middle mouse button over a point (m,n) on the bottom half of the applet, then the yellow point on the top moves to (M(m,n),--). The right yellow buttons control the second coordinate. In case you don't have a 3-button mouse, the applet has a 3-button mouse emulator. It is the control panel w/the word mouse in it.## basic properties

We are mainly interested in the arithmetic graph G(A) when A=p/q is rational and pq is odd. We call thisthe odd caseIn the odd case, G(p/q) is a polygonal arc that starts at (0,0) and ends at (q,-p). The orbit of the point X0=(1/q,1) is periodic, but the map M discussed above is not one-to-one. If we kept iterating the return map, in the odd case, we would generate an infinite embedded polygonal arc. My monograph establishes these basic properties. When pq is even, we observe that G(p/q) is always a closed embedded polygon. When you push thenew kitebutton on the applet, the arithmetic graph is drawn in red on the bottom half of the applet. The boundary of the region is the actual arithmetic graph. We have colored it in to make a pretty picture. The distance from M(m,n) to the origin is proportional to the distance from (m,n) to the blue halfplane. So, if G(p/q) rises far away from this blue halfplane, then the orbit of X0=(1/q,1) goes far away from the origin. Thus, the oscillations of the arithmetic graph are directly related to the problem of finding unbounded orbits for outer billiards on kites.## the demo

1. Select a lattice point (m,n) on the bottom half of the applet. For best effect, choose (m,n) to lie on the arithmetic graph. 2. Click the return map. You will now see a blue square appear near the yellow one on the bottom half. This blue square indicates an adjacent vertex in the arithmetic graph. If necessary, to see the whole orbit you can rescale the window with mouse clicks. 3. On the bottom half of the applet, select the lattice point marked by the blue square. You will see that the correponding yellow point on the top half moves to a point that has the same first coordinate as the blue point. 4. The yellow point might not coincide with the blue point. They could lie on different rays. You can use the yellow buttons to change rays so that the yellow and blue points match. 5. iterate the above 4 steps. This will illustrate how the arithmetic graph is traced out from the dynamics of the outer billiards map.## The Irrational Case

So far we have talked about the arithmetic graph in the rational case. The irrational case is similar but there is one difference. In the rational case, the orbit of the point (t,1) is, from a combinatorial point of view, independent of t as long as t lies in the interval (0,2/q). This is why we choose to focus our attention on the orbit of the point (1/q,1). Setting e(p/q), we have our basic map M(m,n)=(2Am+2n+e,--) In the irrational case, there is typically no canonical choice of e, and so we just pick a value of e and call it theoffsetAll we require is that e is not in 2Z[A]. We call G(A,e) the arithmetic graph relative to the offset e.## Other components

We have described the arithmetic graph G(A,e) as the result of following the orbit of the point X0=M(0,0). However, we could follow the orbit of any point M(m,n). In other words, we can defined an extended version of the arithemtic graph where two points (m1,n1) and (m2,n2) are joined by an edge if and only if there is a choice of numbers y1,y2 in {-1,1} such that (M(m1,n1),y1) and (M(m2,n2),y2) are related by the return map R. This notion never leads to a contradiction, because reflection in the X-axis conjugates R to its inverse. To, we would connect (m1,n1) to (m2,n2) iff we would connect (m2,n2) to (m1,n1). our graph G(A,e) is just the component of the extended graph that contains the origin.## the strips

In the top window we have drawn a certain 4 strips. If you zoom out, you can get a better sense of these strips. If you select a lattice point (m,n) fairly far away from the magenta boundary, the corresponding point M(m,n) lies far from the origin in the top window. Select such a far-away point and doing the return map. You will then understand the significance of the strips.## Another applet

This applet draws nicer pictures of the arithmetic graph, and lets you choose fractions with denominator up to 1000.