The Arithmetic Graph

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
the  new kite  button.  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 the  return map  button
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 this  the odd case 

In 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 the  new kite 
button 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 the  offset 
All 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.