The Embedding Theorem

This demo gives a pictorial proof of
the following result. 
 
 Embedding Theorem   The extended
arithmetic graph is a disjoint union
of embedded polygons and infinite
embedded polygonal arcs 

In order to understand this demo,
you should first read about the
 arithmetic graph  and the
 Master Picture Theorem.

The applet on the right demonstrates 
the Embedding Theorem.  Drag the mouse
over the big blue square at the top to
initiate the applet.

A tour of the applet 
Before we give the proof, we
need to explain how the applet
works.  The bottom half of the
applet shows slices of the two
4 dimensional polytopes that
arise in the Master Picture
Theorem.  We think of these
polytopes as fibering over the
big blue square at the top
left part of the applet.

If you drag the mouse over this
square, you change the
Slice. The vertical direction
selects the kite parameter A,
and the horizontal direction
just selects the third
dimension of the two polyhedra
associated to the parameter A.
In the demo for the Master
Picture Theorem, we only let
you move along the horizontal
direction, with the parameter
fixed.

The little colored checkerboard
next to the slice selector has
no active function in the
applet.  It simply reminds
you of how the Master Picture
Theorem works.  That is, it
reminds you of how the colors
in the polytope partition
determine the local picture
of the arithmetic graph.

The  shift  control
panel controls the placement
of one of the polytopes.  
Try toggling this control
panel and seeing how the
polytope moves.  Our proof of
the Embedding Theorem breaks
into two halves, and each
half uses a different 
positioning.

The  color  control 
panel allows you to simplify
the colorings.  You will see
below how this can be useful
for seeing certain things
better.

In the top right of the applet,
you will see two black 3x3 grids.
These grids are the  piece
selectors    The left piece
selector (with the red bar
just above it) corresponds to
the left polytope. The right piece
selector (with the blue bar
just below it) corresponds to
the right polytope.

Let's concentrate on the left
piece selector.  Try clicking
on some of the buttons.  If you
turn on one of the squares in
the piece selector, the lattice
orbit of the corresponding
piece is drawn in the lower
half of the applet. Recall
that our polyhedron (for each
parameter) is a fundamental
domain for the action of
a certain lattice.  We can
act on each piece by this
lattice and plot the image in 
the plane.  We don't plot
the whole image, which is
infinite, but only enough
for you to see what is going on.

The little red button turns
on and off the display of
the left polytope.  The little
blue button does the same thing
for the right polytope.

 The Immersion Property 
For the first part of the proof
we will explain why every point
of the arithmetic graph has
valence 2.  By definition, every
point of the arithmetic graph has
valence either 1 or 2.  The thing
we are trying to rule out is the
possibility that both edges
incident to a vertex coincide.

Recall from the statement of the
Master Picture Theorem that the
two edges incident to a vertex
are determined by which polyhedra
in the partitions intersect the
lattice associated to the point.
The relevant edge is nontrivial
if and only if the corresponding
lattice point lies on one of the
colored (non-grey) polyhedra.

 Step 1: 
Set the shift to normal
and the coloring to normal.
Turn on the middle squares
of both piece selectors.
You will see that the
lattice orbit of the left
grey piece is the same
as the lattice orbit of
the right grey pieces.
This means that a lattice
point (associated to a
vertex of the arithmetic
graph) lands in the left
grey pieces if and only if
it lands in the right grey
pieces.  This translates
into the statement that
if one edge of the 
extended arithmetic graph
is trivial, so is the 
other one.

 Step 2: 
Set the shift to normal
and the coloring to
simplified.  Turn on the
bottom right pieces
on each piece selector.
You will see that the
lattice of blue pieces
does not overlap with
the lattice of red pieces.
You can tour the whole
4 dimensional picture by
dragging the mouse over
the slice selector.  You
get the same disjointness
result for all slices.
This shows that the
two edges of the arithmetic
graph at a vertex cannot
both point down and to the
right. 

 Step 3:
Now go around and do the same 
thing for each of the other
peripheral squares on the
piece selectors.  As you
go around, you should turn
off everything but the
two pieces you want to
focus on.  What you see
is that it is impossible
for the lattice to intersect
the two polyhedra in pieces
of the same (normal) color.
This translates into the
statement that two nontrivial
edges of the arithmetic graph,
incident to the same vertex,
cannot point in the same
direction.  Hence the
arithmetic graph has valence
2 at every nontrivial point.

 The Embedding Property 
To see that the extended
arithmetic graph is embedded,
we need to rule out the 
possibility that there is a
local crossing, as shown in
the figure below.



This happens if and only
if the grid G(m,n) intersects
one of the polytopes in a
light red piece, and the
grid G(m,n+1) intersects
one of the polytopes in a
pink piece.

For each of these two
points, there are two
ways that the offending
edge could arise. Thus,
there are 4 cases to
consider.  Let L be the
fundamental domain on the
left and let R be the one
on the right.

 Case 1 (R/R) 
Suppose that the relevant edge
associated to (m,n) arises
because a point in G(m,n)
lands in a lt. red polyhedron
in R, and the relevant edge
associated to (m,n+1) arises
because a point in G(m,n+1)
lands in a pink polyhedron in R.
In this case, both points
intersect the same slice
of R.  But you can inspect
the picture and see that 
the lt. red and pink polyhedra
never appear in the same
slice.

 Case 2 (R/L) 
In this case, the relevant
edge associated to (m,n)
arises because a point
in G(m,n) lands in a
red polyhedron in R and
a point in G(m,n+1) lands
in a pink polyhedron
in L.  However, there is
no pink polyhedron i L.


 Case 3 (L/L) 
Again, there is no pink
polyhedron in L.

 Case 4 (L/R) 
In this case, the relevant
edge associated to (m,n)
arises because a point
in G(m,n) lands in a
red polyhedron in L and
a point in G(m,n+1) lands
in a pink polyhedron
in R.  You can see this
case by setting the shift
position to "offset", and
turning on the red pieces
of L and the pink pieces
of R. You will see that the
lattice images of the two
kinds of pieces never
overlap.