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This demo gives a pictorial proof of the following result.Embedding TheoremThe extended arithmetic graph is a disjoint union of embedded polygons and infinite embedded polygonal arcsIn 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. Theshiftcontrol 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. Thecolorcontrol 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 thepiece selectorsThe 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.