| Among the objects taken out of Larry's office after his death was the tetrahedral structure shown here on the right. It represents three dimensional projective space over the field of two elements -- i.e., P3(F2) -- and was built by a student in one of Larry's algebra classes, presumably as homework. The set of "lines" coming into each "point" is held together by electrical tape which is slowly drying out and releasing its grip. In the background, another student's construct can be seen -- with bright red "points" and lines made from coloured pipe cleaners. Though clearly sturdier, it is also smaller and even more confusing to the eye. |
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A very prominent feature of the gadget is the mess of circular wires adorning it. There are ten of them: one on each face of the tetrahedron, and one coming off the centre of each edge and touching the centres of the opposite faces. They represent those "lines" in P3(F2) which do not remain straight in Euclidean 3-space. Let us explain.
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The tetrahedron is embedded in Euclidean 4-space as the set of points
(x,y,z,w) such that x+y+z+w = 1. In the diagram on the left, let the point at the top lie on the
w-axis and the vertices of the base, from left to right, belong to the axes for x,y,z. Then the blue face satisfies the equation z = 0.
Keeping z = 0, the three edges of the face are given by x = 0, y = 0, w = 0, while the lines through the middle come from x = y, |
The extra "line" turns out to contain the three midpoints of the edges on that face. Thus it is an F2-line which is broken in Euclidean 3-space. It is here drawn as a yellow triangle but usually represented by a circle. The points in question have coordinates (#,#,0,#) where exactly one of the # equals 0 and the other two both equal 1/2. What does it mean to say that they satisfy x+y+w = 0 over F2?
| The vertices of the unit cube in E4 (Euclidean 4-space) are mapped bijectively onto the affine 4-space over F2 in the obvious way, with the 15 non-zero ones going to P3(F2). Though not
linear, this map preserves linear dependence: since the latter is just a question of certain
(integral) determinants vanishing, it will survive a fortiori in F2.
On the tetrahedron, the original hypercube vertices are scaled by suitable factors 1/m to make them satisfy x+y+z+w = 1. But this scaling does not affect linear dependence! Hence if points are collinear (or coplanar) in the tetrahedral model, then so are their images in P3(F2). |
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For instance, the plane y = z shown in blue on the preceding diagram has 3 new straight lines (through the middle), which therefore automatically correspond to lines in P3(F2). The broken line -- again shown as a yellow triangle -- connects the points corresponding to the hypercube vertices (1,0,0,1), (1,1,1,0), and (0,1,1,1). Over F2 they again satisfy x+y+w = 0 -- this time under the constraint y = z, of course.
Altogether we now have accounted for the 10 broken lines -- and for 25 straight ones, as a careful count will reveal. That is all: there are 105 pairs of points which fall into 35 blocks of three, one block per line. However, by sticking to equations with at most two non-trivial coefficients, we have only seen 10 of the 15 possible planes. To see another type, consider the equation x+y+w = 0 already encountered above.
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It defines a "pyramid" -- i.e., an irregular tetrahedron with three
long isosceles faces and one equilateral face (the "base"). The only straight lines are the three long edges (shown in blue). The vertices of the base form one of the broken lines.
Of the three others, each of which connects two long edges with the opposite point of the base, only the one given by y = z is shown. If broken lines were represented (as usual)
by ovals instead of triangles, the diagram would be even messier.
There are 4 such folded "planes" of pyramid type. As noted above, they reveal no new lines -- all lines having already been seen in one of the two types of flat planes. |
| There is one more F2-plane to be accounted for, the one corresponding to x+y+z+w = 0. Its points are the midpoints of the six edges,
hence define a regular octahedron. The only straight lines are its three space diagonals,
the four broken lines bound alternate faces.
To finish this story, let us go back and try to replace the determinant argument above with something more elementary. It really was not needed to identify the F2-planes, which were exhibited one by one. The fact to be proved is this: if three points U, V, W are collinear in the tetrahedral model, then so are their F2 counterparts. |
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Suppose that V lies between U and W. Then the pair U, W consists either of a tetrahedral vertex and a point on the opposite face, or of two midpoints of opposite edges. In both cases -- letting U*, V*, W* denote the corresponding hypercube vertices -- it turns out that V* = U*+W*, and this relation carries over modulo 2.