Index
Recall that a generalized quadrangle satisfies the following axioms:
GENERALIZED QUADRANGLE
1. Two distinct points are contained in at most one line.
2. Given a line l and a point p not on l, there is exactly one line k through p that intersects l (in some point q)
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Recall the definition for an incidence graph. The incidence graphs of projective planes are in fact generalized hexagons.
We say that a generalized quadrangle is of order (s,t) if every line contains s + 1 points and every point is contained in exacly t +1 lines. It follows that this geometry has (s + 1)(st + 1) points and (t + 1)(st + 1) lines.
The generalized quadrangle of order (2,2) is the smallest non-trivial quadrangle. It follows that in this quadrangle each line will contain 3 points and each point will contain 3 lines. In addition, there are 15 lines and 15 points. We have seen a plane model for this geometry - it is called a "doily".
Its interesting to prove to yourself that the axioms of a generalized quadrangle actually hold in this case.
Now, lets look at a bit more graph theory. Recall that a circuit that includes each edge of a graph exactly once is called Eulerian. A graph is considered Eulerian if it contains an Eulerian circuit. While each edge is passed through only once when traversing an Eulerian circuit, an Eulerian circuit can pass through each vertex as much as neccessary. However, a graph that contains a circuit that goes through each vertex of G exactly once is called Hamiltonian.
Of course it makes sense to talk about circuits on any geometry, not just graphs, so lets look at circular walks on geometries - specifically generalized quadrangles. The walk will go along lines of the geometry. We consider a point "visited" if it is used to change directions (to get from one line to the next). Points that are simply passed over along a given line do not count as visited. Lines will be considered "visited" if two consecutive points on the walk lie along that line. The goal of a circular walk is to visit every point once and to visit every line once. Let's look at an example on the doily.
Now, recall from Problem 32 that by Singer's Theorem any finite Desarguesian plane has a fixed-point-free automorphism. This automorphism cyclically takes each point to each point and each line to each line. This is really a circular walk! Thus, we know that any finite Desarguesian plane must contain a circular walk. In the case of our projective planes, it is trivial to arrive at the circular walk from any fixed point-free automorphism. Let's look at the Fano Plane. Recall the fixed-point-free automorphism given in problem 32: f: 0 a 4, 1 a 0, 2 a 1, 3 a 2, 4 a 6, 5 a 3, 6 a 5. From this, we can find a circular walk on the Fano plane. Simply use the Fano Plane as usually labeled:
Now start at point 1. In our collineation, 1 a 0, so follow the line from 1 to 0. Then under f, f: 0 a 4, so from 0 go to point 4, then point 6, point 5, point 3, point 2 and back to point 1, as the collineation indicates. We can see that this is indeed a circular walk!
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PROBLEM 33: Find a circular walk on the Projective plane of order 3 as well as the fixed-point-free automorphism that corresponds to it. [Hint: We can construct this plane using the difference set {1, 2, 4, 10}.]
Solution
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References: Beutelspacher & Rosenbaum, Polster, Toth
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