By D. Mumford

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If 8(x, y) +8(y, z) :s: nand projyx -I projyz, then 8(x, z) = 8(x, y) +8(y, z). 7 Lemma. c. If 6(x, y) + 6(y, z) + 6(z, x) :s: 2n, projab i projac, for {a, b, c} = {x, y, z}, with 6(a, b) < n, for all {a, b} <;;; {x, y, z}, then 6(x, y) + 8(y, z) + 8(z, x) = 2n and x, y, z are contained in a unique apartment of r. Proof. This follows immediately from the previous lemma considering the element u opposite x and contained in the path connecting y with z. From the assumptions it indeed follows that u is unique, and there are two different paths of length n joining x and u: one containing the path from x to y; the other one containing the path connecting x with z.

If R. = 1, then r has order (2,2) (and is unique, as follows from our construction). If R. 1 (iii). 7(ii). If e= 3, then we have a generalized quadrangle of order (2,4). Now let e > 3. Let X and Y be two points of Q so that (J"x and (J"y are disjoint, and let Y' E Q be so that (J" y = (J" y'. Let P be any point of P collinear with both Y and y'. Let z and z' be the unique points of Q on PY and PY', respectively (and different from Y and y', respectively). We clearly have (J" z = (J" z', but (J" z is not disjoint with (J"x, for otherwise (J"z(J";1 has order 3, and hence it is an even permutation, as is (J"x(J";I, and (J"y(J";1 is a transposition, which is a contradiction.

Z is a line). If 8(x, z) = n, then f 1 (z) <:;;: f n-l (x). Then double counting of all flags {y, z} with 8(x,y) = n -1, 8(x,z) = n gives the formula (t + l)lfn(x)1 = Ifn-1(x)ls if n is even, and (s + l)lfn(x)1 = Ifn-1(x)lt if n is odd (then s = t). The result now 0 easily follows. Throughout this book, we will make freely use of the above lemma without referring explicitly to it. In fact, we will only need the various values for n = 3,4,6,8 and we write some of them down separately in the next result.