Lectures on Buildings (Perspectives in Mathematics, Vol 7) by Mark Ronan

By Mark Ronan

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F! t nt; 1 if m is odd; for 1n odd, reversing = the roles of sand t gives an alternative proof that s t. If 111 is even show that the total number of chambers is (s + 1)(t + 1)(1 + st + ... + (st) T- 1 ). 16. e. lies on exactly 2 edges). -gon ~o by introducing a. nd taking the vertices of ~o to be the type 2 vertices of ~. 4) that 111. = 2,3,4 or 6 only. 17. A polarity of a generalized 1n-gon is an (ou ter) au tomorphism of order 2 interchanging the t\VO types of vertices. Show that the chanlbers fixed by a polarity are mutually opposite, and if there are no fixed chambers then every chamber is carried to an opposite one.

In this case there is, in W, a minimal gallery frolll 1 to Xl via l-i of reduced type f. Let y be the second term in the unique gallery of type f from 0'(1) to a(xI), and define a(ri) y. Again we need to show that 6(y,a(x)) rix for all X E X, so define {3(x) = ri6(y,a(x)). Since y is i-adjacent to 0'(1), we see that = 6(y,a(x)) = = rix or x, and therefore {3(X) = x or rix. Now, as a Illap froln X to W, {3 is a cOlnposite of three maps: 0, 6(y, ) and ri (left nlultiplication). The first and last of these preserve distances, and the middle one does not increase distances, because it preserves adjacency.

3. Let 0' be a root, and 1r a panel in Ga. If x, y ยข 0' are chambers in St(1r), show that 0' U {x} is isometric to 0' U {y}. Conclude that 0' U {x} lies in an apartment, and show that 0' is the intersection of all apartments containing it. 4. Given any two chambers x and y in a thick building, show that the set of all chambers on Ininimal galleries froln x to y is the saine as the intersection of all apartlnents containing both x and y. 8) and Exercise 3]. 5. Let W be finite, and define two chambers x and y to be opposite if they are opposite in sOlne apartment A containing both.

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