By Claus M. Ringel

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A2,a3,c. t to b 2. a2, Thus there Let y being a sincere positive root of X. be a Since (z), we have However, + Za2 + Za3 + z c for any positive root is the coefficient c F(5). determine t 0 = zt - z z t then c, then to one of a 1.... a2, b 1.... t remains to is a neighbor is a completion If there is no edge from then t . i i, we obtain a contradiction. al,a2,a3, assume that there is no edge from If there is also no edge from type for all is a neighbor both of is a neighbor zb i=I of the root y of F(4), we have Y ~ - Yc ~ 2 Oc(y) at the v e r t e x c).

Must be comparable C(2)), a! < d, we consider {al,d,bl,b2,Cl,C2,C3,C4}. are neighbors c E S, then form a subset thus aI or smaller convex subset in T in case again of type C(5) (since the subset d < b 2. is obtained c i < Ci+l, b2 d < b2, and, of course, S" a I < a2, b I < b2, and This finishes with instead of As above, we see that S' a! < d, or a strictly We argue similarly a subset c4 from Thus, b I S" S, conand a2 by replacing i = 1,2,3, by chains, w h i c h is convex in and T, and the proof.

D, we consider {al,d,bl,b2,Cl,C2,C3,C4}. are neighbors c E S, then form a subset thus aI or smaller convex subset in T in case again of type C(5) (since the subset d < b 2. is obtained c i < Ci+l, b2 d < b2, and, of course, S" a I < a2, b I < b2, and This finishes with instead of As above, we see that S' a! < d, or a strictly We argue similarly a subset c4 from Thus, b I S" S, conand a2 by replacing i = 1,2,3, by chains, w h i c h is convex in and T, and the proof. References The two theorems of Ovsienko (both the statements and the proofs) are taken from [Ov], also using oral cormmunioations by Ovsienko.

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