Two Order Invariants Related to the Fixed Point Property

19 Problem 1: In all the examples seen in this paper, we have (P) = (P) if the latter is even, and (P) = (P) ? 1 otherwise. Could one nd a poset for which this is not the case? Problem 2: In all the examples of the article, the function takes all even values between 0 and (P). Can it be shown that this is always the case, that is, for any nite connected order P, the spectrum of is f0; 2 Problem 3: Are and invariants of the comparability graph as well, that is, given two orders P and Q that have the same comparability graph, do we have (P) = (Q) and (P) = (Q)? References 1] S. Abian and A.B. Brown, A theorem on partially ordered sets with application to xed point theorems, Can. 2] E. Corominas, Sur les ensembles ordonn es projectifs et la propri et e du point xe, C. R. 18 Proposition 6.3 Every minimal automorphic order is-and-minimal. Proof: Let P be minimal automorphic and R be a proper retract of P. By Proposition 6.1, there are retracts R 1 and R 2 of R with respective auto-morphisms f 1 and f 2 , such that (R) = (f 1) and (R) = (f 2). Then R 1 and R 2 are proper retracts of P, so by minimality of P f 1 and f 2 have xed points and (R) = (R) = 0. The following example shows that not all-minimal orders are minimal automorphic. Example 5: We show that for n 1, X 4n+2 (see Section 4) is-minimal. Let R be a retract of X 4n+2 such that (R) = (X 4n+2) = 2n + 2, and let be the corresponding retraction. Then R has diameter 2n + 2 and there exists an endomorphism f of R such that d(x; f(x)) = 2n + 2 for all x 2 R. Since X 4n+2 is a braid, f must be the restriction to R of the antipodal map. Assume without loss of generality that 0 2 R, so f(0) = 4n + 2 2 R and R contains a fence of length 2n + 2 from 0 to 4n + 2. Suppose that fence is n?1 i=0 f4i; 4i+3gf4n; 4n+1; 4n+2g, all other cases being similar. By taking the closure of the fence under f, we get n?1 i=0 f4i; 4i + …