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RICHARD P. STANLEY RICHARD P. STANLEY

RICHARD P. STANLEY - PDF document

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RICHARD P. STANLEY - PPT Presentation

0 1984 30 B For all 0 i fn the linear transformation Ii1 145145lQji ViVni is invertible Let us call a Peck poset P unitary if Qi6 yeF YP x EPi 1 JL Let G be group ID: 394242

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0 1984 30 RICHARD P. STANLEY (B) For all 0 i fn, the linear transformation @?I--i+1 ‘*‘&+lQji: Vi+Vn-i is invertible. Let us call a Peck poset P unitary if Qi6)= yeF+, YP x EPi- (1) J&L Let G be group of automorphisms of a poset P, and let P/G be the quotient poset. The elements of P/G are the orbits of G, and P/G if there exist x E @, x’ E 0’ such thatx ’in P. The purpose of this paper is to give a simple, straightforward proof that if P is unitary Peck then P/G is Peck. A somewhat weaker result was P is a Boolean algebra. Namely, if Vi/G denotes the vector space with basis Pi/G (the orbits of G on Pi), then for 0 i n there are Vj/G such that: (a) if 0 E Pi/G then �tii(O = n,s, c(7 ICC?” forsome co#EC, and (b) c?EPjfG rank $ij=min{ IPi/GI, IPj/GI}. This result implies that P/G has the Sperner property, but it is not If P is a unitary Peck poset then P/G is Peck. NOTE: (a) If P P/G need not be unitary Peck. For instance, if P is the Boolean algebra B, (which is unitary Peck) and G the cyclic group of order 5 (acting in the obvious way), then P/G is given by I (9