4

FRANCESCO BRENTI

Finally, let us remark that even in the case that a? is a natural labeling Conjectures 1.1,

1.2, 2.1 and 2.2 are also still open.

1.3 Partial results on the Poset Conjecture

In this section we present some partial results on the Poset Conjecture. Of course, one of

the first things that come to mind in trying to resolve the conjecture is to see if some of the

familiar operations on posets preserve its validity.

It is immediately clear, for example, that the conjecture holds for a labeled poset (P, u?)

if and only if it holds for its dual (P*,oJ) (where U(x) = \P\ -f 1 — u(x) for x € P ) because

0(P , QJ;Z) = Q,(P*,uJ;z) by the combinatorial definition of the order polynomial. On the

other hand, we can easily convince ourselves that even if the conjecture holds for two labeled

posets (P, a?i) and (Q, 0*2) it is almost impossible to understand if it also holds for their direct

product ( P x § , w i x u?2) or for the power ( P , ^ ) ^ ' ^ , (see [63], §3.2 for the definitions of

these operations).

The only other operations for which we can have any reasonable hope of success are the

ordinal sum © and the disjoint union +. The first operation does indeed preserve the Poset

Conjecture as we will now show. Let (P,u;i) and (Q,u?2) be two labeled posets. If we denote

with p and Q the partial orderings on P and Q then recall that the ordinal sum P © Q

is the poset that has P l+l Q as underlying set partially ordered by letting x y if either

x,y € P and x p y; or x, y € Q and x Q y; or x € P , y € Q. We denote with u?i © u2 the

labeling of P © Q defined by

/

m

v

M

d e f f «l(aO if 3 € P ,

\a,2(*) + |P| iixeQ.

Note that ujt © ^2 is natural if and only if both u^ and c^ are natural.

Propositio n 1.3.1 Let (P,a?i) and ( Q , ^ ) be as above. Then

zWiPeQ^t^^z) = zW(Q®P,W2@wi]z)

= W(P,UJ1;Z)W(Q,LU2;Z)

In particular, W(P © Q, LJ\ © UJ2] Z) has only real zeros if and only if both W(P,UJI\ z) and

W{Q^2] z) have only real zeros.

Proof: From the definitions of P © Q and uo\ © UJ2 it follows easily by direct combinatorial

reasoning that

s

es(P © Q,u) = J2es-i(Q^2)(ei(P,uj1) + e ^ P , ^ ) ) ,

1=0

for all s 0, where LJ = u)\ © UJ2. Hence we have

zE(P9QtU)(z) = ^ e

s

( P © Q ,

W

)

2

s + 1