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
Previous Page Next Page