We prove the following probabilistic result for weak order comparability:
Theorem 1.4.1. Let π, σ ∈ Sn be selected independently and uniformly at random, and let Pn∗ := P(π σ). Then Pn∗ is submultiplicative, i.e. Pn∗1+n2 ≤ Pn∗1Pn∗2. Con- sequently there exists ρ = limpn
Pn∗. Furthermore, there exists an absolute constant c >0 such that
n
Y
i=1
(H(i)/i)≤Pn∗ ≤c(0.362)n, where H(i) :=Pi
j=11/j. Consequently, ρ≤0.362.
The proof of the upper bound is parallel to that of Theorem 1.2.1, lower bound, while the lower bound follows from the non-inversion (resp. inversion) set criterion
described last section. Empirical estimates indicate thatρis close to 0.3. So here too, as in Theorem 1.2.1, the upper bound seems to be qualitatively close to the actual probability Pn∗. And our lower bound, though superior to the trivial bound 1/n!, is decreasing superexponentially fast with n, which makes us believe that there ought to be a way to vastly improve it.
Paradoxically, it is the lower bound that required a deeper combinatorial insight.
Clearly the number of π’s below (or equal to) σ equals e(P), the total number of linear extensions of P = P(σ), the poset induced by σ. (The important notion of P(σ) was brought to our attention by Sergey Fomin [24].) We prove that for any posetP of cardinality n,
e(P)≥n!. Y
i∈P
d(i), (1.1)
where d(i) := | {j ∈ P : j ≤i in P} |. (This bound is an exact value of e(P) if the Hasse diagram is a (directed) rooted tree, Knuth [34, sect. 5.1.4, ex. 20], or a forest of such trees, Bj¨orner and Wachs [8].) The bound (1.1) fore(P(σ)) together with the independence of sequential ranks in the uniform permutation were the key ingredients in the proof of Theorem 1.4.1, lower bound.
Mikl´os B´ona [12] has informed us that this general lower bound for e(P) had been stated by Richard Stanley as a level 2 exercise in [46, p. 312, ex. 1] without a solution. We have decided to keep the proof in the dissertation, since we could not find a published proof anywhere either. The classic hook formula provides an example of a poset P for which (1.1) is markedly below e(P). It remains to be seen whether (1.1) can be strengthened in general, or at least forP(σ). As an illustration,
1 2
3 4
Figure 1.3: The permutation-induced poset P(2143).
P = P(2143) has the Hasse diagram appearing in Figure 1.3. Then e(P) = 4, but (1.1) delivers only
e(P)≥24/9 =⇒ e(P)≥3.
Regarding the lattice properties of (Sn,), note that the identity permutation 12ã ã ãn is the unique minimum, andn(n−1)ã ã ã1 is the unique maximum. Letπ1, . . . , πr ∈Sn be selected independently and uniformly at random. It is natural to ask: “How likely is it that the infimum (resp. supremum) of {π1, . . . , πr} is the unique mini- mal (resp. maximal) element in the weak order lattice?” Equivalently, what is the asymptotic number ofr-tuples (π1, . . . , πr) such that inf{π1, . . . , πr}= 12ã ã ãn (resp.
sup{π1, . . . , πr} =n(n−1)ã ã ã1), n → ∞? It turns out that the answer is the same whether we consider infs or sups, which allows us to focus only on infimums. We prove the following:
Theorem 1.4.2. Let Pn(r) =P (inf{π1, . . . , πr}= 12ã ã ãn). Then 1. As a function of n, Pn(r) is submultiplicative. Hence, there exists
p(r) = lim n q
Pn(r)= inf n q
Pn(r), r ≥1.
2. For eachr ≥1, puthr(z) =P
j≥0 (−1)j
(j!)r zj and Hr(z) = (hr(z))−1. Then, letting P0(r) = 1, we have
Hr(z) =X
n≥0
Pn(r)zn, from which we obtain (Darboux theorem [2])
Pn(r) ∼ − 1 z∗h0r(z∗)
1
(z∗)n, n→ ∞.
Here, z∗ = z∗(r) ∈ (1,2) is the unique simple root of hr(z) = 0 in the disc
|z| ≤2. Consequently, p(r) = 1/z∗.
Unlike our results about comparability in the Bruhat and weak orderings, here we have asharp asymptotic formula. The key to the proof of this theorem is establishing the exact formula
Pn(r)=
n−1
X
k=0
(−1)k X
b1,...,bn−k≥1 b1+ããã+bn−k=n
1
(b1!)rã ã ã(bn−k!)r,
which follows from the principle of inclusion-exclusion. This formula for Pn(r) is, in some sense, an r-analog of that for the Eulerian numbers (B´ona [11], Knuth [34]).
Indeed, it turns out thatPn(r)is the probability that the uniform, independent permu- tations π−11 , . . . , π−1r have no common descents. Introduce the random variable Sn(r),
the number of these common descents, so thatPn(r) =P
Sn(r) = 0
. Another natural question here is:
Problem. What is the limiting distribution of Sn(r)?
We believe that the answer here is “Gaussian”, as it is in the case of the number of descents in a single uniformly random permutation (Sachkov [44]). Our feeling is that the proof will involve use of the bivariate generating function Fr(x, y) = P
n≥1xnE h
(1 +y)Sn(r) i
, which we prove has the simple form
Fr(x, y) = xfr(xy)
1−xfr(xy), fr(z) :=X
j≥0
zj (j + 1)!r.
Interestingly, this generating function is a special case of a more general result proved by Richard Stanley [45], although he was probably unaware of the connections his work had with the weak ordering.
In conclusion we mention several papers that are in the same spirit of this dissertation.
First, [39] and [40] (both by Boris Pittel), where the “probability-of-comparability”
problems were solved for the poset of integer partitions of n under dominance order, and for the poset of set partitions of [n] ordered by refinement. Also, [41] (again by B. Pittel), where the “infimum/supremum” problem was solved for the lattice of set partitions of [n] ordered by refinement. In [16], E. R. Canfield presents an enlightening extension of the inf/sup work done in [41]. Very recently, in [1], R. M.
Adin and Y. Roichman explore the valency properties of a typical permutation in the Bruhat order Hasse diagram.
This work is in large part the result of an intensive collaborative effort with my doctoral advisor, Boris Pittel. Portions of this dissertation have been accepted for publication (2006) in the journalTransactions of the American Mathematical Society (see [28] for availability).
CHAPTER 2
THE PROOF OF THE BRUHAT ORDER UPPER BOUND
In this chapter, we focus on the proof of Theorem 1.2.1, upper bound. The proof divides naturally into three steps, hence the divisions of the sections that follow.
We need to show that
P (π ≤σ) = O n−2 .
The argument is based on the (0,1)-matrix criterion. We assume thatnis even. Only minor modifications are necessary forn odd.