Wachs permutations, Bruhat order and weak order

Abstract

We study the partial orders induced on Wachs and signed Wachs permutations by the Bruhat and weak orders of the symmetric and hyperoctahedral groups. We show that these orders are graded, determine their rank function, characterize their ordering and covering relations, and compute their characteristic polynomials, when partially ordered by Bruhat order, and determine their structure explicitly when partially ordered by right weak order.

Full text

arXiv:2212.04932v1 [math.CO] 9 Dec 2022
Wachs permutations, Bruhat order and weak order
Francesco Brenti ∗and Paolo Sentinelli†
Abstract
We study the partial orders induced on Wachs and signed Wachs permutations
by the Bruhat and weak orders of the symmetric and hyperoctahedral groups. We
show that these orders are graded, determine their rank function, characterize their
ordering and covering relations, and compute their characteristic polynomials, when
partially ordered by Bruhat order, and determine their structure explicitly when par-
tially ordered by right weak order.
1 Introduction
Wachs permutations are a class of permutations first introduced (in the even case)
in [22] to study the signed Eulerian numbers and the signed major index enumerator
of the symmetric groups. This class was extended in [4] to the odd case and to signed
permutations in order to study the enumerators of the odd and even major indices of
classical Weyl groups twisted by their one-dimensional characters. In this work we study
Wachs (and signed Wachs) permutations in their own right. More precisely, we study
the partial orders induced on them by the Bruhat and weak orders of the symmetric
and hyperoctahedral groups. These orders are fundamental objects in algebraic combi-
natorics and have important connections to algebra and geometry (see, e.g., [2, Chaps.2
and 3], [1, Chap.2], [16, Chap.1], [11, Chap.5], [7, Chap.10], [17, Chap.2], and the refer-
ences cited there). Many subsets of the symmetric and hyperoctahedral groups have been
studied as posets under the Bruhat order such as, for example, quotients, descent classes,
and generalized quotients [3], complements of quotients [19], involutions [9, 12, 13, 15, 18],
conjugation-invariant sets of involutions [5, 8], and twisted identities [10]. In this paper
∗Dipartimento di Matematica, Universitá di Roma “Tor Vergata”, Via della Ricerca Scientifica, 1, 00133
Roma, Italy. brenti@mat.uniroma2.it
†Dipartimento di Matematica, Politecnico di Milano, Milan, Italy. paolosentinelli@gmail.com
1

we show that Wachs permutations possess many nice properties when partially ordered by
Bruhat and weak orders. More precisely, we show that they form a graded poset, deter-
mine their rank function, characterize the ordering and covering relations, and compute
the characteristic polynomial, when partially ordered by Bruhat order, and determine their
structure explicitly when partially ordered by right weak order.
The organization of the paper is as follows. In the next section we collect some notation
and results that are used in the sequel. In §3 we study the poset obtained by partially
ordering Wachs permutations with respect to Bruhat order. We show that these posets
are always graded (Theorem 3.9), characterize their order and covering relations (Theorem
3.13 and Proposition 3.12), and compute their characteristic polynomials (Corollary 3.16).
In §4, using some of the results in §3, we obtain analogous results for the poset induced on
signed Wachs permutations by Bruhat order. More precisely, we show that these posets
are graded (Theorem 4.10), characterize their order and covering relations (Theorem 4.13,
Proposition 4.2, Corollary 4.12, and Corollary 4.6), and compute the characteristic poly-
nomials (Corollary 4.16). In §5 we study the posets obtained by partially ordering Wachs
and signed Wachs permutations under right weak order. We show that they are always
isomorphic to the direct product of a Boolean algebra with the weak order on the whole
group in rank n
2(Theorems 5.1 and 5.3). Finally, in §6, we discuss some conjectures and
open problems arising from the present work and the evidence that we have about them.
2 Preliminaries
In this section we recall some notation, definitions, and results that are used in the
sequel. As Nwe denote the set of non-negative integers and as Pthe set of positive integers.
If n∈N,[n] := {1,2, ..., n}; in particular [0] = ∅. For n∈P, in the polynomial ring Z[q]
the q-analogue of nis defined by [n]q:=
n−1
P
i=0
qiand the q-factorial by [n]q! :=
n
Q
i=1
[i]q. For
a set Aand f:A→Nwe define
A(x, f ) := X
w∈A
xf(w)
and f(u, v) := f(v)−f(u), for all u, v ∈A. The cardinality of a set Xwill be denoted by
|X|and the power set of Xby P(X). Given a cartesian product X×Yof two sets Xand
Y, we indicate with π1and π2the projections on Xand Yrespectively.
Let n∈N,i∈Z,q∈Qand J⊆[n]; then we define Je:= {j∈J:j≡0 mod 2},
Jo:= {j∈J:j≡1 mod 2},J+i:= {i+j:j∈J} ∩ [n], and qJ := S
j∈J
{qj} ⊆ Q.
We follow Chapter 3 of [20] for notation and terminology concerning posets. We just
2

recall some definitions. Given two posets Pand Q, their ordinal product P⊗Qis defined
by ordering the set P×Qin the following manner: (x, y)6(x′, y′)if and only if x=x′
and y6y′or x < x′, for all (x, y),(x′, y′)∈P×Q. If µis the Möbius function of a
graded poset Pwith minimum ˆ
0, maximum ˆ
1and rank function ρ, then the characteristic
polynomial of Pin the indeterminate xis defined by
CP(x) := X
z∈P
µ(ˆ
0, z)xρ(z,ˆ
1).
Next recall some basic results in the theory of Coxeter groups which will be useful in
the sequel. The reader can consult [2] or [11] for further details. Let (W, S)be a Coxeter
system. The length of an element z∈Wwith respect of the given presentation is denoted
as ℓ(z). If J⊆S, we let
WJ:= {w∈W:ℓ(ws)> ℓ(w)∀s∈J},
JW:= {w∈W:ℓ(sw)> ℓ(w)∀s∈J},
D(w) := {s∈S:ℓ(ws)< ℓ(w)},
and, more generally, for any A⊆Wwe let AJ:= A∩WJ. The subgroup WJ⊆Wis the
group generated by J. In particular WS=Wand W∅={e}, being ethe identity of W.
We consider on Wthe Bruhat order 6(see, e.g., [2, Chapter 2] or [11, Chapter 5]) and
on any subset we consider the induced order. When the group Wis finite, there exists a
unique maximal element w0of maximal length. We recall this characterizing property of
the Bruhat order, known as the lifting property (see [2, Proposition 2.2.7]):
Proposition 2.1. Let v, w ∈Wbe such that v < w and s∈D(w)\D(v). Then v6ws
and vs 6w.
For J⊆S, each element w∈Whas a unique expression w=wJwJ, where wJ∈WJ
and wJ∈WJ(see [2, Proposition 2.4.4]). Often we consider the projection PJ:W→WJ
defined by PJ(w) = wJ. This map is order preserving ( [2, Proposition 2.5.1]). We let
T:= {wsw−1:w∈W, s ∈S}. The following lemma will be useful in the next section; for
a proof see [2, Lemma 2.2.10].
Lemma 2.2. Suppose that x < xt and y < ty, for x, y ∈W,t∈T. Then, xy < xty.
For w∈Wwe set TL(w) := {t∈T:ℓ(tw)< ℓ(w)}; the right weak order 6Ron
Wis the partial order whose cover relations are defined by letting u⊳Rvif and only if
v−1u∈Sand ℓ(v) = ℓ(u) + 1, for all u, v ∈W(see [2, Chapter 3]). Then the following
characterization holds (see [2, Proposition 3.1.3]):
3

Proposition 2.3. Let (W, S)be a Coxeter system. Then u6Rvif and only if TL(u)⊆
TL(v), for all u, v ∈W.
Now, for any n∈P, let Snbe the group of all bijections of the set [n]. It is well
known that it is a Coxeter group with set of generators {s1, s2, ..., sn−1}, being sithe
simple inversion given, in one line notation, by 12...(i+ 1)i...n. Given a permutation
σ=σ(1)σ(2)...σ(n)∈Sn, the action of sion the right is given by σsi=σ(1)σ(2)...σ(i+
1)σ(i)...σ(n), for all i∈[n−1]. For i, j ∈[n], the action on the right of a transposition
(i, j),i6=j, then is given by σ(i, j) = σ(1)σ(2)...σ(i−1)σ(j)...σ(j−1)σ(i)...σ(n), for
all σ∈Sn. As a Coxeter group, identifying {s1, s2, ..., sn−1}with [n−1], we have that
(see e.g. [2, Propositions 1.5.3 and 1.5.2]) D(σ) = {i∈[n−1] : σ(i)> σ(i+ 1)}and
ℓ(σ) = ℓA(σ), where ℓA(σ) := inv(σ) = |{(i, j)∈[n]2:i < j, σ(i)> σ(j)}|, for all σ∈Sn.
Then, given J⊆[n−1],SJ
n={σ∈Sn:σ(i)< σ(i+ 1) ∀i∈J}.
For i∈[n]and A⊆Sndefine A(i) := {σ∈A: pos(σ) = i}, being pos : Sn→[n]the
function defined by pos(σ) := σ−1(n), for all σ∈Sn. We find it convenient to define the
following involution on [n]:
i∗:= 






i−1,if i≡0 mod 2;
i+ 1,if i≡1 mod 2 and i+ 1 ∈[n];
n, otherwise,
and the simple inversion s∗
i:= (i, i∗), for all i∈[n]. Given a permutation σ∈Sn,
k∈[n]and i∈[k], define σi,k as the i-th element in the increasing rearrangement of
{σ(1), σ(2), ..., σ(k)}<and σ−1
i,k as the position of σ(i)in {σ(1), σ(2), ..., σ(k)}<. So σj,k =
σ(i)if j=σ−1
i,k . The following theorem, known as the “tableau criterion”, characterizes the
Bruhat order on Sn(see [2, Theorem 2.6.3]).
Theorem 2.4. Let n∈Pand σ, τ ∈Sn. The following are equivalent:
1. σ6τ;
2. σi,k 6τi,k for all k∈D(σ)and i∈[k];
3. σi,k 6τi,k for all k∈[n−1] \D(τ)and i∈[k].
We indicate by wnthe maximum of the poset (Sn,6), i.e., in one line notation, wn=
n...321.
The descent number and the major index are the functions des : Sn→Nand maj :
Sn→Ndefined respectively by des(σ) := |D(σ)|and maj(σ) := Pi∈D(σ)i. A famous result
of McMahon asserts that ℓAand maj are equidistribuited over Sn(see, e.g. [20, Proposition
4

1.4.6]), i.e. Sn(x, ℓA) = Sn(x, maj), and this rank-generating function is known to be (see,
e.g [20, Corollary 1.3.10])
Sn(x, ℓA) = [n]x!.(1)
Following [4] we define functions from Snto Nby letting
odes(σ) := |D(σ)o|,emaj(σ) := X
i∈D(σ)e
i
2,
for all σ∈Sn, and call these functions odd descent number and even major index respec-
tively.
Following [4], for n∈P, we let
W(Sn) := {σ∈Sn:|σ−1(i)−σ−1(i∗)|61if i∈[n−1]}
and call the elements of W(Sn)Wachs permutations. It is not hard to see that, if nis even,
W(Sn) = {σ∈Sn:|σ(i)−σ(i∗)|61if i∈[n−1]}.
Let [±n] := {−n, . . . , −1,1,...,n}. We denote by Bnthe group of bijective functions
σ: [±n]→[±n]satisfying −σ(i) = σ(−i), for all i∈[n]. We use the window notation.
So, for example, the element [−2,1] ∈B2represents the function σ: [±2] →[±2] such
that σ(1) = −2 = −σ(−1) and σ(2) = 1 = −σ(−2). We let Neg(σ) := {i∈[n] :
σ(i)<0},neg(σ) = |Neg(σ)|,nsp(σ) := |{(i, j)∈[n]2:i < j, σ(i) + σ(j)<0}|,
sB
j:= (j, j + 1)(−j, −j−1) for j= 1, ..., n −1,s0:= (1,−1), and SB:= {s0, sB
1, ..., sB
n−1}.
It is well known that (Bn, SB)is a Coxeter system and that, identifying SBwith [0, n −1],
the following holds (see, e.g., [2, §8.1]).
Proposition 2.5. Let σ∈Bn. Then ℓB(σ) = ℓA(σ) + neg(σ) + nsp(σ), and D(σ) = {i∈
[0, n −1] : σ(i)> σ(i+ 1)}, where σ(0) := 0.
We denote by 6the Bruhat order of Bnand by σ7→ ˜σthe embedding Bn֒→S±n
(where S±nis the set of all bijections of [±n]) . The following result is [2, Corollary 8.1.9].
Proposition 2.6. We have that σ6τin Bnif and only if ˜σ6˜τin S±n.
The next result, which appears in [14, Theorem 5.5], characterizes the cover relations
in the Bruhat order of Bn. For σ∈Bnand i, j ∈[±n],i < j, we say that (i, j)is a rise for
σif σ(i)< σ(j)(i.e., if (i, j)is not an inversion of σ). Given a rise (i, j)for σwe then say,
following [14], that (i, j)is central if (0,0) ∈[i, j]×[σ(i), σ(j)], that (i, j)is free if there is
no i < k < j such that σ(i)< σ(k)< σ(j), and that it is symmetric if i=−j.
Theorem 2.7. Let n∈Pand σ, τ ∈Bn. Then σ⊳τin Bruhat order if and only if either
5

1. τ=σ(i, j)(−i, −j)where (i, j)is a non-central free rise of σor
2. τ=σ(i, j)where (i, j)is a central symmetric free rise of σ.
Following [4], for n∈P, we let
W(Bn) := {σ∈Bn:|σ−1(i)−σ−1(i∗)|61if i∈[n−1]}
and call the elements of W(Bn)signed Wachs permutations.
Our aim in this work is to study the sets of Wachs and signed Wachs permutations
under the Bruhat order and the weak order.
3 Wachs permutations and Bruhat order
Let (W, S)be a Coxeter system. For any independent set I⊆S(i.e. st =ts for all
s, t ∈I) we define a subgroup of Wby
GI:= {w∈W:Iw=I},
where Iw:= {wsw−1:s∈I}. Note that GIis a subgroup of W, and that, since Iis
independent, WI⊆GI. Furthermore, WIis normal in GI.
Proposition 3.1. PI(GI)is a subgroup of GI, isomorphic to the quotient GI/WI. In
particular we have the group isomorphisms
GI≃PI(GI)⋉WI≃S2≀IPI(GI).
Proof. Let w∈GIand write w=wIwIwhere wI∈WIand wI∈WI. Let s∈I. Then, by
our hypothesis, wsw−1=wIwIs(wI)−1(wI)−1=wIs(wI)−1so wI∈GI. So PI(GI)⊆GI.
Let u, v ∈GIand write u=uIuIand v=vIvIwhere uI, vI∈WIand uI, vI∈WI,
so uI, vI∈PI(GI). Let s∈I. Then there is t∈Isuch that vIs=tvI. Furthermore,
since uI, vI∈WI,ℓ(tvI)> ℓ(vI), and ℓ(uIt)> ℓ(uI). Therefore, by Lemma 2.2 we
obtain that uIvIs=uItvI> uIvI. Hence uIvI∈WI. But, since vI∈GI, there exists
˜uI∈WIsuch that uv =uIvI˜uIvIso (uv)I=uIvI. Hence uIvI∈PI(GI). In particular,
(uI)−1= (u−1)I, so PI(GI)is a subgroup of GI, and PI:GI→PI(GI)is a surjective
homomorphism whose kernel is WI. The last statements follow by the definitions.
Let m∈P,W:= S2m. For any set X, we consider P(X)as an abelian group, the
operation being the symmetric difference +, i.e. A+B:= (A\B)∪(B\A), for all
A, B ∈X. Then it is straightforward to see that if we take I={si:i≡1 mod 2} ⊆ S
6

then W(S2m) = GI. The group isomorphisms PI(W(S2m)) ≃Smand WI≃ P([m]) then
imply
W(S2m)≃Sm⋉P([m]) = S2≀Sm.
Therefore W(S2m)is isomorphic to the hyperoctahedral group.
We consider Snwith the Bruhat order and the subset W(Sn)⊆Snwith the induced
order. It is not difficult to see, using Theorem 2.4 that PI(W(S2m)) ≃Smas posets for
all m > 0, where the set Smis ordered by the Bruhat order. Let m > 0. For u∈Smand
T⊆[m]let φ−1
2m(u, T ) := v, where v∈ W(S2m)is defined by
v(2i−1) = (2u(i)−1,if i6∈ T,
2u(i),if i∈T,
and
v(2i) = (2u(i),if i6∈ T,
2u(i)−1,if i∈T,
for all i∈[m]. The following result follows easily from our definitions and Theorem 2.4,
and its proof is omitted.
Proposition 3.2. Let m∈P. Then
1. φ2mis a bijection;
2. φ2m:W(S2m)→Sm⊗ P([m]) is order preserving;
3. φ−1
2m:Sm× P([m]) → W(S2m)is order preserving.
Moreover ℓ(v) = 4ℓ(τ) + |T|if φ(v) = (τ , T ).
Figure 3 shows the Hasse diagram of (W(S5),6). By the following example we see that
φ2mand φ−1
2mare not poset isomorphisms in general.
Example 3.3.Let m= 3. Then (123,{1,2,3})6(132,∅)in Sm⊗P([m]) but φ−1
2m(123,{1,2,3}) =
214365 125634 = φ−1
2m(132,∅)in W(S2m).
Let m= 2,u= 2143 and v= 3412. Then u, v ∈ W(S4),u < v and φ2m(u) =
(12,{1,2})(21,∅) = φ2m(v)in Sm× P([m]).
We consider now W(S2m+1). For any m > 0it is not difficult to see that the set
W(S2m+1)is not a group. The previous general construction in Coxeter systems gives, for
W=S2m+1, the group GI≃ W (S2m)which, as a set, can be included in W(S2m+1). For
n > 1, define a function χn:W(Sn)→ W(Sn−1)by
7

4321
3421 4312
3412
2143
1243 2134
1234
Figure 1: Hasse diagram of (W(S4),6).
χn(v)(i) = (v(i),if i < pos(v);
v(i+ 1),if i>pos(v),
for all i∈[n−1],v∈ W(Sn). Note that the function χnis not necessarily order preserving.
For example, if m= 2 and u= 21345,v= 51234 then u6vbut χn(u) = 2134 1234 =
χn(v).
Let φ2m+1 :W(S2m+1)→[m+ 1] ×Sm× P([m]) be the function defined by
φ2m+1(v) := ((pos(v) + 1)/2, τ, T ),
if χ2m+1(v) = (τ , T ), for all v∈ W(S2m+1). For example, φ2m+1(4312756) = (3,213,{1}).
As in the even case, noting that u6vimplies pos(v)6pos(u), we have the following
result whose proof follows easily from our definitions and Proposition 3.2.
Proposition 3.4. Let m > 0. Then
1. φ2m+1 is a bijection;
2. φ2m+1 :W(S2m+1)→[m+ 1]∗⊗Sm⊗ P([m]) is order preserving;
3. φ−1
2m+1 : [m+ 1]∗×Sm× P([m]) → W (S2m+1)is order preserving,
where i6jin [m+ 1]∗if and only if j6i, for all i, j ∈[m+ 1]. Moreover ℓ(v) =
4ℓ(τ) + |T|+ 2(m−i+ 1) if φ2m+1 (v) = (i, τ, T ).
Let n > 0and fn:W(Sn)→S⌊n
2⌋be the function defined by the assignment (i, τ , T )7→
τif nis odd, and (τ , T )7→ τif nis even. By Proposition 3.2, if nis even then fnis order
8

54321
53421 54312
43521 53412
52143 34521 43512
51243 52134 34512 43215
21543 51234 34215 43125
12543 21534 34125
12534 21435
12435 21345
12345
Figure 2: Hasse diagram of (W(S5),6).
9

preserving. Corollary 3.8 states that the function fnis order preserving also in the odd
case. We also define a function ℓW:W(Sn)→Nby
ℓW(v) := ℓ(v)−ℓ(fn(v)),
for all v∈ W(Sn). For example, ℓW(342156) = 5 −1 = 4 and ℓW(3472156) = 9 −1 = 8.
The minimum and the maximum of the poset (W(Sn),6)are respectively the identity of
Sn, which corresponds to (e, ∅)in the even case and to (m+ 1, e, ∅)in the odd one, and
the element of maximal length wnof Sn, which corresponds to (wm,[m]) in the even case
and to (1, wm,[m]) in the odd one, where m:= n
2. Moreover ℓW(e) = 0 and
ℓW(wn) = n
2−⌊n
2⌋
2.(2)
Note that if nis even then ℓW(wn) = (3n−2)n
8=(3m−1)m
2. So {ℓW(w2m)}m∈Pis
the sequence of pentagonal numbers (see A000326 in OEIS). If nis odd then ℓW(wn) =
3(n2−1)
8=3m(m+1)
2; so {ℓW(w2m+1)}m∈Pis the sequence of triangular matchstick numbers
(see A045943 in OEIS).
Let n > 0and m:= n
2; for any i, j ∈[m],i < j, we define an involution wA
i,j :
W(Sn)→ W(Sn)by
wA
i,j (v) = ((τ(i, j), T +{i, j}),if nis even and v= (τ, T );
(k, τ (i, j), T +{i, j}),if nis odd and v= (k, τ , T ),
for all v∈ W(Sn). For example, if v= 4312765 ∈ W(S7), then wA
2,3(v) = 4356721.
Lemma 3.5. Let m > 0,v= (τ, T )∈ W(S2m),i, j 6∈ Tand τ(i, j)⊳τ. Then wA
i,j(v)< v
and ℓW(v)−ℓW(wA
i,j (v)) = 1. If u= (σ, S)∈ W (S2m),u < v and σ6τ(i, j)then
u6wA
i,j(v).
Proof. We have that ˆv:= wA
i,j(v) = (τ(i, j), T ∪ {i, j}). By the tableau criterion it is easy
to deduce that ˆv < v and the equality ℓW(ˆv, v) = 1 follows easily from Proposition 3.2
and the definition of ℓW. We now show that u6ˆv. We use Theorem 2.4. Note first that,
since uand ˆvare Wachs permutations and σ6τ(i, j),ul,h 6ˆvl,h for all 16l6hif his
even. Furthermore, since u < v,ul,h 6ˆvl,h for all l∈[h]if h62i−2or h>2j. So let
k∈[m]be such that k6∈ Tand i < k < j. We wish to show that ul,2k−16ˆvl,2k−1,for all
16l62k−1.
Let a:= v(2i−1) and b:= v(2j−1). So a+ 1 = v(2i)and a > b + 1 = v(2j)as well
as ˆv(2i−1) = b+ 1,ˆv(2i) = b,ˆv(2j−1) = a+ 1 and ˆv(2j) = a. Let c:= v(2k−1) and
d:= u(2k−1). Note that, since τ(i, j )⊳τ,c6∈ [b, a + 1] and that c≡1 (mod 2) since
10

k6∈ T. Let r, s ∈[k]be such that v2r−1,2k−1=cand u2s−1,2k−1=d. We have two cases to
consider according as to whether c < b or c > a + 1.
Say c > a + 1.
Let p, q ∈[k]be such that v2q−1,2k−1=aand ˆv2p−1,2k−1=b, so p6q,v2q,2k−1=a+ 1
and ˆv2p,2k−1=b+ 1. Then 2r−1>2qsince c > a + 1. Note that
vl,2k−1= ˆvl,2k−1(3)
if 16l62p−2or 2q+ 1 6l62k−1and that
vl,2k−1= ˆvl+2,2k−1(4)
if 2p−16l62q−2. Moreover, since u6vby our hypothesis,
ul,2k−16vl,2k−1,(5)
for all 16l62k−1.
Note that, by (3) and (5), we have
ul,2k−16vl,2k−1= ˆvl,2k−1
if 16l62p−2or 2q+ 1 6l62k−1. So assume 2p−16l62q.
If x, y ∈P,x, y ≡1 (mod 2), then we find it convenient to write {x, x + 1}6{y, y + 1}
if x6yand similarly for <.
We have two cases to consider.
Let s > q. Then, since σ6τ(i, j), we have that
{u2h−1,2k−1, u2h,2k−1}6{ˆv2h−1,2k−1,ˆv2h,2k−1}
for all p6h6q, so ul,2k−16ˆvl,2k−1for all 2p−16l62q.
Let s6q. If p6h6s−1then, since σ6τ(i, j ), we have that
{u2h−1,2k−1, u2h,2k−1}6{ˆv2h−1,2k−1,ˆv2h,2k−1}
so ul,2k−16ˆvl,2k−1for all 2p−16l62s−2.
We may therefore assume that max{2p−1,2s−1}6l62q. Since σ6τ(i, j), we
have that
ul,2k−26ˆvl,2k−2
11

for 16l62k−2. But
ul,2k−1=ul−1,2k−2
if 2s6l62k−1since {u(1), ..., u(2k−2)}={u(1), ..., u(2k−1)} \ {d}, and similarly
ˆvl,2k−1= ˆvl,2k−2
if 16l62r−2. Therefore
ul,2k−1< ul+1,2k−1=ul,2k−26ˆvl,2k−2= ˆvl,2k−1,
if 2s−16l62r−2, so ul,2k−16ˆvl,2k−1if 2s−16l62q, because 2q62r−2.
The case c < b is similar (and slightly simpler) except that one uses 2krather than
2k−2and obtains that
ul,2k−1=ul,2k6ˆvl,2k= ˆvl−1,2k−1<ˆvl,2k−1
if 2p6l62s−1. We leave the details to the interested reader.
The following lemma can be proved using Lemma 3.5 and the definitions.
Lemma 3.6. Let m > 0,v= (k, σ, S)∈ W(S2m+1 ),i, j, k 6∈ Sand σ(i, j)⊳σ. Then
wA
i,j(v)< v,v(k, k + 2) < v and ℓW(wA
i,j(v)) = ℓW(v(k, k + 2)) = ℓW(v)−1.
The next result is the crucial technical tool for the proof of our main theorem, and
states that elements of W(S2m+1)are “join-irreducible” in a certain sense.
Proposition 3.7. Let m > 0,u, v ∈ W (S2m+1 ),u < v and i:= pos(v). If u < v and
pos(u, v)>0then u6zwhere z∈ W(S2m+1 )is defined by
z=(v(i, i + 2),if v < v(i+ 1, i + 2);
v(i+ 1, i + 2),otherwise.
Proof. Let h:= (i+ 1)/2and consider first the case i+ 1 6∈ D(v). Let a:= v(i+ 1); then
a+ 1 = v(i+ 2) = z(i),a=z(i+ 1) and z(i+ 2) = 2m+ 1. Define c:= u(i),r∈[m+ 1]
be such that u(2r−1) = 2m+ 1,p∈[h]be such that z2p−1,i =a+ 1 and q∈[h]be such
that u2q−1,i =c. Notice that u(i+ 1) ∈ {u(i) + 1, u(i)−1}since pos(u, v)>0. Assume
that q6p. Note that
ul,i 6vl,i =zl,i
12

if 16l62p−2, while
ul,i =ul−1,i−16vl−1,i−1=zl−1,i−1=zl,i
if 2p6l, and
u2p−1,i =






u2p−2,i−16v2p−2,i−1=z2p−2,i−1=z2p−2,i < z2p−1,i,if q < p,
u2p−1,i+1 6v2p−1,i+1 =a < a + 1 = z2p−1,i ,if q=pand u(i)< u(i+ 1),
u2p−1,i+1 + 1 6v2p−1,i+1 + 1 = a+ 1 = z2p−1,i,if q=pand u(i)> u(i+ 1),
since u6v. Moreover ul,i+1 =ul−1,i 6zl−1,i =zl,i+1 if l>2p+ 1 while ul,i+1 6vl,i+1 =
zl,i+1, if l62p−2. Let l∈ {2p−1,2p}. There are some cases to be considered.
1. q < p and u(i)> u(i+ 1): in this case we have that u2p,i+1 =u2p−1,i 6z2p−1,i =
z2p,i+1 and so u2p−1,i+1 6u2p,i+1 −16z2p,i+1 −1 = z2p−1,i+1.
2. q < p and u(i)< u(i+ 1): in this case we have that u2p−1,i+1 =u2p−1,i −16
z2p−1,i −1 = z2p−1,i+1 and so u2p,i+1 =u2p−1,i+1 + 1 6z2p−1,i+1 + 1 = z2p,i+1.
3. q=pand u(i)> u(i+ 1): in this case u2p,i+1 =u2p−1,i 6z2p−1,i =z2p,i+1 and
u2p−1,i+1 =u2p−1,i −16z2p−1,i −1 = z2p−1,i+1.
4. q=pand u(i)< u(i+ 1): in this case u2p−1,i+1 6v2p−1,i+1 =a=z2p−1,i+1 and
u2p,i+1 =u2p−1,i+1 + 1 6z2p−1,i+1 + 1 = z2p,i+1.
So we have proved that u6zwhenever q6p. Consider the case q > p. If l62p−2
or l>2qthe result follows as above. Let 2p6l62q−2. Then ul,i =ul,i+1 6vl,i+1 =zl,i.
Moreover
u2p−1,i =u2p−1,i+1 6v2p−1,i+1 =a < a + 1 = z2p−1,i.
Let u(i)< u(i+ 1). Then
u2q−1,i =u2q−1,i+1 6v2q−1,i+1 =z2q−1,i.
If u(i)> u(i+ 1) we have that u2q−1,i =u2q−1,i+1 + 1 6v2q−1,i+1 + 1 = z2q−1,i and then
we have proved that u6zalso in case q > p.
Let’s consider the case si+1 ∈D(v). Define a:= v(i+ 2) and v2p−1,i+1 := a+ 1. If
l6= 2p−1we have that ul,i+1 6vl,i+1 =zl,i+1. Let l= 2p−1. Since u2p−1,i+1 ≡1 (mod 2),
a≡1 (mod 2) and u2p−1,i+1 6a+ 1, we conclude that u2p−1,i+1 6a=z2p−1,i+1.
Corollary 3.8. Let n∈P. Then fn:W(Sn)→S⌊n
2⌋is order preserving.
13

Proof. If nis even the result was already observed. If nis odd let u, v ∈ W(Sn)and
u6v. We prove the result by induction on pos(u, v). If pos(u, v) = 0 by the tableau
criterion we conclude that χn(u)6χn(v)and then the result follows by the even case.
Let pos(u, v)>0and i:= pos(v). In this case, by Proposition 3.7, u6v(i, i + 2) < v
or u6v(i+ 1, i + 2)(i, i + 2) < v(i+ 1, i + 2) < v. Hence, by the inductive hypothesis,
fn(u)6fn(v(i, i + 2)) = fn(v)or fn(u)6fn(v(i+ 1, i + 2)(i, i + 2)) = fn(v(i+ 1, i + 2)) =
fn(v).
We can now prove the main result of this section.
Theorem 3.9. Let n > 0. Then (W(Sn),6)is graded, of rank n
2−⌊n
2⌋
2, and its rank
function is ℓW.
Proof. Assume first n= 2m,m > 1. Let u, v ∈ W(S2m)with u < v,φ(u) = (σ, S)and
φ(v) = (τ, T ). We prove that, if ℓW(u, v)6= 1, then there exists z∈ W(S2m)such that
u < z < v and ℓW(z, v) = 1. Note that, by Proposition 3.2, σ6τ. We have two cases to
consider.
1. ℓ(σ, τ) = 0: then σ=τso, by Proposition 3.2, S⊆T. Therefore there exists
i∈T\S. So s2i−1∈D(v)\D(u)hence, by the Lifting Property, u < vs2i−1< v,
vs2i−1∈ W(S2m)and ℓW(vsi, v) = 1.
2. ℓ(σ, τ)>0: in this case, since σ < τ , there exist 16x < y 6msuch that
σ6τ(x, y)⊳τ. Let i:= 2x−1and j:= 2y−1. There are three cases to consider.
(a) si∈D(v): let z:= vsiand w:= (τ(x, y), T ). We want to prove that u < z. Let
a+ 1 := v(i) = v2p−1,i,b:= w(i) = v(j) = w2q−1,i and c:= u(i) = u2s−1,i. In
particular q6p. We only have to prove that ul,i 6(vsi)l,i for all l6i. Since
(vsi)l,i =vl,i if l6= 2p−1, it is enough to prove that u2p−1,i 6(vsi)2p−1,i. Note
that ul,i+1 6vl,i+1 = (vsi)l,i+1 for all 16l6i+ 1.
If s6pthen
(vsi)2p−1,i >(vsi)2p−2,i =w2p−1,i =w2p,i+1 >u2p,i+1 >u2p−1,i
because b < a, and σ6τ(x, y).
If s > p then u2p−1,i =u2p−1,i+1 6(vsi)2p−1,i+1 = (vsi)2p−1,i. Since ℓW(vsi, v ) =
1the result follows.
(b) si6∈ D(v)and sj∈D(v): we then claim that u < vsj. Let b+1 = v(j) = v2p−1,j ;
then b= (vsj)(j) = (vsj)2p−1,j . As in the case above it is sufficient to prove
14

that u2p−1,j 6(vsj)2p−1,j . If u(j)> u2p−1,j then u2p−1,j =u2p−1,j+1 =
u2p,j+1 −16v2p,j +1 −1 = b. Let u(j)6u2p−1,j; since σ6τ(x, y)we
have {u2p−1,j −1, u2p,j −1}6{w2p−1,j−1, w2p,j−1}. Then, u2p−1,j < u2p,j =
u2p−1,j−16w2p−1,j −1=b, since a > b + 1.
(c) si, sj6∈ D(v): in this case, by Lemma 3.5, u < wA
x,y(v)< v and wA
x,y(v)∈
W(S2m),ℓW(wA
x,y(v), v) = 1.
Now assume n= 2m+ 1,m > 0. Let u6v,i:= pos(v), and ℓW(u, v)>1. We prove
that there exists z∈ W(S2m+1)such that u < z < v and ℓW(z, v) = 1. If pos(u, v) = 0
then the result follows by the previous point. If pos(u, v)>0we have, by Proposition 3.7,
u < v(i, i + 2) < v if si+1 6∈ D(v)and u < vsi+1 < v otherwise.
Remark 3.10.In general W(Sn)∩SJ
nis not graded; one can see this by considering J={s1}
and the interval [124365,561234] in W(S6)∩SJ
6.
We can now compute the rank-generating function of (W(Sn),6).
Corollary 3.11. Let m > 0. Then
W(S2m)(x, ℓW) = (1 + x)m[m]x3!,
W(S2m+1)(x, ℓW) = (1 + x)m[m+ 1]x2[m]x3!.
Moreover
W(S2m)(x, ℓW) = W(S2m)(x, 3 emaj + odes)
and
W(S2m+1)(x, ℓW) = W(S2m+1)(x, (3 emaj + o des) ◦χ2m+1 + pos).
Proof. The result follows by (1), Theorem 3.2 and the definition of ℓW. In fact, W(S2m+1)(x, ℓW) =
[m+ 1]x2W(S2m)(x, ℓW)and ℓ(v) = odes(v) + 4ℓ(f2m(v)), for all v∈ W(S2m).
From Proposition 3.11 we find that the polynomials W(Sn)(x, ℓW)are reciprocal, i.e.
xℓW(wn)W(Sn)(x−1, ℓW) = W(Sn)(x, ℓW). In fact the poset (W(Sn),6)is self-dual, for
all n∈P, since the map v7→ vwnis an antiautomorphism of (W(Sn),6)such that
ℓW(vwn) = ℓW(wn)−ℓW(v)(see [2, Propositions 2.3.2 and 2.3.4]).
From the combinatorial description of the rank function of W(S2m+1)we can deduce
a description of its cover relations.
Proposition 3.12. Let m∈P,u, v ∈ W(S2m+1),u= (i, σ, S)and v= (j, τ , T ). Then
u⊳vif and only if either
15

1. i=j,σ=τand S⊳T, or
2. j=i−1,σ=τ,i−1∈Sand T=S\ {i−1}, or
3. i=j,σ⊳τ,T∩ {a, b}=∅and S=T∪ {a, b},
where (a, b) := τ−1σ. In particular, if u, v ∈ W (S2m),u= (σ, S)and v= (τ , T ), then
u⊳vif and only if either σ=τand S⊳T, or σ⊳τ,T∩ {a, b}=∅and S=T∪ {a, b}.
Proof. If point 1or 2hold then ℓW(v)−ℓW(u) = 1 and by Theorem 3.4 there follows that
u6v. If point 3holds then the result follows by Lemma 3.6.
Conversely let u⊳v. Then σ6τby Corollary 3.8 and ℓW(u, v) = 1. If i−j>2then,
by Proposition 3.7 there is z∈ W(S2m+1)such that u6z < v and ℓW(z, v ) = 1; so z=u,
which is a contradiction since k−j61, being z= (k, ρ, R). Hence assume i−j61.
If σ < τ then there exists a reflection (r, s)such that σ6τ(r, s)⊳τ. If i−j= 1 then
by Proposition 3.7 u=v(2j−1,2j+1) and σ=τ, which is a contradiction. Therefore i=j.
As in the proof of Theorem 3.9, if r∈Tthen (σ, S)6(τ , T \ {r}); if r6∈ Tand s∈Tthen
(σ, S)6(τ, T \ {s}). Moreover ℓW((τ, T \ {r}),(τ, T )) = 1 and ℓW((τ, T \ {s}),(τ , T )) = 1
in these cases. Then (σ, S) = (τ, T \{r})or (σ, S) = (τ, T \{s}), a contradiction. Therefore
r, s 6∈ Tand, as in the proof of Theorem 3.9, (σ, S)6(τ(r, s), T ∪ {r, s})⊳(τ, T ), and
ℓW((τ(r, s), T ∪ {r, s}),(τ, T )) = 1. So we conclude that S=T∪{r, s}and σ=τ(r, s)⊳τ.
Assume now σ=τ. If i=jthen S⊳T(else ℓW(u, v) = ℓ(u, v)>2). If i−j= 1 then
by Proposition 3.7 we have that u=v(2j−1,2j+ 1) and v < vs2jand the result follows.
The second statement follows from the first one by observing that W(S2m)is isomorphic
to the interval [e, 2m...321(2m+ 1)] in W(S2m+1).
For example the permutation 782156934 ∈ W(S9)covers the Wachs permutations
781256934,782156439 and 652187934, which correspond respectively to cases 1, 2 and 3
in Proposition 3.12.
From Proposition 3.12 we can now prove the second main result of this section, namely
a characterization of the Bruhat order relation on Wachs permutations.
Theorem 3.13. Let m > 0and u, v ∈ W (S2m+1),u= (i, σ, S),v= (j, τ , T ). Then u6v
if and only if
σ6τ,S(u, v)⊆T(u, v), and j6i,
where, for X⊆[m+ 1],X(u, v) := X∩([j−1] ∪[i, m]) ∩F(σ, τ), being F(σ, τ) :=
{k∈[m] : σ(k) = τ(k)}. Moreover ℓW(u) = 3ℓ(σ) + |S|+ 2(m−i+ 1). In particular, if
u, v ∈ W(S2m),u= (σ, S),v= (τ, T ), then u6vif and only if σ6τand S∩F(σ, τ )⊆T,
and ℓW(u) = 3ℓ(σ) + |S|.
16

Proof. Let u6v. We may assume u⊳v. There are three cases to consider.
1. i=j,σ=τand S⊳T: in this case F(σ, τ ) = [m]so the result follows.
2. j=i−1,σ=τand S=T∪ {i−1}, with i−16∈ T: in this case F(σ, τ ) = [m]and
S∩([i−2] ∪[i, m]) ⊆T.
3. i=j,σ⊳τ,T∩ {a, b}=∅and S=T∪ {a, b}, where (a, b) = τ σ−1: we have
F(σ, τ) = [m]\ {a, b}and then S∩F(σ, τ )⊆T.
Now let σ6τ,j < i and S∩([j−1] ∪[i, m]) ∩F(σ, τ )⊆T. In Smthere exists a saturated
chain σ=σ0⊳σ1⊳... ⊳σn=τwith n=ℓ(σ, τ ). Define (ai, bi) := σ−1
iσi−1for all
i∈[n]. We have the following chain in (W(S2m+1),6):
(i, σ, S)6(i, σ, S ∪[j, i −1])
⊳(i−1, σ, (S∪[j, i −2]) \ {i−1})
⊳(i−2, σ, (S∪[j, i −3]) \ {i−2, i −1})
⊳... ⊳(j+ 1, σ, (S∪ {j})\[j+ 1, i −1])
⊳(j, σ, S \[j, i −1])
6(j, σ, (S\[j, i −1]) ∪ {a1, b1})
⊳(j, σ1,(S\[j, i −1]) \ {a1, b1})
6(j, σ1,(S\([j, i −1] ∪ {a1, b1})) ∪ {a2, b2})
⊳(j, σ2,(S\[j, i −1]) \ {a1, b1, a2, b2})
6... 6(j, τ, (S\[j, i −1]) \ {a1, b1, ..., an, bn})6(j, τ , T ),
since {a1, b1, ..., an, bn}= [m]\F(σ, τ). The length formula follows by Proposition 3.4.
The last statement follows immediately noting that the map (σ, S)7→ (m+ 1, σ, S)is
a poset isomorphism between W(S2m)and {(i, σ, S)∈ W(S2m+1) : i=m+ 1}.
We illustrate the preceding theorem with an example. Let u= (4,2431,{1,2,3})∈
W(S9)and v= (3,3421,{2})∈ W(S9). Then we have that 2431 <3421 and S∩([j−1] ∪
[i, m]) ∩F(σ, τ) = {1,2,3} ∩ ({1,2} ∪ {4})∩ {2,4}={2}; hence, by Theorem 3.13, u < v.
The following lemma can be easily deduced by Theorem 3.13 so we omit its verification.
Lemma 3.14. Let m > 0and u, v ∈ W(S2m+1). If u6(i, σ, S1)6vand u6(i, σ, S2)6v
in (W(S2m+1),6)then u6(i, σ, S1∪S2)6v.
The characterization obtained in Theorem 3.13 enables us to give an explicit expression
for the Möbius function of lower intervals in the poset of Wachs permutations partially
ordered by Bruhat order, and shows, in particular, that it has values in {0,1,−1}.
17

Proposition 3.15. Let m > 0, and v= (j, τ, T )∈ W(S2m+1). Then
µ(e, v) = ((−1)|T|,if τ=eand j=m+ 1;
0,otherwise.
In particular, if v= (τ, T )∈ W(S2m)then
µ(e, v) = ((−1)|T|,if τ=e;
0,otherwise.
Proof. We proceed by induction on ℓW(v). If τ=eand j=m+ 1 then, by Theorem 3.13,
the interval [e, v]is isomorphic to a Boolean algebra, so we conclude that µ(e, v) = (−1)|T|,
as desired. So assume that either τ6=eor j < m+1. Then, by Proposition 3.12, ℓW(v)>2,
so, by Lemma 3.14 there exists R⊆[m],R6=∅, such that [e, v]∩ {(k, ρ, U )∈ W(S2m+1 ) :
ρ=e, k =m+ 1}= [e, (m+ 1, e, R)]. Hence
µ(u, v) = −X
x∈[e,v)
µ(e, x)
=−X
x∈[e,(i,σ,R)]
µ(e, x)−X
x∈[e,v)\[e,(i,σ,R)]
µ(e, x)
=−X
x∈[e,v)\[e,(i,σ,R)]
µ(e, x) = 0
by our induction hypothesis, and the fact that |[e, (i, σ, R)]| 6= 1, where [u, v) := {z∈
W(S2m+1) : u6z < v}.
The statement about Wachs permutations in the even case follows from the odd one
as in the proof of Theorem 3.13.
We conclude by computing, using Proposition 3.15, the characteristic polynomial of
the poset of Wachs permutations.
Corollary 3.16. The characteristic polynomial of (W(Sn),6)is
(x−1)⌊n
2⌋x(n
2)−(⌊n
2⌋+1
2),
for all n∈P.
Proof. The result follows from Poposition 3.15 and Theorem 3.9.
The following is a commutative diagram that summarizes the poset morphisms consid-
ered in this section. The function πistands for the canonical projection on the i-th factor
18

of a Cartesian product. Notice that, if Aand Bare posets, the projection π1:A⊗B→A
is order preserving, whereas π2:A⊗B→Bis not order preserving.
[m+ 1]∗×Sm× P([m]) W(S2m+1) ([m+ 1]∗×Sm)⊗ P([m])
Sm× P([m]) W(S2m)Sm⊗ P([m])
Sm
φ−1
2m+1
π2×π3
φ2m+1
f2m+1
φ−1
2m
π1
φ2m
f2mπ1
4 Signed Wachs permutations and Bruhat order
For n > 0recall (see [4]) that the set of signed Wachs permutation is
W(Bn) := {σ∈Bn:|σ−1(i)−σ−1(i∗)|61∀i∈[n−1]}.
So, for example, [−2,−1,4,3] ∈ W(B4)while [3,4,−2,1] /∈ W(B4). In the even case, as in
type A, we have the following group isomorphism (see Proposition 3.1)
W(B2m)≃Bm⋉P([m]) = S2≀Bm.
We define a bijection φ:W(B2m)→Bm× P ([m]) as follows. For σ∈Bmand T⊆[m]
let φ−1(σ, T ) := v, where v∈ W(B2m)is defined by
v(2i−1) = (2σ(i)−χ(σ(i)>0),if i6∈ T,
2σ(i) + χ(σ(i)<0),if i∈T,
and
v(2i) = (2σ(i) + χ(σ(i)<0),if i6∈ T,
2σ(i)−χ(σ(i)>0),if i∈T,
for all i∈[m]. For example, let v:= [−3,−4,1,2,6,5] ∈ W(B6); then φ(v) = ([−2,1,3],{1,3}).
Because of this bijection from now on we freely identify the sets W(B2m)and Bm×P([m]),
so if v∈ W(B2m)and φ(v) = (σ, T ), then we simply write v= (σ, T )and we define
ℓW(v) := ℓB(v)−ℓB(σ).
Recall that we denote by v7→ ˜vthe natural embedding Bn֒→S±n. Note that if
v∈ W(B2m)then ˜v∈ W(S±2m). Indeed, if v= (σ, T )then ˜v= (˜σ, −T∪T). In fact, by
Proposition 2.6, this is an injective group and poset morphism W(B2m)֒→ W(S±2m). For
19

example, for u= [−2,−1,6,5,−3,−4] = ([−1,3,−2],{2,3})∈ W(B6)we have
˜u= (4,3,−5,−6,1,2,−2,−1,6,5,−3,−4) = ((2,−3,1,−1,3,−2),{−3,−2,2,3})∈ W(S±6).
Notice that if nis odd then the image of a signed Wachs permutation is not a Wachs
permutation. For example, if u= [−2,−1,6,5,−3,−4,7] ∈ W(B7)we have
˜u= (−7,4,3,−5,−6,1,2,−2,−1,6,5,−3,−4,7) 6∈ W(S±7).
It is known that 2ℓB(v) = ℓA(˜v) + neg(v)(see e.g. [2, Exercise 8.2]) so we have that
ℓW(v) = ℓW(˜v) + neg(σ)
2,(6)
for all v= (σ, T )∈ W(B2m), because 2 neg(σ) = neg(v).
Proposition 4.1. The function W(B2m)→Bmdefined by the assignment (τ, T )7→ τis
order preserving.
Proof. Let u= (σ, S)∈ W (B2m)and v= (τ , T )∈ W(B2m). We have that u6vimplies
˜u6˜vand then, by Corollary 3.8, ˜σ6˜τ. By Proposition 2.6, this implies σ6τ.
It is easy to characterize, using Theorem 3.13, the Bruhat order relation between signed
Wachs permutations in the even case.
Proposition 4.2. Let u, v ∈ W(B2m),u= (σ, S),v= (τ, T ). Then u6vif and only if
σ6τin Bmand S∩F(σ, τ )⊆T, where F(σ, τ ) := {i∈[m] : σ(i) = τ(i)}.
Proof. We have that ˜u= (˜σ, −S∪S)and similarly ˜v= (˜τ , −T∪T). But, by Theorem
3.13, ˜u6˜vif and only if ˜σ6˜τand (−S∪S)∩F(˜σ, ˜τ)⊆ −T∪T, which in turn happens
if and only if σ6τand S∩F(σ, τ)⊆T.
The next result is the analogue of Proposition 3.2.
Proposition 4.3. Let m > 0. Then
1. φ:W(B2m)→Bm⊗ P([m]) is order preserving;
2. φ−1:Bm× P([m]) → W(B2m)is order preserving.
Moreover ℓB(τ, T ) = 4ℓB(τ) + |T| − neg(τ), for all (τ , T )∈ W(B2m).
20

Proof. Points 1. and 2. are direct consequences of Propositions 2.6 and 3.2. The last
equality follows by the formula of Proposition 3.2; in fact
ℓB(τ, T ) = ℓA(˜τ , −T∪T) + neg(τ, T )
2= 2ℓA(˜τ) + |T|+ neg(τ)
= 4ℓB(τ)−2 neg(τ) + |T|+ neg(τ).
For any (i, j)∈[n]×[±n],i6=j, we find it convenient to define
(i, j)B:= ((i, j)(−i, −j),if i6=|j|;
(i, −i),otherwise.
So the set of reflections of Bnis (see [2, Proposition 8.1.5])
TBn={(i, j)B: 1 6i < |j|6n} ∪ {(i, −i)B:i∈[n]}.
Let m > 0. For any reflection (i, j)B∈TBmwe define an involution wB
i,j :W(B2m)→
W(B2m)by letting
wB
i,j(τ, T ) := (τ(i, j)B, T +{i, |j|}),
for all (τ, T )∈ W(B2m), where X+Ystands for the symmetric difference between two sets
Xand Y. For example, if v= ([−2,1,4,3],{1,4})then wB
3,−3(v) = ([−2,1,−4,3],{1,3,4})
and wB
1,−3(v) = ([−4,1,2,3],{3,4}).
Remark 4.4.We observe that, under the embedding W(B2m)֒→ W(S±2m), we have that
wB
i,j(v)7→ wA
−i,−j(wA
i,j(˜v)), if i6=−j, and wB
i,−i(v)7→ wA
i,−i(˜v).
The next technical result enables us to “lift” some order theoretic properties from B2m
to W(B2m). Its proof relies on the corresponding result in type A, namely Lemma 3.5.
Lemma 4.5. Let m > 0,i∈[m]and j∈[±m]. Assume v= (τ , T )∈ W(B2m),i, |j| 6∈ T
and τ(i, j)B⊳τ. Then wB
i,j(v)< v and ℓW(wB
i,j (v), v) = 1. Moreover if u= (σ, S)∈
W(B2m)and σ6τ(i, j)Bthen u6wB
i,j (v)< v.
Proof. Consider the element ˜v= (˜τ , −T∪T)∈ W(S±2m). Then by our hypothesis and
Theorem 2.7 we have that ˜τ(i, j)(−i, −j)⊳˜τ(i, j)⊳˜τif i6=|j|, and ˜τ(i, −i)⊳˜τif
i=|j|. So, by Lemma 3.5, ˜u6wA
−i,−j(wA
i,j(˜v)) < wA
i,j(˜v)<˜v(since i, j 6∈ Timplies
−i, −j6∈ −T) if i6=|j|, and ˜u6wA
i,−i(˜v)<˜vif i=|j|. By Proposition 2.6 and Remark
4.4 we conclude that u6wB
i,j(v)< v. Note also that, by Theorem 2.7, i6=|j|and
v(i, j)B⊳vin Bnimply neg(v(i, j)B) = neg(v), and that v(i, −i)B⊳vin Bnimplies
21

neg(v(i, −i)B) = neg(v)−1. Therefore, if i, |j| 6∈ Tand τ(i, j)B⊳τ, then by Lemma 3.5
and (6) we have that ℓW(wB
i,j (v), v) = 1.
The following result characterizes the cover relations of the ordering induced by Bruhat
order on the signed Wachs permutations in the even case.
Corollary 4.6. Let m > 0,u, v ∈ W (B2m),u= (σ, S)and v= (τ , T ). Then u⊳vif and
only if either one of the following conditions is satisfied:
1. σ=τand S⊳T;
2. σ⊳τ,T∩ {a, |b|} =∅and S=T∪ {a, |b|}, where (a, b)B:= τ−1σ.
Proof. Let u⊳vin W(B2m). Then σ6τby Corollary 4.1. Moreover, if σ < ω < τ for
some ω∈Bmthen, by Proposition 4.3, u6(ω, S)6v. So u⊳vimplies σ=τor σ⊳τ.
If σ=τthen, by Corollary 4.2, S⊆T; since u⊳v, we have that S⊳T. Assume
now σ⊳τand let (a, b)B:= τ−1σ. If a∈Tor |b| ∈ Tthen, by point 2 of the proof of
Theorem 3.9, ˜u6˜v(a, a + 1)(−a−1,−a)<˜vand ˜u6˜v(|b|,|b|+ 1)(−|b| − 1,−|b|)<˜v,
respectively. Hence u⊳vand σ⊳τimply T∩ {a, |b|} =∅and u=wB
a,b(v), by Lemma
4.5. This implies S=T∪ {a, |b|}.
The converse can be proved by noting that if condition 1 or 2 are satisfied then, by
Proposition 3.12, [˜u, ˜v] = {˜u, w1, w2,˜v}, with w1, w2∈ W(S±2m)\ W(B2m), or [˜u, ˜v] =
{˜u, ˜v}, where the intervals are taken in W(S±2m).
For example [−2,−1,3,4,6,5,−7,−8] ⊳[−2,−1,4,3,6,5,−7,−8] ⊳[−2,−1,5,6,3,4,−7,−8].
By Corollary 4.6 we have that if u⊳vholds in W(B2m)then either u=v(2i−1,2i)Bfor
some i∈[m], or u=wB
i,j(v)for some (i, j)B∈TBm.
We can now prove the even part of the main result of this section.
Theorem 4.7. The poset W(B2m)is graded, with rank function ℓW, and its rank is 3m2.
Proof. Since e, wB2m
0∈ W(B2m), these are the minimum and maximum of the poset
W(B2m), respectively. Let u= (σ, S)∈ W (B2m)and v= (τ , T )∈ W(B2m)be such
that u⊳v. Therefore, by Corollary 4.6, either u=v(2i−1,2i)Bfor some i∈[m]or
u=wB
i,j(v), where σ=τ(i, j)B⊳τ. In both cases ℓW(u, v) = 1, and this proves the
first statement. The rank of the poset W(B2m)is, by (6), ℓW(w0(B2m)) = 3m2, being
ℓW^
w0(B2m)=m(6m−1).
We now investigate the Bruhat order on W(Bn)for nodd. Let v∈ W(B2m+1). Note
that there is a bijection between W(B2m+1)and W(B2m)×[±(m+ 1)] given by v7→
22

(v, (pos(v) + sgn(pos(v)))/2) where
v(i) = (v(i),if i < v−1(2m+ 1),
v(i+ 1),if i>v−1(2m+ 1),
for i∈[2m]. Combining this with the bijection between W(B2m)and Bm× P([m]) ex-
plained at the beginning of this section we obtain a bijection φbetween W(B2m+1)and
[±(m+ 1)] ×Bm× P([m]). If v∈ W(B2m+1)and (i, σ, S)∈[±(m+ 1)] ×Bm× P([m])
correspond under this bijection then we write v= (i, σ, S), and we define
ℓW(v) := ℓB(v)−ℓB(σ),(7)
and ˇv:= (σ, S)(so ˇv∈ W (B2m)). So, for example, if v= [−1,−2,5,6,−7,3,4,]then
v= (−5,[−1,3,2],{−1})so ℓW(v) = (9 + 3 + 7) −2 = 17, and ˇv= [−1,−2,5,6,3,4]. Note
that, if u, v ∈W(B2m+1)are such that u−1(2m+ 1) = v−1(2m+ 1) then
ℓW(u, v) = ℓW(˘u, ˘v).(8)
The next result is the analogue of Proposition 3.4.
Proposition 4.8. Let m > 0. Then
1. φ:W(B2m+1)→[±(m+ 1)]∗⊗Bm⊗ P([m]) is order preserving;
2. φ−1: [±(m+ 1)]∗×Bm× P([m]) → W (B2m+1 )is order preserving.
Moreover ℓB(v) = 4ℓB(τ) + |T| − neg(τ) + 2(m−i+ 1) −3χ(i < 0) = ℓB(˘v) + 2(m−i+
1) −3χ(i < 0), for all v= (i, τ , T )∈ W(B2m+1).
Proof. The first two points follow easily from Proposition 4.3 and the fact that if u6v
then pos(u)>pos(v). The length equality is easy to check using our definitions and the
well known fact (see, e.g., [2, Prop. 8.1.1]) that ℓB(v) = inv(v) + neg(v) + nsp(v)for all v
in the hyperoctahedral group.
For v∈ W (B2m+1 )we define an element c(v)∈ W(B2m+1)by
c(v) := 






v(−1,1)B,if j=−1;
v(j+ 1, j + 2)B,if j6=−1and v(j+ 1) > v(j+ 2);
v(j, j + 2)B,if j6=−1and v(j+ 1) < v(j+ 2);
(9)
where j:= v−1(2m+1). For example, c([−9,4,3,−6,−5,2,1,−8,−7]) = [9,4,3,−6,−5,2,1,−8,−7],
c([4,3,−6,−5,9,2,1,−8,−7]) = [4,3,−6,−5,9,1,2,−8,−7] and c([3,4,−9,1,2,6,5,−7,−8]) =
23

[−9,4,3,1,2,6,5,−7,−8]. Note that ℓW(c(v), v) = 1. The next result implies that c(v)is
the only coatom uof vsuch that v−1(2m+ 1) 6=u−1(2m+ 1), and is the main technical
tool in our proof of the fact that W(Bn), under Bruhat order, is graded.
Theorem 4.9. Let m∈P, and u, v ∈D(B2m+1 ),u < v, be such that v−1(2m+ 1) <
u−1(2m+ 1). Then u6c(v).
Proof. Let j:= v−1(2m+ 1). We distinguish various cases according as to whether j < −1,
j=−1, or j > 0. Let, for brevity, z:= c(v). Given a signed permutation w∈Bn
and j∈[±n]we find it convenient to denote by wjthe increasing rearrangement of
{w(−n), w(−n+ 1),...,w(j)}.
Assume first that j > 0and j+ 1 ∈DR(v). Let a:= v(j+ 2) so v(j+ 1) = a+ 1. Then
vk=zkfor all k∈[±(2m+ 1)] \ {j+ 1,−j−2}. Since u, v, z are all Wachs permutations
there are x1,...,x˜m, y1,...,y˜m+1 ∈[±(2m)], where ˜m:= 2m−1+j
2, such that
vj+1 = (−2m−1, x1, x1+ 1,...,xp, xp+ 1, a + 1, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1,2m+ 1)
zj+1 = (−2m−1, x1, x1+ 1,...,xp, xp+ 1, a, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1,2m+ 1)
and
uj+1 = (−2m−1, y1, y1+ 1,...,y˜m+1, y ˜m+1 + 1),
for some 06p6˜m. Since uj+1 6vj+1 we have that yp+1 6a+ 1. But yp+1 and aare of
the same parity if they have the same sign so yp+1 6aand hence uj+1 6zj+1.
Similarly, there are x1,...,x˜m, y1,...,y˜m∈[±(2m)], where ˜m:= 2m−j−1
2such that
v−j−2= (x1, x1+ 1,...,xp, xp+ 1,−a, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1)
z−j−2= (x1, x1+ 1,...,xp, xp+ 1,−a−1, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1)
and
u−j−2= (−2m−1, y1, y1+ 1,...,y˜m, y ˜m+ 1),
for some 06p6˜m. Since u−j−26v−j−2we have that yp6xp+ 1. But xp+ 1 <−a−1
so yp+ 1 6−a−1and hence u−j−26z−j−2.
Suppose now that (j > 0and) j+ 1 /∈DR(v). Consider first vj, zj, uj, vj+1, zj+1 , uj+1.
Since v, u, and zare all Wachs permutations, and by our definition of z, there are {x1,...,x˜m}<,
{y1,...,y˜m}<⊆[±(2m)], where ˜m:= 2m−1+j
2such that
vj= (−2m−1, x1, x1+ 1,...,x˜m, x ˜m+ 1,2m+ 1)
24

zj= (−2m−1, x1, x1+ 1,...,xp, xp+ 1, a + 1, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1)
uj= (−2m−1, y1, y1+ 1,...,yq, yq+ 1, b∗, yq+1, yq+1 + 1,...,y˜m, y ˜m+ 1)
vj+1 = (−2m−1, x1, x1+ 1,...,xp, xp+ 1, a, xp+1, xp+1 + 1,...,x˜m, x ˜m+1,2m+ 1)
zj+1 = (−2m−1, x1, x1+ 1,...,xp, xp+ 1, a, a + 1, xp+1 , xp+1 + 1,...,x˜m, x ˜m+ 1)
uj+1 = (−2m−1, y1, y1+ 1,...,yq, yq+ 1, b, b + 1, yq+1, yq+1 + 1,...,y˜m, y ˜m+ 1)
where a:= v(j+ 1) (=z(j+ 1)), b∗:= u(j), and p, q ∈[ ˜m]. By our hypothesis u6v
so uj6vjand uj+1 6vj+1 (componentwise). This easily implies that uj6zjand
uj+1 6zj+1 (componentwise) keeping in mind the fact that xrand yshave the same
parity if they have the same sign for all r, s ∈[ ˜m](so ys6xr+ 1 implies ys6xrfor all
r, s ∈[ ˜m]). For example, if q < p, then yp6a(since uj+1 6vj+1), while yk6xk+ 1
(since uj6vj) so yk6xkif p+ 1 6k6˜m. Similarly, if q > p, then yp+1 6a(since
uj+1 6vj+1) while b6xq+1 so b6xq, while yk6xk+ 1 (since uj6vj) so yk6xkif
q+ 1 6k6˜m, and yk6xk−1+ 1 (since uj+1 6vj+1) so yk6xk−1if p+ 2 6k6q. The
case p=qis even simpler, and is therefore omitted.
Consider now u−j−2, z−j−2, v−j−2and u−j−1, z−j−1, v−j−1.Then reasoning as above we
have that there are {x1,...,x˜m}<,{y1,...,y˜m}<⊆[±(2m)] , where ˜m:= m−j+1
2, such
that
v−j−2= (x1, x1+ 1,...,xp, xp+ 1,−a−1, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1)
z−j−2= (−2m−1, x1, x1+ 1,...,x˜m, x ˜m+ 1)
u−j−2= (−2m−1, y1, y1+ 1,...,y˜m, y ˜m+ 1)
v−j−1= (x1, x1+ 1,...,xp, xp+ 1,−a−1,−a, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1)
z−j−1= (−2m−1, x1, x1+ 1,...,xp, xp+ 1,−a, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1)
u−j−1= (−2m−1, y1, y1+ 1,...,yq, yq+ 1,−b, yq+1, yq+1 + 1,...,y˜m, y ˜m+ 1)
where p, q ∈[ ˜m]. As above, the fact that u−j−26v−j−2and u−j−16v−j−1easily implies
that u−j−26z−j−2and u−j−16z−j−1. For example, if q < p, then yk6xk+ 1 (since
u−j−26v−j−2) so yk6xkfor 16k6pand hence −b < yq+1 6xq+1. Similarly, if q>p.
Consider now the case j < −1(so j6−3). Assume first that j+ 1 /∈DR(v). Consider
uj, zj, vj, uj+1, zj+1, vj+1. If u−1(−2m−1) > j then we conclude exactly as in the case
j > 0. So assume that u−1(−2m−1) < j. Then reasoning as above we conclude that
25

there are {x1,...,x˜m}<,{y1,...,y˜m}<⊆[±(2m)], where ˜m:= m+j+1
2, such that
vj= (x1, x1+ 1,...,x˜m, x ˜m+ 1,2m+ 1)
zj= (x1, x1+ 1,...,xp, xp+ 1, a + 1, xp+1 , xp+1 + 1,...,x˜m, x ˜m+ 1)
uj= (−2m−1, y1, y1+ 1,...,y˜m, y ˜m+ 1)
vj+1 = (x1, x1+ 1,...,xp, xp+ 1, a, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1,2m+ 1)
zj+1 = (x1, x1+ 1,...,xp, xp+ 1, a, a + 1, xp+1 , xp+1 + 1,...,x˜m, x ˜m+ 1)
uj+1 = (−2m−1, y1, y1+ 1,...,yq, yq+ 1, c, yq+1, yq+1 + 1,...,y˜m, y ˜m+ 1)
vj+2 = (x1, x1+ 1,...,xp, xp+ 1, a, a + 1, xp+1 , xp+1 + 1,...,x˜m, x ˜m+ 1,2m+ 1)
uj+2 = (−2m−1, y1, y1+ 1,...,yq, yq+ 1, c, c + 1, yq+1, yq+1 + 1,...,y˜m, y ˜m+ 1)
where a:= v(j+ 1) (=z(j+ 1)), c:= u(j+ 1), and p, q ∈[ ˜m]. By our hypothesis u6v
so uj6vj,uj+1 6vj+1, and uj+2 6vj+2 (componentwise). As above, this implies that
uj6zjand uj+1 6zj+1. For example, if p < q, then yp+1 6a+ 1 (so yp+1 + 1 6xp+1),
yp+1 6a,c6xq+1, and yi6xi−1+ 1 (so yi6xi−1) for p+ 2 6i6q(since uj+2 6vj+2),
while yi6xi+ 1 (so yi6xi) for p+ 1 6i6˜m( since uj6vj). Similarly, if q < p, then
yp6aand yi6xi+ 1 (so yi6xi) for p+ 1 6i6˜m(since uj+1 6vj+1). Finally, if q=p,
then c6a+ 1,yp+1 6aand yi6xi+ 1 (so yi6xi) for p+ 1 6i6˜m(since uj+2 6vj+2 ).
Consider now u−j−1, v−j−1, z−j−1, u−j−2, v−j−2, z−j−2. If u−1(−2m−1) > j then we
conclude exactly as in the case j > 0. So assume that u−1(−2m−1) < j. Then as above,
since u, v, and zare all Wachs permutations we conclude that there are {x1,...,x˜m}<⊆
[±(2m)] and {y1,...,y˜m}<⊆[±(2m)] such that
v−j−2= (x1, x1+ 1,...,xp, xp+ 1,−a−1, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1,2m+ 1)
z−j−2= (−2m−1, x1, x1+ 1,...,x˜m, x ˜m+ 1,2m+ 1)
u−j−2= (−2m−1, y1, y1+ 1,...,yq, yq+ 1, c∗, yq+1, yq+1 + 1,...,y˜m, y ˜m+ 1)
v−j−1= (x1, x1+ 1,...,xp, xp+ 1,−a−1,−a, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1,2m+ 1)
z−j−1= (−2m−1, x1, x1+ 1,...,xp, xp+ 1,−a, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1,2m+ 1)
u−j−1= (−2m−1, y1, y1+ 1,...,yq, yq+ 1, c, c + 1, yq+1, yq+1 + 1,...,y˜m, y ˜m+ 1)
v−j−3= (x1, x1+ 1,...,x˜m, x ˜m+ 1,2m+ 1)
26

u−j−3= (−2m−1, y1, y1+ 1,...,y˜m, y ˜m+ 1)
where ˜m:= m−j+3
2,−a:= v(−j−1) (=z(−j−1)), and c∗:= u(−j−2), for some
p, q ∈[ ˜m]. It is then not hard to conclude that u−j−26z−j−2and u−j−16z−j−1. For
example, if p6q, then yi6xi+ 1 (since u−j−26v−j−2) so yi6xifor 16i6p. If
p > q then yi6xi+ 1 ( since u−j−36v−j−3) so yi6xifor 16i6˜mand hence
c+ 1 < yq+1 6xq+1.
Assume now that (j < −1and) j+ 1 ∈DR(v). Then vk=zkif k∈[±(2m+ 1)] \
{−j−2, j + 1}. Assume first that u−1(2m+ 1) >−j. Let a:= v(j+ 2) so v(j+ 1) = a+ 1.
Then there are x1,...,x˜m, y1,...,y˜m∈[±(2m)], where ˜m:= 2m−3−j
2such that
v−j−2= (x1, x1+ 1,...,xp, xp+ 1,−a, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1,2m+ 1) (10)
z−j−2= (x1, x1+ 1,...,xp, xp+ 1,−a−1, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1,2m+ 1) (11)
and
u−j−2= (−2m−1, y1, y1+ 1,...,yq, yq+ 1, c∗, yq+1, yq+1 + 1,...,y˜m, y ˜m+ 1),
for some 06p, q 6˜mwhere c∗:= u(−j−2). It is then easy to see that u−j−26z−j−2.
Indeed, if q < p then, since u−j−26v−j−2,yp6−aso, as above, yp6−a−1and
hence u−j−26z−j−2. If q>pthen yp6xp+ 1 (since u−j−26v−j−2) so yp6xp. But
xp+ 1 <−ahence yp+ 1 6−a−1so u−j−26z−j−2.
Similarly, there are {x1,...,x˜m}<,{y1,...,y˜m}<∈[±(2m)], where ˜m:= 2m+j+1
2, such
that
vj+1 = (x1, x1+ 1,...,xp, xp+ 1, a + 1, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1,2m+ 1) (12)
zj+1 = (x1, x1+ 1,...,xp, xp+ 1, a, xp+1, xp+1 + 1,...,x˜m, x ˜m+ 1,2m+ 1) (13)
and
uj+1 = (−2m−1, y1, y1+ 1,...,yq, yq+ 1, c∗, yq+1, yq+1 + 1,...,y˜m, y ˜m+ 1),
for some 06p,q6˜m, where c∗:= u(j+ 1), and we conclude exactly as in the last case.
Assume now that u−1(2m+ 1) <−j. Then there are x1,...,x˜m, y1,...,y˜m+1 ∈[±2m],
where ˜m:= 2m+1+j
2such that (12) and (13) hold and
uj+1 = (y1, y1+ 1,...,y˜m+1, y ˜m+1 + 1)
27

and the result follows easily. Similarly, there are x1,...,x˜m, y1,...,y˜m∈[±2m], where
˜m:= 2m−3−j
2, such that (10) and (11) hold and
u−j−2= (−2m−1, y1, y1+ 1,...,y˜m, y ˜m+ 1,2m+ 1)
for some 06p6˜m. Hence, since u−j−26v−j−2,yp6xp+ 1 <−a−1so yp+ 1 6−a−1.
Finally, if j=−1then z=v(−1,1) so there are x1,...,xm,y1,...,ym∈[±2m]such
that v−1= (x1, x1+ 1,...,xm, xm+ 1,2m+ 1),z−1= (−2m−1, x1, x1+ 1,...,xm, xm+ 1)
and u−1= (−2m−1, y1, y1+ 1,...,ym, ym+ 1), so the conclusion follows easily. This
concludes the proof.
Recall that if v= (σ, S, i)∈ W(B2m+1 )then we let ˘v= (σ, S)∈ W(B2m). We can now
prove the main result of this section.
Theorem 4.10. W(Bn)is graded, with rank function ℓW, and its rank is n2−n
22.
Proof. If nis even then this follows from Theorem 4.7. So assume that n= 2m+ 1 for
some m > 0. Since W(B2m+1)has both a maximum and a minimum element it is enough
to show that if u, v ∈ W(B2m+1)and u⊳vthen ℓW(u, v) = 1. So let u, v ∈ W (B2m+1)
be such that u⊳v. Let j:= u−1(2m+ 1), and i:= v−1(2m+ 1). Since u < v we have
that i6j. If i < j then by Theorem 4.9 we have that u6c(v)< v so u=c(v)and
hence ℓW(u, v) = 1. If i=jthen ˘u⊳˘vso, by Theorem 4.7, ℓW(˘u, ˘v) = 1 and hence,
by (8), ℓW(u, v) = 1. Finally, it is not difficult to see, by our definition of ℓW, that
ℓW(w0(B2m+1)) = 3m2+ 4m+ 1.
We remark that the sequence {ℓW(w0(B2m))}m∈Pgives the number of edges of the
complete tripartite graph Km,m,m (see A033428 in OEIS), and {ℓW(w0(B2m+1))}m∈Pis
the sequence of octagonal numbers (see A000567 in OEIS). We can now compute the
rank-generating function of (W(Bn),6).
Corollary 4.11. Let m > 0. Then
W(B2m)(x, ℓW) = (1 + x)m[m]x3!
m
Y
i=1
(1 + x3i−1)
and
W(B2m+1)(x, ℓW) = [m+ 1]x2(1 + x2m+1)(1 + x)m[m]x3!
m
Y
i=1
(1 + x3i−1).
Proof. Note first that by our definition and Proposition 4.3 if u∈ W(B2m),u= (σ, S),
28

then ℓW(u) = 3ℓB(σ) + |S| − neg(σ). Therefore
X
u∈W(B2m)
xℓW(u)=X
σ∈BmX
S⊆[m]
x3ℓB(σ)+|S|−neg(σ)= (1 + x)mX
σ∈Bm
x3ℓB(σ)−neg(σ).
Let J:= [m]. Then by [2, Prop. 2.4.4] every element of σ∈Bmmay be expressed
in a unique way as σ=zw where w∈(Bm)Jand z∈(Bm)J. But the elements of
(Bm)Jare permutations of Smand the z∈(Bm)Jare characterized by the fact that
z(1) < z(2) <··· < z(m). Hence these elements zare in bijection with subsets S⊆[m],
where the subset Sis the set of negative values taken by σ. Furthermore, we then have that
ℓB(σ) = inv(σ)+neg(σ)+nsp(σ) = inv(w)+neg(z)+nsp(z) = inv(w)+|S|+Ps∈S(s−1) =
inv(w) + Ps∈Ss. We therefore have that
X
σ∈Bm
x3ℓB(σ)−neg(σ)=X
w∈SmX
S⊆[m]
x3 inv(w)−|S|+3 Ps∈Ss=X
w∈Sm
x3 inv(w)X
S⊆[m]
xPs∈S(3s−1)
and the first equation follows. For the second formula we have, using Proposition 4.8 and
(7),
X
u∈W(B2m+1 )
xℓW(u)= −1
X
i=−m−1
x2m−2i−1+
m+1
X
i=1
x2(m−i+1)!W(B2m)(x, ℓW)
= [m+ 1]x2(1 + x2m+1)W(B2m)(x, ℓW).
Theorem 4.10 enables us to explicitly determine the cover relations in W(Bn)for n
odd.
Corollary 4.12. Let u, v ∈ W (B2m+1),u= (i, σ, S),v= (j, τ , T )(σ, τ ∈Bm,S, T ⊆[m],
i, j ∈[±(m+ 1)]). Then u⊳vif and only if either one of the following conditions is
satisfied:
1. i=j,σ=τ, and S⊳T;
2. j=i−1,σ=τand either |j+χ(j < 0)| ∈ S,T=S\ {|j+χ(j < 0)|} if j6=−1,
or S=Tif j=−1;
3. i=j,σ⊳τ,T∩ {|a|,|b|} =∅and S=T∪ {|a|,|b|}, where (a, b)B=τσ−1.
Proof. Note that, since W(B2m+1)is graded, and its rank function is ℓW,u⊳vif and only
if u6vand ℓW(u, v) = 1. If either 1. or 2. hold then it is easy to check that u6vand
ℓW(u, v) = 1.
29

Assume now that 3. holds. Suppose first that τ=σ(a, b)(−a, −b). Since σ⊳τ
we have from Theorem 2.7 that ab > 0. We may assume that 0< a < b. Again by
Theorem 2.7 we have that {k∈[a+ 1, b −1] : σ(a)< σ(k)< σ(b)}=∅. This, by
3., implies that inv(v)−inv(u) = 4 −2,neg(v) = neg(u), and nsp(v) = nsp(u). Hence
ℓW(u, v) = ℓB(u, v)−ℓB(σ, τ ) = 2 −1 = 1. Suppose now that τ=σ(c, −c). We may
clearly assume that c > 0. Then by Theorem 2.7 {k∈[c−1] : −σ(c)< σ(k)< σ(c)}=∅.
This, again by 3., implies that inv(v)−inv(u) = −2(σ(c)−1) −1,neg(v) = neg(u) + 2,
and nsp(v) = nsp(u) + 2(σ(c)−1) + 1. Hence ℓW(u, v) = ℓB(u, v)−ℓB(σ, τ ) = 2 −1 = 1.
Conversely, assume that u⊳v. If j > i then by Theorem 4.9 u6c(v)< v so u=c(v).
If j=−1then 2. follows easily from (9). If j6=−1then v(pos(v) + 1) < v(pos(v) + 2)
(else, by (9), v−1(2m+ 1) = c(v)−1(2m+ 1) so i=j, a contradiction) and 2. again follows
from (9). If i=jand σ=τthen it follows easily from Proposition 2.6 and Theorem 2.4
that S⊆Tand 1. follows since u⊳v. So assume (i=jand) σ < τ . Then, since i=j,
(σ, S)⊳(τ , T )in W(B2m)so 3. follows from Corollary 4.6.
So, for example, in W(B9)we have that [9,2,1,−3,−4,8,7,−6,−5]⊳[−9,2,1,−3,−4,8,7,−6,−5]
⊳[1,2,−9,−3,−4,8,7,−6,−5]⊳[1,2,−9,−3,−4,8,7,−5,−6]⊳[1,2,−9,−6,−5,8,7,−4,−3],
where the first two covering relations are of type 2. (with j=−1, and j=−2, respectively),
the third one is of type 1., and the fourth one of type 3., with (a, b)B= (2,4)(−2,−4).
The next theorem characterizes the Bruhat order on signed Wachs permutations, in
the odd case. Note that it generalizes and puts in perspective the results of Proposition
4.8.
Theorem 4.13. Let m > 0and u, v ∈ W(B2m+1 ),u= (i, σ, S),v= (j, τ , T ). Then u6v
if and only if
σ6τ,S(u, v)⊆T(u, v), and j6i ,
where, for X⊆[m],X(u, v) := X∩([min{|i|,|j|} − 1] ∪[max{|i|,|j|}, m]) ∩F(σ, τ), being
F(σ, τ) := {i∈[m] : σ(i) = τ(i)}. Moreover ℓW(u) = 3ℓB(σ) + |S| − neg(σ) + 2(m−i+
1) −3χ(i < 0).
Proof. Let u6v. We may assume u⊳v. It is clear from Corollary 4.12 that σ6τand j6
i. Furthermore in case 1 of Corollary 4.12 and in case 2 with j=−1we have that S⊆Tand
then S(u, v)⊆T(u, v). In case 2 with j6=−1we have that iand jhave the same sign and
F(σ, τ) = [m]; hence T=S\ {j}if i > 0, so T(u, v) = (S\ {j})∩([j−1] ∪[j+ 1, m]) =
S(u, v). If i < 0then T=S\ {|i|} and T(u, v) = (S\ {|i|})∩([|i| − 1] ∪[|i|+ 1, m]) =
30

S(u, v). Finally in case 3 we have that F(σ, τ ) = [m]\ {|a|,|b|}; since S=T∪ {|a|,|b|} we
obtain T(u, v) = S(u, v).
Now let σ6τ,j6iand S(u, v)⊆T(u, v). Assume j6i < 0. We then have the
following chain in (W(B2m+1),6):
(i, σ, S)6(i, σ, S ∪[|i|,|j| − 1])
⊳(i−1, σ, (S∪[|i|+ 1,|j| − 1]) \ {|i|})
⊳(i−2, σ, (S∪[|i|+ 2,|j| − 1]) \ {|i|,|i|+ 1})
⊳... ⊳(j, σ, S \[|i|,|j| − 1]) 6(j, τ , T ),
where the last inequality follows from Proposition 4.2 and the facts that (S\[|i|,|j| − 1]) ∩
F(σ, τ) = S(u, v)⊆T(u, v)⊆Tand (j, σ, S \[|i|,|j| − 1]) 6(j, τ , T )if and only if
(σ, S \[|i|,|j| − 1]) 6(τ, T ). If j < 0< i and |j|< i we have the following chain in
(W(B2m+1),6):
(i, σ, S)6(i, σ, S ∪[|j|, i −1])
⊳(i−1, σ, (S∪[|j|, i −2]) \ {i−1})
⊳(i−2, σ, (S∪[|j|, i −3]) \ {i−2, i −1})
⊳... ⊳(−j, σ, S \[|j|, i −1]) 6(j, σ, S \[|j|, i −1]) 6(j, τ , T ),
where the last inequality follows as in the previous case. If j < 0< i and |j|> i have that
(i, σ, S)6(−i, σ, S)6(j, τ , T ), where the second inequality follows by the first case above.
The case i > j > 0is similar and easier, so we omit it. The length formula follows from
Proposition 4.8.
We illustrate the previous theorem with an example. Let m= 4,u= [3,4,−5,−6,1,2,9,−7,−8]
and v= [−3,−4,−9,1,2,−5,−6,−8,−7]. Then u= (4,[2,−3,1,−4],{2,4}), and v=
(−2,[−2,1,−3,−4],{1,3}), so σ= [2,−3,1,−4] 6[−2,1,−3,−4] = τ,F(σ, τ) = {4},
j=−264 = i, and S(u, v) = {2,4}∩{1,4}∩{4}={4},T(u, v) = {1,3}∩{1,4}∩{4}=∅,
so u66vin W(B9).
The following lemma is the analogue of Lemma 3.14 and can be easily deduced from
Theorem 4.13 so we omit its verification.
Lemma 4.14. Let m > 0and u, v ∈ W (B2m+1 ). If u6(i, σ, S1)6vand u6(i, σ, S2)6
vin (W(B2m+1),6)then u6(i, σ, S1∪S2)6v.
The next result gives an explicit expression for the Möbius function of lower intervals
in the poset of signed Wachs permutations partially ordered by Bruhat order, and shows,
31

in particular, that it has values in {0,1,−1}. The proof is similar to the one of Proposition
3.15 and we omit it.
Proposition 4.15. Let m > 0, and v= (j, τ, T )∈ W(B2m+1 ). Then
µ(e, v) = ((−1)|T|,if τ=eand j=m+ 1;
0,otherwise.
In particular, if v= (τ, T )∈ W(B2m)then
µ(e, v) = ((−1)|T|,if τ=e;
0,otherwise.
By Proposition 4.15 and Theorem 4.10 we deduce the following result.
Corollary 4.16. The characteristic polynomial of W(Bn)with the Bruhat order is
(x−1)⌊n
2⌋xn2−⌊n
2⌋2−⌊n
2⌋,
for all n∈P.
5 Weak orders on Wachs permutations and signed Wachs
permutations
In the previous sections we have proved several results concerning the Bruhat order
on Wachs permutations and signed Wachs permutations. The Bruhat order of a Coxeter
group is a refinement of two fundamental orders, the left weak order 6Land the right one
6R(see, e.g. [2, Chapter 3]), whose Hasse diagrams are isomorphic to the Cayley graph
of the group, relative to the considered Coxeter presentation. Then it is natural to ask
when our results hold, and in what terms, for the left and right weak orders on Wachs
permutations and signed Wachs permutations.
Differently from the Bruhat order, for the right weak order the answer is easily described
as a Cartesian product (compare with Propositions 4.3 and 4.8). For reasons that will
become clear in the next two pages we begin with signed Wachs permutations. Recall that
the set of reflections of Bnis {(a, b)(−a, −b) : 1 6a < |b|6n} ∪ {(a, −a) : 1 6a6n}.
Therefore, if v∈Bn,(a, b)(−a, −b)∈TL(v)if and only if b > 0and bis to the left of ain
the complete notation of v, or b < 0and bis to the right of a, while (a, −a)∈TL(v)if and
only if ais to the left of −a.
32

[−1,−2,−3]
[−2,−1,−3]
[2,1,−3] [−3,−1,−2]
[1,2,−3] [−3,−2,−1] [3,−1,−2]
[−3,2,1] [3,−2,−1]
[−3,1,2] [3,2,1] [−1,−2,3]
[3,1,2] [−2,−1,3]
[2,1,3]
[1,2,3]
Figure 3: Hasse diagram of (W(B3),6).
33 32 31 30
34 16 15 14 29
35 17 5 4 13 28
36 18 6 3 12 27
19 7 1 2 11 26
20 8910 25
21 22 23 24
Figure 4: Reading clockwise, the bold numbers on the diagonals of the hexagon are the
sequences {3,12,27,...}={rk(W(B2m))}m>0,{2 rk(W(S2m))}m>0,{rk(W(B2m+1))}m>0
and {2 rk(W(S2m+1))}m>0.
33

Theorem 5.1. Let n > 0; then (W(Bn),6R)≃B⌈n
2⌉,6R× P n
2. In particular,
(W(Bn),6R)is a complemented lattice.
Proof. Let nbe even; for v= (τ , T )∈ W(Bn)we have that
TL(τ, T ) = ]
(a,−a)B∈TL(τ)
{(2a−1,−2a+ 1)B,(2a, −2a)B,(2a−1,−2a)B}
]
(a,b)B∈TL(τ):b>0
{(2a−1,2b−1)B,(2a−1,2b)B,(2a, 2b−1)B,(2a, 2b)B}
]
(a,b)B∈TL(τ):b<−a
{(2a−1,2b+ 1)B,(2a−1,2b)B,(2a, 2b+ 1)B,(2a, 2b)B}
⊎ {(v(2a), v(2a−1))B:a∈T, τ (a)>0}
⊎ {(−v(2a−1),−v(2a))B:a∈T, τ (a)<0}.
Note that in the first group of reflections the non-symmetric ones (i.e., not of the form
(k, −k)for some k∈[n]) are never simple and have odd first element, and even and negative
second one. The only simple reflections in the second line have even first element and odd
positive second one. The ones in the third line are never simple, have negative second
element, and if the first one is odd and the second one even then the difference between
the absolute values of the two is >3(and hence are disjoint from the non-symmetric ones
in the first line). The reflections in the last two lines are always simple, and have odd first
element and even and positive second one, since (v(2a), v(2a−1))B= (2τ(a)−1,2τ(a))B
if τ(a)>0while (−v(2a−1),−v(2a))B= (−2τ(a)−1,−2τ(a))Bif τ(a)<0.
Therefore, if u= (σ, S)∈ W (Bn)and v= (τ , T )∈ W(Bn), we have that TL(u)⊆
TL(v)if and only if TL(σ)⊆TL(τ)and σS⊆τT, where, for X⊆[n/2] and w∈Bn/2,
wX:= {w(i) : i∈X}. Indeed, if TL(u)⊆TL(v)and (a, b)B∈TL(σ)is such that (say)
b < −athen, by the equation written at the beginning of this proof, (2a, 2b)B∈TL(u)so
(2a, 2b)B∈TL(v)which implies, by the remarks above, that there is (c, d)B∈TL(τ)such
that c < −dand (2a, 2b)B= (2c, 2d)B, so (a, b)B= (c, d)B∈TL(τ). Similarly all the other
cases. The converse is clear. Hence, by Proposition 2.3, the map (τ, T )7→ (τ, τT)gives the
desired poset isomorphism.
34

Let nbe odd and v= (j, τ, T )∈ W(Bn). Then similarly
TL(j, τ, T ) = ]
(a,−a)B∈TL(¯τ):a6=⌈n/2⌉
{(2a−1,−2a+ 1)B,(2a, −2a)B,(2a−1,−2a)B}
]
(a,b)B∈TL(¯τ):0<b<⌈n/2⌉
{(2a−1,2b−1)B,(2a−1,2b)B,(2a, 2b−1)B,(2a, 2b)B}
]
(a,b)B∈TL(¯τ):−⌈n/2⌉<b<−a
{(2a−1,2b+ 1)B,(2a−1,2b)B,(2a, 2b+ 1)B,(2a, 2b)B}
]
(a,b)B∈TL(¯τ):b=⌈n/2⌉
{(2a−1, n)B,(2a, n)B}
]
(a,b)B∈TL(¯τ):b=−⌈n/2⌉,a6=−b
{(2a−1,−n)B,(2a, −n)B}
⊎{(n, −n)B: (⌈n/2⌉,−⌈n/2⌉)B∈TL(¯τ)}
⊎ {(2τ(a)−1,2τ(a))B:a∈T, τ (a)>0}
⊎ {(2τ(a)−1,2τ(a))B:−a∈T, τ (a)>0}
where
¯τ(k) := 






τ(k),if k < |j|;
sgn(j)·(n+ 1)/2,if k=|j|;
τ(k−1),if k > |j|,
for all k∈ ⌈n/2⌉. The considerations made in the even case apply line by line also to
this case, with the added remark that the reflections in lines 4, 5, and 6 are the only
ones with nor −nin the second position. Therefore, if u= (i, σ, S)∈ W (Bn)and
v= (j, τ, T )∈ W(Bn), we have that TL(u)⊆TL(v)if and only if TL(¯σ)⊆TL(¯τ)and
σS⊆τS, and we conclude as in the previous case. The last statement follows from the fact
that (B⌈n
2⌉,6R)is a complemented lattice (see, e.g., [2, Cor. 3.2.2]).
Regarding left weak order we have the following observations.
Proposition 5.2. Let m > 0. The posets (W(B2m),6R)and (W(B2m),6L)are isomor-
phic.
Proof. The assignment v7→ v−1defines a bijective function W(B2m)→ W(B2m)and
u6Rvif and ony if u−16Lv−1, for all u, v ∈B2m.
On the other hand, the poset (W(B3),6L)is not graded.
For Wachs permutations the situation is analogous but simpler so we leave to the
interested reader the proof of the following result.
35

Theorem 5.3. Let n > 0; then (W(Sn),6R)≃S⌈n
2⌉,6R× P n
2. In particular,
(W(Sn),6R)is a complemented lattice.
As in the case of signed Wachs permutations, the posets (W(S2m),6R)and (W(S2m),6L
)are isomorphic, for all m > 0. On the other hand, the poset (W(S5),6L)is not graded.
6 Open problems
In this section we collect some open problems and conjectures which arise in this work,
and the evidence that we have in their favor.
We have proved in Propositions 3.15 and 4.15 that the Möbius function of lower intervals
in the posets of Wachs permutations and signed Wachs permutations partially ordered by
Bruhat order always has values in {0,1,−1}. We feel that this is true in general.
Conjecture 6.1. Let n∈P. Then
µ(u, v)∈ {0,1,−1}
for all u, v ∈ W(Sn).
We have verified Conjecture 6.1 for n68.
Conjecture 6.2. Let n∈P. Then
µ(u, v)∈ {0,1,−1}
for all u, v ∈ W(Bn).
We have verified Conjecture 6.2 for n66. Note that since W(Sn)is isomorphic, as a
poset, to the interval [e, [n, . . . , 3,2,1]] in W(Bn), Conjecture 6.2 implies Conjecture 6.1.
Recently Davis and Sagan [6] studied the convex hull of various sets of pattern avoiding
permutations. Following this idea, it is natural to look at the convex hulls c(W(Sn)) and
c(W(Bn)) of Wachs and signed Wachs permutations in Rn. In this respect, we feel that
the following is true.
Conjecture 6.3. Let m∈P. Then c(W(S2m)) is a simple polytope.
We have verified Conjecture 6.3 for m65. According to SageMath [21] c(W(S9)) is
not simple.
Conjecture 6.4. Let m∈P. Then c(W(B2m)) is a simple polytope.
36

We have verified Conjecture 6.4 for m63. According to SageMath [21] c(W(B3)) is
not simple.
Regarding left weak order, we feel that the following might be true.
Problem 1. Is (W(S2m+1),6L)a lattice for all m∈P?
We have verified that the answer to Problem 1 is yes if m64.
Acknowledgments. The first named author was partially supported by the MIUR Ex-
cellence Department Project CUP E83C18000100006.
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