Bruhat-Chevalley Order in Reductive Monoids
MOHAN S. PUTCHA
Department of Mathematics, Box 8205, North Carolina State University, Raleigh, NC 27695-8205 Received January 24, 2003; Revised July 7, 2003; Accepted August 5, 2003
Abstract. LetM be a reductive monoid with unit groupG. Letdenote the idempotent cross-section of theG×G-orbits onM. IfWis the Weyl group ofGande,f ∈withe ≤ f, we introduce a projection map fromWeWtoWfW. We use these projection maps to obtain a new description of the Bruhat-Chevalley order on the Renner monoid ofM. For the canonical compactificationXof a semisimple groupG0with Borel subgroupB0ofG0, we show that the poset ofB0×B0-orbits ofX(with respect to Zariski closure inclusion) is Eulerian.
Keywords: reductive monoid, Renner monoid, Bruhat-Chevalley order, projections
2000 Mathematics Subject Classification: Primary 20G99, 20M99, 06A07
Introduction
Reductive monoidsM are Zariski closures of reductive groupsG. By this we mean that if Gis a closed subgroup ofG Ln(k), then the closureMofGinMn(k) is a reductive monoid.
The Bruhat-Chevalley order inGhas a natural extension to reductive monoids. The Renner monoidRtakes the place of the Weyl groupW. Associated with the Bruhat decomposition of aG×G-orbit ofM is aW×W-orbit ofRthat is a graded poset and has been explicitly determined by the author [7]. The ordering on R is more removed from the ordering on W and hence harder to understand. It is the detailed study of the ordering on Rthat is the purpose of this paper.
TheW×W-orbits are indexed by the cross-section latticeofM. For twoW×W-orbits WeW,WfW withe≤ f, we define an upward projection map p :WeW → WfW. These are order-preserving maps with some pleasing properties. Ifσ ∈ WeW, θ ∈ WfW, then we show thatσ ≤θinRif and only if p(σ)≤θ. Combining with the description of the order on the W ×W-orbits in [7], we obtain a new description of the order on R which enables us to obtain several consequences. In particular we show that any length 2 interval in R is either a chain or a diamond. This leads to a conjecture on the M¨obius function onR.
We go on to study canonical reductive monoids associated with the canonical compact- ification of semisimple groups, cf. [13]. We prove that for a canonical monoid, the poset R∗ = R\{0}is Eulerian. This extends a classical result of Verma [18] thatW is Eulerian and a recent result of the author [7] that theW ×W-orbits in R∗are Eulerian.
1. Preliminaries
Let P be a finite partially ordered set with a maximum and minimum element such that all maximal chains have the same length. ThenPadmits a rank function with the minimal element having rank zero. If X ⊆ P, we will say that X isbalanced if the number of even rank elements of X is equal to the number of odd rank elements of X.P is said to beEulerianif forα,β ∈ P withα < β, the interval [α, β] isbalanced. Eulerian posets were first defined by Stanley [16] and have been extensively studied. The Dehn-Somerville equations are valid in these posets and as stated in [17; Section 3.14], Eulerian posets enjoy remarkable duality properties.
Letkbe an algebraically closed field and letGbe a reductive group defined overk. Let T be a maximal torus contained in a Borel subgroup BofG. LetW = NG(T)/T denote the Weyl group ofGand letSdenote the generating set of simple reflections ofW. Then Ghas the Bruhat decomposition:
G=
w∈W
BwB (1)
As in [1], the Bruhat-Chevalley order onW is defined as:
x≤y ifBxB⊆By B (2)
As is well known, this is equivalent x being a subword of a reduced expression y = α1. . . αm, α1, . . . , αm∈ S. The lengthl(y) is defined to bem. Ifw0is the longest element of W, then B− = w0Bw0 is the Borel subgroup of G opposite to B relative to T. If x1, . . . ,xn∈W, then let
x1∗ · · · ∗xn =
x1. . .xn ifl(x1. . .xn)=l(x1)+ · · · +l(xn) undefined otherwise
Forx,y∈W, letx◦y,xy∈W be defined as:
B(x◦y)B =BxByB,B−(xy)B=B−x By B (3) Lemma 1.1 Let x,y∈W . Then
(i) x◦y=x1∗y=x∗y1for some x1≤x,y1≤ y
(ii) x◦y=max{x y|y≤ y} =max{xy|x≤x} =max{xy|x≤x,y≤ y}
(iii) xy=min{x y|y≤ y} =min{xy|x≥x} =min{xy|x≥x,y≤ y}
(iv) xy=x y1with y1≤y and x =(x y1)∗y1−1 (v) xy=x y if and only if l(x y)=l(x)−l(y)
(vi) (xy)◦y−1=(xy)y−1and(x◦y)y−1=(x◦y)y−1
Proof: (i) follows the Tits axioms and induction on length. (ii) then follows from (i) and (3). (iii) and (iv) follow from (i) and (ii) by noting thatxy=w0((w0x)◦y).
(iv) Letxy=s. Then by (iv)x=sy1−1for somey1≤ y. Then by (i), (ii), x=sy1−1≤s◦y1−1≤s◦y−1=s∗y−1
for somes≤s. Sox=s∗y2−1for somes≤s,y2≤y. So s≥s≥s=x y−21≥xy=s
Hences=sand (xy)◦y−1=(xy)y−1.
Corollary 1.2 Let x,x,y,y∈W . If x ≥xand x∗y=x∗y,then y≤y. Proof: By Lemma 1.1,
x∗y=x∗y=x◦y≤x◦y=x∗y1 for somey1≤ y. By [7; Lemma 2.1],y≤y1. Soy≤y. Lemma 1.3 Let x0,y0,s∈W such that s0=x0y0 ≤s. Then
Y =
y≤ y0−1s◦y=x0 is a balanced subset of W.
Proof: By Lemma 1.1 (iv),s0=x0y1, withy1≤y0andx0 =s0∗y1−1. Lety∈Y. Then y≤y0−1ands◦y=x0. By Lemma 1.1 (i),x0=s∗ywiths≤s. Then
s0=x0y0≤x0y−1=s
Sinces0∗y1−1=x0 =s∗y, we see by Corollary 1.2 thaty≤y−11 . Hence Y =
y∈W |y≤y1−1,x0=s◦y
We prove by induction onl(y1) thatY is balanced. Ifl(y1) = 0, thenx0 = s0 < s and Y = ∅. So letl(y1)>0. Lety1=α∗y2, α∈S. Let
Y1 = {y∈Y | yα >y}
Y2 = {y∈Y | yα <y,yα∈Y} Lety∈Y1. Theny≤y1−1. So
yα=y◦α≤ y1−1◦α=y1−1
and
s◦(yα)=s◦y◦α=x0◦α=s0◦y1−1◦α=s0◦y1−1 =x0
Henceyα∈Y2. ThusY1Y2is balanced. Let Y3 = {y∈Y |yα <y,yα /∈Y}
SoY =Y1Y2Y3. Letx1=s0y2−1=s0∗y2−1. Then by induction hypothesis, Z =
y∈Y |y≤y2−1,s◦y=x1
is balanced. Lety∈ Z. Thenx1=s◦y=s1∗yfor somes1 ≤s. Sincex1<x1α, we see thaty<yα. Then
yα=y◦α≤ y2−1◦α=y1−1
and
s◦(yα)=s◦y◦α=x1◦α=x0 Thusyα∈Y3. Conversely let y∈Y3. Then
(yα)∗α=y≤y−11=y2−1∗α Henceyα≤y2−1. Alsos◦(yα)=x0and
(s◦(yα))◦α=s◦((yα)◦α)=s◦y=x0
Thus
(s◦(yα))∗α=x0 =x1∗α
Sos◦(yα) = x1 andyα ∈ Z. Since Z is balanced, we see thatY3 is balanced. Hence Y =Y1Y2Y3is balanced.
IfI ⊆S, then as usual letWI denote the parabolic subgroup ofW generated byI and let DI = {x∈W |xw=x∗w for all w∈WI}
denote the set of minimal length left coset representatives ofWI.
LetM be areductive monoidhavingGas its unit group. Thus Mis the Zariski closure ofGin someMn(k), whereMn(k) is the monoid of alln×nmatrices overk. We refer to [6, 14] for details. The idempotent set E( ¯T) of ¯T is a finite lattice isomorphic to the face lattice of a rational polytope. As in [5], let
= {e∈E( ¯T)|Be=e Be}
Thenis a cross-section of theG×G-orbits ofMsuch that for alle, f ∈, e≤ f ⇔e∈MfM
Here as usual,e≤ f meanse f =e= f e.is called thecross-section latticeof M. All maximal chains inhave the same length. We note that forMn(k),
=
Ir 0
0 0 0≤r≤n
is the usual set of idempotent representatives of matrices of different ranks.
By [10], the Bruhat decomposition (1) is extended toMas
M =
σ∈R
BσB (4)
whereR=NG(T)/T is theRenner monoidofM.W is the unit group ofRand R=
e∈
W eW (5)
The Bruhat-Chevalley order (2) onW extends toRas:
σ ≤θ ifBσB⊆BθB (6)
Then eachWeW is an interval in R and by [10] all maximal chains in R have the same length. Ris an inverse semigroup. This means that the map,σ →σ−1is an involution of R. Here ifσ =xey∈WeW, thenσ−1=y−1ex−1. Unlike inW, this involution is not order preserving. However by [11], the map
σ →w0σ−1w0 (7)
is an order preserving involution of R. Lete∈. Then as in [8], consider (inR),
λ(e)= {s∈S|se=es} (8)
and
λ∗(e)=
f≥e
λ(f), λ∗(e)=
f≤e
λ(f) (9)
Then
W(e)=Wλ(e)= {w∈W |we=ew}, W∗(e)=Wλ∗(e),
W∗(e)=Wλ∗(e) = {w∈W |we=e=ew}
are parabolic subgroups ofW with
W(e)=W∗(e)×W∗(e) (10)
MoreoverW∗(e) is the Weyl group of the unit group ofeMe. See [6; Chapter 10] for details.
IfI =λ(e),K =λ∗(e), let
D(e)=DI, D∗(e)=DK (11)
Let
W∗I,K =DI ×WI\K ×D−1I (12) Forσ =(x, w,y), σ=(x, w,y)∈WI∗,K, define
σ ≤σ ifw=w1∗w2∗w3 with xw1≤x, w2≤w, w3y≤ y (13) By [7; Theorem 2.5],
W∗I,K is isomorphic to the dual ofWeW (14)
The order onRis more subtle. Letσ ∈ R. Then
σ =xey for unique e∈,x∈ D∗(e),y∈ D(e)−1 (15) This is called thestandard formofσ. Letσ =xey, θ =s f t ∈ Rin standard form. Then by [4],
σ ≤θ⇔e≤ f, x≤sw, w−1t ≤y for somew∈W(f)W∗(e) (16)
Letσ = xeybe in standard form. Let x ≤ x1,y1 ≤ y. Let y1 = uy2,u ∈ W(e),y2 ∈ D(e)−1. Letx1u = x2z,x2 ∈ D∗(e),z ∈ W∗(e). Then x1ey1 = x2ey2 in standard form.
Now
x≤x1=x2zu−1=x2u−1·uzu−1
Sinceuzu−1 ∈ W∗(e) andx ∈ D∗(e),x ≤x2u−1. Alsouy2 = y1 ≤ y. Henceσ ≤ x2ey2. Thus,
σ =xeyin standard form ⇒σ ≤x1ey1for allx1≥x,y1≤y (17) Ife, f ∈withe≤ f, thene∈ f T and so we see directly from (6) that
xey≤x f y for allx,y∈W (18)
The length function onRis defined as follows. Letσ =xeyin standard form. Then
l(σ)=l(x)+l(e)−l(y) (19)
wherel(e) is the length of the longest element inD(e). We refer to [4, 7, 11, 14] for further details. In particular,
length function = rank function onWeW,e∈ (20)
where the rank function is determined from the grading ofWeW. 2. Projections
We wish to better understand the order≤on R given by (6), (16). Fore∈, letzedenote the longest element inW∗(e). Lete, f ∈withe≤ f. Letσ =xey∈WeW in standard form. Letzey=uy1,u∈W(f),y1∈ D(f)−1. We define the projection ofσ inW fWas:
pe,f(σ)=(x u)f y1 (21)
We claim that (21) is in standard form. Let x=x1v,x1 ∈ D(f), v ∈ W(f). Sincex ∈ D∗(e) andW∗(f)⊆W∗(e), we see thatv∈W∗(f). By (10),vu∈W∗(f). Thusxu = x1(vu)∈ D∗(f). Hence (21) is in standard form. Nowzfzey =uy1withu =zfu ∈ W(f) and by the above,vu=vu. Hence we also have
pe,f(σ)=(xu)f y1in standard form (22)
The following result in conjunction with (14) yields a new description of the order onR.
Theorem 2.1 Let e, f ∈,e≤ f.Then
(i) pe,f :WeW→WfW is order preserving andσ ≤ pe,f(σ)for allσ ∈WeW.
(ii) Ifσ ∈WeW,θ∈W fW, thenσ ≤θif and only if pe,f(σ)≤θ.
(iii) If h∈with e≤h ≤ f , then pe,f = ph,f ◦ pe,h. (iv) pe,f is onto if and only ifλ∗(e)⊆λ∗(f).
(v) pe,f is1−1if and only ifλ(f)⊆λ(e).
Proof: Letσ = xey in standard form. Let zey = uy1,u ∈ W(f),y1 ∈ D(f)−1. By Lemma 1.1,xu=xu1withu1≤u. Then
u1y1≤uy1=zey
Henceu1y1=zyfor somez≤z1,y≤y. Thenz∈W∗(e) and (z−1u1)y1=y≤ y. Also sincex∈ D∗(e),
x≤xz=(xu1) u−11 z By (16), (21),
σ =xey≤xu1f y1= pe,f(σ) (23)
Letθ=s f tin standard form such thatσ ≤θ. Then by (16), x≤sw, w−1t ≤y for somew∈W(f)W∗(e)
Sox=s1∗w1for somes1≤s, w1≤w. Sincex ∈D∗(e), w1∈ D∗(e). Noww=w2∗z for somew2∈W(f),z∈W∗(e). Thenw1 ≤w2. Sincet ∈ D(f)−1andy∈ D(e)−1,
w1−1t≤w−21t =zz−1w−21t =zw−1t ≤z◦(w−1t)≤z◦y≤ze◦y=zey=uy1
Sincet,y1 ∈ D(f)−1andw1,u ∈ W(f), we see by [7; Lemma 2.2] thatw1 =w3∗w4
withw−14 ≤u, w3−1t ≤y1. So
xu ≤xw4−1=s1w1w−14 =s1w3≤s◦w3 =s∗w5
for somew5≤w3. Alsow−51t ≤w3−1t ≤y1. Hence by (21),
pe,f(σ)=(xu)f y1≤s f t=θ (24)
So ifσ ∈ WeW withσ ≤ σ, then by (23),σ ≤ σ ≤ pe,f(σ). So by (24), pe,f(σ) ≤ pe,f(σ). This proves (i), (ii).
Lete≤ h ≤ f in. Letσ ∈ WeW, θ = pe,f(σ). Then by (i), (ii), pe,f(σ) ≤ ph,f ◦ pe,h(σ). Letσ =xey, θ =s f tin standard form. Then by (16),x≤sw, w−1t≤ yfor some w∈W(f)W∗(e). Sow=w1∗zfor somew1∈W(f),z∈W∗(e). Then by (17), (18),
σ =xey≤swew−1t =sw1ew1−1t ≤sw1hw−11t≤sw1fw−11t =θ
So ifπ=sw1hw−11 t∈WhW, thenσ ≤π≤θ. By (ii),pe,f(σ)≤πandph,f(π)≤θ. So by (i),
ph,f ◦ pe,f(σ)≤ ph,f(π)≤θ=pe,f(σ) This proves (iii).
(iv) Suppose first thatλ∗(e)⊆ λ∗(f). By (9),λ∗(e) ⊆λ∗(f). Soλ(e) ⊆λ(f). Hence W(e) ⊆ W(f) and W∗(e) = W∗(f). So D∗(e) = D∗(f) and D(f) ⊆ D(e). Thus if θ =x f y ∈WfWis in standard form, thenσ =xeyis in standard form and pe,f(σ)=θ. Thus pe,f is onto.
Assume conversely that pe,f is onto. Letw be the longest element inW∗(f) and let θ =wf ∈W f W. There existsσ =xeyin standard form such thatpe,f(σ)=θ. By (21), w=xufor someu∈W(f). Sow≤xandx∈W(f). By (9), (11),x∈ D∗(e)⊆D∗(f).
So by (10),x ∈ W∗(f). Thusx =w. Sincewis the longest element ofW∗(f)x, α < x for allα∈λ∗(f). Sincex ∈ D∗(e),xα >xfor allα∈λ∗(e). Henceλ∗(e)∩λ∗(f)= ∅. There existsx1ey1in standard form such thatpe,f(x1ey1)= f. By (21),ze∗y1 ∈ W(f).
Hence by (10),ze∈W(f)=W∗(f)×W∗(f). SinceW∗(e)∩W∗(f)= {1},ze∈W∗(f).
Henceλ∗(e)⊆λ∗(f).
(v) Letλ(f)⊆λ(e). ThenW(f)⊆W(e) and D(e)⊆ D(f). Letxey,xey,s f tbe in standard form such that
pe,f(xey)= pe,f(xey)=s f t (25)
Letze=vy1, wherev∈W(f)∩W∗(e) andy1∈ D(f)−1∩W∗(e). Letu∈W(f)⊆W(e).
Then
u(y1y)=(uy1)y
=(u∗y1)y since y1∈ D(f)−1
=(u∗y1)∗y, since uy1∈W(e),y∈ D(e)−1
=u∗(y1y)
Soy1y∈D(f)−1. Similarlyy1y∈ D(f)−1. By (25), y1y=t =y1y, xv=s=xv
Soy=y. Sincev ∈W∗(e),x=xvandx=xv. Sox=x. Thuspe,f is 1−1.
Assume conversely thatpe,f is 1−1. Letve, vf denote the longest elements ofW(e) and W(f) respectively. Thusvew0 andvfw0 are respectively the longest elements of D(e)−1 andD(f)−1. Let
zevew0=vy, v∈W(f),y∈ D(f)−1 (26) Then
pe,f(evew0)= f y= pe,f(v−1evew0)
Since pe,f is 1−1,evew0=v−1evew0. Sov∈W∗(e). Soz=v−1ze∈W∗(e) and by (26), zvew0=z∗(vew0)=y≤vfw0
Sovew0 ≤vfw0. Hencevf ≤ve. ThusW(f)⊆W(e) andλ(f)⊆λ(e). This completes the proof.
Example 2.2 LetG=G L3(k),M =M3(k). Then W is the group of permutation matrices and R is the monoid of partial permutation matrices (rook monoid). Let
e=
1 0 0
0 0 0
0 0 0
,
f =
1 0 0
0 1 0
0 0 0
.
Thenλ(e)=λ∗(e)= {(23)}, λ(f)= {(12)}, λ∗(f)= ∅. Hencepe,fis not 1−1 or onto.pe,f
is given in Table 1. Sinceλ∗(I)=θ,pf,I is onto. pf,I is given in Table 2. Combining with [7; figures 3 and 4], one obtains the Hasse diagram ofR.
Example 2.3 Letφ: Mn(k)→MN(k) be defined as:
φ(A)=A⊗ ∧2A⊗ · · · ⊗ ∧nA
where
N = n r=1
n r
Table 1. Projection from rank 1 to rank 2.
σ pe,f(σ)
1 0 0
0 0 0
0 0 0
1 0 0
0 0 1
0 0 0
0 1 0
0 0 0
0 0 0
0 1 0
0 0 1
0 0 0
0 0 1
0 0 0
0 0 0
0 1 0
0 0 1
0 0 0
0 0 0
1 0 0
0 0 0
0 0 1
1 0 0
0 0 0
0 0 0
0 1 0
0 0 0
0 0 1
0 1 0
0 0 0
0 0 0
0 0 1
0 0 0
0 1 0
0 0 1
0 0 0
0 0 0
0 0 0
1 0 0
0 0 1
0 0 0
1 0 0
0 0 0
0 0 0
0 1 0
0 0 1
0 0 0
0 1 0
0 0 0
0 0 0
0 0 1
0 1 0
0 0 0
0 0 1
LetM denote the Zariski closure ofφ(Mn(k)) inMN(k). ThenW is the symmetric group of degreenandS= {(12),(23), . . . ,(n−1 n)}. Also
= {eI|I⊆S} ∪ {0}
with
eK ≤eI⇔K⊆I
Table 2. Projection from rank 2 to rank 3.
σ pf,I(σ) σ pf,I(σ)
0 0 0 0 1 0 1 0 0
0 0 1
0 1 0
1 0 0
0 0 0
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1
0 0 0
0 0 1
1 0 0
0 1 0
0 0 1
1 0 0
0 0 1
0 0 0
0 1 0
1 0 0
0 0 1
0 1 0
0 0 0
1 0 0
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1
1 0 0
0 0 0
0 1 0
1 0 0
0 0 1
0 1 0
0 0 0
1 0 0
0 1 0
0 0 1
1 0 0
1 0 0
0 0 0
0 0 1
1 0 0
0 1 0
0 0 1
0 0 0
0 0 1
0 1 0
1 0 0
0 0 1
0 1 0
1 0 0
0 1 0
0 0 0
1 0 0
0 1 0
0 0 1
0 1 0
1 0 0
0 0 0
0 1 0
1 0 0
0 0 1
0 0 1
0 1 0
0 0 0
1 0 0
0 1 0
0 0 1
0 0 0
1 0 0
0 0 1
0 1 0
1 0 0
0 0 1
0 1 0
0 0 0
0 0 1
1 0 0
0 1 0
0 0 1
0 0 1
0 0 0
1 0 0
0 1 0
0 0 1
1 0 0
1 0 0
0 0 1
0 0 0
1 0 0
0 1 0
0 0 1
1 0 0
0 0 0
0 1 0
1 0 0
0 0 1
0 1 0
0 1 0
0 0 1
0 0 0
1 0 0
0 1 0
0 0 1
and
λ(eI)=λ∗(eI)=I, λ∗(eI)= ∅, I⊆S So by Theorem 2.1, peK,eI is onto forK⊆I.
Example 2.4 Letφ:S Ln(k)→G LN(k) be defined as:
φ(A)=A⊕ ∧2A⊕ · · · ⊕ ∧nA
where N=2n −1. Let M denote Zariski closure in MN(k) ofkφ(S Ln(k)). Again W is symmetric group of degreenandS= {(12),(23),(n−1 n)}. Then
= {1} ∪ {eI|I⊆S}
with
eI ≤eK⇔K⊆I and
λ(eI)=λ∗(eI)=I, λ∗(eI)= ∅, I⊆S So by Theorem 2.1, peI,eK is 1−1 forK⊆I.
Corollary 2.5 Let e< f in, σ∈WeW, θ∈WfW. Letσ=xey in standard form,zfzey= uy1with u∈W(f),y1∈D(f)−1. Thenθcoversσif and only if f covers e in,pe,f(σ)=θ and l(xu)=l(x)−l(u).
Proof: Ifθcoversσ, then by Theorem 2.1,f coverseinandθ=pe,f(σ). So assume that f coverseinandθ=pe,f(σ). The maximum elements ofWeWandWfWare respectively w0zeeandw0zf f. Since f coverse, we see by (9) and [6; Chapter 10] that
λ∗(f)=λ(f)∩λ∗(e) So by (22),
pe,f(w0zee)=w0zef zfze (27)
coversw0zee. By (19), (20), [σ, w0ze] has length
l(w0ze)−l(x)+l(y) (28)
and [w0f zfze, w0zf f] has length
l(w0zf)−l(w0ze)+l(zfze) (29)
By (27)–(29), [σ, w0zf f] has length
l(w0zf)−l(x)+l(y)+l(zfze)+1 (30) By (22),θ=(xu)f y1. Also
l(u)+l(y1)=l(zfzey)=l(zfze)+l(y) (31)
By (19), (20), [θ, w0zf f] has length
l(w0zf)−l(xu)+l(y1) (32)
By (30)–(32), [σ, θ] has length l(xu)+l(u)−l(x)+1 Henceθcoversσ if and only if
l(xu)=l(x)−l(u)
By Lemma 1.1, this is true if and only ifl(xu)=l(xu). This completes the proof.
Corollary 2.6 Any interval in R of length2has at most4elements.
Proof: Consider an interval [σ, θ] inRof length 2. Letσ∈WeW,θ=WfW. Thene≤ f. Case 1. e= f. By (14), WeW is isomorphic to the dual of W∗I,K where I=λ(e) and K=λ∗(e). NowW∗I,K is a subposet ofW∗I,∅with the same rank function. By [7; Theorem 3.3],W∗I,∅is an Eulerian poset. Hence any interval of length 2 inWI∗,∅has 4 elements. It follows that|[σ, θ]| ≤4 inWeW.
Case 2. e< f and f does not coverein. Then by Corollary 2.5, [e, f] has length 2 in . NowE( ¯T) is the face lattice of a polytope. Hence in,|[e,f]| ≤4. So in,
[e, f]= {e,h,h,f}
withe<h,e<h< f and with the possibility thath=h. So by Theorem 2.1, [σ, θ]= {σ,pe,h(σ),pe,h(σ), θ}
inR.
Case 3. f coverseinandθ=pe,f(σ). Letσ=xeyin standard form. Ifπ∈(σ, θ), then π∈WeWandπ coversσ. So by (14), either Rπ=Rσ or πR=σR. Letπ1, π2∈(σ, θ) such that Rπ1 = Rσ = Rπ2. Thenπ1=x1ey, π2=x2ey in standard form. Letzfzey= uy1,u∈W(f),y1∈D(f)−1. Sinceθcoversπ1andπ2, we see by Corollary 2.5 that
x1u f y1= pe,f(π1)=θ=pe,f(π2)=x2u f y1 (33) in standard form. It follows that x1u=x2u. Hencex1 = x2 andπ1=π2. Dually by (7), π1R=π2Rimplies thatπ1=π2. It follows that|[σ, θ]| ≤4.
Case 4. f covers e in and pe,f(σ)=θ1< θ. Then θ1 covers σ and θ covers θ1. Letπ1, π2∈(σ, θ), π1=π2, π1=θ1, π2=θ1. Thenπ1, π2∈WeWandθcoversπ1, π2. So pe,f(π1)=θ=pe,f(π2). Sinceπ1, π2coverσ, we see by (14) that fori=1,2, πiR=σR, or Rσ=Rπi. Sinceπ1=π2, we can assume by (33) thatRπ1=Rσ,Rπ2=Rσ. Soπ1= xey, π2=xeyin standard form,xcoversxandycoversy. Sinceθcoversπ1, π2,
zfzey=uy1,zfzey=vy1,u, v∈W(f),y1∈D(f)−1 Thenθ1=x1f y1, θ=x1f y1in standard form with
x1=xu, x1 =xv=xu and by Corollary 2.5,
x=x1∗u−1=x1∗v−1, x=x1 ∗u−1
Nowx1 coversx1 and henceu−1 coversv−1by Corollary 1.2. Since xcoversx, this contradicts the exchange condition forW. So|[σ, θ]| ≤4, completing the proof.
Corollary 2.6 leads us to the following conjecture concerning the M¨obius functionµon R. We refer to [17; Chapter 3] for the theory of M¨obius functions on posets:
Conjecture 2.7 Letσ, θ∈R, σ≤θ. Then µ(σ, θ)=
(−1)l[σ,θ] if every interval of length 2 in [σ, θ] has 4 elements
0 otherwise
Herel[σ, θ] denotes the length of the interval [σ, θ].
Theorem 3.4 below establishes Conjecture 2.7 for canonical monoids.
3. Canonical monoids
In this section we will assume thatM is a canonical monoid. This means that∗=\{0}
has a least elemente0withλ(e0)= ∅. Then as in Example 2.3,∗is in 1−1 correspondence with the subsets ofS. So we can write:
∗= {eI|I⊆S} (34)
with
λ(eI)=λ∗(eI)=I, λ∗(eI)= ∅,I⊆S
and
eK ≤eI ⇔K⊆I
See [9, 13] for details. Example 2.3 is an example of a canonical monoid. More generally ifG0is a semisimple group and ifφ:G0→G Ln(k) is an irreducible representation with highest weight in the interior of the Weyl chamber, then the Zariski closure in Mn(k) of kφ(G0) is a canonical monoid. Canonical monoids are closely related to canonical com- pactifications of semisimple groups in the sense of [2]. The connection between reductive monoids and embeddings of homogenous spaces is studied in [12]. See also [19]. Basically the canonical compactification is obtained as the projective variety X=(M\{0})/center.
Then the B×B-orbits of X are indexed byR∗=R\{0}. See [13]. The Bruhat-Chevalley order onR∗corresponds to the Zariski closure inclusion ofB×B-orbits ofX, the geometric properties of which have been studied in [15].
LetMbe a canonical monoid. ForI⊆S, letRI =W eIW=W eID−I1. Then by (5), (34), R∗ =R\{0} =
I⊆S
RI (35)
ForK⊆I, we write pK,I for peK,eI. So pK,I:RK→RI. By [7; Theorem 3.3], eachRI is an Eulerian poset. We will show in this section thatR∗is an Eulerian poset.
Lemma 3.1 Letσ∈R∅,s∈W such that p∅,S(σ)<s. Then[σ,s]∩R∅is balanced.
Proof: Lete=e∅, σ=x0ey0,s0=p∅,S(σ)=x0y0. Thens0<s.Fory≤y0, let Ay=[σ,s]∩W ey =
xey|x0≤x≤sy−1
Thus Ay is a non-trivial interval in R∅ unless x0=sy−1. So Ay is balanced unless x0=sy−1. By Lemma 1.3,
Y =
y≤ y0|x0=sy−1
is balanced. It follows that [σ,s]∩R0is balanced.
Corollary 3.2 Letσ∈R∅, θ∈RI such that p∅,I(σ)< θ. Then[σ, θ]∩R∅is balanced.
Proof: Lete=e∅, f =eI, σ=x0ey0, θ=s f t in standard form. Suppose x0∈/sWI. For t≤y≤y0, let
Ay=[σ, θ]∩W ey By [3],
w=wy=max{u∈WI|u−1t ≤y}
