Primitive orthogonal idempotents for R-trivial monoids
Chris Berg, Nantel Bergeron, Sandeep Bhargava, Franco Saliola
aa r X i v : . [ m a t h . R T ] O c t Primitive orthogonal idempotents for R-trivialmonoids
Chris Berg , Nantel Bergeron , Sandeep Bhargava , andFranco Saliola
Fields Institute, 222 College Street, Toronto, ON, Canada York University, 4700 Keele Street, Toronto, ON, Canada Universit´e du Qu´ebec `a Montr´eal, Montr´eal, QC, CanadaJune 17, 2018
Abstract
We construct a recursive formula for a complete system of primi-tive orthogonal idempotents for any R -trivial monoid. This uses thenewly proved equivalence between the notions of R -trivial monoid andweakly ordered monoid. Recently, Denton [4] gave a formula for a complete system of primitive or-thogonal idempotents for the 0 -Hecke algebra of type A , the first since thequestion was raised by Norton [6] in 1979. A complete system of primitiveorthogonal idempotents for left regular bands was found by Brown [3] andSaliola [9]. Finding such collections is an important problem in representationtheory because they decompose an algebra into projective indecomposablemodules: if { e J } J ∈ I is such a collection for a finite dimensional algebra A ,then A = ⊕ J ∈ I Ae J , where each Ae J is a projective indecomposable module.They also allow for the explicit computation of the quiver, the Cartan in-variants, and the Wedderburn decomposition of the algebra (see [2, 1]). For1xample, in [5], Denton, Hivert, Schilling, and Thi´ery use a construction ofa system of primitive orthogonal idempotents for any J -trivial monoid S toderive combinatorially the Cartan matrix and quiver of S .Schocker [10] constructed a class of monoids, called weakly ordered monoids ,to generalize simultaneously 0-Hecke monoids and left regular bands, withthe broader aim of finding a complete system of orthogonal idempotents forthe corresponding monoid algebras. We achieve this goal here.A key step is to recognize that the notions of weakly ordered monoidand R -trivial monoid are one and the same. This was first pointed outto us by Thi´ery [13] after an intense discussion between the authors andDenton, Hivert, Schilling, and Thi´ery. In Section 2, we fill out an outlineof a proof provided by Steinberg [12], who independently made this sameobservation. In Section 3, we use this equivalence to build a recursive formulafor a complete system of primitive orthogonal idempotents for any R -trivialmonoid. This covers, in particular but not only, the previously known casesof J -trivial monoids [5] and left regular bands. R -trivial monoids Given any monoid S , that is, a set with an associative multiplication and anidentity element, we define a preorder ≤ as follows. Given u, v ∈ S , write u ≤ v if there exists w ∈ S such that uw = v . We write u < v if u ≤ v but u = v . Unless stated otherwise, the monoids throughout the paper areendowed with this “weak” preorder. In the monoid theory literature, the dual of this preorder is known as Green’s R -preorder . Definition 2.1.
A finite monoid S is said to be a weakly ordered monoid if there is a finite upper semi-lattice ( L , (cid:22) ) together with two maps C, D : S → L satisfying the following axioms:1. C is a monoid morphism, i.e. C ( uv ) = C ( u ) ∨ C ( v ) for all u, v ∈ S .2. C is a surjection.3. If u, v ∈ S are such that uv ≤ u , then C ( v ) (cid:22) D ( u ) .4. If u, v ∈ S are such that C ( v ) (cid:22) D ( u ) , then uv = u . emark 2.2. This notion was introduced by Schocker [10] to generalize0-Hecke monoids and left regular bands, with the broader aim of finding acomplete system of orthogonal idempotents for the corresponding monoidalgebras. In his paper, he actually calls these weakly ordered semigroups .However our understanding is that monoids include an identity element andsemigroups do not. So throughout the paper we call these weakly orderedmonoids.
Definition 2.3.
A monoid S is R -trivial if, for all x, y ∈ S , xS = yS implies x = y . We restrict our discussion to finite R -trivial monoids. Example 2.4.
A monoid S is called a left regular band if x = x and xyx = xy for all x, y ∈ S . Left regular bands are R -trivial. Indeed, if xS = yS , then there exist u, v ∈ S such that xu = y and x = yv . But then,since uv = uvu , x = yv = xuv = xuvu = yvu = xu = y. Finitely generated left regular bands are also weakly ordered monoids, seeShocker [10], e.g. 2.4 and Brown [3, Appendix B].
Example 2.5.
Let G be a Coxeter group with simple generators { s i : i ∈ I } and relations: • s i = 1, • s i s j s i s j · · · | {z } m ij = s j s i s j s i · · · | {z } m ij for some positive integers m ij .Then the H G (0) has generators { T i : i ∈ I } and relations: • T i = T i , • T i T j T i T j · · · | {z } m ij = T j T i T j T i · · · | {z } m ij for some positive integers m ij .The weakly ordered monoid H G (0) has maps C and D onto the lattice ofsubsets of I . The map C is the content of an element: C ( T i T i · · · T i k ) = { i , i , . . . , i k } . The map D is the set of right descents of an element: D ( x ) =3 i ∈ I : xT i = x } . Note that the preorder for this monoid coincides with theweak order on the elements of the Coxeter group G .Of particular interest is the case when G is the symmetric group S n .Norton [6] gave a decomposition of the monoid algebra C H S n (0) into leftideals and classified its irreducible representations. She raised the questionof constructing a complete system of orthogonal idempotents for the algebra,which was first answered by Denton [4]. Example 2.6.
Let S be the monoid with identity generated by the followingmatrices: g := and g := . Then S = { , g , g , g g , g g } and S is both an R -trivial monoid and aweakly ordered monoid. For example, we can take L to be usual lattice ofsubsets of { , } , with C : S → L given by C (1) = ∅ , C ( g ) = { } , C ( g ) = { } , C ( g g ) = C ( g g ) = { , } , and D : S → L given by D (1) = ∅ , D ( g ) = { } , D ( g ) = D ( g g ) = { } , D ( g g ) = { , } . The monoid S , however, is neither a left regular band, since g g is notidempotent, nor isomorphic to the 0-Hecke monoid H G (0) on two generators,since the latter always has an even number of elements.The fact that the above examples are all weakly ordered and R -trivial isno coincidence: the purpose of this section is to show that these two notionsare equivalent. Remark 2.7.
A monoid S is R -trivial if and only if the preorder ≤ definedabove is a partial order.Proof. Suppose S is an R -trivial monoid and x, y ∈ S are such that x ≤ y and y ≤ x . Then there exist u, v ∈ S such that xu = y and yv = x . So y ∈ xS and x ∈ yS , implying that yS ⊆ xS and xS ⊆ yS . That is, xS = yS .Since S is R -trivial, x = y .On the other hand, suppose that the given preorder is a partial order,and that xS = yS for some x, y ∈ S . Since x = x · ∈ xS = yS , we havethat x = yu for some u ∈ S . So y ≤ x . Similarly, y ∈ xS implies that x ≤ y .The antisymmetry of ≤ implies then that x = y . So S is R -trivial.4 orollary 2.8. A weakly ordered monoid is an R -trivial monoid.Proof. Let S be a weakly ordered monoid. Lemma 2 . S is a partial order. The result now follows from Proposition 2.7.We will show that any finite R -trivial monoid S is a weakly orderedmonoid using an argument outlined by Steinberg [12]. We must establish theexistence of an upper semi-lattice L and two maps C and D from S to L that satisfy the conditions of Definition 2.1. We gather here the definitionsof L , C and D :1. L is the set of left ideals Se generated by idempotents e in S , orderedby reverse inclusion;2. C : S → L is defined as C ( x ) = Sx ω , where x ω is the idempotent powerof x (see Lemma 2.10);3. D : S → L is defined as D ( u ) = C ( e ), where e is some maximal elementin the set { s ∈ S : us = u } (with respect to the preorder ≤ ).The following lemma is a simple statement about R trivial monoids whichis used frequently throughout the paper. Lemma 2.9.
Suppose S is an R -trivial monoid. If x, y, z ∈ S are such that xyz = x , then xy = x .Consequently, if x, y , y , . . . , y m ∈ S are such that xy · · · y m = x , then xy i = x for all ≤ i ≤ m .Proof. If xyz = x then xyS = xS . Therefore xy = x by the definition of S being R -trivial. The second statement immediately follows from the first.The remainder of this section is dedicated to showing that these objectsare well defined and that they satisfy the conditions of Definition 2.1. Webegin by recalling some classical results from the monoid literature. Thefollowing is [7, Proposition 6.1]. Lemma 2.10. If S is a finite monoid, then for each x ∈ S , there existsa positive integer ω = ω ( x ) such that x ω is idempotent, i.e. ( x ω ) = x ω .Furthermore, if S is R -trivial, then we also have x ω x = x ω . roof. Consider the elements x, x , x , . . . . Since S is finite, there exist pos-itive integers i and p such that x i + p = x i . Then x k + p = x k for all k ≥ i , so ifwe take ω = ip , then ( x ω ) = x ω + ip = x ω .If S is R -trivial, then x ω ≤ x ω x ≤ x ω x ω = x ω , and so x ω x = x ω . Remark 2.11.
In what follows, if x ∈ C S and there exists N such that x N +1 = x N , we sometimes abuse notation by writing x ω in place of x N . Lemma 2.12.
Let S be a finite R -trivial monoid. For all x and y in S ,1. ( xy ) ω x = ( xy ) ω ;2. ( xy ) ω y = ( xy ) ω ;3. ( xy ) ω x ω = ( xy ) ω ;4. ( xy ) ω y ω = ( xy ) ω ; 5. ( x ω y ω ) ω x ω = ( x ω y ω ) ω ;6. ( x ω y ω ) ω = ( x ω y ω ) ω ( xy ) ;7. ( x ω y ω ) ω = ( x ω y ω ) ω ( xy ) ω .Proof. (1) Since ( xy ) ω x ∈ ( xy ) ω S , it follows that ( xy ) ω xS ⊆ ( xy ) ω S . Toshow the reverse inclusion, note that ( xy ) ω = ( xy ) ω ( xy ) = (cid:0) ( xy ) ω x (cid:1) y ∈ ( xy ) ω xS , where the first equality follows from Lemma 2.10. So ( xy ) ω S ⊆ ( xy ) ω xS . Thus ( xy ) ω xS = ( xy ) ω S . Since S is an R -trivial monoid, thedesired result follows.(2) Apply (1) and Lemma 2.10:( xy ) ω = ( xy ) ω ( xy ) = (cid:16) ( xy ) ω x (cid:17) y = ( xy ) ω y. (3) This follows from applying (1) repeatedly.(4) This follows from applying (2) repeatedly.(5) Let u = x ω and v = y ω . Now, by (1), ( uv ) ω u = ( uv ) ω .(6) We compute:( x ω y ω ) ω = ( x ω y ω ) ω − x ω y ω = ( x ω y ω ) ω − x ω y ω y (by Lemma 2.10)= ( x ω y ω ) ω y = ( x ω y ω ) ω x ω y (by (5))= ( x ω y ω ) ω x ω xy (by Lemma 2.10)= ( x ω y ω ) ω xy (by (5))(7) This follows by repeatedly applying part (6).6e are now ready to construct a lattice corresponding to the R -trivialmonoid S . Define L := { Se : e ∈ S such that e = e } . That is, L is the set of left ideals generated by the idempotents of S . Definea partial order on L by Se (cid:22) Sf ⇐⇒ Se ⊇ Sf.
Proposition 2.13. If e, f are idempotents in S , then S ( ef ) ω is the leastupper bound of Se and Sf in L .Proof. First, let us show that S ( ef ) ω is an upper bound for Se and Sf .Since, by Lemma 2.12 (1), ( ef ) ω = ( ef ) ω e , we have that ( ef ) ω ∈ Se . Hence S ( ef ) ω ⊆ Se and S ( ef ) ω (cid:23) Se . Moreover, ( ef ) ω = (cid:16) ( ef ) ω − e (cid:17) f ∈ Sf . So S ( ef ) ω ⊆ Sf and S ( ef ) ω (cid:23) Sf . So S ( ef ) ω is an upper bound for Se and Sf .Next, let us show that S ( ef ) ω is the least upper bound for Se and Sf .Suppose g is an idempotent in S such that Sg is an upper bound for Se and Sf . That is, Sg ⊆ Se and Sg ⊆ Sf . Since Sg ⊆ Se , g = te for some t ∈ S .But then ge = ( te ) e = te = te = g . Similarly, Sg ⊆ Sf implies that gf = g .So g ( ef ) = ( ge ) f = gf = g and it follows that g = g ( ef ) = (cid:16) g ( ef ) (cid:17) ( ef ) = g ( ef ) = (cid:16) g ( ef ) (cid:17) ( ef ) = g ( ef ) = · · · = g ( ef ) ω . Consequently, g ∈ S ( ef ) ω , Sg ⊆ S ( ef ) ω , and Sg (cid:23) S ( ef ) ω . So S ( ef ) ω is theleast upper bound of Se and Sf .As a result, we may define the join of two elements Se and Sf in L by Se ∨ Sf = S ( ef ) ω . That is, L is an upper semi-lattice with respect to this join operation. Thisobservation proves the following. Proposition 2.14.
The map C : S → L defined by C ( x ) = Sx ω is a surjec-tive monoid morphism. roof. Let x, y ∈ S . By Lemma 2.12 (5), we know that ( x ω y ω ) ω = ( x ω y ω ) ω ( xy ) ω .Hence, ( x ω y ω ) ω ∈ S ( xy ) ω and S ( x ω y ω ) ω ⊆ S ( xy ) ω .To show the reverse inclusion, we begin by noting that, by Lemma 2.12 (2),( xy ) ω = ( xy ) ω x ω . So ( xy ) ω ∈ Sx ω and S ( xy ) ω ⊆ Sx ω . That is, S ( xy ) ω (cid:23) Sx ω .Lemma 2.12 (4), implies that ( xy ) ω ∈ Sy ω , which implies that S ( xy ) ω ⊆ Sy ω and S ( xy ) ω (cid:23) Sy ω . In particular, S ( xy ) ω is an upper bound for both Sx ω and Sy ω . So S ( xy ) ω (cid:23) Sx ω ∨ Sy ω = S ( x ω y ω ) ω , that is, S ( xy ) ω ⊆ S ( x ω y ω ) ω .Thus C ( xy ) = S ( xy ) ω = S ( x ω y ω ) ω = Sx ω ∨ Sy ω = C ( x ) ∨ C ( y ), and C is a monoid morphism. Finally, we know that every element of L is of theform Se for some idempotent e in S . But then C ( e ) = Se ω = Se ; that is, C is a surjective morphism.Here is an alternate and useful characterization of C ( x ). Proposition 2.15. C ( x ) = { a ∈ S : ax = a } for all x ∈ S .Proof. Take an arbitrary element in C ( x ) = Sx ω , say tx ω . Since (cid:0) tx ω (cid:1) x = t (cid:0) x ω x (cid:1) = tx ω by Lemma 2.10, we see that tx ω ∈ { a ∈ S : ax = a } . On theother hand, take b ∈ { a ∈ S : ax = a } . Then bx ω = ( bx ) x ω − = bx ω − = ( bx ) x ω − = bx ω − = · · · = bx = b. Therefore, b ∈ Sx ω .We now define the map D : S → L . Given u ∈ S , let D ( u ) = C ( e ), where e is a maximal element in the set { s ∈ S : us = u } . To check that D is welldefined, let e and f be two distinct maximal elements in { s ∈ S : us = u } .Since e ≤ ef and u ( ef ) = ( ue ) f = uf = u , by the maximality of e , e = ef .Similarly, since f ≤ f e and u ( f e ) = u , the maximality of f implies f = f e .Then, by Proposition 2.14, C ( e ) = C ( ef ) = C ( e ) ∨ C ( f ) = C ( f ) ∨ C ( e ) = C ( f e ) = C ( f ) . Note that the maximality of e and ue = u also implies that e = e , that is, e is idempotent.The next proposition shows that the maps C and D interact in preciselythe manner given in conditions 2 and 3 in Definition 2.1. The followinglemma will help us prove this proposition. Lemma 2.16.
Let x, y ∈ S . If x ≤ y , then C ( x ) (cid:22) C ( y ) . roof. If s ∈ C ( y ), then sy = s . Since x ≤ y , there exists t ∈ S such that y = xt . So sxt = s , implying sx ≤ s . That is, s ∈ C ( x ). Hence C ( y ) ⊆ C ( x ),or C ( x ) (cid:22) C ( y ) since s ≤ sx and S is R -trivial. Proposition 2.17.
Let u, v ∈ S . (i) If uv ≤ u , then C ( v ) (cid:22) D ( u ) . (ii) If C ( v ) (cid:22) D ( u ) , then uv = u .Proof. (i) Since u ≤ uv , u = uv . Hence v lies in the set { s ∈ S : us = u } .Let e be a maximal element in this set such that v ≤ e . Then, by Lemma2.16, C ( v ) (cid:22) C ( e ) = D ( u ).(ii) By definition, D ( u ) = C ( e ), where e is a maximal element of { s ∈ S : us = u } . So if C ( v ) (cid:22) D ( u ), then C ( v ) (cid:22) C ( e ). Hence C ( e ) ⊆ C ( v ). Since ue = u , u lies in C ( e ). So u is also a member of C ( v ); that is, uv = u .Propositions 2.14 and 2.17 tell us that an R -trivial monoid is a weaklyordered monoid. Combining this with Corollary 2.8, we have the followingresult. Theorem 2.18.
A finite monoid S is a weakly ordered monoid if and onlyif it is an R -trivial monoid. Definition 3.1.
Let A be a finite dimensional algebra with identity . Wesay that a set of nonzero elements Λ = { e J : J ∈ I} of A is a completesystem of primitive orthogonal idempotents for A if:1. each e J is idempotent : that is, e J = e J for all J ∈ I ;2. the e J are pairwise orthogonal : e J e K = 0 for J, K ∈ I with J = K ;3. each e J is primitive (meaning that it cannot be further decomposedinto orthogonal idempotents): if e J = x + y with x and y orthogonalidempotents in A , then x = 0 or y = 0 ;4. { e J : J ∈ I} is complete (meaning that the elements sum to the iden-tity): P J ∈I e J = 1 . emark 3.2. If Λ is a maximal set of nonzero elements satisfying conditions(1) and (2), then Λ is a complete system of primitive orthogonal idempotents(that is, (3) and (4) also hold). Indeed, e J is primitive, for if e J could bewritten as x + y , then we could replace e J in Λ with x and y , contradicting themaximality of Λ. To see (4), we just note that if P K e K = 1, then 1 − P K e K is idempotent and orthogonal to all other e K . Combining this element withΛ would again contradict the maximality of Λ.Let S denote a finite weakly ordered monoid with C and D being theassociated “content” and “descent” maps from S to an upper semi-lattice L . We let G denote a set of generators of S . The main goal of this paper isto build a method for finding a complete system of orthogonal idempotentsfor the monoid algebra C S . In particular, this solves the problem posed byNorton about the 0-Hecke algebra for the symmetric group.For each J ∈ L , we define a Norton element A J T J . Let us begin bydefining T J : T J = (cid:16) Y g ∈G C ( g ) (cid:22) J g ω (cid:17) ω ∈ S. Remark 3.3.
A different ordering of the set G of generators may producedifferent T J ’s; so we fix an (arbitrarily chosen) order uniformly for all J .We now define the A J in the Norton element A J T J . First we let B J = Y g ∈G C ( g ) J (1 − g ω ) ∈ C S. In the spirit of Lemma 2.10, we would like to raise B J to a sufficientlyhigh power so that it is idempotent. However, B J is not an element of themonoid S , so ( B J ) ω may not be well defined. The following lemma andcorollary shows that it actually is. Definition 3.4.
Given x = P w ∈ S c w w ∈ C S , the coefficient of w in x is c w . We say that w is a term of x if the coefficient of w in x is nonzero. Lemma 3.5.
Let b ∈ S and suppose bx ω = b for some x ∈ G with C ( x ) J .If c is a term of bB J , then c > b . roof. Let D = { x ω : x ∈ G , C ( x ) J, bx ω = b } . By assumption D is notempty. Let g , g , . . . , g m be the generators which appear in the definition of B J . Then B J = X i
For every J ∈ L , there exists an integer N such that y ω B NJ = 0 for all y ∈ G with C ( y ) (cid:14) J .Proof. Let N = ℓ + 1, where ℓ is the length of the longest chain of elementsin the poset ( S, ≤ ).Suppose y ω B NJ = 0. Let c N be a term of B NJ . Then c N is a term of c N − B J for some term c N − in y ω B N − J . Since y ω y ω = y ω , Lemma 3.5 impliesthat y ω is not a term of y ω B kJ for any k ≥
1, so that c N − = y ω g ω · · · g ωm forsome m ≥ g i ∈ G with C ( g i ) J . In particular, c N − g ωm = c N − , andso, again by Lemma 3.5, c N > c N − . Repeated application of this argumentproduces a decreasing chain c N > c N − > c N − > · · · > c of elements in S , contradicting the fact that the length of the longest chainof elements in ( S, ≤ ) is ℓ . Corollary 3.7.
For every J ∈ L there exists an N such that B N +1 J = B NJ .Proof. By Lemma 3.6, ( B J − B NJ = 0 for a sufficiently large N since everyelement of B J − αy ω where α ∈ C , y ∈ G and C ( y ) (cid:14) J . Remark 3.8.
Corollary 3.7 is a special property of an R -trivial monoid, andis not true for a general monoid. For instance if an element x of a semigroup S generates a finite cyclic group of order , then (1 − x ) k = 2 k − − k − x , so (1 − x ) k +1 = (1 − x ) k for all k . This now allows us to define A J = B ωJ .11 emma 3.9. Let J ∈ L . Then:1. T J x = T J for all x such that C ( x ) (cid:22) J ;2. y ω A J = 0 for all y such that C ( y ) J and y ∈ G .Proof. Since J = C ( T J ), C ( x ) (cid:22) J implies C ( x ) ⊇ C ( T J ). We also knowthat T J ∈ C ( T J ) because T J is idempotent. So T J ∈ C ( x ), that is, T J x = T J .The second part follows from Lemma 3.6 since A J = B NJ . Remark 3.10.
Although T J and A J are idempotents individually, theirproduct, the Norton element z J , need not be. For example, take the 0-Heckealgebra H (0) corresponding to the symmetric group S . Let J be the subset { , , } of { , , , , } . Then T J = T T T T , A J = (1 − T )(1 − T )(1 − T )and z J is their product. No power of z J is idempotent. Lemma 3.11.
The coefficient of T J in z J = A J T J is 1. All other terms y in z J have C ( y ) ≻ J .Proof. The coefficient of the identity element 1 in A J is 1. Each term of A J T J is of the form aT J for a term a of A J . If a = 1, then C ( a ) (cid:14) J so C ( aT J ) = C ( a ) ∨ C ( T J ) ≻ C ( T J ) = J . Hence the coefficient of T J in A J T J is 1 and all other terms have content greater than J . Lemma 3.12. If J K then z J z K = 0 .Proof. Since J K , there exists a g ∈ G with C ( g ) (cid:22) J but C ( g ) K .Then, using Lemma 3.9 (1) and Lemma 3.9 (2), z J z K = A J T J A K T K = A J ( T J g ω ) A K T K = A J T J ( g ω A K ) T K = 0 . Lemma 3.13.
For all J ∈ L , there exists an N such that (1 − z J ) N z J = 0 .Proof. To simplify the notation, let us temporarily set T = T J , A = A J and z = z J = AT . We first note that for any integer k ≥ − z ) k z = z (1 − z ) k z = AT (1 − AT ) k AT = A ( T (1 − A ) T ) k AT.
We will show that ( T (1 − A ) T ) N A = 0 for N > ℓ , where ℓ is the length ofthe longest chain in the poset ( S, ≤ ).12et us write 1 − A = P a ∈ S c a a where each term has c a = 0 only if a = g ω · · · g ωk with C ( g i ) J for all i . Therefore T (1 − A ) T = X a ∈ S c a T aT = X a ∈ STaT = Ta c a T a + X a ∈ STaT = Ta c a T aT.
Note that c = 0 since 1 is not a term of (1 − A ). If T aT = T a , then we have
T aT · ( T (1 − A ) T ) = T a (1 − A ) T = T a − T aAT = T a since aA = 0 by Lemma 3.9. Thus,( T (1 − A ) T ) N = X a ∈ STa T = Ta c a T a + X a ∈ STa T = Ta c a T a T ( T (1 − A ) T ) N − = X a ∈ STa T = Ta c a T a + X a ∈ STa T = Ta c a T a T ( T (1 − A ) T ) N − . Next, rewrite the second summand above using the same argument: X a ∈ STa T = Ta c a T a T ( T (1 − A ) T ) N − = X a ∈ STa T = Ta c a T a T X a ∈ S c a T a T ! ( T (1 − A ) T ) N − = X a ,a ∈ STa T = Ta c a c a T a T a T ( T (1 − A ) T ) N − = X Ta T = Ta Ta Ta T = Ta Ta c a c a T a T a + X Ta T = Ta Ta Ta T = Ta Ta c a c a T a T a T ( T (1 − A ) T ) N − . T (1 − A ) T ) N in the form( T (1 − A ) T ) N = X c a T a + · · · + X c a · · · c a N T a · · · T a N ! + X Ta ··· TaiT = Ta ··· Tai ≤ i ≤ N c a · · · c a N T a · · · T a N T. By Lemma 3.9, we have a i A = 0 for all terms a i in 1 − A , and so( T (1 − A ) T ) N · A = X Ta ··· TaiT = Ta ··· Tai ≤ i ≤ N c a · · · c a N T a · · · T a N T A. This summation is 0 as it ranges over an empty set: indeed, if it is not empty,we would have an increasing chain of length
N > ℓ , namely
T a < T a T a < T a T a T a < · · · < T a T a · · · T a N , Therefore, ( T (1 − A ) T ) N A = 0. Definition 3.14.
Let J ∈ L . Let P J := X n,m ≥ (1 − z J ) n + m z J = X k ≥ ( k + 1) (1 − z J ) k z J . (In Remark 3.20 we establish a summation-free formula for P J .) Remark 3.15.
Lemma 3.13 shows there are only finitely many terms inthe summation of P J . Therefore P J is a well defined element of C S for each J ∈ L . Remark 3.16.
A monoid S is called J -trivial if SxS = SyS implies x = y for all x, y ∈ S . When S is J -trivial it suffices to define P K = X n ≥ (1 − z K ) n z K . Lemma 3.17.
The coefficient of T J in P J is and all other terms y of P J have C ( y ) ≻ J . roof. If n + m > T J is idempotent, A J T J A J T J (1 − A J T J ) n + m = A J T J A J ( T J − T J A J T J ) n + m . Each term x in ( T J − T J A J T J ) n + m has C ( x ) ≻ J , so no T J appears in z J (1 − z J ) n + m . The coefficient of T J in z J is 1, by Lemma 3.11. Hence T J appears in z J (1 − z J ) with coefficient 1. By Lemma 3.11, since all of theterms y = T J of z J have C ( y ) ≻ J and P J is a polynomial in z J , all otherterms w of P J must have C ( w ) ≻ J . Remark 3.18.
As polynomials in x we have for any nonnegative integer N : x N X n =0 (1 − x ) n = 1 − (1 − x ) N +1 . Proposition 3.19.
For each J ∈ L , the element P J is idempotent.Proof. Let J ∈ L be fixed and let N be such that (1 − z J ) N z J = 0. Let ustemporarily denote z J by z . We can use Lemma 3.18 to rewrite P J as P J = X n,m ≥ z (1 − z ) n + m = N X n =0 N − n X m =0 z (1 − z ) n + m = N X n =0 (1 − z ) n z N − n X m =0 (1 − z ) m ! = N X n =0 (1 − z ) n (cid:0) z − z (1 − z ) N − n +1 (cid:1) = z N X n =0 (1 − z ) n ! − ( N + 1) z (1 − z ) N +1 = 1 − (1 − z ) N +1 − ( N + 1) z (1 − z ) N +1 . This implies that z P J = z since z (1 − z ) N +1 = 0, and so P J = N X n =0 N − n X m =0 (1 − z ) n + m z ! P J = N X n =0 N − n X m =0 (1 − z ) n + m z = P J . Remark 3.20.
As shown in the calculation above, one could define P J as P J = 1 − (1 + ( N + 1) z J )(1 − z J ) N +1 , where N is the length of the longest chain in the monoid, or even N = | S | .For a J -trivial monoid, it suffices to take P J = 1 − (1 − z J ) N +1 .15 emma 3.21. For all
J, K ∈ L , with J K , P J P K = 0 .Proof. Follows from Lemma 3.12 and the fact that P J is a polynomial in z J with no constant term. Definition 3.22.
For each J ∈ L , let e J := P J − X K ≻ J e K ! . Lemma 3.23. T J occurs in e J with coefficient 1. All other terms y of e J have C ( y ) ≻ J . In particular, e J = 0 .Proof. We proceed by induction. If J is maximal, then e J = P J , so thestatement is implied by Lemma 3.17.Now suppose the statement is true for all M ≻ J . Then e J = P J (1 − P M ≻ J e M ). By induction, all terms x of e M have C ( x ) (cid:23) M ≻ J . So terms y from P J e M have C ( y ) (cid:23) M ≻ J . The only other terms are those from P J ,for which the statement was proved in Lemma 3.17. Lemma 3.24. e K P J = 0 for K J .Proof. The proof is by a downward induction on the semi-lattice. If K ismaximal, then e K = P K , so by Lemma 3.21, e K P J = P K P J = 0.Now suppose that for every L ≻ K , e L P J = 0 for L J , and we willshow that e K P J = 0 for K J . We expand e K P J : e K P J = P K − X L ≻ K e L ! P J = P K P J − X L ≻ K P K e L P J . Since K J , we have P K P J = 0 by Lemma 3.21, and e L P J = 0 by induction,since L ≻ K and K J implies L J . Corollary 3.25. e J is idempotent.Proof. We expand e J e J : e J e J = P J − X M ≻ J e M ! P J − X M ≻ J e M ! = P J P J − X M ≻ J e M P J ! − X M ≻ J e M ! (1) = P J − X M ≻ J e M ! (2) = P J − X M ≻ J e M ! = e J , Lemma 3.26. e J e K = 0 for J = K .Proof. The proof is by downward induction on the lattice L . For a maximalelement M ∈ L , e M = P M , so e M e K = P M P K (1 − P e L ) = 0 by Lemma3.21. Now suppose that for all M ≻ J , e M e K = 0 for M = K and we willshow that e J e K = 0 for J = K . We expand e J e K : e J e K = P J (1 − X L ≻ J e L ) e K = P J ( e K − X L ≻ J e L e K ) (1)If K J , then P L ≻ J e L e K = 0 by our induction hypothesis, so P J ( e K − P L ≻ J e L e K ) = P J e K = P J P K (1 − P M ≻ K e M ) = 0 by Lemma 3.21.If K ≻ J , then P L ≻ J e L e K = e K since e K is idempotent and e L e K = 0for L = K by the inductive hypothesis. Therefore e K − P L ≻ J e L e K = 0 andhence the right hand side of (1) is zero. Theorem 3.27.
The set { e J : J ∈ L} is a complete system of primitiveorthogonal idempotents for C S .Proof. From [10], we know that the maximal number of such idempotentsis the cardinality of L . The rest of the claim is just Lemma 3.23, Corollary3.25 and Lemma 3.26. Appendix: Two examples
We illustrate the above constructions on two examples.
Idempotents for the free left regular band on two gen-erators
Let S be the left regular band freely generated by two elements a, b . Then S = { , a, b, ab, ba } . All elements of S are idempotent. Also aba = ab and bab = ba . The lattice L has four elements: ∅ := S, a := Sa, b := Sb and ab := Sab = Sba , where ∅ ≺ a ≺ ab and ∅ ≺ b ≺ ab , but a and b have no relation.We begin by computing the elements P J .17 = ∅ : Neither of the generators satisfies C ( g ) (cid:22) J , so T ∅ = 1 ∈ S . B ∅ = (1 − a )(1 − b ). Also B ∅ = (1 − a )(1 − b )(1 − a )(1 − b ) = (1 − a − b + ab )(1 − a )(1 − b )= (1 − a − b + ab )(1 − b ) = (1 − a − b + ab ) = B ∅ . Therefore A ∅ = B ∅ = 1 − a − b + ab , so z ∅ = 1 − a − b + ab is idempotent and P ∅ = 1 − a − b + ab.J = a : Then C ( a ) (cid:22) a and C ( b ) a , so T a = a and B a = 1 − b = A a since 1 − b is idempotent. Therefore z a = (1 − b ) a = a − ba . z a = a − ab andone can check that z a = z a , so P a = z a (1 + (1 − z a ) + (1 − z a ) + . . . ) = z a = a − ab. One can check that P a is idempotent. J = b : Similarly, P b = b − ba.J = ab : C ( a ) , C ( b ) (cid:22) ab , so T ab = ab and A ab = 1. z ab = ab is idempo-tent, so P ab = ab. We can now compute the idempotents e J . Since ab is maximal, e ab = ab. Since P a e ab = ( a − ab ) ab = ab − ab = 0, e a = P a (1 − e ab ) = P a = a − ab and similarly, e b = b − ba. Finally, note that P ∅ e a = (1 − a − b + ab )( a − ab ) = 0 and similarly P ∅ e b = 0,so that e ∅ = P ∅ (1 − e a − e b − e ab ) = P ∅ − P ∅ e ab = 1 − a − b + ab − ab + ba = 1 − a − b + ba. One can check that { e ∅ , e a , e b , e ab } is a collection of mutually orthogonalidempotents. 18 dempotents of H S (0) As mentioned above, H S (0) has generators T , T , T , T . In this case, thecorresponding lattice L is the lattice of subsets of { , , , } . The monoid H S (0) is actually a J -trivial monoid, so we can use the simplified formulafrom Remark 3.16. We use the shorthand notation T i ··· i k to denote theelement T i · · · T i k .If J = { , , , } , then T J = T ω = T . Also A J = 1, so z J = A J T J = T J . Also, P J = z J , and since J is maximal, e J = P J , so e { , , , } = T . If J = { , , } , then T J = T and A J = 1 − T . Then z J = (1 − T ) T = T − T . One can check that z J = z J , so P J = z J . Also,one can check that P J is orthogonal to e { , , , } . So e J = P J . Therefore e { , , } = T − T . Similarly, e { , , } = − T + T . Now let J = { , , } . Then T J = T and A J = (1 − T ). Letting z J = A J T J , one can check that z J (1 − z J ) = 0, so P J = z J (1 + (1 − z J )).Again P J is orthogonal to e { , , , } , so e J = P J . Therefore e { , , } = − T + T − T + T . Similarly, e { , , } = − T + T − T + T . When J = { , } , T J = T and A J = (1 − T )(1 − T )(1 − T ). Then z J is already idempotent, so P J = z J . One can check that P J is alreadyorthogonal to e { , , , } , e { , , } , e { , , } . Therefore, e { , } = T − T + T − T − T + T . Similarly, 19 { , } = T − T − T + T + T − T . If J = { , } , T J = T T and A J = (1 − T )(1 − T ). One can checkthat z J (1 − z J ) = 0, and P J = z J (1 + 1 − z J ) is idempotent. P J is or-thogonal to e { , , , } and e { , , } , but not orthogonal to e { , , } . So we define e { , } = P { , } (1 − e { , , } ). Then e { , } = − T + T − T + T + T − T − T + T − T + T − T + T + T − T + T − T . Similarly, e { , } = − T + T + T + T − T − T + T − T − T − T + T + T + T − T − T + T . We continue in this way, constructing all of the idempotents for the alge-bra. For the sake of completeness, the other idempotents are: e { , } = − T + T − T + T − T + T + T − T ; e { , } = − T + T + T − T − T + T + T − T − T + T + T − T ; e { } = − T + T − T + T + T − T + T − T + T − T − T + T − T + T − T + T + T − T + T − T + T − T − T + T ; e { } = T − T + T − T − T + T − T + T − T + T + T − T + T − T + T − T − T + T − T + T + T − T − T + T ; e { } = − T + T − T + T − T + T + T − T − T + T − T + T − T + T + T − T + T − T + T − T + T − T − T + T ; e { } = T − T + T − T − T + T − T + T + T − T − T + T + T − T + T − T − T + T − T +20 − T + T + T − T . Finally, e {} is just the signed sum of all elements, with sign determinedby Coxeter length: e {} = P w ( − ℓ ( w ) T w . One can check (ideally not by hand!) that { e J : J ⊆ { , , , }} is acomplete system of orthogonal idempotents. Acknowledgements
The authors are grateful to Tom Denton, Florent Hivert, Anne Schilling,Benjamin Steinberg and Nicolas M. Thi´ery for useful and open mathemati-cal discussions. We first learned of the equivalence between R -trivial monoidsand weakly ordered monoids from Thi´ery after discussions between the au-thors and Denton, Hivert, Schilling, and Thi´ery. The proof presented inSection 2 was outlined by Steinberg.This research was facilitated by computer exploration using the open-source mathematical software Sage [11] and its algebraic combinatorics fea-tures developed by the
Sage-Combinat community [8]. We are especiallygrateful to Nicolas M. Thi´ery and Florent Hivert for sharing their code withus. This work is supported in part by CRC and NSERC. It is the resultof a working session at the Algebraic Combinatorics Seminar at the FieldsInstitute with the active participation of C. Benedetti, A. Bergeron-Brlek,Z. Chen, H. Heglin, D. Mazur and M. Zabrocki.
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