THE WEAK ORDER ON INTEGER POSETS GRE´GORYCHATEL,VINCENTPILAUD,ANDVIVIANEPONS Abstract. Theweakorderonthesymmetricgroupnaturallyextendstoalatticeonallinteger binaryrelations. Wefirstshowthatthesubposetofthisweakorderinducedbyintegerposets defines as well a lattice. We then study the subposets of this weak order induced by specific families of integer posets corresponding to the elements, the intervals, and the faces of the permutahedron,theassociahedron,andsomerecentgeneralizationsofthose. The weak order is the lattice on the symmetric group S(n) defined as the inclusion order of 7 inversions, where an inversion of σ ∈S(n) is a pair of values aσ−1(b). It 1 is a fundamental tool for the study of the symmetric group, in connection to reduced expressions 0 of permutations as products of simple transpositions. It can also be seen as an orientation of the 2 skeletonofthepermutahedron(theconvexhullofallpermutationsofS(n)seenasvectorsinRn). n The weak order naturally extends to all integer binary relations, i.e. binary relations on [n]. a Namely, for any two integer binary relations R,S on [n], we define in this paper J 7 R(cid:52)S ⇐⇒ RInc ⊇SInc and RDec ⊆SDec, 2 where RInc:= {(a,b)∈R|a≤b} and RDec:= {(b,a)∈R|a≤b} respectively denote the increas- ] inganddecreasingsubrelationsofR. Wecallthisordertheweak orderonintegerbinaryrelations, O see Figure 1. The central result of this paper is the following statement, see Figure 5. C Theorem 1. The weak order on the integer posets on [n] is a lattice. . h Ourmotivationforthisresultisthatmanyrelevantcombinatorialobjectscanbeinterpretedby t a specificintegerposets,andthesubposetsoftheweakorderinducedbythesespecificintegerposets m often correspond to classical lattice structures on these combinatorial objects. To illustrate this, [ westudyspecificintegerposetscorrespondingtotheelements,totheintervals,andtothefacesin 1 theclassicalweakorder,theTamariandCambrianlattices[MHPS12,Rea06],thebooleanlattice, v and other related lattices defined in [PP16]. By this systematic approach, we rediscover and shed 5 light on lattice structures studied by G. Chatel and V. Pons on Tamari interval posets [CP15], 9 by G. Chatel and V. Pilaud on Cambrian and Schröder-Cambrian trees [CP14], by D. Krob, 9 M. Latapy, J.-C. Novelli, H.-D. Phan and S. Schwer on pseudo-permutations [KLN+01], and by 7 0 P. Palacios and M. Ronco [PR06] and J.-C. Novelli and J.-Y. Thibon [NT06] on plane trees. . 1 Part 1. The weak order on integer posets 0 7 1.1. The weak order on integer binary relations 1 : v 1.1.1. Integer binary relations. Our main object of focus are binary relations on integers. An i integer(binary)relationofsizenisabinaryrelationon[n]:={1,...,n},thatis,asubsetRof[n]2. X As usual, we write equivalently (u,v)∈R or uRv, and similarly, we write equivalently (u,v)∈/ R ar or u(cid:54)Rv. Recall that a relation R∈[n]2 is called: • reflexive if uRu for all u∈[n], • transitive if uRv and vRw implies uRw for all u,v,w ∈[n], • symmetric if uRv implies vRu for all u,v ∈[n], • antisymmetric if uRv and vRu implies u=v for all u,v ∈[n]. From now on, we only consider reflexive relations. We denote by R(n) (resp. T(n), resp. S(n), resp. A(n)) the collection of all reflexive (resp. reflexive and transitive, resp. reflexive and symmetric, resp.reflexiveandantisymmetric)integerrelationsofsizen. WedenotebyC(n)thesetof integer congruencesofsizen, thatis, reflexivetransitivesymmetricintegerrelations, andbyP(n) VPiwaspartiallysupportedbytheFrenchANRgrantSC3A(15CE40000401). 1 2 GRE´GORYCHATEL,VINCENTPILAUD,ANDVIVIANEPONS thecollectionofintegerposetsofsizen,thatis,reflexivetransitiveantisymmetricintegerrelations. In all these notations, we forget the n when we consider a relation without restriction on its size. A subrelation of R ∈ R(n) is a relation S ∈ R(n) such that S ⊆ R as subsets of [n]2. We say that S coarsens R and R extends S. The extension order defines a graded lattice structure on R(n) whose meet and join are respectively given by intersection and union. The complemen- tation R(cid:55)→{(u,v)|u=v or u(cid:54)Rv} is an antiautomorphism of (R(n),⊆,∩,∪) and makes it an ortho-complemented lattice. NotethatT(n),S(n)andA(n)areallstablebyintersection,whileonlyS(n)isstablebyunion. Inotherwords,(S(n),⊆,∩,∪)isasublatticeof(R(n),⊆,∩,∪),while(T(n),⊆)and(A(n),⊆)are meet-semisublattice of (R(n),⊆,∩) but not sublattices of (R(n),⊆,∩,∪). However, (T(n),⊆) is a lattice. To see it, consider the transitive closure of a relation R∈R(n) defined by Rtc:=(cid:8)(u,w)∈[n]2 (cid:12)(cid:12)∃v1,...,vp ∈[n] such that u=v1Rv2R...Rvp−1Rvp =w(cid:9). ThetransitiveclosureRtc isthecoarsesttransitiverelationcontainingR. Itfollowsthat(T(n),⊆) is a lattice where the meet of R,S ∈ R(n) is given by R∩S and the join of R,S ∈ R(n) is given by (R∪S)tc. Since the transitive closure preserves symmetry, the subposet (C(n),⊆) of integer congruences is a sublattice of (T(n),⊆). 1.1.2. Weak order. From now on, we consider both a relation R and the natural order < on [n] simultaneously. To limit confusions, we try to stick to the following convention throughout the paper. We use couples (u,v) when we do not know whether uv for the natural order. In contrast, we use couples (a,b) and (b,a) when we know that a≤b for the natural order. Let In:= (cid:8)(a,b)∈[n]2 (cid:12)(cid:12)a≤b(cid:9) and Dn:= (cid:8)(b,a)∈[n]2 (cid:12)(cid:12)a≤b(cid:9). Observe that In∪Dn = [n]2 while I ∩D = {(a,a)|a∈[n]}. We say that the relation R ∈ R(n) is increasing (resp. de- n n creasing) when R ⊆ I (resp. R ⊆ D ). We denote by I(n) (resp. D(n)) the collection of all n n increasing (resp. decreasing) relations on [n]. The increasing and decreasing subrelations of an integer relation R∈R(n) are the relations defined by: RInc:=R∩In =(cid:8)(a,b)∈R(cid:12)(cid:12)a≤b(cid:9)∈I(n) and RDec :=R∩Dn =(cid:8)(b,a)∈R(cid:12)(cid:12)a≤b(cid:9)∈D(n). Inourpictures,wealwaysrepresentanintegerrelationR∈R(n)asfollows: wewritethenumbers 1,...,nfromlefttorightandwedrawtheincreasingrelationsofRaboveinblueandthedecreasing relations of R below in red. Although we only consider reflexive relations, we always omit the relations (i,i) in the pictures (as well as in our explicit examples). See e.g. Figure 1. Besides the extension lattice mentioned above in Section 1.1.1, there is another natural poset structure on R(n), whose name will be justified in Section 2.1. Definition2. The weakorderonR(n)istheorderdefinedbyR(cid:52)SifRInc ⊇SInc andRDec ⊆SDec. The weak order on R(3) is illustrated in Figure 1. Observe that the weak order is obtained by combiningtheextensionlatticeonincreasingsubrelationswiththecoarseninglatticeondecreasing subrelations. Inotherwords,R(n)isthesquareofan(cid:0)n(cid:1)-dimensionalbooleanlattice. Itexplains 2 the following statement. Proposition 3. The weak order (R(n),(cid:52)) is a graded lattice whose meet and join are given by R∧ S=(RInc∪SInc)∪(RDec∩SDec) and R∨ S=(RInc∩SInc)∪(RDec∪SDec). R R Proof. Theweakorderisclearlyaposet(antisymmetrycomesfromthefactthatR=RInc∪RDec). ItscoverrelationsarealloftheformR(cid:52)R(cid:114){(a,b)}foraRIncborR(cid:114){(b,a)}(cid:52)RwithbRDeca. Therefore, the weak order is graded by R (cid:55)→ | RDec | − | RInc |. To check that it is a lattice, consider R,S ∈ R(n). Observe first that R∧ S is indeed below both R and S in weak order. R Moreover,ifT(cid:52)RandT(cid:52)S,thenTInc ⊇RInc∪SInc andTDec ⊆RDec∩SDec,sothatT(cid:52)R∧ S. R This proves that R∧ S is indeed the meet of R and S. The proof is similar for the join. (cid:3) R Remark 4. DefinethereverseofarelationR∈RasRrev:= (cid:8)(u,v)∈[n]2 (cid:12)(cid:12)(v,u)∈R(cid:9). Observe that (Rrev)Inc =(RDec)rev and (Rrev)Dec = (RInc)rev. Therefore, the reverse map R (cid:55)→ Rrev defines an antiautomorphism of the weak order (R(n),(cid:52),∧ ,∨ ). Note that it preserves symmetry, R R antisymmetry and transitivity. THE WEAK ORDER ON INTEGER POSETS 3 3 2 1 3 2 1 3 d. 2 te 3 3 t 1 mi 2 2 3 o 1 1 e 2 r a 3 3 1 n] 2 2 [ 1 3 1 ∈ 2 i 3 3 3 3 r 1 o 2 2 2 2 f 1 1 23 1 1 i,i() 3 3 1 ns 2 2 o 1 3 1 ati 2 el 3 3 3 3 r 1 2 2 2 2 ve 1 1 3 1 1 xi 2 fle 3 3 e 1 r 2 3 2 All 1 1 3 3 2 3 3 3. 1 e 2 2 2 2 z 3 si 1 1 1 1 f 2 o 3 3 3 3 s 1 n 2 2 2 2 o 3 ti 1 1 1 1 a 2 el 3 3 3 3 r 1 y 2 2 2 2 r 1 1 3 1 1 na 2 bi 3 1 3 er 2 2 g 1 3 1 nte 2 i 3 3 3 3 ) 1 e 2 2 3 2 2 xiv 1 1 1 1 e 2 efl 3 3 r 1 ( 2 2 n 3 o 1 1 2 er 3 3 3 3 d 1 r 2 2 2 2 o 3 k 1 1 1 1 a 2 we 3 3 1 e 2 2 h 3 T 1 1 2 . 1 3 3 1 e 2 2 r 3 u 1 1 g 2 i F 1 3 2 1 3 2 1 4 GRE´GORYCHATEL,VINCENTPILAUD,ANDVIVIANEPONS 1.2. The weak order on integer posets In this section, we show that the three subposets of the weak order (R(n),(cid:52)) induced by antisymmetric relations, by transitive relations, and by posets are all lattices (although the last two are not sublattices of (R(n),(cid:52),∧ ,∨ )). R R 1.2.1. Antisymmetric relations. We first treat the case of antisymmetric relations. Figure 2 shows the meet and join of two antisymmetric relations, and illustrates the following statement. Proposition 5. The meet ∧ and the join ∨ both preserve antisymmetry. Thus, the antisym- R R metric relations induce a sublattice (A(n),(cid:52),∧ ,∨ ) of the weak order (R(n),(cid:52),∧ ,∨ ). R R R R Proof. Let R,S ∈ A(n). Let a < b ∈ [n] be such that (b,a) ∈ R∧ S. Since (b,a) is decreasing R and (R∧ S)Dec = RDec ∩SDec, we have bRDec a and bSDec a. By antisymmetry of R and S, R we obtain that a(cid:54)RInc b and a(cid:54)SInc b. Therefore, (a,b) ∈/ RInc ∪SInc = (R∧ S)Inc. We conclude R that (b,a) ∈ R∧ S implies (a,b) ∈/ R∧ S and thus that R∧ S is antisymetric. The proof is R R R identical for ∨ . (cid:3) R R∈A(4) S∈A(4) R∧ S∈A(4) R∨ S∈A(4) R R 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 Figure 2. The meet R∧ S and join R∨ S of two antisymmetric relations R,S. R R Our next two statements describe all cover relations in (A(n),(cid:52)). Proposition 6. All cover relations in (A(n),(cid:52)) are cover relations in (R(n),(cid:52)). In particu- lar, (A(n),(cid:52)) is still graded by R(cid:55)→|RDec|−|RInc|. Proof. Consider a cover relation R(cid:52)S in (A(n),(cid:52)). We have RInc ⊇SInc and RDec ⊆SDec where at least one of the inclusions is strict. Suppose first that RInc (cid:54)= SInc. Let (a,b)∈RInc(cid:114)SInc andT:= R(cid:114){(a,b)}. NotethatTisstillantisymmetricasitisobtainedbyremovinganarcfrom an antisymmetric relation. Moreover, we have R (cid:54)= T and R (cid:52) T (cid:52) S. Since S covers R, this implies that S=T=R(cid:114){(a,b)}. We prove similarly that if RDec (cid:54)=SDec, there exists aa,we haveaRiRbanda

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