A nice bijection for a content formula for skew semistandard Young tableaux

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1 A nic bijction for a contnt formula for skw smistandard Young tablau Martin Rub Institut für Mathmatik Univrsität Win a10410@unt.univi.ac.at Submittd: Ma ; Accptd: April MR Subjct Classifications: 05E10 Abstract Basd on Schütznbrgr s vacuation and a modification of ju d taquin w giv a bijctiv proof of an idntit conncting th gnrating function of rvrs smistandard Young tablau with boundd ntris with th gnrating function of all smistandard Young tablau. This solvs Ercis b of Richard Stanl s book Enumrativ Combinatorics 2. 1 Introduction Th purpos of this articl is to prsnt a solution for Ercis b of Richard Stanl s book Enumrativ Combinatorics 2 [5]. Thr Stanl askd for a nic bijctiv proof of th idntit ( ) q n(r) = q n(p ) (1 q a+c(ρ) ) (1) R rvrs SSYT of shap λ/µ with R ij a + µ i i P SSYT of shap λ/µ ρ λ/µ whr a is an arbitrar intgr such that a + c(ρ) > 0 for all clls ρ λ/µ. 1 Hr and in th squl w us notation dfind blow: Dfinition 1.1. A partition is a squnc λ =(λ 1 λ 2...λ r )withλ 1 λ 2 λ r > 0 for som r. 1 In fact this is th corrctd vrsion of th idntit originall givn in [5] to b found at Stanl took it from [1] Thorm 3.1 whr th formula is statd incorrctl too. th lctronic journal of combinatorics (2002) #R18 1

2 a. Frrrs diagram b. skw Frrrs diagram c. rvrs SSYT Figur 1. Th Frrrs diagram of a partition λ is an arra of clls with r lft-justifid rows and λ i clls in row i. Figur 1.a shows th Frrrs diagram corrsponding to ( ). W lablthcllinthi th row and j th column of th Frrrs diagram of λ b th pair (i j). Also w writ ρ λ ifρ is a cll of λ. A partition µ =(µ 1 µ 2...µ s )iscontaind in a partition λ =(λ 1 λ 2...λ r ) if s r and µ i λ i for i {1 2...s}. Th skw diagram λ/µ of partitions λ and µ whrµ is containd in λ consists of th clls of th Frrrs diagram of λ which ar not clls of th Frrrs diagram of µ. Figur 1.b shows th skw diagram corrsponding to ( )/(2 2 1). Th contnt c(ρ) of a cll ρ =(i j) ofλ/µ is j i. Givn partitions λ and µ atabloid of shap λ/µ is a filling T of th clls of th skw diagram λ/µ with non-ngativ intgrs. T ρ dnots th ntr of T in cll ρ. Th norm n(t ) of a tabloid T is simpl th sum of all ntris of T.Thcontnt wight w c (T )ofa tabloid T is ρ λ/µ T ρ (a + c(ρ)) whr a is a givn intgr such that a + c(ρ) > 0 for all clls ρ λ/µ. A smistandard Young tablau of shap λ/µ shortssyt isatabloidp such that th ntris ar wakl incrasing along rows and strictl incrasing along columns. A rvrs smistandard Young tablau of shap λ/µ is a tabloid R such that th ntris ar wakl dcrasing along rows and strictl dcrasing along columns. In Figur 1.c a rvrs SSYT of shap ( )/(2 2 1) is shown. 2 A Bijctiv proof of Idntit 1 In fact w will giv a bijctiv proof of th following rwriting of Idntit 1: ( ) q n(p ) = q n(r) 1 1 q a+c(ρ) P SSYT of shap λ/µ = R rvrs SSYT of shap λ/µ with R ij a + µ i i (RT ) R rvrs SSYT of shap λ/µ with R ij a + µ i i T tabloid of shap λ/µ q n(r) q wc(t ). ρ λ/µ th lctronic journal of combinatorics (2002) #R18 2

3 So all w hav to do is to st up a bijction that maps SSYT P onto pairs (R T ) whr R is a rvrs SSYT with R ij a + µ i i and T is an arbitrar tabloid such that n(p )=n(r)+w c (T ). Th bijction consists of two parts. Th first stp is a modification of a mapping known as vacuation which consists of a spcial squnc of so calld ju d taquin slids. An in dpth dscription of ths procdurs can b found for ampl in Bruc Sagan s Book Th smmtric group [4] Sctions 3. and W us vacuation to bijctivl transform th givn SSYT P in a rvrs SSYT Q which has th sam shap and th sam norm as th original on. Th scond stp of our bijction also consists of a squnc of modifid ju d taquin slids and bijctivl maps a rvrs SSYT Q onto a pair (R T ) as dscribd abov. This procdur is vr similar to bijctions discovrd b Christian Krattnthalr proving Stanl s hook-contnt formula. [2 3] n(.) =43 n(.) =43 n(.) =1w c (.) =24 Figur 2. A complt ampl for th bijction can b found in th appndi. Thr w chos a = 6 and map th SSYT P of shap ( )/(2 2 1) on th lft of Figur 2 to th rvrs SSYT Q in th middl of Figur 2 which in turn is mappd to th pair on th right of Figur 2 consisting of a rvrs SSYT R whr th ntr of th cll ρ =(i j) is lss or qual to a + µ i i and a tabloid T so that n(q) =n(r)+w c (T ). In th algorithm dscribd blow w will produc a filling of a skw diagram stp b stp starting with th mpt tablau of th givn shap. Thorm 2.1. Th following two maps dfin a corrspondnc btwn SSYT and rvrs SSYT of th sam shap λ/µ and th sam norm: Givn a SSYT P of shap λ/µ produc a rvrs SSYT Q of th sam shap and thsamnormasfollows: Lt Q b th mpt tablau of shap λ/µ. WHILE thr is a cll of P which contains an ntr Lt b th minimum of all ntris of P. Among all clls τ with P τ = lt ρ =(i j) b th cll which is situatd most right. WHILE ρ has a bottom or right nighbour in P that contains an ntr th lctronic journal of combinatorics (2002) #R18 3

4 Dnot th ntr to th right of ρ b and th ntr blow ρ b. W allow also that thr is onl an ntr to th right or blow ρ and th othr cll is missing or mpt. If < or thr is no ntr blow ρ thn rplac b and lt ρ b th cll (i j +1). Othrwis if or thr is no mpt to th right rplac and lt ρ b th cll (i +1j). b Put Q ρ qual to and dlt th ntr of th cll ρ from P. Not that clls of P which contain an ntr still form a SSYT. In th proof blow ρ will b calld th cll whr th ju d taquin slid stops. Givn a rvrs SSYT Q of shap λ/µ produc a SSYT P of th sam shap and thsamnormasfollows: Lt P b th mpt tablau of shap λ/µ. WHILE thr is a cll of Q which contains an ntr Lt b th maimum of all ntris of Q. Among all clls τ with Q τ = lt ρ =(i j) b th cll which is situatd most lft. St P ρ = and dlt th ntr of th cll ρ from Q. WHILE ρ has a top or lft nighbour in P that contains an ntr Dnot th ntr to th lft of ρ b and th ntr abov ρ b. W allow also that thr is onl an ntr to th lft or abov ρ and th othr cll is missing or mpt. If > or thr is no ntr abov ρ thn rplac b and lt ρ b th cll (i j 1). Othrwis if or thr is no ntr to th lft rplac and lt ρ b th cll (i 1j). b th lctronic journal of combinatorics (2002) #R18 4

5 ρ 1 ρ 2 ρ 0 2 ρ Figur 3. Th clls of P which contain an ntr now form a SSYT. In th proof blow ρ will b calld th cll whr th ju d taquin slid stops. Proof. Not that what happns during th cution of th innr loop of ( ) isaju d taquin forward (backward) slid prformd on Q into th cll ρ s Sction 3. of [4]. Firstwhavtoshowthat is wll dfind. I.. w hav to chck that aftr ach ju d taquin forward slid aftr th ntr in th cll ρ is dltd from P th clls of P which contain an ntr form a SSYT as statd in th algorithm. This follows bcaus aftr ithr tp of rplacmnt in th innr loop th onl possibl violations of incras along rows and strict incras along columns in P can onl involv and th ntris to its right and blow. Whn th ju d taquin forward slid is finishd ρ is a bottom-right cornr of P hnc aftr dlting th ntr in ρ no violations of incras or strict incras can occur. Nt w show that indd producs a rvrs SSYT. In fact w vn show that th tabloid dfind b th clls of Q which hav bn filld alrad is a rvrs SSYT at vr stag of th algorithm. Clarl vr cll of Q is filld with an ntr actl onc. Furthrmor at th tim th cll ρ is filld th clls in Q to th right and to th bottom of ρ ifthist ar filld alrad othrwis ρ would not b a bottom-right cornr of P. Bcaus th squnc of ntris chosn is monotonicall incrasing rows and columns of Q ar dcrasing. So it rmains to show that th columns of Q ar in fact strictl dcrasing. Suppos that ρ 1 and ρ 2 ar clls both containing th sam minimal ntr andρ 1 is right of ρ 2. Whn th ju d taquin forward slid in is prformd into th cll ρ 1 th ntr dscribs a path from ρ 1 to th cll whr th slid stops which w will dnot b ρ 1. Similarl w hav a path from ρ 2 to a cll ρ 2. Now suppos ρ 1 is in th sam column as but blow ρ 2 as dpictd in Figur 3. Clarl in this cas th two paths would hav to cross and w had th following situation: th lctronic journal of combinatorics (2002) #R18 5

6 First (th star is a placholdr for an ntr w do not know) c z would b rplacd b z c. In this situation z would hav to b smallr thn. Thn whn th ju d taquin forward slid into th cll ρ 2 is prformd th following situation would aris at th sam four clls: c z would hav to b rplacd b c z. But this cannot happn bcaus thn would hav to b strictl smallr than z. It can b shown in a vr similar mannr that indd producs a SSYT. W lav th dtails to th radr. Finall w want to prov that is invrs to. Suppos that in a ju d taquin forward slid into th cll ρ containing th ntr is prformd on P. Suppos that th slid stoppd in ρ Q ρ is st to and th ntr in ρ is dltd from P. Among th ntris of Q is maimal bcaus smallst ntris ar chosn first in. Furthrmor among thos clls of Q containing th ntr th cll ρ is most lft. This follows bcaus th tabloid dfind b th clls of Q which hav bn filld alrad is a rvrs SSYT and th paths dfind b th ju d taquin slids cannot cross as w hav shown abov. It is straightforward to chck that in this situation th ju d taquin backward slid into ρ prformd on P in stops in th original cll ρ. B induction w find that is invrs to. Th scond stp of th bijction is just as as: Thorm 2.2. Th following two maps dfin a corrspondnc btwn rvrs SSYT Q to pairs (R T ) whr R is a rvrs SSYT with R ij a + µ i i and T is an arbitrar tabloid so that n(q) =n(r)+w c (T ) Q R and T bing of shap λ/µ: Givn a rvrs SSYT Q of shap λ/µ produc a pair (R T ) as dscribd abov as follows: St R = Q and st all ntris of T qual to 0. WHILE thr is a cll τ =(i j) such that R τ >a+ µ i i Lt b maimal so that thr is a cll τ with R τ ( a + c(τ) ) =. Among all clls τ with R τ ( a + c(τ) ) = ltρ =(i j) b th cll which is situatd most bottom. St R ρ =. WHILE <R (ij+1) or R (i+1j) Dnot th ntr to th right of ρ b and th ntr blow ρ b. W allow also that thr is onl a cll to th right or blow ρ and th othr cll is missing. th lctronic journal of combinatorics (2002) #R18 6

7 If 1 > or thr is no cll blow ρ thn rplac b 1 and lt ρ b th cll (i j +1). Othrwis if +1 or thr is no cll to th right rplac b +1 and lt ρ b th cll (i +1j). Incras T ρ b on. Givn a pair (R T ) as dscribd abov produc a rvrs SSYT Q of shap λ/µ as follows: St Q = R. WHILE thr is a cll τ =(i j) such that T τ 0 Lt b minimal so that thr is a cll τ with Q τ = and T τ 0. Among ths clls τ lt ρ =(i j) b th cll which is situatd most right. Dcras T ρ b on. WHILE + a + c(ρ) >Q (ij 1) or + a + c(ρ) Q (i 1j) Dnot th ntr to th lft of ρ b and th ntr abov ρ b. W allow also that thr is onl a cll to th lft or abov ρ and th othr cll is missing. If >+ 1 or thr is no cll abov ρ thn rplac b +1 and lt ρ b th cll (i j 1). Othrwis if 1 or thr is no cll to th lft rplac b 1 and lt ρ b th cll (i 1j). th lctronic journal of combinatorics (2002) #R18 7

8 Incras Q ρ b a + c(ρ). Rmark. Bcaus of th obvious similarit to ju d taquin slids w will call what happns in th innr loop of ( ) amodifid ju d taquin (backward) slid into ρ prformd on R (Q). Lmma 2.3. Th two maps 2.2. and 2.2. ar wll dfind. I.. th tabloid R producd b is indd a rvrs SSYT with R ij a + µ i i and th tabloid Q producd b is indd a rvrs SSYT. Also th quation n(q) =n(r)+w c (T ) holds. Furthrmor th following statmnt is tru: Suppos that prforms a modifid ju d taquin slid on R into a cll ρ 1 with R ρ1 =. Aftr this suppos that anothr modifid ju d taquin slid on R into a cll ρ 2 with th sam ntr is prformd. Lt ρ 1 and ρ 2 b th clls whr th slids stop. Thn ρ 1 is lft of ρ 2 or ρ 1 = ρ 2. A corrsponding statmnt holds for Algorithm Proof. First of all w hav to prov that Algorithm 2.2. trminats. W rquird that a + c(τ) > 0 for all clls τ which implis that vr tim whn w rplac th ntr in cll ρ b (s th bginning of th outr loop of th algorithm) w dcras ma τ=(ij) (R τ a µ i + i). It is as to s that this maimum is nvr incrasd in th subsqunt stps of th algorithm. It is as to chck that aftr vr tp of rplacmnt within th modifid ju d taquin slids th validit of th quation n(q) =n(r)+w c (T ) is prsrvd. So it rmains to show that aftr vr modifid ju d taquin slid of th rsulting filling R of λ/µ is in fact a rvrs SSYT: W hav that Q τ ( a + c(τ) ) = is maimal at th vr lft of λ/µ bcaus rows ar dcrasing in Q. Thrfor whn Q τ >a+ µ i i as rquird for th cution of th outr loop of whav = Q τ ( a + c(τ) ) >a+ µ i i (a + µ i +1 i) = 1 so is non-ngativ. Furthrmor aftr ithr tp of rplacmnt during th modifid ju d taquin slid th onl possibl violations of dcras along rows or strict dcras along columns can involv onl th ntr and th ntris to th right and blow. B induction R must b a rvrs SSYT. Th scond statmnt of th lmma is shown with an argumnt similar to that usd in th proof of Thorm 2.1. Whn th ju d taquin forward slid in is prformd into th cll ρ 1 th ntr dscribs a path from ρ 1 to th cll ρ 1 whr th slid stops. Similarl w hav a path from ρ 2 to ρ 2. W conclud that if ρ 1 wr strictl to th right of ρ 2 that ths paths would hav to cross. (S Figur 4). Hnc w had th following situation: First (th star is a placholdr for an ntr w do not know) z would b rplacd b 1 z. th lctronic journal of combinatorics (2002) #R18 8

9 ρ 2 ρ 1 ρ ρ 0 2 Figur 4. In this situation would hav to b strictl smallr thn z. Thn whn th modifid ju d taquin slid into ρ 2 is prformd th following situation would aris at th sam four clls: 1 z would hav to b rplacd b But this cannot happn bcaus thn would hav to b at last as big as z is. Th corrsponding statmnt for Algorithm 2.2. is shown similarl. Proof of Thorm 2.2. It rmains to show that and ar invrs to ach othr. This is prtt obvious considring th lmma: Suppos that th pair (R T ) is an intrmdiat rsult obtaind aftr a modifid ju d taquin slid into th cll ρ. AftrthisT ρ is incrasd whr ρ is th cll whr th slid stoppd. Thn th ntr in ρ must b among th smallst ntris of R sothatt ρ 0 bcaus th squnc of s in th clls chosn for th modifid ju d taquin slids is monotonicall dcrasing. If thr is mor than on cll ρ which contains a minimal ntr of R and satisfis T ρ 0 th lmma assrts that th right-most cll was th last cll chosn for th modifid ju d taquin slid. Hnc it is crtain that th right-most cll containing a minimal ntr as slctd bfor th modifid ju d taquin slid of is ρ. It is as to chck that th rplacmnts don in ar actl invrs to thos in. For ampl suppos th following rplacmnt is prformd in : z. z z is rplacd b 1. Thn w had 1 >and bcaus of strictl dcrasing columns z>. Thrfor in this is rvrsd and w nd up with th original situation. Similarl w can show that is invrs to too. th lctronic journal of combinatorics (2002) #R18

10 Appndi A: Stp b stp ampl This appndi contains a complt ampl for th algorithms dscribd abov for a SSYT of shap ( )/(2 2 1) and a =6. First th SSYT P on th lft of Figur 2 is transformd into th rvrs SSYT Q in th middl of Figur 2 using Algorithm Th ampl has to b rad in th following wa: Each pair (P Q) in th tabl dpicts an intrmdiat rsult of th algorithm. Th cll of P containing th ncircld ntr is th cll into which th nt ju d taquin slid is prformd. Th ju d taquin path is indicatd b th dottd lin in Q. P Q P Q Th invrs transformation of th rvrs SSYT Q into th SSYT P can b tracd in th sam tabl w onl hav to start at th right bottom whr th tablau P is mpt and work our wa upwards to th top lft of th tabl. Not that th ju d taquin paths ar th sam. In th scond stp of th bijction this rvrs SSYT Q is mappd onto a pair (R T ) whr R is a rvrs SSYT with R ij a + µ i i T is a tabloid and n(q) =n(r)+w c (T ). First th algorithm initialiss R to Q and sts all ntris of T to zro. Using modifid ju d taquin slids R is thn transformd into a rvrs SSYT whr th ntris ar th lctronic journal of combinatorics (2002) #R18 10

11 a. a + µ i i b. Th tabloid with ntris a + c(ρ) Figur 5. boundd as rquird. First th algorithm chcks whthr thr ar still clls in R which ar too larg. For rfrnc w giv th rlvant bounds in Figur 5.a. Thn for slcting th cll into which th modifid ju d taquin slid is prformd w nd to calculat R ρ (a + c(ρ)). Again for rfrnc w displa ths valus for ach cll in Figur 5.b. Each row of th tabl blow dpicts an intrmdiat rsult of Algorithm Th clls containing th ncircld ntr ar th clls into which th modifid ju d taquin slid will b prformd th clls containing th bod ntr indicat whr th last modifid ju d taquin slid stoppd. In th third column th ju d taquin path for th slctd cll is indicatd. R T ju d taquin path Again th invrs transformation can b tracd in th sam tabl starting at th bottom moving upwards. Now th clls containing th bod ntr ar th clls into th lctronic journal of combinatorics (2002) #R18 11

12 which th nt modifid ju d taquin slid will b prformd th clls containing th ncircld ntr indicat whr th last slid stoppd. Of cours th ju d taquin paths ar th sam as for. Appndi B: A complt matchup for SSYT of shap (3 2)/(1) with norm 5 whra =2 In th tabl blow ou find a complt matchup for SSYT of shap (3 2)/(1) with norm 5 whr a = 2. Th first column contains all SSYT of shap (3 2)/(1) and norm 5. In th scond column th corrsponding rvrs SSYT obtaind b vacuation ar displad. Finall in columns thr and four th rsults of Algorithm 2.2. can b found. This tabl was producd with a Common-LISP-implmntation of th algorithms abov which can b found on th author s hompag. 2 P Q R T th lctronic journal of combinatorics (2002) #R18 12

13 P Q R T Rfrncs [1] Sara C. Bill William Jockusch and Richard P. Stanl Som combinatorial proprtis of Schubrt polnomials Journal of Algbraic Combinatorics 2 (13) no [2] Christian Krattnthalr An involution principl-fr bijctiv proof of Stanl s hookcontnt formula Discrt Mathmatics and Thortical Computr Scinc (18) no [3] Anothr involution principl-fr bijctiv proof of Stanl s hook-contnt formula Journal of Combinatorial Thor Sris A (1) no [4] Bruc E. Sagan Th smmtric group Wadsworth & Brooks/Col Pacific Grov California 187. [5] Richard P. Stanl Enumrativ combinatorics vol. 2 Cambridg Univrsit Prss 1. th lctronic journal of combinatorics (2002) #R18 13

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