Inner tableau translation property of the weak order and related results

0.19 MB·English

by  Muge Taskin

INNER TABLEAU TRANSLATION PROPERTY OF THE WEAK ORDER AND
RELATED RESULTS
MU¨GETAS¸KIN
1
1 Abstract. LetSYTn bethesetofallstandardYoungtableauxwithncellsand≤weak beMelnikov’sthe
0 weak order on SYTn. The aim of this paper is to introduce a conjecture on the weak order, named the
2
propertyofinnertableautranslation,anddiscussitssignificance. Wewillalsoprovetheconjectureforsome

n specialcases.
a
J

1
3
1. Introduction

] The weakorderfirstintroducedby Melnikovandwellstudied in[12, 13,14]due its strongconnections to
O
Kazhdan-Lusztig and geometric order on standard Young tableaux, where the latter are induced from the
C
representationtheoryofspeciallinearalgebraandsymmetric group. We havethe followinginclusionamong
.
h all of these three orders:
t
a weak order ( Kazhdan-Lusztig (KL) order ⊆ geometric order .
m
The fact that its definition uses just the combinatorics of tableaux such as Knuth relations and the weak
[
orderon symmetric group, givesthe weak orderan importantplace among these orders. On the other hand

3 the only justification for its well-definedness is induced from above inclusion, in other words there is no self
v
contained proof of this basic fact.
0
Our aim here is to bring the attention to the following conjecture on the weak order (which is first asked
9
in[21]), calledthe propertyofinner tableau translation. This propertyisknownto be satisfiedbyKazhdan-
8
4 Lusztig and geometric orders and its importance on the weak order relies on the fact that it provides a self
. contained proof for the well-definedness of this order.
9
0
Conjecture 1.1. Given two tableau S < T having the same inner tableau R, replacing R with another
0 weak
same shape tableau R′ in S and T still preserves the weak order.
1
:
v In the following we first provide the definitions and related background for the weak order. In the third
i section, by assuming the conjecture we will provide a self contained the proof for the well definedness the
X
weak orderand we close this section with the discussionon how this conjecture plays specific role in studies
r
a of Poirier and Reutenauer Hopf algebra on standard Young tableaux. In the last section, we prove the
conjecture for the case when the inner tableau R has hook shape or a shape which consists of two rows or
two columns.
2. Related background
2.1. Definition of the weak order. ThedefinitionoftheweakorderuseswellknownRobinson-Schensted
(RSK)correspondencewhichbijectivelyassignstoeverypermutationw ∈S apairofsameshapetableaux
n
(I(v),R(w)) ∈ SYT ×SYT , where I(w) and R(w) are called the insertion and recording tableau of w
n n
respectively. Onthe otherhandanequivalence relation∼onS due toKnuth[8]playsa crucialroleinthis
n
K
correspondence. Namely:
u ∼ w ⇐⇒ I(u)=I(w).
K
We will denote the corresponding equivalence classes in S by {Y } .
n T T∈SYTn
Thisresearchformsapartoftheauthor’sdoctoralthesisattheUniv. ofMinnesota,underthesupervisionofVictorReiner,
andpartiallysupportedbyNSFgrantDMS-9877047.
1
2 MU¨GETAS¸KIN
Let us explain these algorithms briefly. Denote by (I ,R ) the same shape tableaux obtained by
i−1 i−1
insertion and recording algorithms on the first i−1 indices of w = w ...w . In order to get I , if w is
1 n i i
greater then the last number on the first row of I , it is concatenated to the right side of the first row of
i−1
I , otherwise, w replaces the smallest number, say a among all numbers in the first row greater then w
i−1 i i
and this time insertion algorithm is applied to a on the next row. Observe that after finitely many steps
the insertion algorithm terminates with a new added cell. The resulting tableau is then I and recording
i
tableau R is found by filling this new cell in R with the number i. We illustrate these algorithms with
i i−1
the following example
Example 2.1. Let w=52413. Then,
1 4 1 3
2 2 4
I =5 ⇒I = ⇒I = ⇒I = 2 ⇒I = 2 4 =I(w)
1 2 5 3 5 4 5
5 5
1 3 1 3
1 1 3
R =1 ⇒R = ⇒R = ⇒R = 2 ⇒R = 2 5 =R(w)
1 2 2 3 2 4 5
4 4
K
Definition 2.2. We say u,w∈S differ by one Knuth relation, written u∼=w, if
n
either w =x ...yxz...x and u=x ...yzx...x
1 n 1 n
or w =x ...xzy...x and u=x ...zxy...x
1 n 1 n
K
for some x<y <z. Two permutations are called Knuth equivalent, written u∼=w, if there is a sequence of
permutations such that
K K
u=u ∼=u ...∼=u =w.
1 2 k
Schu¨tzenberger’s jeu de taquin slides [20] are one of the combinatorial operations on tableaux that we
apply often in the following sextions.
Definition 2.3. Let λ=(λ ,λ ,...,λ ) and µ=(µ ,µ ,...,µ ) be two Ferrers diagrams such that µ⊂λ.
1 2 k 1 2 l
Then the corresponding skew diagram is defined to be the set of cells
λ/µ={c:c∈λ,c∈/ µ}.
Askewdiagramiscallednormalifµ=∅. Apartial skew tableauofshapeλ/µisanarrayofdistinctintegers
elements whose rows and columns increase. A standard skew tableau of shape λ/µ is partial skew tableau
whose elements are {1,2,...,n}.
We next illustrate the forward and backward slides of Schu¨tzenberger’s jeu de taquin [20] without the
definition.
4 2 4
Example 2.4. Let P = 2 5 and Q = 3 5 . Below we illustrate a forward and backward slide
1 3 1
on P and Q through the cells indicated by dots.
• 4 2 4 2 4 2 4
2 5 → • 5 → 3 5 = 3 5
1 3 1 3 1 • 1
4
2 4 2 4 • 4
2 5
3 5 → • 5 → 2 5 =
1 3
1 • 1 3 1 3
THE WEAK ORDER 3
Theothermainingredientoftheweakorderisthe(right)weakBruhatorder,≤ ,onS whichobtained
weak n
by taking the transitive closure of the following relation:
u≤ w if w=u·s and length(w)=length(u)+1
weak i
where s denotes the adjacent transposition (i,i+1) and length(w) measures the size of a reduced word of
i
w. The weak order has analternative characterization[3, Prop. 3.1] in terms of (left) inversion sets namely
u≤ w if and only if Inv (u)⊂Inv (w)
weak L L
where Inv (u):={(i,j):1≤i<j ≤n and u−1(i)>u−1(j)}.
L
Definition 2.5. The weak order (SYT ,≤ ), first introduced by Melnikov [12] under the name induced
n weak
Duflo order, is the partial order induced by taking transitive closure of the following relation:
S ≤ T if there exist σ ∈Y , τ ∈Y such that σ ≤ τ.
weak S T weak
The necessity of taking the transitive closure in the definition of the weak order is illustrated by the
following example (cf. Melnikov [12, Example 4.3.1]).
1 4 5 1 4
1 2 5
Example 2.6. Let R= , S = 2 , T = 2 5 with
3 4
3 3
Y ={31425,34125,31452,34152,34512},
R
Y ={32145,32415,32451,34215,34251,34521},
S
Y ={32154,32514,35214,32541,35241}.
T
Here R < S since 34125< 34215, and S < T since 32145< 32154. Therefore R < T.
weak weak weak weak weak
On the other hand, for every ρ∈Y and for every τ ∈Y we have (2,4)∈Inv (ρ) but (2,4)∈/ Inv (τ).
R T L L
2.2. Some basic properties of the weak order. For u ∈ S and 1 ≤ i < j ≤ n, let u be the word
n [i,j]
obtained by restricting u to the segments [i,j] and std(u ) ∈ S be the permutation obtained from
[i,j] j−i+1
u by subtracting i−1 from each letter.
[i,j]
Similarly for S ∈SYT and 1≤i<j ≤n, let S be the normal shape tableau obtained by restricting
n [i,j]
S to the segments [i,j] and by applying Schu¨tzenberger’s back word jeu-de-taquin slides. Then std(S )∈
[i,j]
SYT be the tableau obtained from S by subtracting i−1 from each letter.
j−i+1 [i,j]
In fact Inv (u)⊂Inv (w) gives Inv (u )⊂Inv (w ) for all 1≤i<j ≤n and hence
L L L [i,j] L [i,j]
(2.1) u≤ w implies std(u )≤ std(w ) for all 1≤i<j ≤n.
weak [i,j] weak [i,j]
The following basic fact about RSK, Knuth equivalence, and jeu-de-taquin are essentially due to Knuth
and Schu¨tzenberger; see Knuth [7, Section 5.1.4] for detailed explanations.
Lemma 2.7. Given u∈S , let I(u) be the insertion tableau of u. Then for 1≤i<j ≤n,
n
std(I(u) )=I(std(u )).
[i,j] [i,j]
Therefore we have following:
Lemma 2.8. The weak order restricts to segments, i.e.,
S ≤T implies std(S )≤std(T ) for all 1≤i<j ≤n.
[i,j] [i,j]
Remark 2.9. Melnikov shows in [12, Page 45] that the geometric order also restricts to segments. On
the other hand the same fact about Kazhdan-Lusztig order was first shown by Barbash and Vogan [1] for
arbitrary finite Weyl groups (see also work by Lusztig [10]) whereas the generalization to Coxeter groups is
due to Geck [4, Corollary 3.4].
Now recall that (left) descent set of a permutation τ is defined by
Des (τ):={i:1≤i≤n−1 and τ−1(i)>τ−1(i+1)}.
L
4 MU¨GETAS¸KIN
1
2
3
4
1
1 23 3412 213 2314
2 4
312 213 3142 2143
12
123 1423 1324 1234
1234
1
2
3
4
5
23415 24513 23514 34512
24153 25143 34152 23154 35142
124
12435 123345 12534 13425 14523 53
12345 14253 12354 13245 13254
12345 41235 31245 15234
12345
Figure 1. The weak on SYT for n=2,3,4,5.
n
On the other hand the descent set of the standard Young tableau T is described intrinsically by
Des(T):={(i,i+1):1≤i≤n−1 and
i+1 appears in a row below i in T}.
As a consequence of a well-known properties of RSK we have the following basic fact:
Lemma 2.10. For any τ ∈Y we have
T
Des (τ)=Des(T)
L
i.e., the left descent set is constant on Knuth classes.
We let (2[n−1],⊆) be the Boolean algebra of all subsets of [n−1] ordered by inclusion.
Lemma 2.11. Let ≤ be any order on SYT which is stronger than the weak order and restricts to segments.
n
Then the map
(SYT ,≤)7→(2[n−1],⊆)
n
sending any tableau T to its descent set Des(T) is order preserving.
We denote by (Par ,≤op ) the set of all partitions of the number n ordered by the opposite (or dual)
n dom
dominance order, that is, λ≤op µ if
dom
λ +···+λ ≥µ +···+µ for all k.
1 k 1 k
The following can be easily deduced from Greene’s theorem [5].
THE WEAK ORDER 5
Lemma 2.12. S ≤ T implies shape(T)≤ shape(S)
weak dom
Recall that for a standard young tableau T, Tt denotes the transpose of T whereas Tevac denotes the
tableau found by applying the Schu¨tzenberger’s [19] evacuation map on T. For any τ =τ τ ...τ ∈Y we
1 2 n T
have
τt =τ τ ...τ ∈Y
n n−1 1 Tt
τevac =(n+1−τn)(n+1−τn−1)...(n+1−τ1)∈YTevac
Proposition 2.13. Suppose S ≤ T in SYT . Then
weak n
(1) Sevac ≤ Tevac.
weak
(2) Tt ≤ St.
weak
Proof. Let w be the longest element in S . Then the maps
0 n
w 7→w w and w 7→ww
0 0
are clearly anti-automorphisms and hence w 7→ w ww is a automorphism of (S ,≤ ). On the other
0 0 n weak
hand I(ww ) is just the transpose tableau of I(w) [18] whereas I(w ww ) is nothing but the evacuation of
0 0 0
I(w) [19]. (cid:3)
2.3. Inner tableau translation property. The dual Knuth relations ∼ on S plays the main role in the
n
K∗
definition of inner tableau translation property. In its most basic form this relation is defined through the
Knuth relations applied on the inverse of permutations. Namely,
σ ∼ τ in S if and olny if σ−1 ∼ τ−1.
n
K∗ K
Anequivalentdefinitioncanbegivenbytakingthetransitiveclosureofthefollowing: Wesayσ andτ differs
by a single dual Knuth relation determined by the triple {i,i+1,i+2} if
either σ =...i+1...i...i+2... and τ =...i+2...i...i+1...
or σ =...i+1...i+2...i... and τ =...i...i+2...i+1...
Since left descent sets are all equal for the permutations lying in the same Knuth class, the dual Knuth
relation defines an action on the standard Young tableaux. In order to present this action let us give the
following definition.
Definition2.14. ForS ∈SYT letA={(i,j)}beacelllyinginshape(S),whereidenotestherownumber
n
counted from the top and j denotes the column number counted from the left. Then
(S,A,ne):={(k,l) | k <i and l≥j}
(S,A,sw):={(k,l) | k ≥i and l<j}
Suppose that σ ∈Y has i∈Des(σ) but i6∈Des(σ). Therefore σ has one of the following form
S
σ =...i+1...i...i+2... or σ =...i+1...i+2...i....
Now denote by C ,C and C the cells labeled by i,i+1 and i+2 in S, respectively. Then
i i+1 i+2
either C ∈(S,C ,ne)∩(S,C ,sw) or C ∈(S,C ,ne)∩(S,C ,sw)
i i+1 i+2 i+2 i+1 i
andthe actionofa single dualKnuth relationdeterminedby the triple {i,i+1,i+2}onS interchangesthe
places of i+1 and i+2 in the fist case and it interchanges the places of i and i+1 in the second case.
The following theorem (see [17, Proposition 3.8.1]) provides an important characterization of the dual
Knuth relation.
Proposition 2.15. Let S,T ∈SYT . Then S∼T if and only if shape(S)=shape(T).
n
K∗
Definition 2.16. Let {α,β}={i,i+1} and SYT[α,β] :={T ∈SYT |α∈Des(T),β 6∈Des(T)}. Then we
n n
have inner translation map
V :SYT[α,β] 7→SYT[β,α]
[α,β] n n
which send every tableau T ∈ SYT[α,β] to a tableau obtained as a result of the action of the single dual
n
Knuth relation determined by the triple {i,i+1,i+2}.
6 MU¨GETAS¸KIN
The inner translationmapis firstintroduced by Voganin [22] where he alsoshows that Kazhdan-Lusztig
order is preserved under this map. For geometric order this result is due to Melnikov [15, Proposition 6.6].
On the other hand the example given below shows that the weak order does not satisfy this property.
Example 2.17.
1 2 4 1 2 5
1 2 4 1 2 3
≤ 3 6 but 6≤ 3 6
3 5 6 weak 4 5 6 weak
5 4
where the latter pair is obtained from the former by applying a single dual Knuth relation on the triple
{3,4,5}.
A weaker version of the inner translation property can be defined in the following manner:
Definition 2.18. For 1 ≤ k < n and R ∈ SYT let SYTR := {T ∈ SYT | T = R}. Then for
k n n [1,k]
R,R′ ∈SYT , having the same shape, we have inner tableau translation map
k
V[R,R′] :SYTnR 7→SYTnR′
which send every T ∈SYTR to the tableau T′ obtained by replacing R with R′.
n
As a consequence of Proposition2.15, one can generate T′ by a sequence of dual Knuth relations applied
on the subtableau R of T. Therefore if a partial order is preserved under inner translation map then it is
alsopreservedunder inner tableautranslationmap. Hence Kazhdan-Lusztigand geometricorders havethis
property. Onthe otherhanditis stillreasonabletoaskwhether the weakorderis preservedunderthe inner
tableau translation property.
Conjecture 2.19. Let S⋖ T be a covering relation in SYTR and R′ be a tableau obtained by applying
weak n
to R a single dual Knuth relation. Then
V[R,R′](S)⋖weakV[R,R′](T) in SYTnR′.
In other words the weak order on standard Young tableau is preserved under the inner tableau translation
map.
Remark 2.20. Recallthat any tableau R′ ∈SYT with the same shape as R, can be obtained by applying
r
to R a sequence of dual Knuth relation by Proposition2.15. Therefore one can generalize the conjecture for
any tableau R and R′ having the same shape.
As it is stated earlier this conjecture is checked by computer programing up to n=9. In the last section
we also show that for a specific case the conjecture is true.
3. Applications of the conjecture
3.1. Well-definedness of the weak order. By assuming Conjecture 2.19, we first prove the following
result.
Theorem 3.1. The weak order on SYT is well defined.
n
Proof. It is enough to show that if S ≤ T and S 6= T then T 6≤ S. By Lemma 2.11 we know that
weak weak
S ≤ T implies Des(S)⊂ Des(T) and if Des(T)−Des(S)6=∅ then clearly T 6≤ S. Now we suppose
weak weak
that Des(T)=Des(S). Let k be the smallest integer satisfying
S =T .
[1,k] [1,k]
So k < n and S and T differ only by the position of the corner cells labeled by k+1. On the
[1,k+1] [1,k+1]
other hand by Lemma 2.8 have
S (cid:12) T
[1,k+1] weak [1,k+1]
and Lemma 2.12 together with the fact that shape(T )=shape(S ) gives
[1,k] [1,k]
(3.1) shape(T )(cid:12) shape(S ).
[1,k+1] dom [1,k+1]
THE WEAK ORDER 7
Let A = {(i,j)} and B = {(i′,j′)} denote the cells labeled by k+1 in S and T respectively.
[1,k+1] [1,k+1]
Then (3.1) implies that
i<i′ and j >j′
i.e., the corner cell B lies below the corner cell A and therefore there exists a corner cell C = (i′′,j′′) of
S =T which satisfies
[1,k] [1,k]
i≤i′′ <i′ and j >j′ ≥j′′.
Let R denote S =T and let R′ be another tableau in SYT such that shape(R)=shape(R′) and
[1,k] [1,k] k
the corner cell C of R′ is labeled by k. Denote also by S′ and T′ the tableaux obtained by replacing R with
R′ in S and T respectively. Now by Conjecture 2.19 we have S′⋖ T′ and moreover
weak
k ∈Des(T′)−Des(S′).
The last argument shows that T′ 6≤ S′ and therefore by Conjecture 2.19 T 6≤ S. (cid:3)
weak weak
3.2. Poirier-Reutenauer Hopf algebra on ZSYT = ⊕ ZSYT . Following the work of Malvenuto
n≥0 n
and Reutenauer on permutations [11], Poirier-Reutenauer construct two graded Hopf algebra structures on
Z module of all plactic classes {PC } , where PC = u. The product structure of the one
T T∈SYT T PP(u)=T
that concerns us here is given by
(3.2) PCT ∗PCT′ = X shf(u,w)
P(u)=T
P(w)=T′
where w is obtained by increasing the indices of w by the length of u and shf denotes the shuffle product.
Then the bijection sending eachplactic class to its defining tableaugives us a Hopf algebrastructure onthe
Z module of all standard Young tableaux, ZSYT =⊕ ZSYT .
n≥0 n
In [16] Poirier and Reutenauer explain this product using jeu de taquin slides. Following an analogous
resultof Loday andRonco [9, Thm. 4.1]onpermutations, the author shows the following resultin [21]: For
S ∈ SYT , T ∈ SYT where k+l = n, let T the tableau which is obtained by increasing the indices of T
k l
by k. Denote by S/T the tableau whose columns are obtained by concatenating the columns of T over S
below and by S\T the tableau whose rows are obtained by concatenating the rows of T over S from the
right. Then by [21, Thm. 4.2]
S∗T = X R
R∈SYTn:
S\T≤weakR≤weakS/T
Namely the product structure can be read on the weak order poset of standard Young tableaux.
1 2
1 2 1 1 2 4 3
Example 3.2. Let S = and T = . Then S\T = , S/T = . Then
3 2 3 5 4
5
PC ∗ PC =shf(312,54)+shf(132,54)
1 2 1
3 2
=PC +PC +PC +PC .
1 2 4 1 2 4 1 2 1 2
3 5 3 3 4 3
5 5 4
5
On the other hand one can check from Figure 1 that the product S∗T is equal to the sum of all tableaux
in the interval [S\T,S/T].
By using the facts that (S\T)evac =Tevac\Sevac and (S/T)evac =Tevac/Sevac and Proposition2.13, one
can easily deduce the following corallary to Conjecture 2.19.
Corollary 3.3. Let S,S′,T,T′ be standard Young tableaux satisfying
shape(S)=shape(S′) and shape(T)=shape(T′).
8 MU¨GETAS¸KIN
Then the intervals of the weak order [S\T,S/T] and [S′\T′,S′/T′] are isomorhic. Equivalently, the shuffle
product S∗T is determined by the shapes of the tableaux rather than the tableaux itself.
4. The cases where the conjecture holds
Lemma 4.1. Suppose that S ≤ T is a covering relation in SYTR and R,R′ ∈ SYT has the same
weak n k
shape. If std(S[k+1,n])(cid:12)weak std(T[k+1,n]) then V[R,R′](S)⋖weakV[R,R′](T).
Proof. It is enough to consider the case when R and R′ differ by only one dual Knuth relation determined
by the triples {i,i+1,i+2} for some i≤k−2. Since S < T is a covering relation, there exist σ ∈Y
weak S
and τ ∈Y such that for some i<n,
T
σ =a ...a a ...a ≤a ...a a ...a =τ, where a <a
1 j j+1 n 1 j+1 j n j j+1
i.e., σ <τ is also covering relation the right weak order on S . On the other hand by Lemma 2.8 we have
n
I(σ )=S and T =I(τ )
[k+1,n] [k+1,n] [k+1,n] [k+1,n]
Furthermore the assumption S (cid:12) T yields that σ (cid:12)τ and σ =τ .
[k+1,n] weak [k+1,n] [k+1,n] [k+1,n] [1,k] [1,k]
Now applying the dual Knuth relation determined by the triple {i,i+1,i+2} on σ and τ gives two new
permutations say σ′ and τ′ such that σ′⋖τ′ in the right weak Bruhat order and therefore
V[R,R′](S)=I(σ′) ≤weak I(τ′)=V[R,R′](T).
Now if there exist a tableau Q∈SYTnR′ satisfying V[R,R′](S)(cid:12)weak Q(cid:12)weak V[R,R′](T) then we have
S (cid:12)weak V[R′,R](Q)(cid:12)weak T
which is clearly a contradiction. Hence V[R,R′](S)⋖weakV[R,R′](T). (cid:3)
Lemma 4.2. Suppose
R= 1 3 and R′ = 1 2 .
2 3
Then S⋖weakT in SYTnR if and only if V[R,R′](S)⋖weakV[R,R′](T) in SYTnR′.
Proof. Since S < T there exist σ ∈Y and τ ∈Y such that for some 1≤j <n we have
weak S T
σ =a ...a a ...a and τ =a ...a a ...a , where a <a .
1 j j+1 n 1 j+1 j n j j+1
Observe that since S and T have the same inner tableau R, we have {σ ,τ }⊂Y ={213,231}.
[1,3] [1,3] R
If{a ,a }6={1,3}thenapplyingthedualKnuthrelationdeterminedby{1,2,3}onσ andτ yieldstwo
j j+1
permutations σ′ ∈V[R,R′](S) and τ′ ∈V[R,R′](T) which still have σ′⋖τ′ in the right weak order. Therefore
V[R,R′](S)≤weak V[R,R′](T) and it must be a covering relation.
Suppose {a ,a } = {1,3}. Since Y = {213,231}, the insertion taleau I(a ...a ) must have the
j j+1 R 1 j−1
number 2 its first row left most position. Let for some i≤j−1
b ...b 2 b ...b
1 i−1 i+1 j−1
betherowwordofI(a ...a )obtainedbyreadingnumbersineachrowofI(a ...a )fromlefttoright,
1 j−1 1 j−1
starting from last row. Therefore the sequence 2 b ...b labels the first row and moreover
i+1 j−1
b ...b 2 b b ...b 1 3 a ...a ∈Y
1 i−1 i+1 i+2 j−1 j+2 n S
b ...b 2 b b ...b 3 1 a ...a ∈Y .
1 i−1 i+1 i+2 j−1 j+2 n T
On the other hand since 2<b <...<b we have
i+1 j−1
2 b b ...b 1 3 ∼ 2 b 1 3 b ...b
i+1 i+2 j−1 i+1 i+2 j−1
K
2 b b ...b 1 3 ∼ b b b 3 1 2...b
i+1 i+2 j−1 i−1 i+1 i+2 j−1
K
and moreover
b ...b 2 b 1 3 b ...b a ...a ∈Y
1 i−1 i+1 i+2 j−1 j+2 n S
b ...b b 2 1 3 b ...b a ...a ∈Y .
1 i−1 i+1 i+2 j−1 j+2 n T
THE WEAK ORDER 9
On the other hand applying dual Knuth relation determined by {1,2,3} we get
b ...b 3b 12b ...b a ...a ∈Y and b ...b b 312b ...b a ...a ∈Y
1 i−1 i+1 i+2 j−1 j+2 n V[R,R′](S) 1 i−1 i+1 i+2 j−1 j+2 n V[R,R′](T)
which are clearly the generator of V[R,R′](S)⋖weakV[R,R′](T).
Suppose now S < T is a covering relation in SYTR′. Since Rt =R′, by Proposition 2.13 we have
weak n
Tt⋖ St in SYTR.byProposition2.13
weak n
Now by the previous result we have V[R,R′](Tt)⋖weakV[R,R′](St) and therefore
V[R′,R](S)=(V[R,R′](St))t⋖weak(V[R,R′](Tt))t =V[R,R′](T).
(cid:3)
Proposition 4.3. Suppose that S⋖ T in SYTR where R∈SYT has exactly two rows. If R′ is another
weak n k
tableau in SYTk having the same shape with R then V[R,R′](S)⋖weakV[R,R′](T) in SYTnR′.
Proof. Suppose that S ≤ T is a covering relation in SYTR and R ∈ SYT has two rows. When k < 3
weak n k
there is nothing to prove. For k = 3 the only case that needs to be explored is when R has non vertical or
non horizontal shape, hence Lemma 4.2 gives the desired result.
So we suppose the statement is true for k−1 and let R∈SYT . It is enough to consider the case when
k
R and R′ differ by only one dual Knuth relation determined by the triple {i,i+1,i+2} where i+2=k. If
i+2 <k then S = T of R has still two rows and induction gives the desired result. If i+2= k
[1,i+2] [1,i+2]
then we have the following classes of possibilities for the tableau R:
∗ ∗ k−2 k ∗ ∗ k−2k−1 ∗ ∗ ∗ k−2 ∗ ∗ ∗ k−1
(4.1) ∗ ∗ k−1 ∗ ∗ k ∗ k−1 k ∗ k−2 k
(a) (b) (c) (d)
Observethatin the lasttwoclassesthe dualKnuth relationdeterminedby {k−2,k−1,k}interchangesthe
placesof k−1 andk−2 andso they refer to the caseswith smallerinner tableauS =T andthe
[1,k−1] [1,k−1]
induction argument gives the required result.
For the first two classes we have the following analysis: Since S⋖ T there exist σ ∈ Y and τ ∈ Y
weak S T
such that σ <τ is also a covering relation the right weak order on S i.e., for some 1≤j <n we have
n
σ =a ...a a ...a and τ =a ...a a ...a , where a <a .
1 j j+1 n 1 j+1 j n j j+1
If {a ,a }6={k,k−2} then applying the dual Knuthrelationdetermined by {k,k−1,k−2} onσ and
j j+1
τ yieldstwopermutationsσ′ ∈V[R,R′](S)andτ′ ∈V[R,R′](T)whichstillhaveσ′⋖τ′ intherightweakorder.
Therefore V[R,R′](S)≤weak V[R,R′](T) and it must be a covering relation.
Now let {a ,a }={k,k−2}, i.e.,
j j+1
(4.2) σ =a ... a (k−2) k a ...a and τ =a ...a k (k−2) a ...a
1 j−1 j+2 n 1 j−1 j+2 n
Case 1. We firstconsider the case illustratedin (4.1)-(a), where k−1 comes before k in everypermutations
in the Knuth classes of S and T. Therefore the tableau I(a ...a ) must have the number k−1 and it
1 j−1
must be located in the first row, since otherwise the number k drops to the second row of T at the end of
the insertion of τ and that is clearly a contradiction. Let for some r ≤j−1
b ...b (k−1) b b ...b
1 r−1 r+1 r+2 j−1
betherowwordofI(a ...a )obtainedbyreadingnumbersineachrowofI(a ...a )fromlefttoright,
1 j−1 1 j−1
startingfrom lastrow. Therefore (k−1) b b ...b lies on the firstrow andso k−1<b <b <
r+1 r+2 j−1 r+1 r+2
...<b .
j−1
Now it is easy to see that
b ...b (k−1)b b ...b (k−2)k a ...a ∼ b ...b (k−1)b (k−2)k ...b a ...a
1 r−1 r+1 r+2 j−1 j+2 n 1 r−1 r+1 j−1 j+2 n
K
lies in the Knuth class S where as
b ...b (k−1)b b ...b k(k−2)a ...a ∼ b ...b b (k−1)(k−2)k...b a ...a
1 r−1 r+1 r+2 j−1 j+2 n 1 r−1 r+1 j−1 j+2 n
K
10 MU¨GETAS¸KIN
lies in the Knuth class of T. Moreover applying dual Knuth relation determined by {k,k−1,k−2} on the
latter permutations we get
b ...b k b (k−2) (k−1) ...b a ...a ∈Y
1 r−1 r+1 j−1 j+2 n V[R,R′](S)
b ...b b k (k−2) (k−1)...b a ...a ∈Y
1 r−1 r+1 j−1 j+2 n V[R,R′](T)
and therefore V[R,R′](S)≤weak V[R,R′](T).
Case 2. For the case illustrated in (4.1)-(b), We have
Tt ≤ St ∈SYTRt
weak N
where Rt is a tableau of at most two columns having k−1 in its first column and k in its second column.
Therefore we obtained σt ∈Y and τt ∈Y by reversing σ and τ of (4.2) .i.e.,
St Tt
σt =a ...(k−1)...a k (k−2) a ...a and τt =a ...(k−1)...a (k−2) k a ...a
n j+2 j−1 1 n j+1 j−1 1
Now consider the tableau I(a ...(k −1)...a ). Suppose first that the left most cell in its first row is
n j+2
labeledbyanumber,sayx,whichissmallerthenk−1. Thisimpliesthatinsertionofthesequence(k−2)kin
toI(a ...(k−1)...a ) placesthe sequence (k−2)k to the rightofx butthis contradictsto the factthat
n j+2
the inner tableauRt has atmosttwo columns. Thereforewe havex=k−1. Now let for some r≤n−j−1
b ...b (k−1) b ...b
1 r−1 r+1 n−j−1
be the row word of I(a ...(k−1)...a ). Therefore (k−1) b b ...b lies on the first row and
n j+2 r+1 r+2 n−j−1
so k−1<b <b <...<b which yields
r+1 r+2 n−j−1
b ...b (k−1)b ...b k (k−2)a ...a ∼b ...b b (k−1)(k−2)k ...b a ...a
1 r−1 r+1 n−j−1 j−1 1 1 r−1 r+1 n−j−1 j−1 1
K
lies in Y whereas
St
b ...b (k−1)b ...b (k−2)k a ...a ∼b ...b (k−1)b (k−2)k...b a ...a
1 r−1 r+1 n−j−1 j−1 1 1 r−1 r+1 n−j−1 j−1 1
K
lies in Y . Now reversing and then applying dual Knuth relation determined by {k,k−1,k−2} on the
Tt
latter permutations we get
a ...a b ...(k−1)(k−2)k b b ...b ∈Y
1 j−1 n−j−1 r+1 r−1 1 V[R,R′](S)
a ...a b ...(k−1)(k−2)b k b ...b ∈Y
1 j−1 n−j−1 r+1 r−1 1 V[R,R′](T)
and therefore V[R,R′](S)≤weak V[R,R′](T).
Lastly the fact that resulting relations are in fact covering relations follows directly.
(cid:3)
Corollary 4.4. Suppose that S ≤ T is a covering relation in SYTR where R ∈ SYT has exactly two
weak n k
colums. If R′ is another tableau in SYTk having the same shape with R then V[R,R′](S)≤weak V[R,R′](T) is
also a coving relation in SYTR′.
n
Proof. By Proposition 2.13 we have Tt ≤ St in SYTRt where Rt has exactly two rows. Now (R′)t has
weak n
the same shape with Rt and by previous theorem V[Rt,(R′)t](Tt)≤weak V[Rt,(R′)t](St). Therefore
V[R,R′](S)=(V[Rt,R′t](St))t ≤weak (V[Rt,R′t](Tt))t =V[R,R′](T).
(cid:3)
Definition 4.5. For T ∈SYT and A is a corner cell of T, denote by
n
T↑A and η(T↑A)
thetableauobtainedbyapplyingreverseinsertionalgorithmtoT throughthecornercellAandrespectively
the number which leaves the tableau at the end.
The following result on the hook shape tableaux is easy to deduce by using reverse RSK algorithm.