A correction to the paper `A new approach to the representation theory of the symmetric groups, III'

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by  Minwon Na

A BIJECTIVE PROOF OF VERSHIK’S RELATIONS FOR THE
KOSTKA NUMBERS
MINWON NA
6
1
0
2
Abstract. Wegiveabijective proofofVershik’srelationsforthe Kostkanumbers
n
using insertion and reverse insertion algorithms.
u
J

8
1. Introduction

O] Throughout this paper, n will denote a positive integer. We write λ (cid:15) n if λ is a
C composition of n, that is, a sequence λ = (λ1,λ2,...,λh) of nonnegative integers such
. that |λ| = h λ = n. In particular, if a sequence λ is non-increasing and λ > 0
h i=1 i i
t for all 1 ≤ i ≤ h, then we write λ ⊢ n and say that λ is a partition of n.
ma Let µ ⊢ nPand λ (cid:15) n. We denote by h(λ) the height of λ, and by Dµ the Young
diagram of µ. For example, let µ = (4,3,1) ⊢ 8. Then the Young diagram D is
[ µ

2
v
5
8
3
0 The rows and the columns are numbered from top to bottom and from left to right,
0 like the rows and the columns of a matrix, respectively. A semistandard Young
.
1 tableau (SSYT) of shape µ and weight, or content, λ is a filling of the Young diagram
0
D with the numbers 1,2,...,h(λ) in such a way that
6 µ
1 (i) i occupies λ boxes, for i = 1,2,...,h(λ),
: i
v (ii) the numbers are strictly increasing down the columns and weakly increasing
i
X along the rows.
r We denote by SSYT(µ,λ) the set of all semistandard tableaux of shape µ and weight
a
λ. In particular, if weight λ = (1n), then such a tableau is called a standard Young
tableau(SYT)ofshapeµ. TheKostkanumber K(µ,λ)isdefinedtobethecardinality
of SSYT(µ,λ).
We denote by λ˜ the partition obtained by rearranging components of a composition
λ, andbyλ(i) thecompositionofn−1 definedbyλ(i) = λ −1, andλ(i) = λ otherwise.
i i j j
For λ = (λ ,...,λ ) ⊢ n and γ ⊢ n − 1, we write γ (cid:22) λ if γ ≤ λ for all i with
1 h i i
1 ≤ i ≤ h, and define
C(λ,γ) = |{i | 1 ≤ i ≤ h, λ(i) = γ}|.
Vershik’s relations for the Kostka numbers is as follows:
f
Date: June 08, 2016.
2010 Mathematics Subject Classification. 05A19, 05E10,20C30.
Key words and phrases. insertion tableau, RSK correspodence, Vershik’s relation, Kostka num-
ber, symmetric group.
1
2 MINWONNA
Theorem 1 ([1, p.143, Theorem 3.6.13] and [5, Theorem 4]). For any λ ⊢ n and
ρ ⊢ n−1, we have
K(µ,λ) = C(λ,γ)K(ρ,γ).
µ⊢n γ⊢n−1
X X
µ(cid:23)ρ γ(cid:22)λ
Theorem 1 arises from restricting a permutation representation of the symmetric
groupS toS andthen applying Young’s ruletobothsides. Aspreviously stated,
n n−1
since K(µ,λ) = |SSYT(µ,λ)|, it is natural to expect a bijective proof of Theorem 1.
In fact, Vershik [5, Theorem 4] claims to give a bijection from
L = SSYT(µ,λ)
µ⊢n
[
µ(cid:23)ρ
to
R = SSYT(ρ,λ(x)),
1≤x≤h
[
where λ = (λ ,λ ,...,λ ) ⊢ n and ρ ⊢ n − 1. In order to explain his proof, we
1 2 h
call a tableau in SSYT(µ,λ) a µ-tableau, and a tableau in SSYT(ρ,λ(x)) a ρ-tableau.
Since µ-tableaux have one more box than ρ-tableaux, Vershik [5, Theorem 4] claims
that removable of one box from µ-tableaux gives a bijection from L to R. Vershik
[5, Section 4] gives examples, each of which comes with a bijection. However, if
λ = (3,3,2) ⊢ 8 and ρ = (4,3) ⊢ 7 then there is no bijection from L to R arising
from removable of one box. More precisely, we consider two tableaux in L as follows:
1 1 1 3
1 1 1 3 3
A = , E = 2 2 2 .
2 2 2
3
The only ρ-tableau obtainable from A by removing one box is
1 1 1 3
Q = .
2 2 2
Similarly, the only ρ-tableau obtainable from E by removing one box is Q.
Let [h] = {1,2,...,h}, and let SSYT (µ) be the set of all SSYT’s of shape µ and
[h]
taking values in [h]. For a partition ρ ⊢ n − 1, Loehr [3, p.399, 10.60] shows that
insertion I and reverse insertion R give mutually inverse bijections
I : SSYT (ρ)×[h] → SSYT (µ),
[h] [h]
µ⊢n
[
µ(cid:23)ρ
R : SSYT (µ) → SSYT (ρ)×[h].
[h] [h]
µ⊢n
[
µ(cid:23)ρ
given by I(T,x) = T ← x and R(S) is the result of applying reverse insertion to S
starting attheunique boxofS notinρ. Inthispaper, wedescribe a bijectionbetween
R and L using the restriction I|R′ and R|L, where R′ = 1≤x≤h(SSYT(ρ,λ(x))×{x}).
This paper is organized as follows. After giving preliminaries in Section 2, we prove
S
Theorem 1inSection3. InSection 4, another bijective proofofVershik’s relationscan
A BIJECTIVE PROOF OF VERSHIK’S RELATIONS FOR THE KOSTKA NUMBERS 3
be given using the Robinson-Schensted-Knuth correspondence. Finally, in section 5,
we give two examples and compare Vershik’s bijection with ours.
2. Prelimilaries
Throughout this paper, let h ≥ 1, x ≥ 1 and n ≥ 1 be integers. Let µ ⊢ n and
λ = (λ ,λ ,...,λ ) (cid:15) n, and T ∈ SSYT(µ,λ). First of all, we need a fundamental
1 2 h
combinatorialalgorithmontableauxcalledrow-insertion, orbumping (see[2, Chapter
1]). We define a insertion tableau, denoted T ← x, by the following procedure.
Algorithm 1. Input: Let T ∈ SSYT(µ,λ) and x be a positive integer.
Output: T ← x
Initialization: S := T, y := x and i := 1
while | {j | y < S(i,j)} |> 0 do
z := min{j | y < S(i,j)}.
x′ := S(i,z).
if (p,q) = (i,z) then U(p,q) := y
else
U(p,q) := S(p,q)
end if
S ← U, y ← x′ and i ← i+1.
end
T ← x(i,µ +1) := y
i
Otherwise, T ← x(p,q) := S(p,q)
Output T ← x.
For example, let µ = (4,3,2,1,1) ⊢ 11 and λ = (1,3,2,2,1,2)(cid:15) 11. Suppose x = 2
and
1 2 2 3
2 4 4
T = 3 6 .
5
6
The x = 2 bumps the 3 from the first row, which then bumps the first 4 from the
second row, which bumps the 6 from the third row, which can be put at the end of
the fourth row. Then
1 2 2 2
2 3 4
T ← x = 3 4 .
5 6
6
It is clear that T ← x is a SSYT of shape ν = (4,3,2,2,1) ⊢ 12 and weight
(1,4,2,2,1,2) (cid:15) 12, where µ (cid:22) ν.
This algorithm is invertible. Given a partition µ ⊢ n and insertion tableau T ← x,
there is the unique box of T ← x not in D (In above example, it is {(4,2)}). From
µ
this box, we can construct the reverse insertion algorithm, so we can recover the
original tableau T.
4 MINWONNA
3. Vershik’s relations for the Kostka numbers
Theorem 2. Let ρ ⊢ n−1 and λ = (λ ,...,λ ) (cid:15) n and, set
1 h
R′ = (SSYT(ρ,λ(x))×{x}),
1≤x≤h
[
L = SSYT(µ,λ).
µ⊢n
[
µ(cid:23)ρ
Then I|R′ and R|L give mutually inverse bijections between R′ and L.
Proof. It is obvious that
R′ ⊂ SSYT (ρ)×[h],
[h]
L ⊂ SSYT (µ).
[h]
µ⊢n
[
µ(cid:23)ρ
For each x ∈ [h], we have
I(SSYT(ρ,λ(x))×{x}) ⊂ L
by the definition of insertion. This implies I(R′) ⊂ L. Conversely, for each µ ⊢ n
with µ (cid:23) ρ, there exists ℓ such that D = D ∪{(ℓ,µ )}. Applying reverse insertion
µ ρ ℓ
at (ℓ,µ ) for each tableau in SSYT(µ,λ), we find R(SSYT(µ,λ)) ⊂ R′. This implies
ℓ
R(L) ⊂ R′. Since I and R are mutually inverse bijections, we have
R′ = RI(R′) ⊂ R(L),
L = IR(L) ⊂ I(R′).
Therefore, I(R′) = L and R(L) = R′. (cid:3)
From Theorem 2, we also construct a bijection from R to L given by T 7→ T ← x.
Corollary 3 (Vershik’s relations for the Kostka numbers). For ρ ⊢ n− 1 and λ =
(λ ,λ ,...,λ ) ⊢ n, we have
1 2 h
K(µ,λ) = C(λ,γ)K(ρ,γ).
µ⊢n γ⊢n−1
X X
µ(cid:23)ρ γ(cid:22)λ
Proof.
K(µ,λ) = |SSYT(µ,λ)|
µ⊢n µ⊢n
X X
µ(cid:23)ρ µ(cid:23)ρ
= |SSYT(ρ,λ(x))| (by Theorem 2)
1≤x≤h
X
= |SSYT(ρ,λ(x))|
γ⊢n−11≤x≤h
Xγ(cid:22)λ λgX(x)=γ
= |SSYT(ρ,γ)| (by [1, Lemma 3.7.1])
γ⊢n−11≤x≤h
Xγ(cid:22)λ λgX(x)=γ
A BIJECTIVE PROOF OF VERSHIK’S RELATIONS FOR THE KOSTKA NUMBERS 5
= |{x | 1 ≤ x ≤ h, λ(x) = γ}||SSYT(ρ,γ)|
γ⊢n−1
X
γ(cid:22)λ
g
= C(λ,γ)K(ρ,γ).
γ⊢n−1
X
γ(cid:22)λ
(cid:3)
4. The Robinson-Schensted-Knuth correspondence
Another bijective proof of Vershik’s relations can be given using the Robinson-
Schensted-Knuth (RSK) correspondence (see [2, Chapter 4] and [4, Section 4.8]).
Definition 4 ([4, Definition 4.8.1]). A generalized permutation is a two-line array
of positive integers
i i ··· i
π = 1 2 m
j j ··· j
1 2 m
(cid:18) (cid:19)
whose columns are in lexicographic order, with the top entry taking precedence. Let
πˆ and πˇ stand for the top and bottom rows of π, respectively. We denote by contπˆ
and contπˇ the composition, where the i-th component equals the number of i’s in πˆ
and πˇ, respectively.
For example, let
1 1 1 2 2 3
π =
2 3 3 1 2 1
(cid:18) (cid:19)
be a generalized permutation, contπˆ = (3,2,1) and contπˇ = (2,2,2).
For ν (cid:15) n and λ (cid:15) n, let GP(ν,λ) be the set of all generalized permutations π such
that contπˆ = ν and contπˇ = λ. The RSK correspondence is a bijection
RSK : GP(ν,λ) → SSYT(µ,λ)×SSYT(µ,ν)
µ⊢n
[
given by RSK(π) = (P(π),Q(π)), where P(π) and Q(π) is the insertion tableau and
the recoding tableau for π, respectively. We consider a bijection
Φ : GP((1n),λ) → GP((1n−1),λ(x))
1≤x≤h(λ)
[
given by
1 2 ··· n−1 n 1 2 ··· n−1
Φ( ) = .
j j ··· j x j j ··· j
1 2 n−1 1 2 n−1
(cid:18) (cid:19) (cid:18) (cid:19)
We define Ψ by the following diagram:
GP((1n),λ) RSK // SSYT(µ,λ)×SYT(µ)
µ⊢n
OO
Φ S Ψ
(cid:15)(cid:15)
GP((1n−1),λ(x)) RSK// SSYT(ρ,λ(x))×SYT(ρ)
1≤x≤h(λ) 1≤x≤h(λ) ρ⊢n−1
S S S
6 MINWONNA
For µ ⊢ n and ρ ⊢ n−1 with ρ (cid:22) µ, we let T(µ,ρ) be the set of all T ∈ SYT(µ)
such that
n if D \D = {(i,j)},
µ ρ
T(i,j) =
T′(i,j) otherwise
(
for some T′ ∈ SYT(ρ).
Let ρ ⊢ n−1 and λ (cid:15) n and, set
R′′ = SSYT(ρ,λ(x))×SYT(ρ),
1≤x≤h(λ)
[
L′ = SSYT(µ,λ)×T(µ,ρ).
µ⊢n
[
µ(cid:23)ρ
Now, we claim that Ψ maps R′′ to L′. Indeed, if (P′,Q′) ∈ R′′, then
1 2 ··· n−1 1 2 ··· n−1 n
(P′,Q′) 7→ 7→ 7→ (P,Q),
j j ··· j j j ··· j x
1 2 n−1 1 2 n−1
(cid:18) (cid:19) (cid:18) (cid:19)
where P = P′ ← x and Q(i,j) = n if D \ D = {(i,j)}, and Q(i,j) = Q′(i,j)
µ ρ
otherwise. Thus Ψ(R′′) ⊂ L′. Conversely, if (S,T) ∈ L′, from the definition of
T (µ,ρ), there is a box {(i,j)} = D \ D , so there exists (P′,Q′) ∈ R′′ such that
µ ρ
Ψ((P′,Q′)) = (S,T). This proves Ψ(R′′) = L′. Since
|R′′| = K(ρ,λ(x))·|SYT(ρ)|,
1≤x≤h(λ)
X
|L′| = K(µ,λ)·|T(µ,ρ)|,
µ⊢n
X
µ(cid:23)ρ
and |T(µ,ρ)| = |SYT(ρ)|, we obtain another bijective proof of Corollary 3.
5. Examples
In this section, we compare Vershik’s bijection with ours.
Example 5 ([5, Example 1]). Let λ = (3,2,1) ⊢ 6 and ρ = (4,1) ⊢ 5. Then
1 1 1 2 2 1 1 1 2 3 1 1 1 2
µ-tableaux : A = , B = , C = ,
3 2 2 3
1 1 1 2
1 1 1 3
D = , E = 2 ;
2 2
3
1 1 2 2 1 1 2 3 1 1 1 2
ρ-tableaux : L = , M = , N = ,
3 2 3
1 1 1 3 1 1 1 2
P = , Q = .
2 2
We remove one box from the first row in A and B, one box from the second row in
C and D, and one box (3,1) in E in order to obtain ρ-tableaux. Then we have a
bijection as follows:
A ↔ L; B ↔ M; C ↔ N; D ↔ P; E ↔ Q.
A BIJECTIVE PROOF OF VERSHIK’S RELATIONS FOR THE KOSTKA NUMBERS 7
The bijection given by Theorem 2 is:
L ↔ L ← 1 = E; M ↔ M ← 1 = D;
N ↔ N ← 2 = A; P ↔ P ← 2 = C;
Q ↔ Q ← 3 = B.
We give an example, for which there is no bijection arising from removable of one
box.
Example 6. Let λ = (3,3,2) ⊢ 8 and ρ = (4,3) ⊢ 7. Then
1 1 1 3 3 1 1 1 2 3 1 1 1 2 2
µ-tableaux : A = , B = , C = ,
2 2 2 2 2 3 2 3 3
1 1 1 3 1 1 1 2
1 1 1 2
D = , E = 2 2 2 , F = 2 2 3 ;
2 2 3 3
3 3
1 1 2 3 1 1 2 2 1 1 1 3
ρ-tableaux : L = , M = , N = ,
2 2 3 2 3 3 2 2 3
1 1 1 2 1 1 1 3 1 1 1 2
P = , Q = , R = .
2 3 3 2 2 2 2 2 3
As mentioned in Section 1, µ-tableaux A and E result in ρ-tableau Q, so there is no
bijection between µ-tableaux and ρ-tableaux arising from removable of one box. The
bijection given by Theorem 2 is:
L ↔ L ← 1 = E; M ↔ M ← 1 = F;
N ↔ N ← 2 = D; P ↔ P ← 2 = C;
Q ↔ Q ← 3 = A; R ↔ R ← 3 = B.
References
[1] T. Ceccherini-Silverstein, F. Scarabotti and F. Tolli, Representation Theory of the Symmetric
Groups, Cambridge University Press, 2010.
[2] W. Fulton, Young Tableaux, London Mathematical Society Student Texts 35, 1997.
[3] Nicholas A. Loehr, Bijective Combinatorics, Chapman and Hall/CRC Press, 2011.
[4] B. Sagan, The Symmetric Group, Springer, 2001.
[5] A. M. Vershik, A new approach to the representation theory of the symmetric groups, III:
Induced representations and Frobenius-Young correspondence, Mosc. Math. J. 6 (2006), no. 3,
567–585.
Research Center for Pure and Applied Mathematics, Graduate School of Infor-
mation Sciences, Tohoku University, Sendai 980–8579, Japan
E-mail address: [email protected]