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INTRODUCTION TO ALGEBRAIC COMBINATORICS:
(INCOMPLETE) NOTES FROM A COURSE TAUGHT BY
JENNIFER MORSE
GEORGE H. SEELINGER
These are a set of incomplete notes from an introductory class on algebraiccombinatorics I took with Dr. Jennifer Morse in Spring 2018. Especiallyearly on in these notes, I have taken the liberty of skipping a lot of details,since I was mainly focused on understanding symmetric functions when writ-ing. Throughout I have assumed basic knowledge of the group theory of thesymmetric group, ring theory of polynomial rings, and familiarity with settheoretic constructions, such as posets. A reader with a strong grasp onintroductory enumerative combinatorics would probably have few problemsskipping ahead to symmetric functions and referring back to the earlier sec-tions as necessary.
I want to thank Matthew Lancellotti, Mojdeh Tarighat, and Per Alexan-dersson for helpful discussions, comments, and suggestions about these notes.Also, a special thank you to Jennifer Morse for teaching the class on whichthese notes are based and for many fruitful and enlightening conversations.
In these notes, we use French notation for Ferrers diagrams and Youngtableaux, so the Ferrers diagram of (5, 3, 3, 1) is
We also frequently use one-line notation for permutations, so the permuta-tion σ = (4, 3, 5, 2, 1) ∈ S5 has
σ(1) = 4, σ(2) = 3, σ(3) = 5, σ(4) = 2, σ(5) = 1
0. Prelimaries
This section is an introduction to some notions on permutations and par-titions. Most of the arguments are given in brief or not at all. A familiarreader can skip this section and refer back to it as necessary.
Date: May 2018.
1
0.1. Some Statistics.
0.1. Definition. A statistic on permutations is a function stat : Sn → N.
0.2. Definition. Given a permutation σ = (σ1, . . . , σn) in one-line notation,we have the following definitions.
(a) The length of σ, denoted `(σ), is the number of pairs (i, j) such thati < j and σi > σj .
(b) The sign of σ is (−1)`(σ).(c) The inversion statistic on σ is defined by
inv(σ) = `(σ)
(d) A descent of σ is an index i ∈ [n − 1] such that σi > σi+1. The setof all descents on σ is denoted Des(σ).
(e) The major statistic or major index is given by
maj(σ) =∑
i∈Des(σ)
i
0.3. Example. Let σ = (3, 1, 4, 2, 5). Then, it has
• Inversions (1, 2), (1, 4), (3, 4). Thus, inv(σ) = 3 = `(σ).• Descents 1, 3 and thus maj(σ) = 1 + 3 = 4.
0.4. Remark. Some readers may be more familiar with the definition of `(σ)to be the length of the reduced word of σ in terms of simple transpositionsof Sn. These two notions are equivalent. This can be made explicit using“Rothe diagrams.”
0.5. Definition. Any statistic which is equidistributed over Sn with theinversion statistic, inv, is called Mahonian.
0.6. Theorem. The major index, maj, is Mahonian.
0.7. Example. Given that
S3 = (123), (213), (132), (231), (312), (321) in one-line notation
we have the corresponding lengths 0, 1, 1, 2, 2, 3 and corresponding descentsets , 1, 2, 1, 2, 1, 2, giving major indexes 0, 1, 2, 1, 2, 3. Thus,on S3, inv and maj are equidistributed.
To prove that maj is Mahonian in general, we will use the characterizationof Mahonian given below.
0.8. Theorem. A statistic stat is Mahonian if and only if it satisfies theidentity ∑
σ∈Sn
qstat(σ) =∑σ∈Sn
qinv(σ)
2
0.9. Remark. Note that∑σ∈Sn
qinv(σ) = [n]q! := (1)(1 + q)(1 + q + q2) · · · (1 + q + · · ·+ qn−1)
which can be proved using an inductive argument by noting∑σ∈Sn
qinv(σ) =∑
σ∈Sn−1
qinv(σ) +∑σ∈Snσ(n)6=n
qinv(σ) = [n− 1]q! +n−1∑j=1
∑σ∈Snσ(n)=j
qinv(σ)
but we will not need this in what follows.
To prove maj is Mahonian, we will use the following theorem, althoughother bijections exist, such as the Carlitz bijection.
0.10. Theorem (Foata Bijection). There exists a bijection ψ : Sn → Sn
such that maj(σ) = inv(ψ(σ)).
Proof. Let w = w1w2 · · ·wn ∈ Sn in one-line notation. Then, set ψ(w1) =w1. Next, if ψ(w1 · · ·wi−1) = v1 . . . vi−1, then, we define ψ(w1 · · ·wi) by
(a) Add wi on the right.(b) Place vertical bars in w1 · · ·wi following the rules
(i) If wi > vi−1, then place a bar to the right of any vk with wi > vk.(ii) If wi < vi−1, then place a bar to the right of any vk with wi < vk.
(c) Cyclically permute one step to the right in any run between two bars.
0.11. Example. Let σ = 31426875 which has maj(σ) = 17. Then, we iterate
3
|3|1|3|1|43|14|2→ 3412
3|4|1|2|63|4|1|2|6|8|341268|7→ 8341267
|8|34126|7|5→ 86341275 = ψ(σ)
Indeed, inv(ψ(σ)) = 17
To invert the Foata bijection, one does the following
(a) For w = w1 · · ·wi, place a dot between wi−1 and wi.(b) Place vertical bars in w1 . . . wi following the rules
(i) If wi > w1, place a bar to the left of any wk with wk < wi andalso before wi.
(ii) If wi < w1, place a bar to the left of any wk with wk > wi andalso before wi.
(c) Cyclically permute every run between two bars to the left.
3
(d) wi will be the ith entry of the resulting permutation.
0.12. Example. For our example, if we start with 86341275, we carry out
|8|63412|7.5→ 8341267 5
|834126.7→ 341268 7
|3|4|1|2|6.8→ 34126 8
|3|4|1|2.6→ 3412 6
|3|41.2→ 314 2
|3|1.4→ 31 4
3.1→ 3 1
.3 3
and thus we recover σ = 31426875.
0.13. Proposition. The Foata bijection preserves the inverse descent set ofσ.
0.14. Definition. The charge statistic on a permutation σ is given by
(a) Write the permutation in one-line notation counterclockwise arounda circle with a ? between the beginning and the end.
?
σn
σn−1. . .
σ2
σ1
(b) To each entry σk, assign an “index” Ik recursively as follows
Ik =
Ik−1 + 1 if ? is passed going from σk−1 to σk in a clockwise path around the circle.
Ik−1 otherwise.
(c) Define charge(σ) =∑n
k=1 Ik, that is, the sum of the indices.
0.15. Example. The permutation (6, 7, 1, 4, 2, 3, 5) has
?6
7
14
2
3
53
4
02
1
2
3
charge 3 + 4 + 0 + 2 + 1 + 2 + 3 = 15.
4
0.16. Definition. We define the cocharge of a permutation σ = (σ1, . . . , σn)to be
cocharge(σ) :=
(n
2
)− charge(σ)
We also definecomaj(σ) =
∑i∈Des(σ)
(n− i)
0.17. Remark. We can also show cocharge can be computed like charge,but with the “opposite” recurrence relation of charge, which is what we willuse to show the proposition below.
0.18. Proposition. Given a permutation σ,
cocharge(σ) = comaj(σ−1)
Idea of Proof. Let σ be a permutation with a descent k, that is, σk > σk+1.This adds n− k to comaj(σ). Then, observe
σ =
(1 · · · k k + 1 · · · nσ1 · · · σk σk+1 · · · σn
)−1→ σ−1 =
(· · · σk+1 · · · σk · · ·· · · k + 1 · · · k · · ·
)That is, k + 1 is to the left of k in σ−1. So, our cocharge diagram will looklike
?. . .
k + 1
. . .
k
. . .
Ik + 1
Ik
Therefore, the k + 1 occurring before the k adds 1 to the index for everynumber greater than k, which adds n− k to the cocharge(σ−1).
0.2. Dominance Order on Partitions.
0.19. Definition. Let S be the set of partitions of degree n, ie whose partssum to n. Then, for λ, µ ∈ S, the dominance order on S is given by
λ E µ⇐⇒∑i=1
λi ≤∑i=1
µi ∀`
where λ = (λ1, . . . , λm) and µ = (µ1, . . . , µk).
0.20. Example. (1, 1, 1, 1) E (2, 2) E (4) and (2, 2, 1) D (2, 1, 1, 1). Further-more, (3, 3) and (4, 1, 1) are incomparable since 3 < 4 but 3 + 3 = 6 > 5 =4 + 1.
0.21. Definition. An partition µ covers λ, denoted µ m λ, when µ B λ(that is, µ D λ, µ 6= λ) and there is no partition ν such that µ B ν B λ.
5
0.22. Remark. Covering relations are sufficient to define a partial order ingeneral.
0.23. Proposition. Given partitions λ, µ of degree n,
λ E µ⇐⇒ λ′ D µ′
where λ′ is the conjugate partition to λ, and similarly for µ′.
Proof. See [Mac79, 1.11].
0.24. Definition. The weak order on Sn is defined by, for σ, γ ∈ Sn,
σ lw γ ⇐⇒ γ = σsi and `(γ) = `(σ) + 1
where si is the transposition switching i and i+ 1.
0.25. Example. In one-line notation,
(123) l (213) = (123)s1 l (231) = (213)s2 = s1s2
Note that
s1 = (213) 6l (312) = s2s1
0.26. Definition. The strong (Bruhat) order on Sn is defined by, for σ, γ ∈Sn,
σ ls γ ⇐⇒ γ = στij and `(γ) = `(σ) + 1
where τij is the transposition switching i and j.
0.27. Theorem (Strong Exchange Property). Given σ ls γ and γ = si1si2 · · · si`,then σ = si1si2 · · · sik · · · si` where the hat means sik is deleted from the ex-pression.
0.3. Young’s Poset on Partitions.
0.28. Definition. The Young order on partitions is given by
λ ≤ µ⇐⇒ λ ⊆ µ
when viewed as Ferrers diagrams. Equivalently,
λ l µ⇐⇒ µ = λ+ an addable corner
0.29. Definition. Given a rectangle (mn), the order ideal generated by (mn)is λ ⊆ (mn) with induced containment order. The rank generating func-tion is given by ∑
λ⊆(mn)
q|λ|
where |λ| =∑λi.
6
0.30. Example. The induced sub-poset of (23) = (2, 2, 2) is given by
∅and thus the rank generating function is
1 + q + 2q2 + 2q3 + 2q4 + q5 + q6
0.31. Definition. We define the quantum binomial or Gaussian polynomialto be [
m+ nn
]q
:=[m+ n]q!
[m]q![n]q!
0.32. Example. [21
]q
=1 + q
1 · 1= 1 + q
0.33. Proposition. Given m,n ∈ N, we have∑λ⊆(mn)
q|λ| =
[m+ nn
]q
Proof. The proof is given by an inductive argument on m + n. For theinductive step, one breaks up the set λ ⊆ (mn) into the set of all λ’scontaining a top left cell and the set of all λ’s that do not, thus giving∑
λ⊆(mn)
q|λ| =∑
λ⊆((m−1)n)
qnq|λ| +∑
λ⊆(mn−1)
q|λ|
1. Tableaux
1.1. Tableaux Basics.
1.1. Proposition. A saturated chain in Young’s Poset is given by
∅ l λ1 l λ2 l · · · l λ`
where λi/λi−1 is one cell.
7
1.2. Definition. A standard Young tableau of shape λ is given by a saturatedchain in Young’s poset where each λi/λi−1 is filled with the number i.
Alternatively, one can define a standard Young tableau of shape λ ` n tobe a filling of λ with 1, 2, . . . , n such that rows and columns are strictlyincreasing.
1.3. Example.
∅ l l l l l ↔ 3 5
1 2 4
1.4. Definition. Given a partition λ, the hook length of cell x = (i, j) ∈ λ,denoted hλ(x), is given by the number of cells in row i easy of x plus thenumber of cells in column j north of x plus 1.
1.5. Example.
λ =
The black shaded cell above has hook length 2 + 2 + 1 = 5. The followingdiagram is filled in with the hook length of each cell.
2 1
4 3 1
1.6. Proposition. Let fλ be the number of standard Young tableaux of shapeλ ` n. Then,
fλ =n!∏
x∈λ hλ(x)
1.2. Robinson-Schensted-Knuth (RSK) Correspondence. We will dis-cuss a bijection between matrices with entries in N0 with finite support andpairs of semistandard Young tableaux (P,Q) of the same shape.
1.7. Definition. Let T = (Ti,j) be a tableau and x be a positive integer.We define row insertion, denoted T ← x to be the algorithm, that takes Tand adds a cell labeled x as follows.
(1) Go to row 1. Pick r1 such that T1,r1−1 ≤ x (if T1,1 > x, r1 = 1).(2) If T1,r1 does not exist, add cell labeled x to the end of the row.(3) Otherwise, replace T1,r1 with x to get T′. Then, do T′ ← T1,r1 on
row 2.
8
(4) Repeat this process until either the bumped entry can be put intothe end of the row into which it is bumped or until it is bumped outof the final row, in which case it will form a new row with one entry.
1.8. Example. Let
T = 8
5 6 6
1 1 2 4
Then,
T← 3 =8
5
4 6 6
1 1 2 3
and T← 5 =8
5 6 6
1 1 2 4 5
Explicitly, in the first case, we get the steps
8
5 6 6
1 1 2 4 ← 3
,8
5 6 6
1 1 2 3
← 4,
8
4 6 6
1 1 2 3
← 5,
5
4 6 6
1 1 2 3
← 8
,8
5
4 6 6
1 1 2 3
Notice that, in both cases, we ended up with a semistandard Young tableau.
1.9. Lemma. If T is a semistandard Young tableau, then so is T← x.
Proof. From the algorithm, it is immediate that the rows will still be weaklyincreasing, so we need only check that the columns are strictly increasing.To do this, we will define the notion of an “insertion path.”
1.10. Definition. The insertion path or bumping route of the row-insertionT← x is the collection of the cells (1, r1), (2, r2), . . . , (k, rk) where (r1, . . . , rk)come from the row-bumping algorithm.
1.11. Example. In
8
5 6 6
1 1 2 4
← 3 =8
5
4 6 6
1 1 2 3
9
the bumping path is
8
5
4 6 6
1 1 2 3
or ((1, 4), (2, 1), (3, 1), (4, 1)).
We want to show that the insertion path (starting at row 1) always movesin a weakly westward direction.
Let us say an entry b, originally in Tn,m, is bumped into Tn+1,m′ . Ifm′ > m, then m′ is the largest integer such that Tn+1,m′−1 ≤ b by theconstruction of the algorithm. So, we have the following picture
Tn+1,m · · · Tn+1,m′
Tn,m · · · Tn,m′
↔> b · · · > b
b · · · ≥ b
where we have Tn,m′ ≥ b by the semistandardness of T, which also givesTn+1,m′ > b. However, this would force Tn+1,m′−1 > b, which contradictsthe inequality Tn+1,m′−1 ≤ b above.
1.12. Lemma. Let T be a semistandard Young tableau. If j ≤ k, then theinsertion path T← j is strictly to the left of (T← j)← k.
1.13. Remark. This lemma can be strengthened, but we are content withthe above. See [Ful97, p 9] for a more complete version. Our proof is adaptedfrom the proof in [Ful97].
Proof of Lemma. Suppose j ≤ k and j bumps an element x from the firstrow. The element y bumped by k from the first row must lie strictly to theright of the cell where j bumped x by the semistandardness of T and theconstruction of the row-bumping algorithm. Furthermore, we get x ≤ y,and so the argument continues from row to row.
x y
j k
10
1.14. Remark. An important point that underlies the RSK correspondenceis that this row-bumping process is invertible provided one knows the inser-tion path. For instance,
8
5
4 6 6
1 1 2 3
Can be undone by simply shifting every entry in the path down.
8
5 6 6
1 1 2 4 3
→ 8
5 6 6
1 1 2 4
, 3
In fact, one can do this only knowing the location of the cell that has beenadded to the diagram, simply running the algorithm backwards
8
5
4 6 6
1 1 2 3
,5
4 6 6
1 1 2 3
← 8,8
4 6 6
1 1 2 3
← 5,8
5 6 6
1 1 2 3
← 4,8
5 6 6
1 1 2 4
, 3
In RSK, we will use another tableau to record this information about whenboxes were added.
1.15. Definition (Robinson-Schensted). Given a word w = w1w2 . . . w`, letP(w) be the tableau
P (w) := ((. . . (( w1 ← w2)← w3)← · · · )← w`−1)← w`
and let Q(w) be the recording tableau or insertion tableau whose entries areintegers 1, . . . , ` and where k is placed in the box that is added at the kthstep in the construction of P (w).
1.16. Example. Let w = 5482. Then, we get successive pairs
5 1 5
4
2
1
5
4 8
2
1 3
5
4
2 8
4
2
1 3
where the last pair is (P(w),Q(w)).
The fact that this is a correspondence between words and pairs of tableauxcomes from the reversibility of the row-bump combined with knowing theinserted cell. In the example above, Q is a standard tableaux. We will provea more general correspondence, which will get us what we need about Q.
11
1.17. Definition. Given a (possibly infinite) matrix A with entries in N0
with finite support and nonzero entries , we define
wA :=
(i1 · · · imj1 · · · jm
)in which, for any pair (i, j) that indexes an entry Ai,j of A, there are Ai,jcolumns equal to
(ij
), and all columns satisfy the following two conditions.
• i1 ≤ · · · ≤ im and• If ir = is and r ≤ s, then jr ≤ js.
1.18. Example.
A =
1 0 20 3 10 0 1
→ wA =
(1 1 1 2 2 2 2 31 3 3 2 2 2 3 3
)1.19. Definition (Robinson-Schensted-Knuth). Given a matrix A with en-tries in N0 with finite support, construct
wA =
(i1 . . . imj1 . . . jm
)Then, we will construct a sequence of pairs of tableau were
(P0,Q0) = (∅,∅),Pt+1 = P← jt+1,
and Qt+1 = Qt with it+1 added in the cell that appeared by doing theinsertion for Pt+1. Then, we say that A corresponds to (Pm,Qm).
1.20. Example. Continue with A,wA as in the example above. Then,
(P0,Q0) = (∅,∅)
(P1,Q1) =
(1,
1)
(P2,Q2) =
(1 3
,1 1
)(P3,Q3) =
(1 3 3
,1 1 1
)
(P4,Q4) =
3
1 2 3, 2
1 1 1
(P5,Q5) =
3 3
1 2 2,
2 2
1 1 1
(P6,Q6) =
3 3
1 2 2 2,
2 2
1 1 1 2
12
(P7,Q7) =
3 3
1 2 2 2 3,
2 2
1 1 1 2 2
(P8,Q8) =
3 3
1 2 2 2 3 3,
2 2
1 1 1 2 2 3
1.21. Lemma. Qt constructed above is a semistandard Young tableau.
Proof. The rows of Qt are certainly weakly increasing since the first row ofwA is weakly increasing by construction. For columns strictly increasing, weobserve that if ik = ik+1, then jk ≤ jk+1, and so by lemma 1.12, we get thatthe insertion path M ← jk is strictly to the left of (M ← jk)← jk+1 and sothe columns must be strictly increasing.
1.22. Theorem. The RSK correspondence is a actual correspondence be-tween matrices with entries in N0 with finite support and pairs of semistan-dard Young tableaux of the same shape.
Proof. To show we have a correspondence, we will construct an inverse al-gorithm. Given (P,Q) = (Pm,Qm), we do the following.
(1) Consider Qr,s, the rightmost entry of the largest element in Q (sinceQ can have repeated entries). Then, set Qm−1 = Qm \ Qr,s.
(2) Pr,s is the last spot in the insertion path of Pm−1 ← jm.(3) We need Pr−1,t, the rightmost element of row r − 1 that is smaller
than Pr,s. Then, do reverse bumping until you get jm.(4) Repeat this process until wA is recovered.
1.23. Example. (a)
Q = 2 2
1 1 1=⇒ i5 = 2
(b)
P =3 3
1 2 2
(c)
3
1 2 2← 3, 3
1 2 3and j5 = 2
We also want to show that this inverse is well-defined, that is, if ik = ik+1,then jk ≤ jk+1. If ik = Qr,s and ik+1 = Qu,v, then r ≥ u and s < v. Then,it must be that the inverse insertion path of Pr,s is strictly to the left of theinsertion path of Pu,v. Thus, we get, in the first row,
jk · · · jk+1 =⇒ jk ≤ jk+1
13
2. Symmetric Polynomials and Symmetric Functions
2.1. Introduction to Symmetric Polynomials. Let R be a commutativering with unity and let x1, x2, . . . , xn be indeterminates.
2.1. Definition. The ring of polynomials in n variables with coefficients inR is
R[x1, x2, . . . , xn] =∑
cαxα | α ∈ Nn
where xα = xα1
1 xα22 · · ·x
αnn
with multiplication xαxβ = xα+β.
2.2. Definition. We say f ∈ R[x1, . . . , xn] has homogeneous degree d if
f =∑|α|=d
cαxα
2.3. Example. 3x21 + 19x1x3 has degree 2.
2.4. Proposition. Let
Rd[x1, . . . , xn] := f ∈ R[x1, . . . , xn] | f has degree dThen, R[x1, . . . , xn] is a graded ring with
R[x1, . . . , xn] =⊕d≥0
Rd[x1, . . . , xn]
Proof. Let f ∈ Ra[x1, . . . , xn] and g ∈ Rb[x1, . . . , xn]. Then, fg ∈ Ra+b[x1, . . . , xn].
2.5. Proposition. There is a degree preserving Sn-action on R[x1, . . . , xn]given by
σ.f(x1, . . . , xn) = f(xσ(1), . . . , xσ(n))
for σ ∈ Sn.
2.6. Example. Let σ = (1, 3, 2) in one-line notation. Then, for
g(x) = 3x21 + 19x1x3 − 7x1 + 18x2x43 ∈ Z[x1, x2, x3]
we haveσg = 3x21 + 19x1x2 − 7x1 + 18x42x3
2.7. Definition. The ring of symmetric polynomials on n indeterminates is
Λn := f ∈ R[x1, . . . , xn] | σf = f ∀σ ∈ SnThis is a subring of R[x1, . . . , xn] and is sometimes denoted R[x1, . . . , xn]Sn
(meaning Sn-action fixed points).
2.8. Example. x1 + x2 + x− 3 and 3x21 + 3x22 + 3x23 − 9x1x2x3 are both inΛ3. However, x1 + x2 is not in Λ3 since the permutation (1, 3, 2) (one-line)yields x1 + x3.
14
2.9. Corollary.
Λn =⊕d≥0
Λdn where Λdn := f ∈ Λn | f has degree d
Furthermore, if R is a field, say k, then Λn is a graded algebra since k[x1, . . . , xn]is a graded algebra.
2.10. Remark. In particular, this makes Λn a vector space over k. Thus,we can ask about various k-bases of Λn.
2.11. Definition. For λ ` d, the monomial symmetric polynomial is
mλ(x1, . . . , xn) :=∑α∈Nnα∼λ
xα
2.12. Example. • m1(x1, x2) = x1 + x2 since (1, 0) ∼ (0, 1).• m211(x1, x2, x3) = x21x2x3 + x1x
22x3 + x!x2x
23 since (211) ∼ (121) ∼
(112).• m211(x1, x2) = 0 because there is no partition with 2 parts that can
be rearranged to (2, 1, 1).
2.13. Proposition. mλ(x1, . . . , xn)λ`d is a basis for Λdn.
Proof. Each element of our set is symmetric since every monomial coefficientis 1 and every permutation of xλ occurs in mλ. Also, linear independence isimmediate; each mλ has unique monomials. Thus, it suffices to show thatmonomial symmetric polynomials span Λdn.
Given f ∈ Λdn, we write
f =∑|α|=d
cαxα where cα = cσ(α), ∀σ ∈ Sn
In particular, cα = cλ for α ∼ λ. In otherwords, λ is a partition rearrange-ment of α Thus,
f =∑λ`d
cλ
(∑α∼λ
xα
)=∑λ`d
cλmλ(x1, . . . , xn)
2.14. Corollary. An immediate corollary to this result is that any basis forΛdn can be indexed by λ ` d.
2.15. Definition. For α a partition, we establish the convention
xα := xα11 xα2
2 · · ·xαnn
2.16. Definition. For r ≥ 0, elementary symmetric polynomial
er(x1, x2, . . . , xn) :=∑
1≤i1<i2<···<ir≤nxi1xi2 · · ·xir
15
and, for λ ` d with λ = (λ1, . . . , λ`),
eλ(x1, . . . , xn) = eλ1(x1, . . . , xn)eλ2(x1, . . . , xn) · · · eλ`(x1, . . . , xn)
2.17. Example.
e1(x1, x2, x3) =x1 + x2 + x3 = m1(x1, x2, x3)
e2(x1, x2, x3) =x1x2 + x1x3 + x2x3 = m11(x1, x2, x3)
e21(x1, x2, x3) =(x1x2 + x1x3 + x2x3)(x1 + x2 + x3)
=x21x2 + x1x22 + x1x2x3+
x21x3 + x1x2x3 + x1x23+
x1x2x3 + x22x3 + x2x23
=m21(x1, x2, x3) + 3m111(x1, x2, x3)
2.18. Proposition. er = m1r and er(x1, . . . , xn)r∈N cannot form a basisof Λ.
Proof. By definition, the elementary symmetric polynomials er are symmet-ric polynomials with degree r monomials with no variable occurring morethan once in each monomial. Thus, it must be that er = m1r . Furthermore,by the corollary to the monomial symmetric polynomials forming a basis,it must be that err∈N cannot form a basis because each er ∈ Λrn and sothere simply are not enough in each Λdn to form a basis.
2.19. Proposition. The set eλ(x1, . . . , xn)λ`d is a basis for Λdn.
Proof. Since this set is indexed by λ ` d, it suffices to show that the setspans Λdn. Since mλ is a basis, it suffices to show
mλ =∑
cλ,µeµ
for some cλ,µ. Instead, we will show that
eη =∑
Mη,µmµ and (Mη,µ)η,µ is a unitriangular matrix
To do this, we will need the combinatorics of (0, 1)-matrices.
2.20. Definition. Given a matrix (0, 1)-matrix A, let the row sum be
row(A) := (r1, f2, . . . , rn), ri = number of 1’s in row i
and column sum be
col(A) := (c1, . . . , cn), cj = number of 1’s in row j
Finally, let Bαβ be the number of (0, 1) matrices with row sum vector α andcolumn sum vector β.
16
2.21. Example. Some matrices with row(A) = (2, 1, 1, 0)1 1 0 01 0 0 01 0 0 00 0 0 0
,
1 1 0 00 1 0 01 0 0 00 0 0 0
,
1 1 0 01 0 0 00 0 1 00 0 0 0
,
1 1 0 01 0 0 00 1 0 00 0 0 0
The second and fourth matrices have col(A) = (2, 2, 0, 0).
Then, we claim
eν′ = mν +∑νBµ
Bν′µmµ
To do this, first consider one term. Let A∗ be the (0, 1)-matrix with rowsum ν ′ where all the one’s are left justified.
2.22. Example. If ν ′ = (2, 1, 1, 0), then
A∗ =
1 1 0 01 0 0 01 0 0 00 0 0 0
Note that col(A∗) = (3, 1) = ν
Such an element has col(A∗) = ν and it is the unique (0, 1)-matrix withrow(A) = ν ′ and col(A) = ν. In particular, if A has ν1 ones in columnone and row(A) = ν;, the first `(ν ′) = ν1 rows have ones. Thus, all onesin column 1 are north justified. Iterating this process gives the uniquenessresult. Any other matrix A with row(A) = ν ′ will have column sum smallerin dominance than ν ′ since the ones in A∗ are pushed all the way to the left.
Thus, let η = ν ′. Then,
eη = eη1(x1, . . . , xn)eη2(x1, . . . , xn) · · · eη`(x1, . . . , xn)
=
∑i1<i2<···<iη1
xi1 · · ·xiη1
∑j1<j2<···<jη2
xj1 · · ·xjη2
· · · ∑z1<···<zη`
xz1 · · ·xzη`
=∑
α obtained by choosingi1<i2<···<iη1j1<j2<···<jη2
...z1<z2<···<zη`
xcol(X) where X =
x1 x2 · · · xn
x1 x2 · · · xn...
......
x1 x2 · · · xn
=
0 1 · · · 11 1 · · · 0...
......
0 1 · · · 0
17
where the circled entries are the i1, i2, . . . , iη1 entries of row 1, the j1, j2, . . . , jη2entries of row 2, etc. and thus
eη =∑
A∈(0,1)-matrices withrow(A)=η
xcol(A) =∑µ`d
∑(0,1)-matrices Awith row(A)=η,
col(A)=µ
mµ =∑µ`d
Bη,µmµ
2.23. Example.
e211 = (x1x2 + x1x3 + x2x3)(x1 + x2 + x3)(x1 + x2 + x3) =∑
α=col(A)
xα
So for instance, x1 x2 x3
x1 x2 x3
x1 x2 x3
xcol(A) = x21x2x3
0 1 11 0 01 0 0
2.24. Corollary (Fundamental Theorem of Symmetric Polynomials). Λnwith coefficients in R is a polynomial ring in the n elementary symmetricpolynomials, that is, Λn ∼= R[e1, . . . , en] as commutative rings.
Proof. By the above, for f ∈ Λdn, we write
f =∑µ`d
cµeµ(x1, . . . , xn)
However, since eµ1 , eµ2 , . . . , eµn commute, define eα := eα11 eα2
2 · · · eαnn .
2.25. Example. e3222 = e3e2e2e2 = e01e32e
13 = e(031).
Then, we get that
f =∑
cµeα
where α1 is the number of 1’s in µ, α2 is the number of 2’s in µ, etc. Thisshows that f ∈ R[e1, . . . , en]. Thus, Λdn ⊆ R[e1, . . . , en]. However, sincee1, . . . , en are all symmetric, it must be that R[e1, . . . , en] ⊆ Λn. Thus, weget the desired isomorphism.
2.26. Definition. For r ∈ N, the complete symmetric polynomial is
hr(x1, . . . , xn) :=∑
1≤i1≤···≤ir≤nxi1 · · ·xir
18
and, for λ a partition, the homogeneous symmetric polynomials are
hλ := hλ1hλ2 · · ·hλ`2.27. Example.
h1(x1, x2, x3) = x1 + x2 + x3
h2(x1, x2, x3) = x21 + x1x2 + x22 + x2x3 + x23 + x1x3
= m2(x1, x2, x3) +m11(x1, x2, x3)
2.28. Proposition. hr(x1, . . . , xn) =∑
λ`rmλ
Proof. This follows immediately by grouping the terms of hr appropriately.
We now wish to show the following.
2.29. Proposition. hλ(x1, . . . , xn)λ`d is a basis for Λdn.
We will give a proof by generating functions.
2.30. Lemma. We have the following generating functions
E(t) =∏i
(1 + xit) =∑k≥0
tkek
H(t) =1∏
i(1− xit)=∑k≥0
tkhk
Proof of Lemma. Observe first
E(t) =∏i
(1 + xit)
= (1 + x1t)(1 + x2t) · · · (1 + xnt)
=∑
1≤i1<···<ik≤ntjxi1xi2 · · ·xik
=∑k≥0
tkek
For H(t), recall the geometric series identity
1
1− xit=∑k≥0
xki tk
and thus ∏ 1
1− xit=∏i
∑k≥0
xki tk
=∑k≥0
tkhk
19
Proof of Proposition 2.29. Since H(t) = 1E(−t) , we get
H(t)E(−t) = 1 =⇒
(∑i
tihi
)∑j
(−t)jej
= 1 +∑k≥1
0 · tk
=⇒n∑j=0
(−1)jhn−jej = 0, ∀n > 0
So, we check
n = 1: h1 − e1 = 0 =⇒ e1 = h1
n = 2: h2 − e1h1 + e2 = 0 =⇒ e2 = e1h1 − h2 = h21 − h2and so we can iterate to show that er is a sum of hµ’s. Thus, we get thathµ spans Λn.
2.2. Symmetric Functions. Notice that we can always get a homomor-phism of rings
Λn+1 Λnby setting xn+1 = 0.
2.31. Example.
e21(x1, x2, x3, x4) = (x1x2 + x1x3 + x1x4 + · · · )(x1 + x2 + x3 + x4)
= m21(x1, x2, x3, x4) + 3m111(x1, x2, x3, x4)
7→ e21(x1, x2, x3) = (x1x2 + x1x3 + x2x3)(x1 + x2 + x3)
= m21(x1, x2, x3) + 3m111(x1, x2, x3)
In fact, one can check that mλ(x1, . . . , xn+1) 7→ mλ(x1, . . . , xn) which isnontrivial provided `(λ) > n. Thus,
2.32. Proposition. Λdn∼= Λdn+1 as R-modules provided n ≥ d.
Thus, the number of variables we work in is largely irrelevant providedthere are sufficiently many. In order to avoid this formal annoyance, we willoften work with “infinite symmetric polynomials”, which we will constructin this following section.
2.33. Definition. Let R be a commutative ring with unity. Then, for aninfinite collection of indeterminates x1, x2, . . ., we say R[[x1, x2, . . .]] is thering of formal series over R with elements given as (formal) infinite sums ofmonomials in the indeterminates, addition given by componentwise addition,and multiplication given by extension of polynomial multiplication.
Note that multiplication is well-defined since, given a product of infiniteseries ( ∑
α∈N∞cαx
α
) ∑β∈N∞
dβxβ
,
20
the coefficient at xγ for γ ∈ N∞ has only finitely many contributions fromthe cα and dβ’s.
2.34. Definition. Given α, β ∈ N∞, we say α ∼ β if the underlying multisetsof entries are equal (not counting 0).
2.35. Example. (2, 0, 1, 0) ∼ (1, 2, 0)
2.36. Definition. We say f ∈ R[[x1, x2, . . .]] is a symmetric infinite seriesif the coefficient of xα is equal to the coefficient of xβ when α ∼ β.
2.37. Example. The element∑d≥0
∑i≥1
xdi is a symmetric infinite series.
2.38. Definition. A symmetric polynomial in infinitely many variables (usu-ally called a symmetric function) is a symmetric infinite series with boundeddegree. The set of these symmetric polynomials forms a ring of symmetricfunctions, denoted Λ.
2.39. Example. Let b ∈ N. Then,b∑
d≥0
∑i≥1
xdi is a symmetric function.
2.40. Remark. For the categorically inclined, the symmetric functions canalso be constructed as an inverse limit by taking
Λd = lim←n
Λdn
in the category of Z-modules and then setting
Λ =⊕k≥0
Λk
This requires setting up the necessary maps, but this is not so hard giventhe above for those who wish to use inverse limits. Note, however, that Λis not the inverse limit of Λn in the category of rings; such an inverse limitwould be the ring of symmetric infinite series. However, Λ would be theinverse limit of Λn in the category of graded rings.
We wish to extend our definitions for bases of symmetric polynomials toour ring of symmetric functions.
2.41. Definition. Let x = (x1, x2, . . .). Then, we define
(a) for λ a partition,
mλ(x) =∑α∼λ
xα
(b) for r ∈ N,
er(x) =∑
1≤i1<i2<···<ir
xi1xi2 · · ·xir
21
(c) for r ∈ N,
hr(x) =∑|λ|=r
mλ =∑
1≤i1≤i2≤···≤ir
xi1xi2 · · ·xir
2.42. Example. h2(x) = m(2)(x)+m(1,1)(x) = x21 +x22 + · · ·+x1x2 +x1x3 +· · ·+ x2x3 + · · ·
Finally, we wish to introduce one more family of symmetric functions.
2.43. Definition. For any r ∈ N, we define the power symmetric functions
pr :=∑i
xri
and, for any partition λ, we define
pλ =∏i
pλi
2.44. Proposition. The generating function for the pr is
P (t) =∑ xi
1− xitand also satisfies the identities
P (t) =H ′(t)
H(t)P (−t) =
E′(t)
E(t)
Proof. By direct computation, we see
P (t) :=∑r≥1
prtr−1 =
∑i≥1
∑r≥1
xri tr−1 =
∑i≥1
xi1− xit
Furthermore, if we are clever at manipulating generating functions, we see
P (t) =∑i≥1
d
dtlog
(1
1− xit
)=
d
dtlog
∏i≥1
1
1− xit
=d
dtlog(H(t)) =
H ′(t)
H(t)
and
P (−t) =∑i≥1
d
dtlog(1 + xit) =
d
dtlog
∏i≥1
(1 + xit)
=d
dtlogE(t) =
E′(t)
E(t)
2.45. Corollary (Newton’s Identities). We have the following identities
nhn =
n∑r=1
prhn−r, nen =
n∑r=1
(−1)r−1pren−r
22
Proof. These come from the proposition above by rearranging and expand-ing. For instance,
H ′(t) = P (t)H(t) =⇒∑i≥1
ihiti−1 =
∑r≥1
prtr−1
∑j≥1
hjtj
and so, fixing a specific n, we have
nhn =∑r+j=n
prhj
giving us the desired identity.
2.46. Theorem. The set pλall partitions λ is a basis for Λ over a field ofcharacteristic 0.
Proof. By Newton’s identities,
pn =n∑j=1
ejpn−j
if we rewrite this recurrence, we get
en =1
n
n∑j=1
(−1)j−1en−jpj
Thus, each en can be written as a rational linear combination of the powersums, giving that Λn ⊆ Q[p1, . . . , pn]. Therefore, the pλ’s form a basis andbecause every power sum polynomial is symmetric, Λn ∼= Q[p1, . . . , pn].
2.47. Remark. An alternative proof is just to let C = (cλ,µ) be a matrixexpressing pλ in terms of the monomial basis. One argues that
pλ = cλλmλ +∑µBλ
cλµmµ
with cλλ 6= 0. Then, if xµ11 · · ·xµmm appears in
pλ = (xλ11 + xλ22 + · · · )(xλ21 + xλ22 + · · · ) + · · ·then each µi must be a sum of λj ’s, thus making µ larger in dominanceorder than λ. This proof is probably simpler than the one above but hasthe disadvantage of not introducing Newton’s identities.
Finally, we note how pn’s are explicitly related to the hn’s and en’s.
2.48. Proposition. Given λ ` n, let mi be the multiplicity of i in λ and setzλ =
∏i imimi!. Then,
hn =∑λ`n
pλzλ
anden =
∑λ`n
(−1)n−`(λ)pλzλ
23
Proof. We show this by writing each hr as a linear combination of pλ’s.Observe,
∞∑k=0
hk(~x)tk =
∞∏i=1
1
1− xit
=∏
exp (− log(1− xit))
=∏
exp
( ∞∑k=1
(xit)k
k
)
= exp
( ∞∑k=1
∞∑i=1
(xit)k
k
)
= exp
( ∞∑k=1
pk(x)tk
k
)
=
∞∏k=1
exp
(pk(x)tk
k
)
=
∞∏k=1
∞∑mk=0
(pk(x)tk)mk
mk! · kmk
=∑λ
1
z(λ)pλt|λ|
Thus, we see that hn =∑
λ`npλzλ
. The relation for en is a similar manipula-
tion of the generating function E(t).
2.49. Remark. Note that z(λ) is equal to the cardinality of the centralizerof an element of Sn with cycle type λ. Indeed, an i-cycle (1, 2, . . . , i) is fixedby i elements and a product of mi disjoint i-cycles can be permuted in mi!ways. So then, why
does this num-ber show uphere?
2.3. Schur functions.
2.50. Definition. The Vandermonde determinant is
∆ := det
xn−11 xn−21 · · · x1 1xn−12 xn−22 · · · x2 1
......
......
xn−1n xn−2n · · · xn 1
2.51. Theorem. ∆ =
∏1≤i<j≤n
(xi − xj)
Proof. Note first that deg ∆ =(n2
)and deg
∏=(n2
)for∏
the right handside of the equality. Also, observe that if xi = xj , then ∆ = 0 since then
24
two rows of the matrix would be identical. Therefore, (xi − xj) divides ∆,so
∆ = c ·∏
1≤i<j≤n(xi − xj)
for some c in our base field. Without too much more work, one can verifythat c = 1.
2.52. Definition. For α ∈ Nn, we define Jacobi’s generalized Vandermondeas
∆α := det
xα11 xα2
1 · · · xαn1xα12 xα2
2 · · · xαn2...
......
xα1n xα2
n · · · xαnn
2.53. Remark. We let ρ := (n − 1, n − 2, . . . , 1, 0). Then, ∆ρ = ∆, theVandermonde determinant.
2.54. Proposition. Let α = (α1, . . . , αn) ∈ Nn. Then, for Jacobi’s general-ized Vandermonde
(a) αi = αj =⇒ ∆α = 0(b) If α has distinct parts, then ∆α is “alternating,” that is σ∆α =
(−1)`(σ)∆α for any permutation σ of [n].(c) ∆ρ divides ∆α.
Proof. (a) If αi = αj , then our matrix has two equal columns, so itsdeterminant is 0.
(b) This is immediate since si∆α swaps the ith and i+ 1st rows.(c) If xi = xj , then ∆α = 0 since our matrix has two equal rows. Thus,
(xi − xj) must divide ∆α. However, since i, j were arbitrary, we getthat every such difference must divide ∆α, so ∆ρ divides ∆α.
Thus, now our following definition makes sense
2.55. Definition. Given a partition λ, Jacobi’s bialternant formula for aSchur function is
sλ(x1, . . . , xn) := ∆λ+ρ/∆ρ
which is a polynomial by the above proposition.
2.56. Remark. Both ∆λ+ρ and ∆ρ are alternating, so ∆λ+ρ/∆ρ is symmet-ric.
2.57. Theorem (Littlewood’s Combinatorial Definition). For a partitionλ ` n,
sλ(x1, . . . , xn) =∑
T∈SSYT(λ) in alphabet [n]
xwt(T)
25
In infinite variables,
sλ(x1, x2, . . .) =∑
T∈SSYT(λ)
xwt(T)
Proof. Postponed.
One can instead take the statement of the theorem to be the definitionof the sλ, if so inclined, and prove that it implies the bialternate formula;in fact, we will do this. See 2.74. We wish to show, using this construction,that the Schur functions form a basis for Λ.
2.58. Example. Consider s21(x1, x2, x3). The possible semistandard Youngtableaux are
2
1 1,
3
1 1,
3
2 2,
2
1 2,
3
1 3,
3
2 3,
2
1 3,
3
1 2
and so
s21(x1, x2, x3) = x21x2 + x21x3 + x22x3 + x1x22 + x1x
23 + x2x
23 + 2x1x2x3
2.59. Proposition. Given λ a partition of n,∑T∈SSYT(λ)
xwt(T)
is a symmetric function.
To prove this fact, we will need the following definition.
2.60. Definition. Let the Kostka number, Kλα be the number of semistan-dard Young tableaux of shape λ with weight α. In other words, |SSYT(λ, α)| =Kλα.
Proof of Proposition. For si a simple reflection in Sn, we will show that∑T∈SSYT(λ)
xwt(T) = si∑
T∈SSYT(λ)
xwt(T)
We observe that∑T∈SSYT(λ)
xwt(T) =∑α∈N∞
∑T∈SSYT(λ,α)
xα =∑α∈N∞
Kλαxα
Therefore, we need only show that Kλ,α = Kλ,siα. To show this, we will usethe Bender-Knuth Involution, defined below. Namely, there is an involution(thus a bijection) between SSYT(λ, α) and SSYT(λ, siα), so
Kλ,α = | SSYT(λ, α)| = | SSYT(λ, siα)| = Kλ,siα
2.61. Definition. The Bender-Knuth Involution is a map φi : SSYT(λ, α)→SSYT(λ, siα) given as follows:
26
(a) We define that entries i and i+ 1 in T ∈ SSYT(λ, α) are married ifthey are in the same column. Otherwise, we say they are single.
2.62. Example. For i = 3, in
T =
3 4 4 4
2 2 2 3 3 4 4
1 1 1 1 2 2 3 3 3
the red entries are married and the pink entries are married. Allother entries are single.
(b) φi(T) is obtained by replacing single entries
iii · · · i︸ ︷︷ ︸a
i+ 1 · · · i+ 1︸ ︷︷ ︸b
in each row by
iii · · · i︸ ︷︷ ︸b
i+ 1 · · · i+ 1︸ ︷︷ ︸a
2.63. Example. If T is the example above, then
φ3(T) =
3 3 4 4
2 2 2 3 3 4 4
1 1 1 1 2 2 3 4 4
2.64. Proposition. φi(T) ∈ SSYT(λ, siα).
Proof. We note that the shape λ is preserved by construction, so we needonly show that
(a) Columns strictly increase(b) Rows weakly increase(c) The weight is siα
Note that single entries in a row are contiguous and between married entries,so rows necessarily must be weakly increasing. That is, in a semistandardtableau T, for φx, y = x+ 1, and z > y > x > w, a sufficiently generic rowmight look like
y y z z . . .
x x x x y y y y
. . . w w x x
Thus, if x is single, it lies below z > x+1 and if y is single, it lies above w < x.Therefore, the replacement is still strictly column increasing. Finally, thereare an equal number of married i and i+ 1 entries, and the number of singlei’s in each row is replaced by the number of single i + 1’s in that row, andvice-versa. Thus, the weight is permuted.
27
Thus, we have shown that Littlewood’s combinatorial definition of Schurfunctions gives a symmetric polynomial. To see the equivalence with Jacobi’sbialternate formula, we will actually prove the more general result
2.65. Theorem. [Ste02, p 2]
∆λ+ρsµ/ν =∑
T∈SSYT(µ/ν)T is λ−Yamanouchi
∆λ+wt(T)+ρ
where sµ/ν and λ-Yamanouchi are defined below.
2.66. Definition. A skew Schur function for skew-shape µ/ν is given by∑T∈SSYT(µ/ν)
xwt(T)
and is symmetric by the Bender-Knuth involution.
2.67. Definition. We say a (skew) semistandard tableau T is a Yamanouchitableau if, for all j, T≥j (only columns east of j − 1) has partition weight.This is also sometimes called a reverse lattice.
2.68. Example. (a)
4 4
2 2 2 3
1 1 1 1 1
is not Yamanouchi; it fails at the second column since wt(T≥4) =(2, 0, 1) is not a partition.
(b) The unique semistandard tableau of shape λ and weight λ is Ya-manouchi. In fact, it is the only semistandard “straight shape” thatcan be Yamanouchi, sometimes called the super-semistandard tableauof shape λ.
2
1 1,
2 2
1 1,
2 2
1 1 1,
3
2 2
1 1 1
(c) Here are some Yamanouchi skew-tableau of the following shape.
:
2
1 1
1
,
2
1 2
1
,
3
1 2
1
2.69. Definition. A λ-Yamanouchi tableau T has, for all j, wt(T≥j) + λ isa partition. (Addition is given by vector addition.)
28
2.70. Example. Take λ = (3, 2). Then,
2 2 3
1 1 1 1
is λ-Yamanouchi since
wt(T≥4) + λ = (4, 2)
wt(T≥3) + λ = (5, 2, 1)
wt(T≥2) + λ = (6, 3, 1)
wt(T≥1) + λ = (7, 4, 1)
Proof of 2.65; [Ste02]. We observe that
∆λ+ρsµ/ν =∑
T∈SSYT(λ)T λ-Yamanouchi
∆λ+wt(T)+ρ
=∑w∈Sn
(−1)`(w)xw(λ+ρ)∑
T∈SSYT(µ/ν)
xwt(T)
=∑w∈Sn
∑T∈SSYT(µ/ν)
(−1)`(w)xw(λ+ρ+wt(T))
=∑
T∈SSYT(µ/ν)
∆λ+ρ+wt(T)
However, we want to show the result is true summing only over T that areλ-Yamanouchi. To do this, we will show that all the non λ-Yamanouchitableaux cancel each other out by finding an appropriate sign-reversing in-volution on them.
To that effect, say T is a “Bad Guy” if T is not λ-Yamanouchi. For sucha T, there exists a j, k ∈ [n] such that
λk + wtk(T≥j) < λk+1 + wtk+1(T≥j)
Pick the largest such j and then pick the pair (j, k) with the smallestsuch k. Then, λ + wt(T>j) is a partition. Since, by column strictness,wtk(T≥j) − wtk+1(T≥j) can change by at most one when j is incrementedor decremented, there must be a k + 1 in column j of T, but no k.
2.71. Example. For an easy example, take λ = ∅ and
T =
3 4 4 4
2 2 2 3 3 4 4
1 1 2 2 3 3 3
2 2 2
1 1 1
29
The shaded part of the tableau is Yamanouchi, but it fails to be Yamanouchiat T≥6. Furthermore, the minimal such k so that our inequality fails is 1,that is
3 = 0 + 3 = λ1 + wt1(T≥6) < λ2 + wt2(T≥6) = 0 + 4 = 4
and λ+ wt(T>6) = (3, 3, 3, 1) is a partition. Indeed also, in column 6, thereis a k + 1 = 2 entry, but not a k = 1 entry.
Thus, we have
λk + wtk(T≥j) + 1 = λk+1 + wtk+1(T≥j)
which, since ρk = ρk+1 + 1, also gives
sk(λ+ wt(T≥j) + ρ) = λ+ wt(T≥j) + ρ
where sk ∈ Sn swaps k and k + 1. Now, we apply the Bender-Knuthinvolution φk to T<j and leave the remainder of T fixed. Let us call theresulting tableau T∗.
2.72. Example.
T =
3 4 4 4
2 2 2 3 3 4 4
1 1 2 2 3 3 3
2 2 2
1 1 1
We apply the Bender-Knuth involution φ1 to the red part only to get
T∗ =
3 4 4 4
1 1 2 3 3 4 4
1 1 2 2 3 3 3
2 2 2
1 1 1
Note that the blue part has an equal number of 1’s and 2’s.
T∗ is still an SSYT since the Bender-Knuth involution only swaps someks and k + 1s in T<j in a semistandard way and column j has a k + 1, butcontains no ks. Thus, by construction, we have the following two properties
(T∗)≥j = T≥j since left unchanged by involution
wt(T∗<j) = sk(wt(T<j)) by construction of Bender-Knuth.
So, T∗ is a Bad Guy and, in fact, the map T 7→ T∗ is an involution on BadGuys (Note, however, it still has fixed points, but we need not worry aboutthose. See 2.73). Finally,
λ+ wt(T∗) + ρ = λ+ wt(T∗≥j) + wt(T∗<j) + ρ
30
= λ+ wt(T≥j) + sk(wt(T<j)) + ρ
= sk(λ+ wt(T≥j) + ρ) + sk(wt(T<j))
= sk(λ+ wt(T) + ρ)
and
∆λ+wt(T∗)+ρ = ∆sk(λ+wt(T))+ρ = −∆λ+wt(T)+ρ
since sk acts by swapping the k and k + 1st rows of ∆, which changes thedeterminant by a sign. Thus, we can cancel the contributions of the BadGuys from our desired sums.
2.73. Remark. Note, there is still some work to do. A concern is that some“Bad Guys” will actually have T<j fixed by the BK involution. Consider,for λ = ∅,
T =
2 3 3
2 2 2
1 1 1
which fails to be Yamanouchi at column 1. However, such a Bad Guy willhave
wt(T) + ρ = (3, 4, 2) + (2, 1, 0) = (5, 5, 2)
and so ∆(5,5,2) = 0 by 2.54(a). In general, if the BK involution fixes T<j ,then T<j necessarily has an equal number of cells labeled k and k + 1, thatis
wtk(T<j) = wtk+1(T<j)
Furthermore, since T is a Bad Guy, our choice of (j, k) is such that
λk + wtk(T≥j) + 1 = λk+1 + wtk+1(T≥j)
So, adding these two expressions together, we get
λk + wtk(T) + 1 = λk+1 + wtk+1(T)
and thus, importantly,
(λ+ wt(T) + ρ)k = (λ+ wt(T) + ρ)k+1 =⇒ ∆λ+wt(T)+ρ = 0
2.74. Corollary (Corollaries to Theorem 2.65). In the following specialcases, we get the following corollaries.
(a) If ν = ∅, λ = ∅, then
∆ρsµ =∑
T∈SSYT(µ),T Yamanouchi
∆wt(T)+ρ
=⇒ ∆ρsµ = ∆µ+ρ
Since there is a uniqueYamanouchi SSYT of shape µ
by 2.68
=⇒ sµ = ∆µ+ρ/∆ρ
31
(b) If ν = ∅, then
∆λ+ρsµ =∑
T∈SSYT(µ)T λ-Yamanouchi
∆λ+wt(T)+ρ
=⇒ sλsµ =∑
T of shape µT λ-Yamanouchi
sλ+wt(T) from dividing by ∆ρ
2.75. Remark. The corollaries above have significance outside of combina-torics.
(a) In representation theory, the product rule for Schur functions givesthe decomposition of tensor products of Specht modules for Sn intoirreducibles.
(b) In Schubert calculus, this is useful for understanding the intersectionof Grassmanian subvarieties. Make this more
precise.We can rephrase the corollary above to get the following major theorem.
2.76. Theorem (Littlewood-Richardson Rule). Given shapes λ, µ,
sλsµ =∑ν
cνλ,µsν
where cνλ,µ, called the Littlewood-Richardson coefficients, counts the number
tableaux of skew shape ν/λ of weight µ which are Yamanouchi . Check this re-moval of λ
2.77. Remark. cνλ,µ is also the number of tableaux of the form
µ→
← λ
where the bottom right connected component must be super semi-standardand the whole skew shape is Yamanouchi of weight ν.
2.78. Example. Naturally, we expect that, since s∅ = 1, we get s∅sµ = sµ.Indeed, cν∅,µ is the number of tableaux of shape ν and weight µ, so
cν∅,µ = | SSYT(ν, µ)| = Kν,µ
recovering sµ from the Littlewood-Richardson rule.
32
For something less trivial, take λ = (2, 1) and µ = (1). Then,
c(3,1)(2,1),(1) = 1, c
(2,2)(2,1),(1) = 1, c
(2,1,1)(2,1),(1) = 1
since these are the only skew-shapes such that ν/(2, 1) can have one box.Therefore, all the other coefficients are 0 and we get
s21s1 = s31 + s22 + s211
2.79. Theorem (Pieri Rule). We have
hrsλ =∑
ν=λ+ horizontal r-strip
sν
2.80. Remark. If one already has the Littlewood-Richardson rule, the Pierirule is actually a specialization since s(r) = hr and it turns out that cν(r),µ is
always 0 or 1.
Proof of 2.79. Recall
hrsλ =
∑T of shape (r)
xwt(T)
∑S of shape λ
xwt(S)
and we want to get a sum of the form∑
ν=λ+ horizontal r-strip
sν =∑
ν=λ+ horizontalr-strip
∑tableaux Iof shape ν
xwt(I)
In other words, we want to find a bijection φrλ from pairs of tableaux (T, S)where T has shape (r) and S has shape λ↔ Tableaux I of shape λ+horizontalr-strip such that wt(T) + wt(S) = wt(I).
Let φrλ be RSK. That is, φrλ(T,S) = (I = S← T) with T = x1 x2 . . . xr .
Since x1 ≤ x2 ≤ · · · ≤ xr, then, by Lemma 1.12, the bumping route of x1 isto the left of the bumping route of x2 and so on.
x1 x2
For the inverse, it must be that ν/λ is a horizontal r-strip. Therefore, theinverse is straightforward to compute directly.
Finally, another important relationship between sλ’s and hn’s is givenbelow without proof.
33
2.81. Theorem (Jacobi-Trudi Identities). Let λ = (λ1, . . . , λ`) be a partitionof n. Then,
sλ = det(hλi−i+j)1≤i,j≤` = det
hλ1 hλ1+1 · · · hλ1+`−1
hλ2−1 hλ2 · · ·...
.... . .
hλ`−(`−1) · · · hλ`
where hn = 0 for all n < 0. Similarly,
sλ′ = det(eλi−i+j)1≤i,j≤`
2.82. Example. If λ = (4, 3, 3, 1, 1), then
s43311 = det
h4 h5 h6 h7 h8h2 h3 h4 h5 h6h1 h2 h3 h4 h50 0 h0 h1 h20 0 0 h0 h1
2.4. Inner Product Structure of Λ.
2.83. Proposition. Λ is an inner-product space with Hall-inner product
〈mλ, hµ〉 = δλ,µ =
1 if λ = µ
0 otherwise
2.84. Remark. This is a symmetric form, although this is not obvious fromthe definition.
A natural question about this inner product is how might it behave withrespect to other bases we have encountered. One significant basis with thisform is the Schur basis.
2.85. Theorem. Schur functions are an orthonormal basis with respect to〈·, ·〉. In other words,
〈sλ, sµ〉 = δλ,µ
There are multiple ways to show this, each of which reveals a differentperspective.
2.86. Lemma. Given a shape λ,
hλ =∑µ
Kµ,λsµ
where Kλ,α = |SSYT(λ, α)| is the Kostka number (see 2.60).
Proof. From the Pieri rule, we see that
hr1s∅ = s(r1)
34
Then,
hr2hr1s∅ = hr2s(r1) =∑
ν=(r1)+horizontal r2-strip
sν
2.87. Example. For instance, if r1 = 5, r2 = 3, we get
hr2s(r1) = s + s + s + s
which, if you fill the gray boxes with 1 and the white boxes with 2, you getan SSYT of weight (5, 3). Thus, the sum is precisely
h(5,3) =∑µ,(5,3)
Kµ,(5,3)sµ
In general, we see that the process of multiplying many hri ’s to s∅ willbe equivalent to building different SSYTs of partition weight (r1, . . . , rk). Infact, since the weight is partition weight, this will exactly get us the desiredequality. I know this is
not the bestproof; the ex-ample reallyillustrates thepoint, though.
2.88. Remark. Note, if we encode the Kostka numbers is a matrix, calledthe Kostka matrix, then we have change of bases given by
(Kλ,µ)µ,λ(sµ)µ = (hλ)λ
and thus also the other change of basis given by taking the inverse of theKostka matrix.
A proof of 2.85. Based on our lemma above and Littlewood’s combinatorialdescription of Schur functions, we observe that
sλ =∑µ
K−1µ,λhµ and sλ =∑µ
Kλ,µmµ
where K−1µ,λ is the (µ, λ)-entry of inverse Kostka matrix (Kα,β)−1α,β. Therefore,
〈sλ, sµ〉 = 〈∑α
Kλ,αmα,∑β
K−1β,µhβ〉
=∑α
Kλ,α
∑β
K−1β,µ〈mα, hβ〉
=∑α
Kλ,αK−1α,µ since 〈mα, hβ〉 = δα,β
= (Id)λ,µ
= δλ,µ
2.89. Corollary. The Schur functions can be characterized as the uniqueorthonormal basis for Λ that is unitriangularly related to the basis mλλ.
In fact, one could take the corollary to be the definition of the Schurfunctions.
35
2.90. Example. Imagine one knows only the above corollary. Then, we haveby the unitriangularity that
s111 = m111
s21 = m21 + u21,111m111
s3 = m3 + u3,21m21 + u3,111m111
Then, since
s111 = m111 = h111 − 2h21 + h3
we get
〈s111, s21〉 = 0 =⇒ 0 = 〈h111 − 2h21 + h3,m21 + u21,111m111〉 = u21,111 − 2
and so
s21 = m21 + 2m111
One could then continue this process to get the unknown coefficients in s3,but this would be quite tedious.
Another proof technique to see the orthonormality of Schur functions isusing “Cauchy kernels.” This process requires us to know more about howthe power sum basis behaves with respect to the Hall inner product.
2.91. Definition. The Cauchy kernel in x = (x1, x2, . . .) and y = (y1, y2, . . .)is defined by
Ω(x, y) =∏i,j≥1
1
1− xiyi
2.92. Proposition. We have the following facts about the Cauchy kernel.
(a) Ω(x, y) =∑
λ hλ(x)mλ(y)(b) Ω(x, y) =
∑λ sλ(x)sλ(y) Add that
Ω(x, y) =∑λ pλ(x)pλ(y).
Only problemis that I do notknow how toprove this yet...
Proof. To see part (a), observe∏i,j≥1
1
1− xjyj=∏j
(∑n
hn(~x)ynj
)and so, expanding out our product and collecting terms, we see∏
j
(∑n
hn(~x)ynj
)= 1+h1(~x)(y1 + y2 + · · · )︸ ︷︷ ︸
`1
+h2(~x) · (y21 + y22 + · · · ) + h1(~x)h1(~x) · (y1y2 + y1y3 + · · ·+ y2y3 + · · · )︸ ︷︷ ︸`2
+ · · ·
Therefore, we have our desired result.
To see part (b), we make use of the RSK correspondence. Note that∏i,j≥1
1
1− xiyj=∏i,j≥1
∞∑k=0
(xiyj)k
36
when expanded out into an infinite series has monomials (each with co-efficient 1) indexed by matrices with nonnegative integer entries. On therighthand side, if we expand
∑λ
sλ(x)sλ(y) =∑λ
∑T∈SSYT(λ)
xwt(T)
∑S∈SSYT(λ)
ywt(S)
=∑λ
∑T,S∈SSYT(λ)
xwt(T)ywt(S)
the monomials are indexed by pairs of SSYTs of the same shape, each withcoefficient 1. Thus, by the RSK correspondence, these two infinite sums areequal.
2.93. Theorem. Two bases bλ and bλ for Λ are dual if and only if∑λ bλ(x)bλ(y) = Ω(x, y).
Proof. Write bλ(x) =∑
µ aλ,µmµ and bγ(x) =∑
ν cγ,νhν . Then,
〈bλ, bγ〉 =∑µ
∑ν
aλ,µcγ,ν〈mµ, hν〉 =∑µ
aλ,µcγ,µ
Thus, the two bases are orthonormal if and only if∑µ
aλ,µcγ,µ = δλ,γ ⇐⇒ ACT = I ⇐⇒ CTA = I ⇐⇒ ATC = I ⇐⇒∑µ
aµ,λcµ,γ = δλ,γ
where A = (aλ,µ) and C = (cγ,ν). However, this happens if and only if∑λ,γ
(∑µ
aµ,λcµ,γ
)hλ(x)mγ(y) =
∑λ,γ
δλ,γhλ(x)mγ(y) =∑λ
hλ(x)mλ(y) = Ω(x, y)
Alternative proof of 2.85. Since Ω(x, y) =∑
λ sλ(x)sλ(y), then the Schurbasis is self-dual by the theorem above.
3. LLT Polynomials
3.1. Some Notions on Semistandard Young Tableaux. Note that wecan consider a semistandard Young tableau T of shape λ on n letters as afunction
T : boxes of λ → 1, . . . , nand it T is a standard Young tableau of shape λ on n letters, then thisfunction is bijective.
3.1. Definition. Let T be a standard Young tableau and let T(x) = a andT(y) = a+ 1. If x is strictly south of y and weakly to the east of y (that is,c(x) > c(y)), we say that a is a descent of T and we say that
des(T) := a | a is a descent of T ⊆ 1, . . . , n− 1
37
3.2. Example.
T =
7 9
2 5 8
1 3 4 6
has des(T) = 1, 4, 6, 8
3.3. Proposition. Every semistandard Young tableau T of shape λ has aunique “standardization” S, that is, a unique standard Young tableau S suchthat
(a) T S−1 is a weakly increasing function and(b) The contents of the “standardization” corresponding to the same let-
ter have increasing contents. Formally, if TS−1(j) = TS−1(j+1) =· · · = T S−1(k) = a, then j, . . . , k − 1 ∩ des(S) = ∅.
Proof. Consider that if λ is a horizontal strip, there is a unique labeling ofthe cells of λ to form a standard tableau (with no descents). Similarly, if λis a vertical strip, there is a unique standard tableau on λ with descents atevery position. To get a weakly increasing sequence, one must simply replacethe numbers of the SSYT in increasing order. To resolve the multiple waysone can relabel boxes with the same number, one starts from lowest contentto highest content. One then gets the desired standardization.
T =
3 4
2 2 4
1 1 3 5 6
has standardization S =
5 7
3 4 8
1 2 6 9 10
since
(T S−1)(1) = 1
(T S−1)(2) = 1
(T S−1)(3) = 2
(T S−1)(4) = 2
(T S−1)(5) = 3
(T S−1)(6) = 3
(T S−1)(7) = 4
(T S−1)(8) = 4
(T S−1)(9) = 5
(T S−1)(10) = 6
and des(S) = 2, 4, 6
3.4. Definition. The content of the cell in row i and column j in λ is j− i.
38
3.5. Example. The diagram is filled such that the number in the cell indi-cates its content.
−3
−2 −1 0
−1 0 1 2
0 1 2 3 4
3.6. Remark. A semistandard Young tableau of weight α can be createdfrom a standard Young tableau on |α| letters where the sequences of cells(1, 2, . . . , α1), (α1 + 1, α1 + 2, . . . , α2), . . . each have increasing content byreplacing the labels 1, 2, . . . , α1 by 1, α1 + 1, α1 + 2 . . . , α2 by 2, etc.This is essentially inverse to the process of “standardization.”
3.7. Example. Take (123)(456)(78)(9)(10), that is, take weight α = (3, 3, 2, 1, 1),so |α| = 10. Then, from the standard Young tableau below (which has thedesired content increasing condition), we get the following semistandardtableau using this method:
10
9
4 5 6
1 2 3 7 8
7→
5
4
2 2 2
1 1 1 3 3
Below, we will seek to construct so-called “semistandard ribbon tableau”from “standard ribbon tableau” in an analogous manner.
3.2. Rim Hook Tabloids. Warning: In this section, partitions λ =(λ1, . . . , λk) will be in increasing order, that is λ1 ≤ λ2 ≤ · · · ≤ λk.
3.8. Definition. (a) A rim hook H of a partition λ is a consecutivesequence of cells on Fλ’s northeast rim such that any two adjacentcells of H have a common edge and removal of H from Fλ leaves a“legal” Ferrers diagram.
(b) A special rim hook is a rim hook which starts on the first column ofλ, that is, it starts in cell (λ1, 1).
(c) The number of rows of the rim hook is called its height, denotedht(H).
(d) The number of cells of the rim hook is called its size, denoted |H|.
3.9. Example. The green shaded boxes form a rim hook. The one on theright is a special rim hook.
39
3.10. Definition. A special rim hook tabloid T of shape µ and type λ isa filling of Fµ repeatedly with rim hooks of sizes λ1, λ2, . . . , λk such thateach rim hook is special.
Note, in this context, rim hooks may be called “ribbons” since they arenot rim hooks of the original diagram.
3.11. Definition. An n-ribbon R is a connected skew shape of n squareswith no subshape of 2× 2 squares.
To get a special rim hook tabloid, one follows the following algorithm
(a) Pick a special rim hook H1 in Fµ and remove the cells of H1 to getFµ(1).
(b) Pick a special rim hook H2 in Fµ(1) and remove it to get Fµ(2).(c) Proceed analogously until there is nothing to remove.
3.12. Definition. The type of a special rim hook tabloid T is λ if (|H1|, |H2|, . . . , |H`|)arranged in weakly increasing order produces partition λ
3.13. Example. Take µ = (3, 3, 3, 4). The only special rim hook tabloids oftype λ = (2, 2, 4, 5) are
I)
|H1| = 4
|H2| = 2
|H3| = 5
|H4| = 2
II)
|H1| = 5
|H2| = 2
|H3| = 2
|H4| = 4
3.14. Definition. The sign of a rim hook H is taken to be (−1)ht(H)−1. Thesign of T is the product of signs of rim hooks of T .
3.15. Example. The sign of rim hook tabloid I) in the example above is
(−1)2−1(−1)1−1(−1)2−1(−1)1−1 = 1At some pointwe switch backto standard con-vention.
3.16. Theorem (Murnaghan-Nakayama Rule). Given a shape λ and r ∈ N,we have that
prsλ =∑µ
(−1)ht(µ/λ)−1sµ
where the sum is over all partitions µ such that µ/λ is a rim-hook of size r.
Proof. Let us work in Λn. First, consider that
∆λ+ρpr =n∑i=1
∆λ+rεi+ρ
40
Then, arrange λ+ rεi + ρ in descending order. If two terms are equal, then∆λ+rεi+ρ = 0, otherwise there will be some p ≤ q such that
λp−1 + ρp−1 > λq + ρq + r > λp + ρp
⇐⇒λp−1 + n− p+ 1 > λq + n− q + r > λp + n− p
and so ∆λ+rεi+ρ = (−1)q−p∆ξ where
ξ = (λ1+n−1, · · · , λp−1+n−p+1, λq + n− q + r, λp + n− p, . . . , λq−1 + n− q + 1︸ ︷︷ ︸q−p rows
, λq+1+n−q−1, . . . , λn)
Thus, µ := ξ − ρ is a partition of the form
µ = (λ1, . . . , λp−1, λp + p− q + r, λp + 1, . . . , λq−1 + 1, λq+1, . . . , λn)
Such a partition is exactly such that µ/λ has a rim-hook of size r. In theexample below, r = 5, p = 2, q = 4
λ =
λ+5ε4+ρ = 7→ ξ =
µ =
To complete the proof, divide both sides by ∆ρ and let n→∞.
3.17. Definition. The type of a rim hook tabloid is α = (α1, . . . , α`) if theribbon labeled i has αi cells.
3.18. Definition. A rim hook tabloid of type α is a sequence of partitions
λ0 ⊆ λ1 ⊆ · · · ⊆ λ`
such that λi/λi−1 is a rim hook with αi-cells and the cells of λi/λi−1 arelabeled with the letter i
41
3.19. Example. A rim hook tabloid with type α = (3, 4, 5) is
2 3 3 3
2 2 2 3
1 1 1 3
3.20. Definition. Fix n, a > 0. A standard ribbon tableau (sometimes calledan n-ribbon tableau) of degree a is a rim hook tabloid of type (n, . . . , n︸ ︷︷ ︸
a
) =
(na).
3.21. Example. A 3-ribbon tableau of degree 6 is given by
4
4 4
3 3
1 3 5 5 5 6
1 1 2 2 2 6 6
3.22. Definition. A ribbon head is the southeastern-most cell in a ribbonR. A ribbon tail is the northwestern-most cell.
3.23. Example. A ribbon is given as the colored cells combined. The tailis the dark colored cell (olive), the head is the light colored cell (lime).
3.24. Definition. Fix n > 0. A semistandard ribbon tableau of weight α iscreated from an n-ribbon tableau on len(α) letters such that the contentsof the ribbon heads labeled 1, 2, . . . , α1 are increasing, the contents of theribbon heads labeled α1+1, α1+2, . . . , α2 are increasing, etc., by replacingthe labels 1, 2, . . . , α1 by 1, α1 + 1, α1 + 2, . . . , α2 by 2, etc.
3.25. Example. Let α = (2, 3, 1, 1). Then, the semistandard ribbon tableauis given on the right
7
7
7
3 3 3 4 6 6
1 2 2 4 5 6
1 1 2 4 5 5
4
4
4
2 2 2 2 3 3
1 1 1 2 2 3
1 1 1 2 2 2
Ribbons with same numbers have increasing content heads.
42
3.26. Definition. Let R be a ribbon. Then, we define the spin of a ribbonto be
spin(R) := ht(R)− 1
If T is a ribbon tableau, we define
spin(T ) :=∑
ribbon R∈Tspin(R)
3.3. p-quotients Of Integer Partitions. For later use in understandingn-ribbon tableaux, we will need to understand a certain quotienting oper-ation on partitions. This exposition will follow problem 8 in chapter 1 ofMacDonald’s book.
3.27. Proposition. Let λ, µ be partitions of length ≤ m such that µ ⊆ λand such that λ−µ is a rim hook of length p. Let ρ = (m−1,m−2, . . . , 1, 0)and set
ξ = λ+ ρ, η = µ+ ρ
Then, η is obtained from ξ by subtracting p from some part of ξi of ξ andrearranging in descending order.
Proof. Consider that, due to the shift by ρ, ξ and µ both have distinct partsand our rim-hook will no longer be connected vertically.
λ = (4, 4, 2, 2)
µ = (4, 1, 1)7→
Furthermore, our rim hook (before the shift) can only move down verticallyif the two rows have the same length and, when it moves horizontally, itmust move to the end of the row. Therefore, after the shift, deletion ofthe rim-hook corresponds to subtracting p from a part of ξ and shifting,resulting in η either way.
7→ ←
3.28. Definition. Given a partition λ = (λ1, . . . , λ`), let mk(λ) be the num-ber of parts of λ of size k. In other words mk(λ) = |λi ∈ λ | λi = k|.
3.29. Definition. Let λ be a partition of length ≤ m, let ξ = ρm + λ whereρm = (m− 1,m− 2, . . . , 1, 0), and let p ≥ 2. Then, we define the p-quotientof λ as follows.
43
Let mr = mr(ξ) = |ξi ∈ ξ | ξi ≡ r mod p|, that is, the number of parts
of ξ congruent to r mod p. Then, ξi = pξ(r)k + r for 1 ≤ k ≤ mr where
ξ(r)1 > ξ
(r)2 > · · · > ξ(r)mr ≥ 0
and let λ(r)k = ξ
(r)k − mr + k so that λ(r) = (λ
(r)1 , . . . , λ
(r)mr) is a partition.
Then, the collection λ = (λ(0), . . . , λ(p−1)) is called the p-quotient of thepartition λ.
3.30. Example. Let λ = (4, 4, 2, 2). Then, ζ = λ + ρ4 = (7, 6, 3, 2) hasdistinct parts. If p = 3, then
m0 = 2,m1 = 1,m2 = 1
andζ2 = 6 = 2 · 3 + 0
ζ3 = 3 = 1 · 3 + 0
ζ1 = 7 = 2 · 3 + 1
ζ2 = 2 = 0 · 3 + 2
=⇒
ζ(0)1 = 2
ζ(0)2 = 1
ζ(1)1 = 2
ζ(2)1 = 0
=⇒
λ(0)1 = 2− 2 + 1 = 1
λ(0)2 = 1− 2 + 2 = 1
λ(1)1 = 2− 1 + 1 = 2
λ(2)1 = 0− 1 + 1 = 0
Thus, we get the 3-quotient λ∗ = ((1, 1), (2), (0)), ie
7→
(, ,∅
)
Note that, for a different choice of m (here we picked m = 4), we would get acyclic rearrangement of the 3-quotient. Thus, p-quotients are defined up tocyclic rearrangement. This fact leads [Mac79] to suggest that a p-quotientshould really be thought of as a “necklace of partitions”.
3.31. Proposition. The p-core of a partition is unique and may be obtainedby removing border strips of length p in any order until it is no longer pos-sible.
Proof. By above, removing a border strip of length p corresponds to sub-tracting p from some part of ζ = λ + ρ and then rearranging the sequencein descending order. Simply subtracting p from parts of ζ until it is nolonger possible is certainly independent of choice, and will result in a uniquep-core.
3.4. LLT Polynomials.
3.32. Definition. Given n > 0 and a shape λ, we define the (spin-squared)LLT polynomial (named after Lascoux, Leclerc, and Thibon) to be
G(n)λ (x, t) =
∑T a ss ribbon tableau
of shape λ
tspin(T)xwt(T)
44
where every ribbon of T is of size n.
3.33. Example. If we take n = 3 and λ = (3, 3), we have the followingpossible ribbon tableaux of shape λ:
1 2 2
1 1 2
α = (1, 1)
spin = 2
2 2 2
1 1 1
α = (1, 1)
spin = 0
1 1 1
1 1 1
α = (2, 0)
spin = 2
2 2 2
2 2 2
α = (0, 2)
spin = 2
Thus giving us
G(3)(3,3) = t2x1x2 + x1x2 + t2x21 + t2x22
Also useful is to rewrite this in terms of the monomial and Schur bases:
G(3)(3,3) = (t2 + 1)m11 + t2m2 = t2s2 + s11
Notice, however, our definition is not clear for something like
G(3)
=???
because we cannot tile with 3-ribbons. To deal with this, we will haveto define the following:
3.34. Definition. An n-core is a partition with no removable n-rim-hook.Equivalently, an n-core is a shape where none of its cells have a hook lengthof n.
3.35. Example. The diagram on the left is not a 3-core, the one on theright is.
3.36. Proposition. For any partition γ, successively removing all removablen-rim hooks in any order yields a unique n-core.
3.37. Definition. Fix n > 0. For any shape λ, let λ be its n-core. We define
G(n)λ := G
(n)
λ/λ
3.38. Example. Now, we can deal with our issue above. Namely,
G(3)
= G(3)
3.39. Proposition. For any skew shape λ, G(n)λ is a symmetric function.
3.40. Proposition. G(n)λ is “Schur-positive”, that is,
G(n)λ =
∑µ
cµλ(t)sµ
such that cµλ(t) ∈ N[t].
45
3.41. Proposition. When t = 1, G(n)λ is a product of Schur functions.
3.42. Example. From above
G(3)(3,3)(x; 1) = s2 + s11 = s1 · s2
3.43. Definition. We define the statistic cospin(T) = (spin max(µ)−spin(T))/2where T is a (semistandard) ribbon tableaux of skew-shape µ and spin max(µ) =maxspin(T) | T is a ribbon tableau of skew-shape µ.3.44. Definition. We define the (cospin) LLT polynomial to be
LLT(n)µ (x, t) =
∑T
xwt(T)tcospin(T)
3.45. Example. To see the difference (and similarity) with the spin-squaredLLT family defined above, we see
LLT(3)(3,3)(x, t) = (t+ 1)m11 +m2
since, in this case, spin = 2 =⇒ cospin = 0 and spin = 0 =⇒ cospin = 1.
3.46. Proposition. For any skew shape µ, LLT(n)µ (x, t) is a symmetric func-
tion.
To show this, we will need the another statistic.
3.5. Bylund and Haiman’s inversion statistic. We define a quotientmap which takes a ribbon tableau T to a tuple T of SSYT’s as follows.
3.47. Definition. Recall the content of a ribbon R is the content of its head.Write the content of each ribbon R in T as rn+ p where 0 ≤ p < n. Then,send R to a square of content r in the p+ 1st SSYT. Why does this
work?! Proba-bly follows fromthe “quotient-ing” aspect. Thep-quotient sec-tion probablyresolves this.
3.48. Example. We send
6
5 5
3
3
2 2 3
1 1
7→
(5 5
2 3,
3 3
1 1,
6
2
)
where the coloring of the SSYT’s are to indicate from which ribbon thenumber comes. For example, the orange 6 has content −1 = −1 · 3 + 2, andso it is sent to the 3rd tableau in a cell with content 1.
3.49. Definition. We define the Bylund and Haiman inversion statistic ona tuple of SSYTs T = (T1, . . . , Tn) to be the number of inversion pairs in
T and denote this inv(T). We define an inversion pair to be a pair (c, d)
of squares containing numbers σ(c) and σ(d) respectively, with c ∈ Ti and
d ∈ Tj where 1 ≤ i < j ≤ n where either
46
(a) σ(c) > σ(d) and c and d have the same content, or(b) σ(c) < σ(d) and the content of c is one more than the content of d
(that is, c is one diagonal below d).
Alternatively, for x ∈ Ti, define the adjusted content Note that this isa modificationthe definitionin [HHL+04]:c(x) = nc(x)+si.
c(x) := nc(x) + (i− 1)
where c(x) is the content of x. Then, an inversion pair is a pair (x, y) suchthat σ(x) < σ(y) and 0 < c(x)− c(y) < n.
3.50. Example. Below the tuple above is filled in with numbers representingthe adjusted contents. (
−3 0
0 3,−2 1
1 4,−1
2
)To use the first definition of inversion pairs, redraw our tuple like so
6
2
3 3
1 1
5 5
2 3
so the contents line up. Then, we see the inversion pairs of the first type(5 , 3
),(
5 , 6),(
5 , 1),(
5 , 3),(
5 , 2),(
2 , 1),(
3 , 1)
that is to say, the 5 in the upper left hand corner of T1 forms an inversionpair with the 3 in the upper left hand corner of T2, the 5 in the upper righthand corner of T1 forms an inversion pair with the 1 and 3 on the samediagonal in T2 and with the 2 in T3, etc. The inversion pairs of the secondtype are (
2 , 3),(
2 , 6),(
1 , 6)
so, for instance, the 2 in the bottom left hand corner of T1 forms an inver-sion pair with the 3 in the upper left hand corner of T2.
Thus, inv(T ) = 9.
3.51. Lemma (Spin Lemma). Let |µ| = 2n. Show that if µ can be tiled bytwo n-ribbons whose cells have at least one content in common, then thereare exactly two tilings of µ and their spins differ by 2.
47
Proof. Below is a sufficiently generic example illustrating the situation.
1
2
2
1
In this situation, there can only be one contiguous “section” of the shapethat has common contents, otherwise the shape cannot be filled with twon-ribbons. Consider such a contiguous section:
We can fill the white boxes two different ways:
In the left configuration, each ribbon has height 1 and in the right config-uration, each ribbon has height 2. Importantly, in the contiguous region,the two ribbons must have the same height since the swap trades a verticalmove for a horizontal move on both ribbons.
Any addition of cells with unique contents will add to the overall height ofthe ribbon tableau, but the contributed height will not be affected by theswap. Thus, the total spins will differ by 2.
3.52. Lemma (Bylund, Haiman). Given a skew shape µ and n > 0, thereis a constant e such that for every standard n-ribbon tableau T of shape µ,we have cospin(T) + e = inv(T)
Proof. We say that standard tableaux S,S′ of shape µ = (µ1, . . . , µn) (thatis, S,S′ are a disjoint collection of n standard tableaux of shapes µ1, . . . , µn)differ by a switch if they are identical except for the positions of two con-secutive entries a, a+ 1. Then, two tableaux S, S′ differing by a switch can
48
only have inv(S), inv(S′) differing by 1.
Now, assume that T,T′ are standard n-ribbon tableau such that T and T′
differ by a switch. We wish to show that cospin(T)− cospin(T′) = inv(T)−inv(T′). Since T,T′ are identical except for the ribbons labeled a and a+ 1,the problem reduces to the case where µ has only two non-empty SSYTsand |µ| = 2n; in other words, T,T′ are n-ribbon tableaux tiled by only 2ribbons each with a distinct content. We now reduce to 2 cases. This is going to
be impossibleto present with-out illustrativediagrams.
• (inv(T) − inv(T′) = 0). For this situation, our switched cells x, ycannot have the same content, nor can they be in two distinct SSYTswith one a column higher than the other. Therefore, µ cannot betiled with two ribbons whose contents differ by less than n. Thus, inthis case, the shape µ is the union of two ribbons whose cells haveno contents in common, otherwise the shape could not be connected.Therefore, these ribbons are unique and so spin(T) = spin(T′) andthus cospin(T) = cospin(T′).
• (| inv(T) − inv(T′)| = 1). This situation is possible if and only if|c(y) − c(x)| < n, which forces the ribbon heads to have contentsdiffering by less than n. Therefore, the two ribbons must have cellswith at least one content in common. Thus, by our lemma above,there are exactly 2 tilings of µ whose spin differ by 2, say T,T′ withspin(T) = spin(T′) + 2 Therefore,
cospin(T) =1
2(spin max(µ)−spin(T)) =
1
2(spin max(µ)−spin(T′)−2) = cospin(T′)−1
3.53. Lemma (Bylund, Haiman). The lemma above also holds for semis-tandard ribbon tableaux.
3.54. Definition. If S is a tuple of standard tableaux of shape µ, we call aa descent of S if S(x) = a, S(y) = a+ 1, and c(x) > c(y) for boxes x, y in S.
3.55. Proposition. Given a tuple of semistandard tableaux T, there is aunique “standardization” S such that T S−1 is weakly increasing and, ifTS−1(j) = TS−1(j+1) = · · · = TS−1(k), then des(S)∩j, . . . , k−1 = ∅where a is a descent if there is an x, y such that S(x) = a, S(y) = a+ 1 andc(x) > c(y).
Proof of Proposition. Observe that if µ is a horizontal strip, there is a unique I find this proofrelatively uncon-vincing, but theexample makesit somewhatclear. Any otherattempt at stan-dardization willforce an unde-sired descent.
standard tableau of shape µ with no descents, and conversely, such a tableauexists only on a horizontal strip.
1 2 3 4 5 6 7
49
Thus, each semistandard tableau T has a unique standardization meetingthe properties above.
T =
(5 5
2 3,
3 3
1 1,
6
2
)has standardization S =
(8 9
3 7,
5 6
1 2,
10
4
)
with des(S) = 2, 4, 7, 9
Proof of Lemma. By the definition of an inversion pair, equal entries T(x) =
T(y) = a contribute nothing to inv(T). In the standardization S, equal en-tries are replaced with entries labeled in increasing order of adjusted contentc, contributing nothing to inv(S). On the other hand, unequal entries of T
give rise to entries ordered in the same way in S, so inv(T) = inv(S).
Given a semistandard n-ribbon tableau T, we define its standardization tobe the unique standard n-ribbon tableau S such that S is the standardizationof T. Then, T and S have the same underlying ribbon tiling and so spin(T) =spin(S). Thus, by the previous lemma,
inv(T) = inv(S) = cospin(S) + e = cospin(T) + e
3.56. Theorem (Bylund, Haiman). For any skew shape µ that can be tiledby n-ribbons, there is a constant e depending on µ such that
te LLT(n)µ (x, t) =
∑T
tinv(T)xwt(T)
where the sum is over all SSYT of shape µ.
Proof. Since cospin(T)+e = inv(T) for all T of shape µ by the lemma above
and xwt(T) = xwt(T) by definition, we have that
te LLT(n)µ (x, t) =
∑T
te+cospin(T)xwt(T) =∑T
tinv(T)xwt(T)
4. Hall-Littlewood Polynomials
Throughout our discussions on symmetric functions, most of our exampleshave been stated and worked out for symmetric functions over reasonablysimple fields, such as Q, or some other relatively structured ring, such as aPID like Z. However, there is no reason we cannot consider Λ over otherrings. In this section, we will primarily be interested in symmetric functionsover Q(t).
50
4.1. Symmetric polynomials over Q(t).
4.1. Example. Consider Λ2 over Q(t). The polynomial
tx21 + tx22 +1− 3t
1 + t2x1x2
would be in Λ2.
4.2. Definition. Let λ be a partition. We define mi(λ) to be the numberof parts in λ of size i. Equivalently, this is the number of rows with i boxesif λ is represented as a Ferrers diagram.
4.3. Example. Let λ = (4, 4, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1). Then,
m1(λ) = 4, m2(λ) = 3, m3(λ) = 2, m4(λ) = 3
4.4. Definition. Let λ be a partition and let n = `(λ). Then, we define Sλn
to be a subgroup of Sn such that
σ ∈ Sλn ⇐⇒ σ(λ) = λ
where σ(λi) = λσ(i).
4.5. Example. Let σ = (1, 3, 2, 5, 4, 8, 7, 6, 12, 11, 10, 9) in one line notation.Then,
Sλn∼= SmN × · · · ×Sm1
σ 7→ (1, 3, 2)× (2, 1)× (3, 2, 1)× (4, 3, 2, 1)
all in one line notation.
4.6. Remark. Recall for σ ∈ Sn, (i, j) is an inversion pair if i < j butσ(i) > σ(j). Thus, if σ ∈ Sλ
n, say σ = w1 × w2 × · · · × wN , then
inv(σ) =
N∑i=1
inv(wi)
since blocks do not contribute inversions outside of the block.
4.7. Definition. For n ∈ N, let
Vn(t) :=
n∏i=1
1− ti
1− t=
n∏i=1
(1 + t+ · · ·+ ti−1)
and, for a partition λ, let
Vλ(t) =∏i≥0
Vmi(λ)(t)
4.8. Remark. One should exercise caution since, for a ∈ N,
V(a)(t) = 1 but Va(t) =a∏i=1
(1 + t+ · · ·+ ti−1)
51
4.9. Example. Continuing with λ = (4, 4, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1), we have
Vλ(t) = V 23 (t)V2(t)V4(t)
4.10. Proposition. Given n ∈ N,
Vn(t) =∑σ∈Sn
tinv(σ)
Proof. The proof is a straightforward application of the quantum factorial.
4.11. Corollary. By the above proposition,
Vλ(t) =∑σ∈Sλn
tinv(σ)
Proof. We check
Vλ(t) =∏i≥0
Vmi(λ)(t)
=∏i≥0
∑wi∈Smi
tinv(wi) by Proposition
=∑
(w1,··· ,wN )∈Sm1×···×SmN
tinv(w1)+inv(w2)+···+inv(wN )
=∑σ∈Sλn
tinv(σ) by Remark 4.6
4.12. Definition. Let λ be a partition. We define
Rλ(x1, . . . , xn; t) :=∑w∈Sn
w
xλ∏i<j
xi − txjxi − xj
∈ Q(t)[x1, . . . , xn]
4.13. Remark. There are many things to notice about this Rλ polynomial,and things to be suspicious of, too.
(a) The product should be reminiscent of the Vandermonde determinant.Recall
∆ = det(xji ) =∏i<j
(xi − xj) =∑w∈Sn
sgn(w)w(xρ)
and, more generally, for α = (α1, . . . , α`) with distinct parts,
∆α = det(xαji ) =
∑w∈Sn
sgn(w)w(xα)
Thus, this Rλ polynomial is of a similar form.
52
(b) In the definition of Rλ, we are dividing the expression by xi − xj ’s,but these elements are not in our ring. Thus, we must argue that thenumerator is somehow divisible by this term. However, recall that aSchur function sλ is
sλ =∆λ+ρ
∆=
∆λ+ρ∏i<j xi − xj
∈ Λ
and so it is not unreasonable for this idea to work out.(c) In fact, when t = 0,
R(x1, . . . , xn; 0) =∑w∈Sn
w
xλ∏i<j
xixi − xj
and, since
∏1≤i<j≤n xi = xρ and w
(∏i<j xi − xj
)= sgn(w)∆, we
get
Rλ(x1, . . . , xn; 0) =∑w∈Sn
sgn(w)
∆w(xλ+ρ)
=1
∆
∑w∈Sn
sgn(w)w(xλ+ρ)
=∆λ+ρ
∆ρ
= sλ
(d) When t = 1, we get
R(x1, . . . , xn; 1) =∑w∈Sn
w
xλ∏i<j
xi − xjxi − xj
=∑w∈Sn
w(xλ)
However, since∑
w∈Sn w(xλ) =∑
w∈Sn xw(λ), then we have
R(x1, . . . , xn; 1) = |StabSn(λ)|∑α∈Nnα∼λ
xα = |Sλn|mλ(x1, . . . , xn)
4.14. Example. We compute the small examples ofR11(x1, x2) andR2(x1, x2).Thus, we are working over S2 = id, (12) (in cycle notation). First,
R11(x1, x2) = x1x2
(x1 − tx2x1 − x2
)+ x1x2
(x2 − tx1x2 − x1
)=
x1x2x1 − x2
(x1 − tx2 − x2 + tx1)
=x1x2x1 − x2
(x1 − x2 + t(x1 − x2))
= x1x2(1 + t)
53
which is (1 + t)s11 in terms of Schur functions. Also, note that
V(11)(t) = V2(t) =1− t2
1− t= 1 + t
Second,
R2(x1, x2) = x21x1 − tx2x1 − x2
+ x22x2 − tx1x2 − x1
=1
x1 − x2(x31 − tx21x2 − x32 + tx1x
22)
=1
x1 − x2(x31 − x32 − tx1x2(x1 − x2))
= x21 + x1x2 + x22 − tx1x2which is s2 − ts11 in terms of Schur functions. Thus, we see that Rλ’s arenot necessarily Schur positive. Also, of note,
V(2) = V1(t) = 1
Thus, in both our examples, Rλ has been divisible by Vλ(t).
4.15. Proposition. Rλ(x1, . . . , xn; t) is a symmetric polynomial.
Proof. Observe that
∏i<j
(xi − txj) = (x1 − tx2) (x1 − tx3) (x1 − tx4) · · ·(x2 − tx3) (x2 − tx4) · · ·
(x3 − tx4) · · ·and so, expanding this product into a sum, each monomial will be of theform ∏
i<j
xriji (−txj)sij , rij , sij ∈ N
However, there are some restrictions we must place on rij and sij . In fact,from each term in initial product above, we are picking either an xi or a−txj ; we cannot pick both. So, the combinatorics of picking monomialsamounts to circling either the (i, j) or (j, i) entry of the following matrix,but not both.
0 x1 x1 x1 · · ·−tx2 0 x2 x2 · · ·−tx3 −tx3 0 x3 · · ·
.... . .
−txn −txn −txn · · · 0
Thus, if we take U = (ui,j) to be a (0, 1)-matrix with ui,i = 0 and ui,j+uj,i =1, we get ∏
i<j
(xi − txj) =∑U
∏i<j
xuiji (−txj)uji
54
where the sum is over all such (0, 1)-matrices U . Thus, the symmetry follows. Elaborate
4.16. Proposition.
Rλ(x1, . . . , xn; t) = Vλ(t)
sλ +∑λBµ
cλ,µ(t)sµ
for some cλ,µ(t) ∈ Q(t).
4.17. Corollary. Rλλ is a basis of symmetric polynomials.
4.18. Remark. If we let
Pλ(x1, . . . , xn; t) =1
Vλ(t)Rλ(x1, . . . , xn; t)
thensλ(x1, . . . , xn) =
∑µ
cλ,µ(t)Pµ(x1, . . . , xn; t)
with cλ,µ(t) ∈ N[t], that is, each cλ,µ(t) is a polynomial in t with no negativecoefficients! This is incredibly remarkable since there are no t’s on the lefthand side, so they must all cancel somehow. Furthermore, it is known that
cλ,µ(t) = Kλ,µ(t) =∑
T∈SSYT(λ,µ)
tcharge(T)
where charge(T) is appropriately defined. Some of this material will beelaborated in the next section.
4.2. t-Generalization of Hall Inner Product. Following the remarkconcluding the previous section, we define the following.
4.19. Definition. Let λ be a partition. Then, we define the Hall-Littlewoodpolynomials to be
Pλ(x; t) =1
Vλ(t)
∑w∈Sn
w
xλ∏i<j
xi − txjxi − xj
4.20. Proposition. (a) Pλ(x; 0) = sλ
(b) Pλ(x; 1) = mλ
(c) Pλ(x; t) is a basis for Λ[t], that is, Λ over Q(t).
4.21. Definition. Let 〈·, ·〉t : Λ[t]× Λ[t]→ Q(t) be given by
〈pλ, pµ〉t = δλ,µzλ∏i
(1− tλi)
where zλ =∏i imimi!. (See 2.48).
4.22. Proposition. The set Pλ(x; t) is characterized by
(a) Orthogonality with respect to 〈·, ·〉t.
55
(b) Unitriangularity with respect to Schur basis. That is,
Pλ(x; t) = sλ +∑λBµ
cλ,µ(t)sµ
4.23. Corollary. If we set t = 1, this gives us that
mλ = sλ + lower order terms
where the lower order terms come from the combinatorics of special rim hooktabloids.
4.24. Definition. Let Q′µ(x; t) in Λ[t] be the dual basis to Pλ(x; t) withrespect to 〈·, ·〉t.
4.25. Proposition. Since sλ =∑
µKλµ(t)Pµ by 4.18, we get
Q′µ =∑λ
Kλ,µ(t)sλ
4.26. Theorem.
Kλ,µ(t) =∑
T∈SSYT(λ,µ)
tcharge(T)
Of course, to understand this theorem, we must define the charge of asemistandard Young tableau.
4.27. Definition. The reading word of a tableau T is the word created fromreading the entries of T from left to right, top to bottom.
4.28. Example. Let
T =
5
4
3 6 8
1 2 7 9
Then, the reading word is 543681279.
4.29. Definition. Recall that the charge of a permutation σ is given bywriting σ counterclockwise on a circle and labeling the letters 1, 2, 3, . . . withan “index” clockwise starting at 0 and increasing if you cross ?. Finally, thecharge is the sum of the indices. Then, for T a standard Young tableau, wedefine charge(T) to be the charge of the reading word of T. Alternatively,one can assign indices I = (I1, I2, . . . , I`) by setting
I1 = 0, Ix =
Ix−1 + 1 if x is east of x− 1
Ix−1 else
56
4.30. Example. Let T be as above so it has reading word σ = 543681279.Then, we compute
?5
4
36 8 1
2
79
111
2 3 013
4
giving charge(T) = 16.
4.31. Definition. We define the charge of a semistandard Young tableau asfollows. Given a word (tableau) of partition weight,
(a) Extract the “standard” (permutation) subwords.(b) Compute charge on each of these subwords.(c) Sum up each of the separate charges to get the charge.
4.32. Example. Let
T =
5
3 4 4 4
2 2 3 3 3
1 1 1 1 2 2 4
Then, we draw the reading word on a charge circle and find the standardsubwords, labeled in different colors and done successively for clarity.
?53
4
4
4
2
23 3 3
1
1
1
1
2
24
→
?53
4
4
4
2
23 3 3
1
1
1
1
2
24
→
?53
4
4
4
2
23 3 3
1
1
1
1
2
24
57
And thus giving us
?53
4
4
4
2
23 3 3
1
1
1
1
2
24
7→
?5
32
1
410
001
?
4
2 3
11
0 1
0
?
4
3 1
21
1 0
1
?
4
3 1
21
1 0
1
Thus, the charge would be 2 + 2 + 3 + 3 = 10.
4.33. Example. Our theorem says that
s21 =∑
partition µ
∑T∈SSYT((2,1),µ)
tcharge(T)
Pµ(x; t)
=
∑T∈SSYT((2,1),(2,1))
tcharge(T)
P21(x; t) +
∑T∈SSYT((2,1),(1,1,1))
tcharge(T)
P111(x; t)
Thus, we can now use our theorem to compute
s21 = t0P21(x; t) + (t2 + t)P111(x; t)
since
2
1 1has charge 0
3
1 2has charge 1
2
1 3has charge 2
4.34. Remark. It should be noted that all Hall-Littlewood polynomials areactually special cases of LLT polynomials. However, I do
not know how toprove this.References
[Ful97] W. Fulton, Young Tableaux, 1997.[Hag08] J. Haglund, The q, t-Catalan Numbers and the Space of Diagonal Harmonics,
2008.
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[HHL+04] J. Haglund, M. Haiman, N. Loehr, J. B. Remmel, and A. Ulyanov, A combi-natorial formula for the character of the diagonal coinvariants (2004).
[Mac79] I. G. Macdonald, Symmetric Functions and Hall Polynomials, 1979. 2nd Edi-tion, 1995.
[Ste02] J. R. Stembridge, A Concise Proof of the Littlewood-Richardson Rule, TheElectronic Journal of Combinatorics (2002).
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