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 algebraic combinatorics I took with Dr. Jennifer Morse in Spring 2018. Especially early 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 the symmetric group, ring theory of polynomial rings, and familiarity with set theoretic constructions, such as posets. A reader with a strong grasp on introductory enumerative combinatorics would probably have few problems skipping 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 which these notes are based and for many fruitful and enlightening conversations. In these notes, we use French notation for Ferrers diagrams and Young tableaux, 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 familiar reader 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 that i < 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 set of all descents on σ is denoted Des(σ). (e) The major statistic or major index is given by X maj(σ) = i i∈Des(σ) 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 transpositions of 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 the inversion 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 descent sets {{}, {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 characterization of Mahonian given below. 0.8. Theorem. A statistic stat is Mahonian if and only if it satisfies the identity X X qstat(σ) = qinv(σ)
σ∈Sn σ∈Sn
2 0.9. Remark. Note that X inv(σ) 2 n−1 q = [n]q! := (1)(1 + q)(1 + q + q ) ··· (1 + q + ··· + q )
σ∈Sn which can be proved using an inductive argument by noting n−1 X inv(σ) X inv(σ) X inv(σ) X X inv(σ) q = q + q = [n − 1]q! + q
σ∈Sn σ∈Sn−1 σ∈Sn j=1 σ∈Sn σ(n)6=n σ(n)=j but we will not need this in what follows. To prove maj is Mahonian, we will use the following theorem, although other 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|4 3|14|2 → 3412 3|4|1|2|6 3|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 and also before wi. (ii) If wi < w1, place a bar to the left of any wk with wk > wi and also 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 of σ. 0.14. Definition. The charge statistic on a permutation σ is given by (a) Write the permutation in one-line notation counterclockwise around a circle with a ? between the beginning and the end. ? σn σ1
σn−1 σ2 ...
(b) To each entry σk, assign an “index” Ik recursively as follows ( Ik−1 + 1 if ? is passed going from σk−1 to σk in a clockwise path around the circle. Ik = Ik−1 otherwise. Pn (c) Define charge(σ) = k=1 Ik, that is, the sum of the indices. 0.15. Example. The permutation (6, 7, 1, 4, 2, 3, 5) has ? 6 5 3 3 7 4 2 3 0 1 1 2 2 4 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 n cocharge(σ) := − charge(σ) 2 We also define X comaj(σ) = (n − i) i∈Des(σ) 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 will use 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 1 ··· k k + 1 ··· n −1 ··· σ ··· σ ··· σ = → σ−1 = k+1 k σ1 ··· σk σk+1 ··· σn ··· k + 1 ··· k ··· That is, k + 1 is to the left of k in σ−1. So, our cocharge diagram will look like ? ...
k + 1 Ik + 1
Ik ...... k Therefore, the k + 1 occurring before the k adds 1 to the index for every −1 number greater than k, which adds n − k to the cocharge(σ ). 0.2. Dominance Order on Partitions. 0.19. Definition. Let S be the set of partitions of degree n, ie whose parts sum to n. Then, for λ, µ ∈ S, the dominance order on S is given by ` ` X X λ E µ ⇐⇒ λi ≤ µi ∀` i=1 i=1 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 in general. 0.23. Proposition. Given partitions λ, µ of degree n,
0 0 λ E µ ⇐⇒ λ D µ where λ0 is the conjugate partition to λ, and similarly for µ0.
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) l6 (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 = si1 si2 ··· si` , then σ = si1 si2 ··· sˆik ··· 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 X q|λ| λ⊆(mn) P 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 polynomial to be m + n [m + n] ! := q n q [m]q![n]q! 0.32. Example. 2 1 + q = = 1 + q 1 q 1 · 1 0.33. Proposition. Given m, n ∈ N, we have X m + n q|λ| = n λ⊆(mn) q Proof. The proof is given by an inductive argument on m + n. For the inductive step, one breaks up the set {λ ⊆ (mn)} into the set of all λ’s containing a top left cell and the set of all λ’s that do not, thus giving X X X q|λ| = qnq|λ| + q|λ| λ⊆(mn) λ⊆((m−1)n) λ⊆(mn−1) 1. Tableaux 1.1. Tableaux Basics. 1.1. Proposition. A saturated chain in Young’s Poset is given by 1 2 ` ∅ l λ l λ 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 saturated chain 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 to be a filling of λ with {1, 2, . . . , n} such that rows and columns are strictly increasing. 1.3. Example. 3 5 ∅ l l l l l ↔ 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 the number 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 following diagram is filled in with the hook length of each cell. 2 1 4 3 1
1.6. Proposition. Let f λ be the number of standard Young tableaux of shape λ ` n. Then, λ n! f = Q 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 and pairs 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 T and 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. 0 0 (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 into the end of the row into which it is bumped or until it is bumped out of 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,
8 8 T ← 3 = and T ← 5 = 5 5 6 6 4 6 6 1 1 2 4 5 1 1 2 3
Explicitly, in the first case, we get the steps
← 8 8 8 8 ← 5 5 8 , , , , 5 6 6 5 6 6 ← 4 4 6 6 4 6 6 5 1 1 2 4 ← 3 1 1 2 3 1 1 2 3 1 1 2 3 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 weakly increasing, 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-insertion T ← 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 8 ← 3 = 5 6 6 5 1 1 2 4 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 moves in a weakly westward direction.
Let us say an entry b, originally in Tn,m, is bumped into Tn+1,m0 . If 0 0 m > m, then m is the largest integer such that Tn+1,m0−1 ≤ b by the construction of the algorithm. So, we have the following picture
Tn+1,m ··· Tn+1,m0 > b ··· > b ↔
Tn,m ··· Tn,m0 b ··· ≥ b
where we have Tn,m0 ≥ b by the semistandardness of T, which also gives Tn+1,m0 > b. However, this would force Tn+1,m0−1 > b, which contradicts the inequality Tn+1,m0−1 ≤ b above. 1.12. Lemma. Let T be a semistandard Young tableau. If j ≤ k, then the insertion path T ← j is strictly to the left of (T ← j) ← k.
1.13. Remark. This lemma can be strengthened, but we are content with the above. See [Ful97, p 9] for a more complete version. Our proof is adapted from the proof in [Ful97].
Proof of Lemma. Suppose j ≤ k and j bumps an element x from the first row. The element y bumped by k from the first row must lie strictly to the right of the cell where j bumped x by the semistandardness of T and the construction 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 correspondence is 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 , 3 8 5 6 6 5 6 6 1 1 2 4 1 1 2 4 3 In fact, one can do this only knowing the location of the cell that has been added to the diagram, simply running the algorithm backwards 8 5 8 8 8 , ← 8, ← 5, ← 4, , 3 5 4 6 6 4 6 6 5 6 6 5 6 6 4 6 6 1 1 2 3 1 1 2 3 1 1 2 3 1 1 2 4 1 1 2 3 In RSK, we will use another tableau to record this information about when boxes were added.
1.15. Definition (Robinson-Schensted). Given a word w = w1w2 . . . w`, let P(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 are integers 1, . . . , ` and where k is placed in the box that is added at the kth step in the construction of P (w). 1.16. Example. Let w = 5482. Then, we get successive pairs 5 1 5 2 5 2 5 4 4 1 4 8 1 3 4 2 2 8 1 3 where the last pair is (P(w), Q(w)).
The fact that this is a correspondence between words and pairs of tableaux comes from the reversibility of the row-bump combined with knowing the inserted cell. In the example above, Q is a standard tableaux. We will prove a 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 i1 ··· im wA := j1 ··· jm in which, for any pair (i, j) that indexes an entry Ai,j of A, there are Ai,j i columns equal to j , and all columns satisfy the following two conditions.
• i1 ≤ · · · ≤ im and • If ir = is and r ≤ s, then jr ≤ js. 1.18. Example. 1 0 2 1 1 1 2 2 2 2 3 A = 0 3 1 → w = A 1 3 3 2 2 2 3 3 0 0 1 1.19. Definition (Robinson-Schensted-Knuth). Given a matrix A with en- tries in N0 with finite support, construct i1 . . . im wA = j1 . . . jm Then, we will construct a sequence of pairs of tableau were 0 0 t+1 (P , Q ) = (∅, ∅), P = P ← jt+1, t+1 t and Q = Q with it+1 added in the cell that appeared by doing the insertion 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, 0 0 (P , Q ) = (∅, ∅) 1 1 (P1, Q1) = , 1 3 1 1 (P2, Q2) = , 1 3 3 1 1 1 (P3, Q3) = , 4 4 3 2 (P , Q ) = , 1 2 3 1 1 1 5 5 3 3 2 2 (P , Q ) = , 1 2 2 1 1 1 6 6 3 3 2 2 (P , Q ) = , 1 2 2 2 1 1 1 2
12 7 7 3 3 2 2 (P , Q ) = , 1 2 2 2 3 1 1 1 2 2 8 8 3 3 2 2 (P , Q ) = , 1 2 2 2 3 3 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 of wA is weakly increasing by construction. For columns strictly increasing, we observe that if ik = ik+1, then jk ≤ jk+1, and so by lemma 1.12, we get that the insertion path M ← jk is strictly to the left of (M ← jk) ← jk+1 and so the 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 (since m−1 m Q can have repeated entries). Then, set Q = Q \ Qr,s. m−1 (2) Pr,s is the last spot in the insertion path of P ← 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) 2 2 Q = =⇒ i5 = 2 1 1 1 (b) 3 3 P = 1 2 2 (c) 3 3 ← 3, and j5 = 2 1 2 2 1 2 3
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 the insertion 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 commutative ring with unity and let x1, x2, . . . , xn be indeterminates. 2.1. Definition. The ring of polynomials in n variables with coefficients in R is n o X α n α α1 α2 αn R[x1, x2, . . . , xn] = cαx | α ∈ N where x = x1 x2 ··· xn with multiplication xαxβ = xα+β.
2.2. Definition. We say f ∈ R[x1, . . . , xn] has homogeneous degree d if X α f = cαx |α|=d 2 2.3. Example. 3x1 + 19x1x3 has degree 2. 2.4. Proposition. Let d R [x1, . . . , xn] := {f ∈ R[x1, . . . , xn] | f has degree d}
Then, R[x1, . . . , xn] is a graded ring with M d R[x1, . . . , xn] = R [x1, . . . , xn] d≥0
a b a+b Proof. Let f ∈ R [x1, . . . , xn] and g ∈ R [x1, . . . , xn]. Then, fg ∈ R [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 2 4 g(x) = 3x1 + 19x1x3 − 7x1 + 18x2x3 ∈ Z[x1, x2, x3] we have 2 4 σg = 3x1 + 19x1x2 − 7x1 + 18x2x3 2.7. Definition. The ring of symmetric polynomials on n indeterminates is
Λn := {f ∈ R[x1, . . . , xn] | σf = f ∀σ ∈ Sn}
Sn This is a subring of R[x1, . . . , xn] and is sometimes denoted R[x1, . . . , xn] (meaning Sn-action fixed points). 2 2 2 2.8. Example. x1 + x2 + x − 3 and 3x1 + 3x2 + 3x3 − 9x1x2x3 are both in Λ3. However, x1 + x2 is not in Λ3 since the permutation (1, 3, 2) (one-line) yields x1 + x3.
14 2.9. Corollary. M d d Λn = Λn where Λn := {f ∈ Λn | f has degree d} d≥0
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 X α mλ(x1, . . . , xn) := x n α∈N α∼λ
2.12. Example. • m1(x1, x2) = x1 + x2 since (1, 0) ∼ (0, 1). 2 2 2 • m211(x1, x2, x3) = x1x2x3 + x1x2x3 + x!x2x3 since (211) ∼ (121) ∼ (112). • m211(x1, x2) = 0 because there is no partition with 2 parts that can be rearranged to (2, 1, 1). d 2.13. Proposition. {mλ(x1, . . . , xn)}λ`d is a basis for Λn. Proof. Each element of our set is symmetric since every monomial coefficient λ is 1 and every permutation of x occurs in mλ. Also, linear independence is immediate; each mλ has unique monomials. Thus, it suffices to show that d monomial symmetric polynomials span Λn.
d Given f ∈ Λn, we write X α f = cαx where cα = cσ(α), ∀σ ∈ Sn |α|=d
In particular, cα = cλ for α ∼ λ. In otherwords, λ is a partition rearrange- ment of α Thus, ! X X α X f = cλ x = cλmλ(x1, . . . , xn) λ`d α∼λ λ`d 2.14. Corollary. An immediate corollary to this result is that any basis for d Λn can be indexed by λ ` d. 2.15. Definition. For α a partition, we establish the convention
α α1 α2 αn x := x1 x2 ··· xn 2.16. Definition. For r ≥ 0, elementary symmetric polynomial X er(x1, x2, . . . , xn) := xi1 xi2 ··· xir 1≤i1 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) 2 2 =x1x2 + x1x2 + x1x2x3+ 2 2 x1x3 + x1x2x3 + x1x3+ 2 2 x1x2x3 + x2x3 + x2x3 =m21(x1, x2, x3) + 3m111(x1, x2, x3) 2.18. Proposition. er = m1r and {er(x1, . . . , xn)}r∈N cannot form a basis of Λ. Proof. By definition, the elementary symmetric polynomials er are symmet- ric polynomials with degree r monomials with no variable occurring more than once in each monomial. Thus, it must be that er = m1r . Furthermore, by the corollary to the monomial symmetric polynomials forming a basis, r it must be that {er}r∈N cannot form a basis because each er ∈ Λn and so d there simply are not enough in each Λn to form a basis. d 2.19. Proposition. The set {eλ(x1, . . . , xn)}λ`d is a basis for Λn. Proof. Since this set is indexed by λ ` d, it suffices to show that the set d spans Λn. Since {mλ} is a basis, it suffices to show X mλ = cλ,µeµ for some cλ,µ. Instead, we will show that X 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 and column sum vector β. 16 2.21. Example. Some matrices with row(A) = (2, 1, 1, 0) 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 , , , 1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The second and fourth matrices have col(A) = (2, 2, 0, 0). Then, we claim X eν0 = mν + Bν0µmµ νBµ To do this, first consider one term. Let A∗ be the (0, 1)-matrix with row sum ν0 where all the one’s are left justified. 2.22. Example. If ν0 = (2, 1, 1, 0), then 1 1 0 0 ∗ 1 0 0 0 A = 1 0 0 0 0 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 with 0 row(A) = ν and col(A) = ν. In particular, if A has ν1 ones in column 0 one and row(A) = ν;, the first `(ν ) = ν1 rows have ones. Thus, all ones in column 1 are north justified. Iterating this process gives the uniqueness result. Any other matrix A with row(A) = ν0 will have column sum smaller in dominance than ν0 since the ones in A∗ are pushed all the way to the left. Thus, let η = ν0. Then, eη = eη1 (x1, . . . , xn)eη2 (x1, . . . , xn) ··· eη` (x1, . . . , xn) X X X = x ··· x x ··· x ··· x ··· x i1 iη1 j1 jη2 z1 zη` i1 17 where the circled entries are the i1, i2, . . . , iη1 entries of row 1, the j1, j2, . . . , jη2 entries of row 2, etc. and thus X col(A) X X X eη = x = mµ = Bη,µmµ A∈{(0,1)-matrices with µ`d (0,1)-matrices A µ`d row(A)=η} with row(A)=η, col(A)=µ 2.23. Example. X α e211 = (x1x2 + x1x3 + x2x3)(x1 + x2 + x3)(x1 + x2 + x3) = x α=col(A) So for instance, x1 x2 x3 x x x col(A) 2 1 2 3 x = x1x2x3 x1 x2 x3 0 1 1 1 0 0 1 0 0 2.24. Corollary (Fundamental Theorem of Symmetric Polynomials). Λn with coefficients in R is a polynomial ring in the n elementary symmetric ∼ polynomials, that is, Λn = R[e1, . . . , en] as commutative rings. d Proof. By the above, for f ∈ Λn, we write X f = cµeµ(x1, . . . , xn) µ`d α α1 α2 αn However, since eµ1 , eµ2 , . . . , eµn commute, define e := e1 e2 ··· en . 0 3 1 (031) 2.25. Example. e3222 = e3e2e2e2 = e1e2e3 = e . Then, we get that X α f = cµe where α1 is the number of 1’s in µ, α2 is the number of 2’s in µ, etc. This d shows that f ∈ R[e1, . . . , en]. Thus, Λn ⊆ R[e1, . . . , en]. However, since e1, . . . , en are all symmetric, it must be that R[e1, . . . , en] ⊆ Λn. Thus, we get the desired isomorphism. 2.26. Definition. For r ∈ N, the complete symmetric polynomial is X hr(x1, . . . , xn) := xi1 ··· xir 1≤i1≤···≤ir≤n 18 and, for λ a partition, the homogeneous symmetric polynomials are hλ := hλ1 hλ2 ··· hλ` 2.27. Example. h1(x1, x2, x3) = x1 + x2 + x3 2 2 2 h2(x1, x2, x3) = x1 + x1x2 + x2 + x2x3 + x3 + x1x3 = m2(x1, x2, x3) + m11(x1, x2, x3) P 2.28. Proposition. hr(x1, . . . , xn) = λ`r mλ Proof. This follows immediately by grouping the terms of hr appropriately. We now wish to show the following. d 2.29. Proposition. {hλ(x1, . . . , xn)}λ`d is a basis for Λn. We will give a proof by generating functions. 2.30. Lemma. We have the following generating functions Y X k E(t) = (1 + xit) = t ek i k≥0 1 X k H(t) = Q = t hk (1 − xit) i k≥0 Proof of Lemma. Observe first Y E(t) = (1 + xit) i = (1 + x1t)(1 + x2t) ··· (1 + xnt) X j = t xi1 xi2 ··· xik 1≤i1<··· 1 X k k = xi t 1 − xit k≥0 and thus Y 1 Y X k k X k = xi t = t hk 1 − xit i k≥0 k≥0 19 1 Proof of Proposition 2.29. Since H(t) = E(−t) , we get ! X i X j X k H(t)E(−t) = 1 =⇒ t hi (−t) ej = 1 + 0 · t i j k≥1 n X j =⇒ (−1) hn−jej = 0, ∀n > 0 j=0 So, we check n = 1: h1 − e1 = 0 =⇒ e1 = h1 2 n = 2: h2 − e1h1 + e2 = 0 =⇒ e2 = e1h1 − h2 = h1 − h2 and so we can iterate to show that er is a sum of hµ’s. Thus, we get that {hµ} spans Λn. 2.2. Symmetric Functions. Notice that we can always get a homomor- phism of rings Λn+1 Λn by 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 is nontrivial provided `(λ) > n. Thus, d ∼ d 2.32. Proposition. Λn = Λn+1 as R-modules provided n ≥ d. Thus, the number of variables we work in is largely irrelevant provided there are sufficiently many. In order to avoid this formal annoyance, we will often work with “infinite symmetric polynomials”, which we will construct in this following section. 2.33. Definition. Let R be a commutative ring with unity. Then, for an infinite collection of indeterminates {x1, x2,...}, we say R[[x1, x2,...]] is the ring of formal series over R with elements given as (formal) infinite sums of monomials 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 infinite series ! X α X β cαx dβx , α∈N∞ β∈N∞ 20 γ ∞ the coefficient at x for γ ∈ N has only finitely many contributions from the cα and dβ’s. ∞ 2.34. Definition. Given α, β ∈ N , we say α ∼ β if the underlying multisets of 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 series if the coefficient of xα is equal to the coefficient of xβ when α ∼ β. X X d 2.37. Example. The element xi is a symmetric infinite series. d≥0 i≥1 2.38. Definition. A symmetric polynomial in infinitely many variables (usu- ally called a symmetric function) is a symmetric infinite series with bounded degree. The set of these symmetric polynomials forms a ring of symmetric functions, denoted Λ. b X X d 2.39. Example. Let b ∈ N. Then, xi is a symmetric function. d≥0 i≥1 2.40. Remark. For the categorically inclined, the symmetric functions can also be constructed as an inverse limit by taking Λd = lim Λd ← n n in the category of Z-modules and then setting M Λ = Λk k≥0 This requires setting up the necessary maps, but this is not so hard given the above for those who wish to use inverse limits. Note, however, that Λ is not the inverse limit of Λn in the category of rings; such an inverse limit would be the ring of symmetric infinite series. However, Λ would be the inverse limit of Λn in the category of graded rings. We wish to extend our definitions for bases of symmetric polynomials to our ring of symmetric functions. 2.41. Definition. Let x = (x1, x2,...). Then, we define (a) for λ a partition, X α mλ(x) = x α∼λ (b) for r ∈ N, X er(x) = xi1 xi2 ··· xir 1≤i1 21 (c) for r ∈ N, X X hr(x) = mλ = xi1 xi2 ··· xir |λ|=r 1≤i1≤i2≤···≤ir 2 2 2.42. Example. h2(x) = m(2)(x)+m(1,1)(x) = x1 +x2 +···+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 X r pr := xi i and, for any partition λ, we define Y pλ = pλi i 2.44. Proposition. The generating function for the pr is X xi P (t) = 1 − xit and also satisfies the identities H0(t) E0(t) P (t) = P (−t) = H(t) E(t) Proof. By direct computation, we see X r−1 X X r r−1 X xi P (t) := prt = xi t = 1 − xit r≥1 i≥1 r≥1 i≥1 Furthermore, if we are clever at manipulating generating functions, we see X d 1 d Y 1 d H0(t) P (t) = log = log = log(H(t)) = dt 1 − xit dt 1 − xit dt H(t) i≥1 i≥1 and X d d Y d E0(t) P (−t) = log(1 + x t) = log (1 + x t) = log E(t) = dt i dt i dt E(t) i≥1 i≥1 2.45. Corollary (Newton’s Identities). We have the following identities n n X X r−1 nhn = prhn−r, nen = (−1) pren−r r=1 r=1 22 Proof. These come from the proposition above by rearranging and expand- ing. For instance, 0 X i−1 X r−1 X j H (t) = P (t)H(t) =⇒ ihit = prt hjt i≥1 r≥1 j≥1 and so, fixing a specific n, we have X nhn = prhj r+j=n giving us the desired identity. 2.46. Theorem. The set {pλ}all partitions λ is a basis for Λ over a field of characteristic 0. Proof. By Newton’s identities, n X pn = ejpn−j j=1 if we rewrite this recurrence, we get n 1 X e = (−1)j−1e p n n n−j j j=1 Thus, each en can be written as a rational linear combination of the power sums, giving that Λn ⊆ Q[p1, . . . , pn]. Therefore, the pλ’s form a basis and ∼ because 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 matrix expressing pλ in terms of the monomial basis. One argues that X pλ = cλλmλ + cλµmµ µBλ µ1 µm with cλλ 6= 0. Then, if x1 ··· xm appears in λ1 λ2 λ2 λ2 pλ = (x1 + x2 + ··· )(x1 + x2 + ··· ) + ··· then each µi must be a sum of λj’s, thus making µ larger in dominance order than λ. This proof is probably simpler than the one above but has the 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 set Q mi zλ = i i mi!. Then, X pλ hn = zλ λ`n and X n−`(λ) pλ en = (−1) zλ λ`n 23 Proof. We show this by writing each hr as a linear combination of pλ’s. Observe, ∞ ∞ X k Y 1 hk(~x)t = 1 − xit k=0 i=1 Y = exp (− log(1 − xit)) ∞ k ! Y X (xit) = exp k k=1 ∞ ∞ k ! X X (xit) = exp k k=1 i=1 ∞ k ! X pk(x)t = exp k k=1 ∞ k Y pk(x)t = exp k k=1 ∞ ∞ k m Y X (pk(x)t ) k = m mk! · k k k=1 mk=0 X 1 = p t|λ| z(λ) λ λ P pλ Thus, we see that hn = . The relation for en is a similar manipula- λ`n zλ tion of the generating function E(t). 2.49. Remark. Note that z(λ) is equal to the cardinality of the centralizer of an element of Sn with cycle type λ. Indeed, an i-cycle (1, 2, . . . , i) is fixed by i elements and a product of mi disjoint i-cycles can be permuted in mi! ways. So then, why does this num- 2.3. Schur functions. ber show up 2.50. Definition. The Vandermonde determinant is here? n−1 n−2 x1 x1 ··· x1 1 n−1 n−2 x x ··· x2 1 ∆ := det 2 2 . . . . . . . . n−1 n−2 xn xn ··· xn 1 Y 2.51. Theorem. ∆ = (xi − xj) 1≤i