Symmetric Functions

Symmetric functions

Contents

1 Preliminaries 1

2 Monomial symmetric functions 2 2.1 Multiplying Monomials ...... 2 2.2 Dominance order ...... 3

3 Elementary symmetric functions 3

4 Homogeneous symmetric functions: hλ 3 4.1 Generating Functions ...... 4

5 Power symmetric functions: pλ 4 5.1 Generating Function: ...... 4

6 Schur functions 5 6.1 Skew Schur functions ...... 5 6.2 Symmetric group action ...... 5 6.3 Schur functions ...... 6 6.4 The Pieri Rule ...... 6

7 Littlewood-Richardson coeﬃcients 6 7.1 The bi-alternant formula ...... 7 7.2 Littlewood-Richardson coeﬃcients ...... 8 7.3 RSK-Correspondence and the Cauchy identity ...... 8

8 Jacobi-Trudi Determinants 9

1 Preliminaries

Let R be a commutative ring and x1, . . . , xn a set of indeterminants. Let α = (α1, . . . , αn) be a vector of non-negative integer coeﬃcients. The set of all α form a monoid under addition which is isomorphic to the monoid of all monomials xα under multiplication:

α β α+β α α αn x x = x with x = x1 ··· xn .

The corresponding algebra is the ring of polynomials, denoted R[x1, . . . , xn], consisting of all polynomials in n variables with coeﬃcients in R. α We write |α| = deg x = α1 + ··· + αn. An element of R[x1, . . . , xn] of the form

X α f(x1, . . . , xn) = cαx |α|=d is called a homogeneous polynomial of degree d. With set of all such polynomials denoted by d R[x1, . . . , xn] , R[x1, . . . , xn] is a graded ring:

M d R[x1, . . . , xn] = R[x1, . . . , xn] d≥0

1 There is a natural degree-preserving Sn action on polynomials, where elements of the symmetric group act by permuting the variables. That is, given P (x) ∈ R[x1, . . . , xn] and σ ∈ Sn:

σP (x1, x2, . . . , xn) = P (xσ1 , xσ2 , . . . , xσn ) .

2 2 For example, σ = (1 3 2) =⇒ σ x1 + 2x1x3 + x2 = x1 + 2x1x2 + x3 . A polynomial is called symmetric if it is invariant under the action of Sn. The set of symmetric d polynomials of degree d in n-variables is denoted Λn and we set,

M d Λn = Λn . d

Λn is a subring of R[x1, . . . , xn], called the ring of symmetric polynomials. Examples include:

x1 + 2x1x2x3 + x2 + x3 ∈ Λ3

3 x1x2x3 + x1x2x4 + x2x3x4 + x1x3x4 ∈ Λ4

2 Monomial symmetric functions

The most obvious symmetric polynomials are constructed by symmetrizing the monomial term xλ, for a partition λ = (λ1, . . . , λn) of weakly decreasing non-negative integers.

The monomial symmetric functions: mλ

X β β β1 β2 βn mλ = x where x = x1 x2 ··· xn β:β∗=λ over all distinct β where β∗ is the partition rearrangement of β.

For example, with λ = (2, 1, 1) and n = 4:

(2, 2, 1, 0) (2, 2, 0, 1) (1, 2, 2, 0) (1, 2, 0, 2) (2, 1, 2, 0) (2, 1, 0, 2) (0, 2, 2, 1) (0, 2, 1, 2) (0, 1, 2, 2)

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 m2,1,1 = x1x2x3 + x1x2x4 + x1x2x3 + x1x2x4 + x1x2x3 + x1x2x4 + x2x3x4 + x2x3x4 + x2x3x4

Furthermore, the mλ are clearly independent and if the homogeneous polynomial

X α f = cαx |α|=d is symmetric, then the cα must remain constant as α ranges over the Sn orbit. Thus, the set of all mλ d with |λ| = d spans Λn. Note if `(λ) > n, we don’t have enough variables and mλ = 0. To avoid this issue, we instead work in the vector space spanned by all the mλ called the ring of symmetric functions:

Λ = R[mλ] .

d Λ is closed under product, and is a graded ring. If Λ is the space spanned by all mλ of degree d, M Λ = Λd . d≥0

d Theorem 1. {mλ : |λ| = d} is a basis of Λ Corollary 2. The dimension of Λd is the number of partitions of d.

2 2.1 Multiplying Monomials The study of the product of certain monomial symmetric functions will later help us understand other bases for Λ. More precisely, we are interested in determining the possible terms in the right hand side of

X ν k m1k mλ = cλ1k mν where 1 = (1,..., 1) . |ν|=|λ|+k

Note that since the product of symmetric functions is symmetric, ν ν cλ1k = the coefficient of mν in mλm1k = the coefficient of x in mλm1k α ν Since the monomial terms in mλ are x where α rearranges to λ, cλ1k is the number of ways to fill rows of ν with at most one row of λ and of (1,..., 1). For example, filling shapes with (2,1) and (1,1) gives

−→ −→ −→ −→

m1,1m2,1 = 2m2,2,1 + m3,2 + 2m3,1,1 + 3m2,1,1,1 It turns out that the only shapes with at least one ﬁlling of λ and 1k are those obtained by adding a vertical k-strip to the diagram of λ. + a vertical 2-strip = , , ,

To convince yourself of this fact, note ﬁrst that since ν is obtained from the parts of λ and 1k, each row of ν must be λi or λi + 1 for some i. Thus, the length of the longest row will be at most λ1 + 1 and at least λ1 implying ν1 contains λ1. Similarly, the length of the second longest row will be λ2 + 1 or λ2 implying ν2 contains λ2, etc.

2.2 Dominance order

We put a partial order C on the set of all partitions (and consequently on the set of monomial symmetric functions mµ C mλ when µ C λ) called dominance order by declaring that

λ B µ ⇐⇒ λ1 + ··· + λj ≥ µ1 + ··· + µj for all j

Many bases are triangularly related with respect to B. Intuitively, λ is greater than µ in dominance order if the Ferrers diagram of λ is wider than µ. 0 0 Remark 3. λ C µ ⇐⇒ λ B µ We can use dominance order to reﬁne our result on the product of monomials:

X ν X ν m1k mλ = cλ,1k mν = mλ+1k + cλ,1k mν (1) k ν=λ+vertical k-strip ν C λ+1

k where λ + 1 = (λ1 + 1, . . . , λk + 1, λk+1,...).

3 Elementary symmetric functions

The elementary symmetric functions are eλ = eλ1 eλ2 ··· eλ` where X ek(x1, . . . , xn) = xi1 xi2 ··· xik and e0 = 1 .

1≤i1

For example, e1 = x1 + x2 + ··· and e2 = x1x2 + x1x3 + ··· . Clearly m1k = ek. Note the degree(ek) = k and degree(eλ = eλ1 ··· eλ` ) = |λ|. Further, when k > n, ek = 0.

Theorem 4. {eµ} forms a basis for Λ.

3 Proof. Homework.

a1 a2 ∞ Since the set {eλ} consists of all monomials e1 e2 ··· where ai ∈ N , and any element of Λ can be expressed as a linear combination of eλ, every element of Λ is uniquely expressible as a polynomial in the er and the er are algebraically independent. ∼ Theorem 5. Λ = Q[e1, e2,...]

4 Homogeneous symmetric functions: hλ

Deﬁnition: Let h0 = 1 and hλ = hλ1 hλ2 ··· hλ` where X X hk(x1, . . . , xn) = mλ = xi1 xi2 ··· xik

|λ|=k 1≤i1≤i2≤···≤ik≤n

Note that h1 = e1. Sometimes hk is called the complete symmetric function since it is the sum over all P P 2 P 2 2 monomials: h1 = xi and h2 = xi + xixj = x1 + x2 + x1x2 + ··· . The homogeneous functions are not triagulary related to the monomials. We shall thus appeal to the use of generating functions to show that the homogeneous symmetric functions provide a basis for Λ.

4.1 Generating Functions Deﬁne X k X k E(t) = ek t H(t) = hk t k≥0 k≥0 Theorem 6. Y Y 1 E(t) = (1 + xit) H(t) = 1 − xit i≥1 i≥1 Proof. Y X X 2 (1 + xit) = (1 + tx1)(1 + tx2) ··· = 1 + txi + t xixj + ··· i≥1 i≥1 i

d Theorem 7. {hλ : λ ` d} is a basis for Λ .

Proof. Since the number elements hλ is the number of partitions of d, it suﬃces to show that they generate the eµ. Further, since both hλ and eλ are multiplicative we shall simply show that ek = f(h1, . . . , hk) for 1 some polynomial f. Our identies reveal that H(t) = E(−t) , or H(t)E(−t) = 1. Substituting hk and e` in the summations gives ! ! X k X l t hk (−t) el = 1 , k l and by comparing coeﬃcients of tn in both sides we see that

X ` (−1) hke` = 0 n 6= 0 (2) k+l=n

Thus, en = h1en−1 − h2en−2 + · · · ± hn−1e1 which is a polynomial in the h’s by induction on n.

4 5 Power symmetric functions: pλ

X k pk = xi and pλ = pλ1 ··· pλ` i r r p1 = x1 + x2 + ··· = e1 = h1 pr = x1 + x2 + ··· = m(r)

d Theorem 8. {pλ : λ ` d} is a basis of Λ P Proof. Let pλ = µ cλµmµ. If we can show that cλµ = 0 for all µ ≥ λ and cλλ 6= 0 then the transition µ1 µ2 µn λ1 λ1 λ2 matrix is invertible and then pλ must be a basis. If x1 x2 ··· xn appears in pλ = (x1 +x2 +··· )(x1 + λ2 x2 + ··· ) ··· , then each µi is a sum of λj’s. Since adding together parts of a partition makes it become larger in dominance order, mλ must be the smallest term that occurs.

5.1 Generating Function: Studying the generating function for the power sums reveals: Proposition 9. def X tn P (t) p = ln H(t) = n n n≥1

1 Proof. Using the Taylor expansion of ln 1−x ,

n n Y 1 X 1 X X (xit) X t X n ln = ln = = xi (1 − xit) (1 − xit) n n i≥1 i≥1 i≥1 n≥1 n≥1 i≥1

Q ri Proposition 10. Let zλ = i i r!

X 1 X (−1)|λ|−`(λ) hn = pλ and en = pλ zλ zλ λ`n λ`n Proof. It suﬃces to show X n X 1 |λ| H(t) = hnt = pλt zλ n≥0 λ Using the proposition,

   k n n r1 2r2 X t X 1 X t X 1 k p1t p2t H(t) = exp  pn  =  pn  = ··· n k! n k! r1, r2,... 1 2 n≥1 k≥0 n≥1 k≥0 r1+r2+···=k 1 pr1 pr2 ··· 1 X 1 2 r1+2r2+··· X |λ| = r r t = pλt r1!r2! ··· 1 1 r 2 ··· zλ r1,r2,... λ where λ = (··· , 3r3 , 2r2 , 1r1 ).

6 Schur functions 6.1 Skew Schur functions Semi-standard skew tableaux are deﬁned on any skew shape will the condition that the ﬁllings are weakly increasing in rows and strictly increasing in columns. The evaluation(content) of any such tableau T is (m1, m2,...) where mi is the number of time i occurs in T .

5 The skew Schur function is deﬁned

X ev(T ) sλ/µ = x , T ∈T (λ/µ) where T (λ/µ) is the set of all semi-standard skew tableaux of shape λ/µ.

The ﬁrst job is to prove these functions are symmetric. As such, let τi denote the simple transposition of letters i and i + 1. Since τi ∈ Sn where τi(α1, . . . , αi, αi+1,...) = (α1, . . . , αi+1, αi,...), it suﬃces to show that there is an involution σi on T (λ/µ) that permutes letters in the tableau so that

ev(σi(T )) = τi(ev(T )) . (3)

6.2 Symmetric group action Any two entries i and i + 1 of a tableau T are married if they occur in the same column. The conditions on rows and columns of skew tableaux imply that if i is married, then its spouse i + 1 occurs in the cell directly above it. And, if any two entries i on the same row are both married, then all entries occuring between these cells are part of a married couple.

i+1 i+1 . i i Furthermore, since columns are strictly increasing, any unmarried entries i or i + 1 in a row must lie to the right of all married i, and similarly any unmarried entries i or i + 1 in a row must lie to the left of all married i + 1. Thus, the set of all unmarried i and i + 1 in any row lie in a contiguous sequence such any cell above these contains on letters strictly larger than i + 1 and those below this sequence have letters strictly smaller than i. Thus, σi is deﬁned to act on a tableaux by replacing every such sequence of r ≥ 0 unmarried i’s and s ≥ 0 unmarried i + 1’s with s i’s followed by r i + 1’s. Since the number of married i’s and i + 1’s is the same, this involution meets the condition (3).

6.3 Schur functions

An important special case of skew Schur functions is the Schur function, sλ = sλ/∅. More precisely,

X ev(T ) sλ = x , T ∈T (λ) where T (λ) is the set of all semi-standard tableaux of shape λ = λ/∅. In three variables, we consider tableaux with ﬁllings on the set {1, 2, 3} and n o 2 3 2 3 3 3 3 2 T (2, 1) = 1 1 1 1 1 2 1 3 2 2 2 3 1 2 1 3 → (2, 1, 0) (2, 0, 1) (1, 2, 0) (1, 0, 2) (0, 2, 1) (0, 1, 2) (1, 1, 1) (1, 1, 1) ev 2 2 2 2 2 2 =⇒ s2,1(x1, x2, x3) = x1x2 + x1x3 + x1x2 + x1x3 + x2x3 + x2x3 + x1x2x3 + x1x2x3 = m2,1 + 2m1,1,1 In the case that λ = (k) or (1k), the Schur functions reduce to the kth complete or elementary symmetric function. That is,

X T T m1 m2 s1k = ek = x where x = x1 x2 ··· T ∈T (1k)

Since the set T (1k) contains only vertical k-strips, the multiplicity of any entry in T must be at most 1. Thus, any term xT in the sum must be a monomial with variables to the ﬁrst power only. On the other hand, X T sk = hk = x T ∈T (k)

6 where we sum over all horizontal k-strips with entries from {1, . . . , n}. Since a horizontal k-strip may contain any k subset of n, i.e. anything from k 1’s to {1, 2, . . . , k}, we see a term xT is a monomial in variables to any power. Since the action σi on tableaux implies the Schur functions are symmetric, we can expand any Schur function in terms of the monomial basis. For any partition λ, X sλ = Kλµmµ µ`|λ| where the Kostka number Kλµ is the number of tableau of shape λ and evaluation µ. Note that the coeﬃcient of µ = 1n will simply be the number of standard tableaux of shape λ since evaluation (1,..., 1) implies each entry occurs only once.

Theorem 11. {sλ} is an integral basis for Λ

Proof. It suffices to show Kλµ = 0 for all µ > λ and Kλλ = 1. If Kλµ is non-zero then there must exist a tableau of shape λ with µ1 ones, µ2 twos, .... Assume µ > λ. This implies λ1 +···+λi−1 = µ1 +···+µi−1 and λi < µi for some i. The column strict condition implies the first row must have all µ1 ones, the second row the µ2 twos, .... In row i of size λi, since µi > λi, the µi i’s will not all fit. However, a leftover i cannot go above this row without violating the increasing column condition and all rows below have already been filled. Therefore, for a λ shaped tableau to exist with evaluation µ, µ ≤ λ. Moreover, 0 our argument shows that if λ = µ, our only option is to fill each row λj with µj j s. Thus Kλλ = 1.

6.4 The Pieri Rule Recall we showed that the formal sum of tableaux

7 Littlewood-Richardson coeﬃcients

More generally is the question of the Schur function expansion of a product of two Schur functions.

7.1 The bi-alternant formula The deﬁnition for Schur functions was originally given by a ratio determinants called alternants:

βj X σβ aβ = det(xi ) = sgn(σ) x . σ∈Sn

Note that for any σ ∈ Sn we have aσβ = sgn(σ)aβ. Letting ρ = (n − 1, n − 2,..., 0), the Schur functions were then deﬁned by the “bi-alternant” formula:

aλ+ρ s˜λ = aρ

As the ratio of determinants, the functions are clearly symmetric. However, the proof thats ˜λ = sλ is equivalent to the functions deﬁned by tableux follows from an interesting identity.

Our point of departure is to consider subtableaux of T ∈ T (λ/µ). let T≥j denote the tableau formed by the columns j, j + 1,... of T . Similarly deﬁned are T>j,Tj, and T

7 Corollary 13. For any partition µ, s˜µ = sµ Proof. The symmetric group action on T (µ/ν) implies that the number of tableaux of evaluation e equals the number with evaluation σe for any σ ∈ Sn. Thus σ(λ + ρ) + ev(T ) and σ(λ + ρ + ev(T )) occur the same number of times as T ranges over T (µ/ν). Therefore

X X σ(λ+ρ) ev(T ) X X σ(λ+ρ+ev(T )) X aλ+ρ sµ/ν = sgn(σ)x x = sgn(σ)x = aλ+ev(T )+ρ

σ∈Sn T ∈T (µ/ν) σ∈Sn T ∈T (µ/ν) T ∈T (µ/ν) We thus are left to prove that the alternants die when T is not λ-yamanouchi. The idea is as follows: suppose λi = λi+1 and T has no i’s and exactly one i + 1. T is not λ-yamanouchi since λ + ev(T ) has λi+1 + evi+1(T ) = λi + evi(T ) + 1 – illegal. In this case, the corresponding alternant aλ+ev(T )+ρ is zero since ρi+1 = ρi − 1 and λi+1 + evi+1(T ) + ρi+1 = λi + evi(T ) + ρi imply that two columns in the determinant are equal. In the general case, we shall prove that there is a unique pairing of all tableaux that are not λ- yamanouchi such that the corresponding alternants cancel out. To this end, suppose T is not λ-yamanouci. Then λ + ev(T≥j) is not a partition for some j, we choose the largest such j. Let k denote the ﬁrst row that is larger then its successor. This given,

λk + evk(T≥j) < λk+1 + evk+1(T≥j) and λk + evk(T>j) ≥ λk+1 + evk+1(T>j) . Since columns of T are strictly increasing, there is at most one k and at most one k + 1 in any column. Thus, the inequality on the right is forced to be an equality and the jth column of T contains k + 1 but no k: λk+1 + evk+1(T ) = λk + evk(T ) + 1. The entry (in column j+1) southwest of the k+1 in column j is strictly smaller than k implying all entries to the west of this are also smaller than k. Thus the tableau Tˆ obtained by acting with the symmetric group on T

7.2 Littlewood-Richardson coeﬃcients

Now, dividing the identity (4) by aρ and using the bi-alternant formula gives X sλ sµ/ν = sλ+ev(T ) . T ∈T (µ/ν) T =λ−yamanouchi When λ = ∅, the yamanouchi tableaux describe the Schur function expansion of a skew Schur function. X sµ/ν = sev(T ) , T ∈T (µ/ν) T =yamanouchi and taking ν = ∅ gives the formula for the Schur function expansion of a product of two Schur functions: X sλ sµ = sλ+ev(T ) . T ∈T (µ) T =λ−yamanouchi Note that the Pieri formulas are the special case when µ = r or 1r.

7.3 RSK-Correspondence and the Cauchy identity We have seen that there is a bijection between standard tableaux and permutations. The RS-correspondence can be generalized to pinpoint a precise relation between pairs of tableaux and generalized permutations. Such a permutation is a two-line array of positive integers whose columns are in lexicographic order with the top entry taking precedence: u u ··· u w = 1 2 m v1 v2 ··· vm

8 1 Lexicographic order implies u1 ≤ u2 ≤ · · · ≤ um and vk−1 ≤ vk if uk−1 = uk . The Robinson-Schensted-Knuth correspondence takes a generalized permutation π to a pair of tableaux by the following algorithm: 1 1 2 2 2 3 w = 3 4 1 1 2 1

∅ ← 3 = 4 1

4 ← 4 = 3 4 1 1

3 2 3 4 ← 1 = 1 4 1 1

3 3 4 2 2 1 4 ← 1 = 1 1 1 1

3 4 3 4 2 2 1 1 ← 2 = 1 1 2 1 1 2

3 3 3 4 2 4 2 2 1 1 2 ← 1 = 1 1 1 1 1 2

Note that when the top row is (1, . . . , n) and the bottom row is a permutation of these letters, a two-rowed array is a permutation in two line notation. As with the RS-correspondence, we refer to the resulting tableaux as the insertion and recording tableau respectively. Theorem 14. The RSK-correspondence is a bijection between two-rowed arrays in lexicographic order and pairs of same-shaped tableaux. Proof. Let P and Q denote the insertion and recording tableau respectively. The insertion algorithm ensures that P is a tableau. Now Q has weakly increasing rows since the cells are added to the periphery at each step and entries are taken from the top of the two-rowed array – a non-decreasing sequence. When two entries of the top row of this array are equal, the corresponding entries in the bottom row are non-decreasing by lexicographic order. Thus, the insertion process of these consecutive entries x ≥ x0 gives rise to cells that do not lie in the same column by the row bumping lemma and therefore the columns of Q are increasing. On the other hand, given a corner and a tableau we have the deletion algorithm that is the precise inverse of Schensted insertion. Choosing corners from largest to smallest, where the rightmost of repeated entries is the largest, we can construct a two-rowed array with lexicographic ordering. Theorem 15. Cauchy Identity

Y 1 X = sλ(x)sλ(y) (5) (1 − xiyj) i,j λ

α β Proof. The coeﬃcient of x y in the right hand side of () is KλαKλβ, or the number of possible pairs of λ-shaped tableaux (P,Q) such that the evaluation of P is α and ev(Q) = β. On the other hand,

1Note: the ordering is lexicographic as an ordering on pairs „u« „u0« ≤ if u < u0 or if u = u0 and v ≤ v0 v v0

9 the left hand side can be expanded using a product of geometric series, 1 Y Y X ai,j = (xiyj) (6) (1 − xiyj) i,j i,j ai,j ≥0    

X a1,1 X a1,2 =  (x1y1)   (x1y2)  ··· (7)

a1,1≥0 a1,2≥0    

X a2,1 X a2,2  (x2y1)   (x2y2)  ··· (8)

a2,1≥0 a2,2≥0

By our recent bijection, we have proved the coeﬃcients of xαyβ in both sides of the identity are equivalent.

8 Jacobi-Trudi Determinants

Recall that sn = hn and s1n = en. More generally, there are simple determinantal expressions for sλ in terms of the homogeneous and elementary bases. For any partition λ, the Jacobi-Trudi formulas are:

0 sλ = hλi−i+j and sλ = eλi−i+j (9)