Symmetric functions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 04 March 2012

Symmetric functions

The symmetric group Sn acts on the vector space n= -span{x1,...,xn} by wxi=xw(i), for wSn, 1in. This action induces an action of Sn on the polynomial ring [Xn] = [x1,...,xn] by ring automorphisms. For a seq uence γ=(γ1,...,γn) of non-negative integers let xγ= x1γ1xnγn, so that [x1,...,xn] = -span{xγ  |  γ0n}. The ring of symmetric functions is [Xn]Sn = {f[Xn]  |  wf=f  for all   wSn}. (Sf 1.1) Define the orbit sums, or monomial symmetric functions, by mλ = γSnλ xγ, for   λ0n, where Snλ is the orbit of λ under the action of Sn. Let P+ = {λ=λ1,...,λn)0n  |  λ1λn} (Sf 1.2) so that {mλ  |  λP+} is a  -basis of   [Xn]Sn. (Sf 1.3)

Partitions

A partition is a collection μ of boxes in a corner where the convention is that gravity goes up and to the left. As for matrices, the rows and columns of μ are indexed from top to bottom and left to right, respectively. The  parts  of  μ  are μi= ( the number of boxes in row  i of μ ), the  length  of  μ  is l(μ)= ( the number of rows of μ ), (Sf 1.4) the  size  of  μ  is |μ|= μ1++μl(μ)= ( the number of boxes of μ ). Then μ is determined by (and identified with) the sequence μ=(μ1,...,μl) of positive integers such that μ1μ2μl>0, where l=l(μ). For example, (5,5,3,3,1,1)= . A partition of k is a partition λ with k boxes. Write λ&rline;k if λ is a partition of k. Make the convention that λi=0 if i>l(λ). The dominance order is the partial order on the set of partitions of k, P+(k) = {partitions of  k} = {λ=(λ1,...,λl)  |  λ1λl>0,  λ1++λl=k}, given by λμ if λ1+λ2++λi μ1+μ2++μi for all   1imax{l(λ),l(μ)}. PUT THE PICTURE OF THE HASSE DIAGRAM FOR k=6 HERE.

Tableaux

Let λ be a partition and let μ=(μ1,...,μn) 0n be a sequence of nonnegative integers. A column strict tableau of shape λ and weight μ is a filling of the boxes of λ with μ1 1s, μ2 2s, ... , μn ns, such that

  1. the rows are weakly increasing from left to right,
  2. the columns are strictly increasing from top to bottom.
If p is a column strict tableau write shp(p) and wt(p) for the shape and the weight of p so that shp(p) = (λ1,...,λn), where λi=  number of boxes in row  i  of  p,  and wt(p) = (μ1,...,μn), where μi=  number of  is  in  p. For example, p= 1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 4 3 3 3 4 4 4 5 4 5 5 6 6 7 7 has shp(p) = (9,7,7,4,2,1,0) and wt(p) = (7,6,5,5,3,2,2). For a partition λ and a sequence μ= (μ1,...μn) 0 of nonnegative integers write B(λ) = {column strict tableaux  p  |  shp(p)=λ}, B(λ)μ = {column strict tableaux  p  |  shp(p)=λ  and   wt(p)=μ}. (Sf 1.5)

Elementary symmetric functions

Define symmetric functions er, 0rn, via the generating function i=1n (1-xiz) = r=0n (-1)rerzr. Then e0=1 and, for 0rn, er=m(1r) = 1i1<i2<<irn xi1xi2xir = shp(p)=(1r) xwt(p), where the last sum is over all column strict tableaux p of shape (1r).

If f(t) is a polynomial in t with roots γ1...γn then the coefficient of   tr   in   f(t)   is   (-1)n-r er(γ1,...,γn). If A is an n×n matrix with entries in 𝔽 with eigenvalues γ1...γn then the trace of the action of A on the rth exterior power of the vector space 𝔽n is tr(A,r𝔽n) = er(γ1,...,γn),   so that   Tr(A) = e1(γ1,...,γn), det(A) = en(γ1,...,γn), and the characteristic polynomials of A is chart(A) = r=0n (-1)n-r en-r (γ1,...,γn)tr.

Let γ=(γ1,...,γn) be a partition. Then eλ' = μλ aλ'μmμ, where aλ'μ is the number of matrices with entries from {0,1} with row sums λ' and column sums μ. Furthermore, aλ'λ=1 and aλ'μ=0 unless μλ.

Proof.
If A is an l×n matrix with entries from {0,1} let xA = i=1n (xi)aij and define rs(A) = (ρ1,...,ρn), cs(A) = (γ1,...,γn), where ρi = j=1l aij  and  γj = i=1n aij, so that rs(A) and cs(A) are the sequences of row sums and column sums of A, respectively. If λ'= (λ1',...,λl') then eλ' = j=1l eλj' = rs(A)=λ' xA = γ0n rs(A)=λ' cs(A)=γ xγ = μ aλ'μ mμ. Since there is a unique matrix A with rs(A)=λ' and cs(A)=λ, aλ'λ=1. If A is a 0,1 matrix with rs(A)=λ' and cs(A)=mu then μ1++μi λ1++λi since there are at most λ1++λi nonzero entries in the first i columns of A. Thus aλ'μ=0 unless μλ.

  1. The set {eλ  |  l(λ')n} is a basis of [Xn]Sn.
  2. [Xn]Sn = [e1,...,en].

Complete symmetric functions

Define symmetric functions hr, r0, via the generating function i=1n 1 1-xiz = r0 hrzr. Then h0=1 and, for r>0, hr = λr mλ = 1i1i2irn xi1xi2xir = sh(p)=(r) xwt(p), where the last sum is over all column strict tableaux p of shape (r).

There is an involutive automorphism ω of [Xn]Sn defined by ω: [Xn]Sn [Xn]Sn ek hk

Proof.
Comparing coefficients of zk on each side of 1= i=1n (1-xiz) i=1n 1 1-xiz yields 0= r=1k (-1)r erhn-r.

  1. The set {hλ  |  l(λ')n} is a basis of [Xn]Sn.
  2. [Xn]Sn = [h1,...,hn].

The monomials in {x1ε1x2ε2xnεn  |  0εin-i} form a basis of [x1,...,xn] as a [x1,...,xn] Sn module.

Proof.
Let I= e1,...,en be the ideal in [x1,...,xn] generated by e1,...,en. Since (1-x1t)(1-xnt) = 0 modI, (1-xi-1t)(1-xnt) = 1 (1-x1t)(1-xit)  modI, and so r=0n-i (-1)r er (xi+1,...,xn)tr = l0 hl(x1,...,xi)tl modI. Comparing coefficients of tn-i+1 on each side gives that, for all 1in, 0=hn-i+1(x1,...,xi) = r=0n-i+1 xn-i+1-r hr(x1,...,xi-1) modI, and thus xin-i+1 = - r=1n-i+1 xn-i+1-r hr(x1,...,xi-1) modI. (Sf 1.11) This identity shows (by induction on i) that xin-i+1 can be rewritten, modI, as a linear combination of monomials in x1,...,xi with the exponent of xi being n-i. In particular, 0=hn-1+1(x1) = x1n  modI and it follows that any polynomial can be written, modI, as a linear combination of monomials x1ε1x2ε2xnεn with 0εin-i. (Sf 1.12) If Sk is the set of homogeneous degree k polynomials in S= [x1,...,xn] and (SW)k is the set of homogeneous degree k polynomials in SW = [e1,...,en] = [x1,...,xn]Sn the Poincaré series of S and SW are 1 (1-t)n = k0 dim(Sk)tk and i=1n ( 1 1-ti ) = k0 dim((SW)k)tk. Then the Poincaré series of S/I is i=1n 1-ti 1-t = [n]! = 1(1+t)(1+t++tn-1). There are n(n-1)21=n! monomials in (???) and thus the monomials in (*) form a basis of S as an SW module. The relations (???) provide a way to expand any polynomial in terms of this basis (with coefficients in SW).

The groups Gr,p,n

Let r and n be positive integers. The group Gr,1,n is the group of n×n matrices with

  1. exactly one non zero entry in each row and each column,
  2. the nonzero entries are rth roots of 1.
Let p be a positive integer (not necessarily prime) such that p divides r. The group Gr,p,n is defined by the exact sequence {1} Gr,p,n Gr,1,n φ /p {1}, where φ(g) = gij0 gij p is the pth power of the product of the nonzero entries of g, and /p is identified with the group of pth roots of unity. Thus Gr,p,n = kerφ is a normal subgroup of Gr,1,n of index p. Examples are
  1. G1,1,n = Sn = WAn-1 is the symmetric group (the Weyl group of type An-1),
  2. G2,1,n = On() = WBn is the hyperoctahedral group of orthogonal matrices with entries in (the Weyl group of type Bn),
  3. G2,2,n = WDn is the group of signed permutations with an even number of negative signs (the Weyl group of type Dn),
  4. Gr,1,1 = /r is the cyclic group of order r of rth roots of unity, and
  5. Gr,r,2 = WI2(r) is the dihedral group of order 2r.

Let ξ=e 2πi r be the primitive rth root of unity and let 𝔬=[ξ]. If x1,...,xn is a basis of 𝔬n then the natural action of Gr,p,n extends uniquely to an action of Gr,p,n on the polynomial ring 𝔬[x1,...,xn] by ring automorphisms. The invariant ring is 𝔬[x1,...,xn]Gr,p,n = {f𝔬[x1,...,xn]  |  wf=f   for all   wGr,p,n}.

Let fi (x1,...,xn) = ei (x1r,...,xnr), for   1in-1   and fn (x1,...,xn) = en (x1rp,...,xnrp).

  1. 𝔬[x1,...,xn]Gr,p,n = 𝔬[f1,...,fn].
  2. 𝔬[x1,...,xn] is a free 𝔬[x1,...,xn]Gr,p,n-module with basis {x1ε1x2ε2xnεn  |  0ε1 r p -1   and   0ε1ir-1,  for   2in}.

Proof.
To show: f1,...,fn generate 𝔬[Xn]W and they are algebraically independent.

Each element wGr,1,n can be written uniquely in the form w= t1γ1tnγnσ, where ti = diag(1,...,1,ξ,1,...,1), σSn, 0γir-1, so that ti is the diagonal matrix with 1s on the diagonal except for ξ in the ith diagonal entry. The element wGr,p,n if γ1++γn = 0 modp, and thus Gr,p,n = {w=t1γ1tnγnσ  |  σSn,  0γn r p -1,   and   0γir-1   for   1in-1}. For each wGr,p,n define a monomial xw = j=1n (xσ(1)xσ(j))γj i  such that σ(i)>σ(i+1) (xσ(1)xσ(i)) .

The polynomial ring 𝔬[x1,...,xn] is a free 𝔬[x1,...,xn]Gr,p,n-module with basis {xw  |  wGr,p,n}.

Proof.

Notes and References

Taken from lecture notes on symmetric functions by Arun Ram.

References

References?

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