## Interpolating symmetric functions

Last update: 21 July 2012

## Interpolating symmetric functions

Define ${q}_{r}\left({X}_{n};q,t\right)={q}_{r}\left({x}_{1},...,{x}_{n};q,t\right)$ by the generating function $∏i=1n 1-xitz 1-xiqz = (q-t) ∑r≥0 qr( x1,...,xn;q,t ) zr.$ Note that with this definition ${q}_{0}=\frac{1}{\left(q-t\right)}.$ Define the elementary symmetric functions, the complete symmetric functions and the power symmetric functions by the formulas $(-t)r-1 er(Xn) = qr( Xn;0,t ), qr-1 hr(Xn) = qr( Xn;q,0 ), and qr-1 pr(Xn) = qr( Xn;q,q ), respectively. (ISF 1)$ The elementary symmetric functions have special importance because of the following ways in which they appear naturally.

1. If $f\left(t\right)$ is a polynomial in $t$ with roots ${\gamma }_{1},...,{\gamma }_{n}$ then
2. If $A$ is an $n×n$ matrix with entries in $𝔽$ with eigenvalues ${\gamma }_{1},...,{\gamma }_{n}$ then the trace of the action of $A$ on the ${r}^{\mathrm{th}}$ exterior power of the vector space ${𝔽}^{n}$ is $tr( A, ⋀r𝔽n ) = er( γ1,...,γn ), so that Tr(A) = e1( γ1,...,γn ), and det(A) = en(γ1,...,γn), (ISF 3)$ and the characteristic polynomial of $A$ is $chart(A) = ∑r=0n (-1)n-r en-r( γ1,...,γn ) tr. (ISF 4)$

Expanding $\frac{1-{x}_{i}tz}{1-{x}_{i}qz}=1+\left(q-t\right)\sum _{l>0}{q}^{l-1}{x}_{i}^{l}{z}^{l}$ and multiplying out $∏i=1n 1-xitz 1-xiqz = 1-x1tz 1-x1qz ⋯ 1-xntz 1-xnqz$ gives $qr = ∑ 1≤i1 ≤⋯≤ ir≤n (q-t) Card( {j | ij from which it follows that $qr = ∑λ⊢r (q-t) l(λ)-1 q r-l(λ) mλ( x1,...,xn ).$ For an $n×n$ matrix $a=\left({a}_{ij}\right)$ with entries from ${ℤ}_{\ge 0}$ let $xa = ∏i=1n (xi)aij, and wt(a) = q |λ|-l(a) (q-t) l(a)-l(λ) ,$ where $l\left(a\right)$ is the number of nonzero entries in $a,$ $l\left(\lambda \right)$ is the number of nonzero entries in $\lambda ,$ and $|\lambda |$ is the sum of the entries of $\lambda .$ Define $rs(a) = ( ρ1,...,ρn ), cs(a) = ( γ1,...,γn ), where ρi = ∑j=1laij and γj = ∑i=1naij,$ so that $\mathrm{rs}\left(a\right)$ and $\mathrm{cs}\left(a\right)$ are the sequences of row sums and column sums of $a,$ respectively.

For a sequence of nonnegative integers $\lambda =\left({\lambda }_{1},...,{\lambda }_{l}\right)$ define $qλ( Xn;q,t ) = qλ1( Xn;q,t ) ⋅ qλ2( Xn;q,t ) ⋅⋯⋅ qλl( Xn;q,t ).$ Then $qλ = ∑μ aλμ (q,t) mμ, where aλμ (q,t) = ∑ a∈Aλμ wt(a),$ and the sum is over partitions $\mu$ such that $|\mu |=|\lambda |.$

 Proof. If $\lambda =\left({\lambda }_{1},...,{\lambda }_{l}\right)$ then $qλ = ∏j=1lqλj = ∑ rs(a)=λ wt(a)xa = ∑γ∈ ℤ≥0n ∑ rs(a)=λ cs(a)=γ wt(a)xγ = ∑μ aλμ (q,t) mμ.$ $\square$

Multiplying out $∏i=1n 1-xitz 1-xiqz = 1 1-x1qz ⋅ 1 1-x2qz ⋅⋯⋅ 1 1-xnqz ( 1-xntz ) ( 1-xn-1tz ) ⋅⋯⋅ ( 1-x1tz )$ gives $qr = ∑ i1≤i2 ≤⋯≤ik >ik+1 >⋯>ir qk-1 (-t)r-k xi1 ⋯ xik xik+1 ⋯ xir. (ISF 6)$ The bijection ???? between sequences ${i}_{1}\le {i}_{2}\le \cdots \le {i}_{k}>{i}_{k+1}>\cdots >{i}_{r}$ and column strict tableaux of shape $\left(k{1}^{r-k}\right)$ yields $qr = ∑k=1r (-t)r-k qk-1 s(k1r-k) (Xn). (ISF 7)$

For each positive integer $k$ define $[k]q,t = qk-tk q-t = qk-1 + tqk-2 +⋯+ tk-2q + tk-1.$

Comparing coefficients of ${z}^{r}$ on each side of $( ∏i=1n 1-xisz 1-xitz ) ( ∏i=1n 1-xitz 1-xiqz ) = ∏i=1n 1-xisz 1-xiqz$ gives $(t-s) qr(Xn;t,s) + (q-t) (t-s) ( ∑j=1r-1 qj( Xn;q,t ) qr-j( Xn;t,s ) ) + (q-t) qr( Xn;q,t ) = (q-s) qr( Xn;qs ). (ISF 8)$

Example. Putting and $t=0$ in (???) yield, respectively, $qr( Xn;q,t ) + ( ∑j=1r-1 hj(Xn) tj qr-j( Xn;q,t ) ) - hr(Xn) [r]q,t = 0 qr( Xn;q,t ) + ( ∑j=1r-1 ej(Xn) (-q)j qr-j( Xn;q,t ) ) + (-1)r er(Xn) [r]q,t = 0 ∑j=0r (-t)r-j [j]q,t hj(Xn) er-j(Xn) = (q-t) qr( Xn;q,t ).$ Then putting ????? (???) gives $rqr( Xn;q,t ) - ( ∑j=1r-1 pj(Xn) ( qj-tj ) qr-j( Xn;q,t ) ) - pr(Xn) [r]q,t = 0,$ and the Newton identities $khk = ∑i=1k pihk-i and kek = ∑i=1k (-1)i-1piek-i,$ are obtained by putting ??? in (???).

Let $\lambda =\left({\lambda }_{1},...,{\lambda }_{n}\right)$ be a partition. Then

1. ${e}_{\lambda \prime }=\sum _{\mu \le \lambda }{a}_{\lambda \prime \mu }{m}_{\mu },$ where ${a}_{\lambda \mu }$ is the number of matrices with entries from $\left\{0,1\right\}$ with row sums $\lambda \prime$ and column sums $\mu .$
2. ${a}_{\lambda \prime \lambda }=1$ and ${a}_{\lambda \prime \mu }=0$ unless $\mu \le \lambda .$
3. $\left\{{e}_{\lambda }\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}l\left(\lambda \prime \right)\le n\right\}$ is a $ℤ-$basis of $ℤ{\left[{X}_{n}\right]}^{{S}_{n}}.$
4. $ℤ{\left[{x}_{1},...,{x}_{n}\right]}^{{S}_{n}}=ℤ\left[{e}_{1},...,{e}_{n}\right]=ℤ\left[{h}_{1},...,{h}_{n}\right]$ and $ℚ{\left[{x}_{1},...,{x}_{n}\right]}^{{S}_{n}}=ℚ\left[{p}_{1},...,{p}_{n}\right].$
5. The set $\left\{{h}_{\lambda }\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}l\left(\lambda \prime \right)\le n\right\}$ is a basis of $ℤ{\left[{X}_{n}\right]}^{{S}_{n}}.$

 Proof. Follows by putting $q=0$ in (???). Since there is a unique matrix $A$ with $\mathrm{rs}\left(A\right)=\lambda \prime$ and $\mathrm{cs}\left(A\right)=\lambda ,$ ${a}_{\lambda \prime \lambda }=1.$ If $A$ is a 0, 1 matrix with $\mathrm{rs}\left(A\right)=\lambda \prime$ and $\mathrm{cs}\left(A\right)=\mu$ then ${\mu }_{1}+\cdots +{\mu }_{i}\le {\lambda }_{1}+\cdots +{\lambda }_{i}$ since there are at most ${\lambda }_{1}+\cdots +{\lambda }_{i}$ nonzero entries in the first $i$ columns of $A.$ Thus ${a}_{\lambda \prime \mu }=0$ unless $\mu \le \lambda .$ This is a consequence of (b) and the fact that $\left\{{m}_{\lambda }\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}l\left(\lambda \right)\le n\right\}$ is a basis of $ℤ{\left[{X}_{n}\right]}^{{S}_{n}}.$ The first equality is an immediate consequence of (c). The second equality follows from the identity (???), which allows one to, inductively, expand ${h}_{r}$ in terms of ${e}_{r},{e}_{r-1},...,{e}_{1}.$ Similarly, the third equality follows from the Newton identity (???) which allows one to, inductively, expand ${p}_{r}$ in terms of ${e}_{r},{e}_{r-1},...,{e}_{1}$ (with coefficients in $ℚ$). $\square$

1. There is an involutive automorphism $\omega$ of $ℤ{\left[{X}_{n}\right]}^{{S}_{n}}$ defined by $ω: ℤ[Xn]Sn → ℤ[Xn]Sn ek ↦ hk .$
2. $\omega \left({q}_{r}\left({X}_{n};q,t\right)\right)={q}_{r}\left({X}_{n};-t,-q\right)$ and $\omega \left({p}_{k}\right)={\left(-1\right)}^{k-1}{p}_{k}.$

 Proof. The map $\omega$ is a well defined ring homomorphism since $ℤ{\left[{X}_{n}\right]}^{{S}_{n}}=ℤ\left[{e}_{1},...,{e}_{n}\right]$ is a polynomial ring. Comparing coefficients of ${z}^{k}$ on each side of $1 = ( ∏i=1n (1-xiz) ) ( ∏i=1n 1 1-xiz ) yields 0 = ∑r=1k (-1)r erhn-r.$ Thus ${e}_{1}={h}_{1},$ and $hk = ∑i=1k(-1) eihk-i and ek = (-1)-k ∑i=0k(-1) eihk-i = ∑j=1k (-1)-j-1 ek-j hj. (ISF 9)$ From the first of these relations, by induction on $k,$ $ω(hk) = ∑i=1k (-1)i+1 hiek-i,$ and, by comparing this identity with the second relation in (???) shows that $\omega \left({h}_{k}\right)={e}_{k}.$ Hence ${\omega }^{2}=\mathrm{id}.$ ???? $\square$

For a partition $\lambda =\left({1}^{{m}_{1}}{2}^{{m}_{2}}\cdots \right)$ of $k$ define $zλ = 1m1 m1! 2m2 m2! ⋯ so that n! zλ = Card( {w∈Sk | w has cycle type λ} ) (ISF 10)$ is the size of the conjugacy class indexed by $\lambda$ in the symmetric group ${S}_{k}.$ Recalling that $ln( 1-xiyj ) = ∑k≥1 xikyjk k since ln(1-t) = ∫11-tdt = ∫( 1+t+t2+⋯ )dt,$ we have $∏i,j 1 1-xiyj = expln( ∏i,j 1 1-xiyj ) = exp ∑i,j ln( 1-xiyj ) = exp( ∑k∑i,j xikyjk k ) = exp( ∑k pk(x)pk(y) k ) = ∏kexp( pk(x) pk(y) k ) = ∏k ∑mk≥0 ( pkmk(x) pkmk(y) kmkmk! ) = ∑m1,m2,... ( p1m1(x) p2m2(x) ⋯ p1m1(y) p2m2(y) ⋯ 1m1 m1! 2m2 m2! ⋯ ) = ∑λ pλ(x) pλ(y) zλ .$

## Notes and References

[Mac] I.G. Macdonald, Symmetric functions and Hall polynomials, Second edition, Oxford University Press, 1995.