## Notes on Schubert PolynomialsChapter 6

Last update: 2 July 2013

## Double Schubert Polynomials

Let $x=\left({x}_{1},\dots ,{x}_{n}\right),$ $y=\left({y}_{1},\dots ,{y}_{n}\right)$ be two sequences of independent indeterminates, and recall (5.8) that

$Δ(x,y)= ∏i+j≤n (xi-yj).$

For each $w\in {S}_{n},$ we define the double Schubert polynomial ${𝔖}_{w}\left(x,y\right)$ to be

$(6.1) 𝔖w(x,y)= ∂w-1w0 Δ(x,y)$

where ${\partial }_{{w}^{-1}{w}_{0}}$ acts on the $x$ variables.

Since $\Delta \left(x,0\right)={x}^{\delta }$ we have

$(6.2) 𝔖w(x,0)= 𝔖w(x),$

the (single) Schubert polynomial indexed by $w\text{.}$

From the Cauchy formula (5.10) we have

$𝔖w(x,y)= ∑v∈Sn ∂w-1w0 𝔖vw0(x) 𝔖v(-y)$

and by (4.2)

$∂w-1w0 𝔖vw0(x)= 𝔖vw(x)$

if $\ell \left(vw\right)=\ell \left(v{w}_{0}\right)-\ell \left({w}^{-1}{w}_{0}\right),$ i.e. if $\ell \left(vw\right)=\ell \left(w\right)-\ell \left(v\right),$ and

$∂w-1w0 𝔖vw0(x)=0$

otherwise. Hence

$(6.3) 𝔖w(x,y)= ∑u,v𝔖u (x)𝔖v(-y)$

summed over all $u,v\in {S}_{n}$ such that $w={v}^{-1}u$ and $\ell \left(w\right)=\ell \left(u\right)+\ell \left(v\right)\text{.}$

From (6.3) it follows that ${𝔖}_{w}\left(x,y\right)$ is a homogeneous polynomial of degree $\ell \left(w\right)$ in ${x}_{1},\dots ,{x}_{n-1},$ ${y}_{1},\dots ,{y}_{n-1}\text{.}$ We have

(6.4)
 (i) ${𝔖}_{{w}_{0}}\left(x,y\right)=\Delta \left(x,y\right),$ (ii) ${𝔖}_{1}\left(x,y\right)=1,$ (iii) ${𝔖}_{{w}^{-1}}\left(x,y\right)={𝔖}_{w}\left(-y,-x\right)=\epsilon \left(w\right){𝔖}_{w}\left(y,x\right)$ for all $w\in {S}_{n},$ (iv) ${𝔖}_{w}\left(x,x\right)=0$ for all $w\in {S}_{n}$ except $w=1\text{.}$

 Proof. (i) is immediate from the definition (6.1). (ii) and (iii) follow from (6.3). (iv) follows from (5.20), since ${𝔖}_{w}\left(x,x\right)=\theta \left({\partial }_{{w}^{-1}{w}_{0}}\Delta \right)=0$ if $w\ne 1\text{.}$ $\square$

(6.5) (Stability) If $m>n$ and $i$ is the embedding of ${S}_{n}$ in ${S}_{m},$ then

$𝔖i(w)(x,y) =𝔖w(x,y)$

for all $w\in {S}_{n}\text{.}$

 Proof. This again follows from (6.3) and the stability of the single Schubert polynomials (4.5). $\square$

From (6.5) it follows that the double Schubert polynomials ${𝔖}_{w}\left(x,y\right)$ are well defined for all permutations $w\in {S}_{\infty }\text{.}$

For any commutative ring $K,$ let $K\left({S}_{\infty }\right)$ denote the $K\text{-module}$ of all functions on ${S}_{\infty }$ with values in $K\text{.}$ We define a multiplication in $K\left({S}_{\infty }\right)$ as follows: for $f,g\in K\left({S}_{\infty }\right),$

$(fg)(w)= ∑u,vf(u) g(v)$

summed over all $u,v\in {S}_{\infty }$ such that $uv=w$ and $\ell \left(u\right)+\ell \left(v\right)=\ell \left(w\right)\text{.}$ For this multiplication, $K\left({S}_{\infty }\right)$ is an associative (but not commutative) ring, with identity element $\underset{\underset{_}{_}}{1},$ the characteristic function of the identity permutation $1\text{.}$ It carries an involution $f↦{f}^{*},$ defined by

$f*(w)= f(w-1)$

which satisfies

$(fg)*= g*f*$

for all $f,g\in K\left({S}_{\infty }\right)\text{.}$

(6.6) Let $f,g\in K\left({S}_{\infty }\right)\text{.}$

 (i) If $fg=f$ and $f\left(1\right)$ is not a zero divisor in $K,$ then $g=\underset{\underset{_}{_}}{1}\text{.}$ (ii) If $fg=\underset{\underset{_}{_}}{1},$ then $gf=\underset{\underset{_}{_}}{1}\text{.}$ (iii) $f$ is a unit (i.e. invertible) in $K\left({S}_{\infty }\right)$ if and only if $f\left(1\right)$ is a unit in $K\text{.}$

 Proof. (i) We have $f\left(1\right)=f\left(1\right)g\left(1\right)$ and hence $g\left(1\right)=1\text{.}$ We shall show by induction on $\ell \left(w\right)$ that $g\left(w\right)=0$ for all $w\ne 1\text{.}$ So let $r>0$ and assume that $g\left(v\right)=0$ for all $v\in {S}_{\infty }$ such that $1\le \ell \left(v\right)\le r-1\text{.}$ Let $w$ be a permutation of length $r\text{.}$ We have $(1) f(w)=(fg) (w)=f(w) g(1)+f(1) g(w)+∑u,v f(u)g(v)$ where the sum on the right is over $u,v\in {S}_{\infty }$ such that $u\ne 1,$ $v\ne 1,$ $uv=w$ and $\ell \left(u\right)+\ell \left(v\right)=\ell \left(w\right),$ so that $1\le \ell \left(v\right)\le r-1$ and therefore $g\left(v\right)=0\text{.}$ Hence (1) reduces to $f\left(1\right)g\left(w\right)=0$ and therefore $g\left(w\right)=0$ as required. (ii) We have $f\left(1\right)g\left(1\right)=1$ so that $f\left(1\right)$ is a unit in $K\text{.}$ Also $f\left(gf\right)=\left(fg\right)f=f,$ whence $gf=\underset{\underset{_}{_}}{1}$ by (i) above. (iii) Suppose $f$ is a unit in $K\left({S}_{\infty }\right),$ with inverse $g\text{.}$ Since is $fg=\underset{\underset{_}{_}}{1}$ we have $f\left(1\right)g\left(1\right)=1,$ whence $f\left(1\right)$ is an unit in $K\text{.}$ Conversely, if $f\left(1\right)$ is an unit in $K$ we construct an inverse $g$ of $f$ as follows. We define $g\left(1\right)=f{\left(1\right)}^{-1}$ and proceed to define $g\left(w\right)$ by induction on $\ell \left(w\right)\text{.}$ Assume that $g\left(v\right)$ has been defined for all $v$ such that $\ell \left(v\right)<\ell \left(w\right)$ and set $g(w)=-f (1)-1 ∑u,vf(u) g(v)$ summed over $u,v$ such that $uv=w,$ $v\ne w$ and $\ell \left(u\right)+\ell \left(v\right)=\ell \left(w\right)\text{.}$ This definition gives $\left(fg\right)\left(w\right)=0$ as required. $\square$

Now let $𝔖\left(x\right)$ (resp. $𝔖\left(x,y\right)\text{)}$ be the function on ${S}_{\infty }$ whose value at a permutation $w$ is ${𝔖}_{w}\left(x\right)$ (resp. ${𝔖}_{w}\left(x,y\right)\text{).}$ (The coefficient ring $K$ is now the ring $ℤ\left[x,y\right]$ of polynomials in the $x\text{'s}$ and $y\text{'s.)}$ Since ${𝔖}_{1}\left(x\right)={𝔖}_{1}\left(x,y\right)=1,$ it follows from (6.6)(iii) that $𝔖\left(x\right)$ and $𝔖\left(x,y\right)$ are units in $K\left({S}_{\infty }\right)\text{.}$

(6.7)
 (i) $𝔖\left(x,0\right)=𝔖\left(x\right),$ (ii) $𝔖\left(x,x\right)=\underset{\underset{_}{_}}{1},$ (iii) $𝔖{\left(x,y\right)}^{*}=𝔖\left(-y,-x\right),$ (iv) $𝔖{\left(x\right)}^{-1}=𝔖\left(0,x\right),$ (v) $𝔖{\left(x\right)}^{*}=𝔖{\left(-x\right)}^{-1},$ (vi) $𝔖\left(x,y\right)=𝔖{\left(y\right)}^{-1}𝔖\left(x\right)=𝔖{\left(y,x\right)}^{-1}\text{.}$

 Proof. (i)-(iii) follow directly from (6.2) and (6.4). From (6.3) and (6.4) we have $𝔖w(x,y)= ∑u,v 𝔖u-1 (-y)𝔖v(x)= ∑u,v𝔖u (0,y)𝔖v(x)$ summed over $u,v\in {S}_{\infty }$ such that $uv=w$ and $\ell \left(u\right)+\ell \left(v\right)=\ell \left(w\right)\text{.}$ In other words, $(1) 𝔖(x,y)=𝔖 (0,y)𝔖(x).$ In particular, when $y=x$ we obtain $𝔖\left(0,x\right)𝔖\left(x\right)=𝔖\left(x,x\right)=\underset{\underset{_}{_}}{1}$ by (ii) above, and hence $𝔖\left(0,x\right)=𝔖{\left(x\right)}^{-1}\text{.}$ This establishes (iv); part (v) now follows from (iv) and (iii), and (vi) from (iv) and (1) above. $\square$

From (6.7) (vi) we have

$𝔖(x)=𝔖(y) 𝔖(x,y)$

or explicitly

$𝔖w(x)=∑u,v 𝔖u(y)𝔖v (x,y)$

summed over $u,v$ such that $uv=w$ and $\ell \left(u\right)+\ell \left(v\right)=\ell \left(w\right),$ so that $u=w{v}^{-1}$ and ${𝔖}_{u}={\partial }_{v}{𝔖}_{w}$ by (4.2). Hence

$𝔖w(x)=∑v 𝔖v(x,y)∂v 𝔖w(y)$

(where the operators ${\partial }_{v}$ act on the $y$ variables). The sum here may be taken over all permutations $v,$ since ${\partial }_{v}{𝔖}_{w}=0$ unless $\ell \left(w{v}^{-1}\right)=\ell \left(w\right)-\ell \left(v\right)\text{.}$ By linearity and (4.13) it follows that

(6.8) (Interpolation Formula) For all $f\in {P}_{n}=ℤ\left[{x}_{1},\dots ,{x}_{n}\right]$ we have

$f(x)=∑w𝔖w (x,y)∂wf(y)$

summed over permutations $w\in {S}^{\left(n\right)}\text{.}$

(The reason for the restriction to ${S}^{\left(n\right)}$ in the summation is that if $w\notin {S}^{\left(n\right)}$ we shall have $w\left(m\right)>w\left(m+1\right)$ for some $m>n,$ and hence ${\partial }_{w}={\partial }_{v}{\partial }_{m}$ where $v=w{s}_{m}\text{;}$ but ${\partial }_{m}f=0$ for all $f\in {P}_{n},$ since $m>n,$ and therefore ${\partial }_{w}f=0\text{.)}$

Remarks. 1. By setting each ${y}_{i}=0$ in (6.8) we regain (4.14).

2. When $n=1,$ the sum is over ${S}^{\left(1\right)},$ which consists of the permutations ${w}_{p}={s}_{p}{s}_{p-1}\dots {s}_{1}$ $\left(p\ge 0\right)\text{;}$ ${w}_{p}$ is dominant, of shape $\left(p\right),$ so that (see (6.15) below) ${𝔖}_{{w}_{p}}\left(x,y\right)=\left(x-{y}_{1}\right)\dots \left(x-{y}_{p}\right)\text{.}$ Hence the case $n=1$ of (6.8) is Newton's interpolation formula

$f(x)=∑p≥0 (x-y1)… (x-yp)fp (y1,…,yp+1)$

where ${f}_{p}={\partial }_{p}{\partial }_{p-1}\dots {\partial }_{1}f,$ or explicitly

$fp(y1,…,yp+1)= ∑i=1p+1 f(yi) ∏j≠i (yi-yj) .$

For any integer $r,$ let ${𝔖}_{w}\left(x,r\right)$ denote the polynomial obtained from ${𝔖}_{w}\left(x,y\right)$ by setting ${y}_{1}={y}_{2}=\dots =r\text{.}$ Since

$𝔖w0(x,r) = Δ(x,r)= ∏i=1n-1 (xi-r)n-i = 𝔖w0(x-r)$

where $x-r$ means $\left({x}_{1}-r,{x}_{2}-r,\dots \right),$ it follows from the definitions (6.1) and (4.1) that

$𝔖w(x,r)=𝔖w (x-r)$

for all permutations $w\text{.}$ Hence, by (6.7)(vi),

$𝔖(x-r)=𝔖 (r)-1𝔖 (x)$

and in particular, for all integers $q,$

$𝔖(q-r)=𝔖 (r)-1𝔖(q)$

from which it follows that

$(6.9) 𝔖(r)=𝔖(1)r$

for all $r\in ℤ\text{.}$

Since ${𝔖}_{w}\left(x\right)$ is a sum of monomials with positive integral coefficients (4.17), ${𝔖}_{w}\left(1\right)$ is the number of monomials in ${𝔖}_{w}\left(x\right)$ (each monomial counted the number of times it occurs). By homogeneity, we have

$(6.10) 𝔖w(r)= rℓ(w) 𝔖w(1).$

From (6.7)(v) and (6.9) we obtain

$𝔖(1)*=𝔖 (-1)-1= 𝔖(1)$

so that we have another proof of the fact (4.30) that ${𝔖}_{w}\left(1\right)={𝔖}_{{w}^{-1}}\left(1\right)\text{.}$

Now consider the function $F=𝔖\left(1\right)-\underset{\underset{_}{_}}{1},$ whose value at $w\in {S}_{\infty }$ is

$F(w)= { numbers of monomials in 𝔖w, if w≠1, 0, if w=1.$

For each positive integer $p$ we have

$(1) Fp = (𝔖(1)-1__)p = ∑r=0p (-1)r (pr) 𝔖(1)r = ∑r=0p (-1)r (pr) 𝔖(1)$

by (6.9). The value of (1) at a permutation $w$ of length $p$ is by (6.10) equal to

$( ∑r=0p (-1)r (pr) rp ) 𝔖w(1)$

which is equal to $p!{𝔖}_{w}\left(1\right)$ (consider the coefficient of ${t}^{p}$ in ${\left({e}^{t}-1\right)}^{p}\text{).}$ On the other hand, ${F}^{p}\left(w\right)$ is by definition equal to

$(2) ∑w1,…,wp F(w1)…F(wp)$

summed over all sequences $\left({w}_{1},\dots ,{w}_{p}\right)$ of permutations such that ${w}_{1}\dots {w}_{p}=w,\ell \left({w}_{1}\right)+\dots +\ell \left({w}_{p}\right)=\ell \left(w\right)=p,$ and ${w}_{i}\ne 1$ for $1\le i\le p\text{.}$ It follows that each ${w}_{i}$ has length $1,$ hence ${w}_{i}={s}_{{a}_{i}}$ say, and that $\left({a}_{1},\dots ,{a}_{p}\right)$ is a reduced word for $w\text{.}$ Since

$𝔖sa=x1+…+ xa$

by (4.4). we have $F\left({w}_{i}\right)={𝔖}_{{s}_{{a}_{i}}}$ $\text{(1)}={a}_{i},$ and hence the sum (2) is equal to $\sum {a}_{1}{a}_{2}\dots {a}_{p}$ summed over all $\left({a}_{1},\dots ,{a}_{p}\right)\in R\left(w\right)\text{.}$

We have therefore proved that

(6.11) The number of monomials in ${𝔖}_{w}$ is

$𝔖w(1)=1p! ∑a1a2…ap$

summed over all $\left({a}_{1},\dots ,{a}_{p}\right)\in R\left(w\right),$ where $p=\ell \left(w\right)\text{.}$

Remarks. 1. The reduced words for ${1}_{m}×w$ $\left(m\ge 1\right)$ are $\left(m+{a}_{1},\dots ,m+{a}_{p}\right)$ where $\left({a}_{1},\dots ,{a}_{p}\right)\in R\left(w\right)\text{.}$ Hence from (6.11) and homogeneity we have

$𝔖1m×w(1m) =1p!∑ (1+a1m)… (1+apm)$

summed over $R\left(w\right)$ as before. Letting $m\to \infty ,$ we deduce that

$(6.12) Card R(w)=p! limm→∞ 𝔖1m×w (1m).$

2. If $w$ is dominant of length $p,$ then ${𝔖}_{w}$ is a monomial by (4.7). and hence in this case

$∑R(w)a1 …ap=p!$

3. Suppose that $w$ is vexillary of length $p\text{.}$ Then by (4.9) we have

$𝔖w=sλ (Xϕ1,…,Xϕr)$

where $\lambda$ is the shape of $w$ and $\varphi =\left({\varphi }_{1},\dots ,{\varphi }_{r}\right)$ the flag of $w\text{.}$ Hence

$𝔖1m×w=sλ ( Xϕ1+m,…, Xϕr+m )$

for each $m\ge 1\text{.}$ If we now set each ${x}_{i}=\frac{1}{m}$ and then let $m\to \infty ,$ we shall obtain in the limit the Schur function ${s}_{\lambda }$ for the series ${e}^{t}$ ([Mac1979], Ch. I. §3. Ex. 5). which is equal to $h{\left(\lambda \right)}^{-1},$ where $h\left(\lambda \right)$ is the product of the hook-lengths of $\lambda \text{.}$ Hence it follows from (6.12) that if $w$ is vexillary of length $p,$ then

$(6.13) Card R(w)= p!h(λ)$

where $\lambda$ is the shape of $w\text{.}$ In other words. the number of reduced words for a vexillary permutation of length $p$ and shape $\lambda ⊢p$ is equal to the degree of the irreducible representation of ${S}_{p}$ indexed by $\lambda \text{.}$

4. It seems likely that there is a $q\text{-analogue}$ of (6.11). Some experimental evidence suggests the following conjecture:

$(6.11q?) 𝔖w(1,q,q2,…) =∑qϕ(a) (1-qa1)… (1-qap) (1-q)… (1-qp)$

summed as in (6.11) over all reduced words $a=\left({a}_{1},\dots ,{a}_{p}\right)$ for $w,$ where

$ϕ(a)=∑ {i:ai

When $w$ is vexillary the double Schubert polynomial ${𝔖}_{w}\left(x,y\right)$ can be expressed as a multi-Schur function, just as in the case of (single) Schubert polynomials (Chap. IV). We consider first the case of a dominant permutation:

(6.14) If $w$ is dominant of shape $\lambda ,$ then

$𝔖w(x,y) = ∏(i,j)∈λ (xi-yj) = sλ ( X1-Yλ1,…, Xm-Yλm )$

where $m=\ell \left(\lambda \right)$ and ${X}_{i}={x}_{1}+\dots +{x}_{i},$ ${Y}_{i}={y}_{1}+\dots +{y}_{i}$ for all $i\ge 1\text{.}$

 Proof. As in (4.6) we proceed by descending induction on $\ell \left(w\right),w\in {S}_{n}\text{.}$ The result is true for $w={w}_{0},$ since ${w}_{0}$ is dominant of shape $\delta$ and $𝔖w0(x,y)= Δ(x,y)= ∏(i,j)∈δ (xi-yj).$ Suppose $w\ne {w}_{0}$ is dominant of shape $\lambda \text{.}$ Then $\lambda \subset \delta$ (and $\lambda \ne \delta \text{).}$ Let $r\ge 0$ be the largest integer such that ${\lambda }_{i}^{\prime }=n-i$ for $1\le i\le r,$ and let $a={\lambda }_{r+1}^{\prime }+1\le n-r-1\text{.}$ Then $w{s}_{a}$ is dominant, $\ell \left(w{s}_{a}\right)=\ell \left(w\right)+1,$ and $\lambda \left(w{s}_{a}\right)=\lambda \left(w\right)+{\epsilon }_{a},$ and therefore $𝔖w(x,y) = ∂a𝔖wsa (x,y) = ∂a ( (xa-yr+1) ∏(i,j)∈λ (xi-yj) )$ by the inductive hypothesis; since ${\lambda }_{a}={\lambda }_{a+1}$ it follows that $𝔖w(x,y)= ∏(i,j)∈λ (xi-yj)$ which is equal to ${s}_{\lambda }\left({X}_{1}-{Y}_{{\lambda }_{1}},\dots ,{X}_{m}-{Y}_{{\lambda }_{m}}\right)$ by (3.5). $\square$

(6.15) If $w$ is Grassmannian of shape $\lambda$ then

$𝔖w(x,y)=sλ ( Xm-Yλ1+m-1, …, Xm-Yλm ) .$

 Proof. This follows from (6.14) just as (4.8) follows from (4.7). $\square$

Finally, let $w$ be vexillary with shape

$λ(w)= ( p1m1,…, pkmk )$

and flag

$ϕ(w)= ( f1m1,…, fkmk )$

as in Chapter IV. Then ${w}^{-1}$ is also vexillary, with shape

$λ(w-1)=λ (w)′= ( q1n1,…, qknk )$

the conjugate of $\lambda \left(w\right),$ and flag

$ϕ(w-1)= ( g1n1,…, gknk )$

where by (1.41)

$gi+qi= fk+1-i+ pk+1-i (1≤i≤k).$

With this notation recalled, we have

$(6.16) 𝔖w(x,y)=sλ ( (Xf1-Ygk)m1 ,…, (Xfk-Yg1)mk ) .$

 Proof. The proof is essentially the same as that of (4.9) (which is the case $y=0\text{).}$ By (4.10) the dominant permutation ${w}_{k}$ constructed from $w$ in the proof of (4.9) has shape $μ= ( gkm1, gk-1m2,…, g1mk )$ and therefore by (6.15) we have $𝔖wk(x,y)=sμ ( X1′,…, Xm′ )$ where $m={m}_{1}+\dots +{m}_{k}=\ell \left(\lambda \right)$ and the sequence $\left({X}_{1}^{\prime },\dots ,{X}_{m}^{\prime }\right)$ is obtained by subtracting the sequence $\left({\left({Y}_{{g}_{k}}\right)}^{{m}_{1}},\dots ,{\left({Y}_{{g}_{1}}\right)}^{{m}_{k}}\right)$ term by term from the sequence $\left({X}_{1},\dots ,{X}_{m}\right)\text{.}$ Hence the same argument as in (4.9) establishes (6.17). $\square$

Remark. From (6.16) and (6.4)(iii) we obtain

$sλ ( Z1m1,…, Zkmk ) = (-1)|λ| sλ′ ( (-Zk)n1,…, (-Z1)nk )$

where ${Z}_{i}={X}_{{f}_{i}}-{Y}_{{g}_{k+i-1}}$ so that (if $rk \left({x}_{i}\right)=rk \left({y}_{i}\right)=1$ for each $i\ge 1\text{)}$

$rk (Zi+1-Zi) = fi+1-fi+ gk+1-i- gk-i = mi+1- nk+1-i$

by (1.41). Hence (6.4)(iii) reduces to the duality theorem (3.8'') (with $\mu =0\text{)}$ when $w$ is vexillary.

Let ${\tau }_{x}$ (resp. ${\tau }_{y}\text{)}$ be the shift operator (4.21) acting on the $x$ (resp. $y\text{)}$ variables. Then we have

$(6.17) τxrτyr𝔖w (x,y)= 𝔖1r×w(x,y)$

for all $r\ge 1$ and all permutations $w\text{.}$

 Proof. By (6.3) and (4.21) we have $τxrτyr𝔖w (x,y)=∑u,v ε(v)𝔖1r×u (x)𝔖1r×v(y)$ summed over $u,v$ such that ${v}^{-1}u=w$ and $\ell \left(u\right)+\ell \left(v\right)=\ell \left(w\right)\text{.}$ By (6.3) again, the right-hand side is equal to ${𝔖}_{{1}_{r}×w}\left(x,y\right)\text{.}$ $\square$

In particular, suppose that $w$ is vexillary. With the notation of (6.16), the flag of ${1}_{r}×w$ (resp. ${1}_{r}×{w}^{-1}\text{)}$ is obtained from that of $w$ (resp. ${w}^{-1}\text{)}$ by replacing each ${f}_{i}$ by ${f}_{i}+r$ (resp. each ${g}_{i}$ by ${g}_{i}+r\text{).}$ Hence by (6.16) we have

$𝔖1r×w(x,y) =sλ ( (Xf1+r-Ygk+r)m1,…, (Xfk+r-Yg1+r)mk )$

and hence

$(6.18) ρr(x) ρr(y) 𝔖1r×w (x,y)=sλ (Xr-Yr)$

for all $r\ge 1,$ where ${\rho }_{r}^{\left(x\right)}$ (resp. ${\rho }_{r}^{\left(y\right)}\text{)}$ is the homomorphism ${\rho }_{r}$ of (4.25) acting on the $x$ (resp. $y\text{)}$ variables.

(6.19) Let ${\pi }_{x}$ (resp. ${\pi }_{y}\text{)}$ denote ${\pi }_{{w}_{0}^{\left(r\right)}}$ acting on the $x$ (resp. $y\text{)}$ variables. Then if $w$ is vexillary of shape $\lambda ,$ we have

$πxπy𝔖w (x,y)=sλ (Xr-Yr).$

 Proof. By (4.24) we have ${\pi }_{x}={\rho }_{r}^{\left(x\right)}{\tau }_{x}^{r}$ and ${\pi }_{y}={\rho }_{y}^{\left(r\right)}{\tau }_{y}^{r}\text{.}$ Hence (6.19) follows from (6.17) and (6.18). $\square$

In particular, suppose that $w$ is dominant of shape $\lambda ,$ so that by (6.14)

$𝔖w(x,y)= ∏(i,j)∈λ (xi-yj)= fλ(x,y) say.$

In this case (6.19) gives

$πxπyfλ (x,y)=sλ (Xr-Yr)$

for all $r\ge 1,$ which is Sergeev's formula (3.12').

## Notes and References

This is a typed excerpt of the book Notes on Schubert Polynomials by I. G. Macdonald.