## Notes on Schubert PolynomialsChapter 3

Last update: 27 June 2013

## Multi-Schur functions

For the time being we shall work in an arbitrary $\lambda \text{-ring}$ $R,$ but we shall use the notation of symmetric functions [Mac1979] rather than that of $\lambda \text{-rings.}$ Thus for $X\in R$ we shall write ${e}_{r}\left(X\right)$ in place of ${\lambda }^{r}\left(X\right)$ for the ${r}^{\text{th}}$ exterior power, and ${h}_{r}\left(X\right)$ in place of ${\sigma }^{r}\left(X\right)$ $\left(={\left(-1\right)}^{r}{\lambda }^{r}\left(-X\right)\right)$ for the ${r}^{\text{th}}$ symmetric power of $X\text{.}$ We have ${e}_{0}\left(X\right)={h}_{0}\left(X\right)=1\text{;}$ ${e}_{1}\left(X\right)={h}_{1}\left(X\right)=X\text{;}$ and ${e}_{r}\left(X\right)={h}_{r}\left(X\right)=0$ if $r<0\text{.}$

Recall that if $\lambda ,\mu$ are partitions and $X\in R,$ the skew Schur function ${s}_{\lambda /\mu }\left(X\right)$ is defined by the formula

$sλ/μ(X)=det (hλi-μj-i+j(X)) 1≤i,j≤n$

where $n\ge \text{max}\left(\ell \left(\lambda \right),\ell \left(\mu \right)\right)\text{.}$ It is zero unless $\lambda \supset \mu \text{.}$

We generalize this definition as follows: let ${X}_{1},\dots ,{X}_{n}\in R$ and let $\lambda ,\mu$ be partitions of length $\le n\text{;}$ then the multi-Schur function ${s}_{\lambda /\mu }\left({X}_{1},\dots ,{X}_{n}\right)$ is defined by

$(3.1) sλ/μ (X1,…,Xn) =det (hλi-μj-i+j(X)) 1≤i,j≤n .$

We also define

$(3.1') sα(X1,…,Xn) =det (hαi-i+j(X)) 1≤i,j≤n$

for any sequence $\alpha =\left({\alpha }_{1},\dots ,{\alpha }_{n}\right)$ of integers of length $n\text{.}$

Remark. In the definition (3.1) the argument ${X}_{i}$ is constant in each row of the determinant. We might therefore also define

$sˆλ/μ (X1,…,Xn)= det (hλi-μj-i+j(X)) 1≤i,j≤n$

with arguments constant in each column. However, we get nothing essentially new: if we define partitions $\stackrel{ˆ}{\lambda },\stackrel{ˆ}{\mu }$ by

$λˆi=N- λn+1-i, μˆi=N- μn+1-i (1≤i≤n)$

where $N\ge \text{max}\left({\lambda }_{1},{\mu }_{1}\right)$ (so that $\stackrel{ˆ}{\lambda }$ and $\stackrel{ˆ}{\mu }$ are the respective complements of $\lambda$ and $\mu$ in the rectangle $\left({N}^{n}\right)\text{),}$ then we have

$sˆλ/μ (X1,…,Xn)= sμˆ/λˆ (Xn,…,X1)$

as one sees by replacing $\left(i,j\right)$ by $\left(n+1-j,n+1-i\right)$ in the determinant (3.1).

(3.2) We have

$sλ/μ (X1,…,Xn)=0$

unless $\lambda \supset \mu \text{.}$

 Proof. If $\lambda \not\supset \mu$ then ${\lambda }_{r}<{\mu }_{r}$ for some $r\le n,$ and hence $λi-μj-i+j ≤λr-μr<0$ whenever $i\ge r$ and $j\le r\text{.}$ It follows that the matrix $\left({h}_{{\lambda }_{i}-{\mu }_{j}-i+j}\left(X\right)\right)$ has an $\left(n-r+1\right)×r$ block of zeros in the south-west corner, and hence its determinant vanishes. $\square$

(3.3) If $\lambda \supset \mu$ and $\ell \left(\lambda \right)=r then

$sλ/μ (X1,…,Xn)= sλ/μ (X1,…,Xr)$

 Proof. We have ${\lambda }_{s}={\mu }_{s}=0$ for $r+1\le s\le n\text{.}$ Hence for each $s>r$ the ${s}^{\text{th}}$ row of the matrix $\left({h}_{{\lambda }_{i}-{\mu }_{j}-i+j}\left(X\right)\right)$ has zeros in the first $s-1$ places, and $1$ in the ${s}^{\text{th}}$ place. $\square$

An element $x\in R$ is said to have finite rank if ${e}_{n}\left(X\right)=0$ for all sufficiently large $n\text{.}$ We then define the rank $rk\left(X\right)$ of $X$ to be the largest $r$ such that ${e}_{r}\left(X\right)\ne 0\text{.}$ If $X,Y$ both have finite rank, the formula

$er(X+Y)= ∑p+q=r ep(X) eq(Y)$

shows that $X+Y$ has finite rank, and that

$rk(X+Y)≤rk (X)+rk(Y).$

(3.4) Let ${X}_{1},\dots ,{X}_{n},{Y}_{1},\dots ,{Y}_{n}\in R$ with $rk\left({Y}_{j}\right)\le j-1$ $\left(1\le j\le n\right)$ (so that ${Y}_{1}=0\text{).}$ Then for all $\alpha \in {ℤ}^{n},$

$sα (X1,…,Xn)= det ( hαi-i+j (Xi-Yj) ) .$

 Proof. We have $hαi-i+j (Xi-Yj)= ∑k=1j hαi-i+k (Xi) hj-k (-Yj),$ since ${h}_{r}\left(-{Y}_{j}\right)={\left(-1\right)}^{r}{e}_{r}\left({Y}_{j}\right)=0$ if $r\ge j\text{.}$ Hence the matrix $( hαi-i+j (Xi-Yj) ) 1≤i,j≤n$ is the product of the matrix $( hαi-i+j (Xi) ) 1≤i,j≤n$ and the matrix $( hi-j (-Yj) ) 1≤i,j≤n$ which is unitriangular. Now take determinants. $\square$

So far the ${X}_{i}$ have been arbitrary elements of the $\lambda \text{-ring}$ $R\text{.}$ But it seems that ${s}_{\alpha }\left({X}_{1},\dots ,{X}_{n}\right)$ is mainly of interest when ${X}_{1},\dots ,{X}_{n}$ is an increasing sequence in $R,$ in the sense that $rk\left({X}_{i+1}-{X}_{i}\right)<\infty$ for $1\le i\le n-1\text{.}$

(3.5) Let ${x}_{i},{y}_{i}$ $\left(i\ge 1\right)$ be elements of $R,$ each of rank $\le 1,$ and let

$Xi=x1+…+xi, Yi=y1+…+yi$

for each $i\ge 0\text{.}$ Then for all $\alpha \in {ℕ}^{n}$ we have

$sα ( X1-Yα1,…, Xn-Yαn ) =∏i=1n ∏j=1αi (xi-yj).$

In particular, if $\lambda$ is a partition of length $\le n,$

$sλ ( X1-Yλ1,…, Xn-Yλn ) =∏(i,j)∈λ (xi-yj).$

 Proof. From (3.4) we have $(*) sα ( X1-Yα1,…, Xn-Yαn ) =det ( hαi-i+j (Xi-Yαi-Xj-1) ) 1≤i,j≤n .$ If $j>i,$ then $hαi-i+j (Xi-Yαi-Xj-1) =±eαi-i+j (Yαi+Xj-1-Xi)$ which is $0$ because $rk(Yαi+Xj-1-Xi) ≤αi+(j-1)-i< αi-i+j.$ Hence the determinant at $\left(*\right)$ is triangular, with diagonal elements $hαi ( Xi-Yαi- Xi-1 ) = hαi(xi-Yαi) = ∑r≥0hr (-Yαi) hαi-r (xi) = ∑r≥0(-1)r er(Yαi) xiαi-r = ∏j=1αi (xi-yj).$ The formula (3.5) now follows. $\square$

In particular, when all the ${y}_{i}$ are zero we have

$(3.5') sα(X1,…,Xn) =∏i=1n xiαi=xα$

for all $\alpha \in {ℕ}^{n}\text{.}$ Also, when all the ${x}_{i}$ are zero (and $\alpha$ is a partition $\lambda \text{)}$ we have

$sλ ( -Yλ1,…, -Yλn ) = (-1)|λ| ∏(i,j)∈λ yj = (-1)|λ| yλ′.$

If we replace the $y\text{'s}$ by $x\text{'s,}$ and $\lambda$ by $\lambda \prime ,$ this becomes

$(3.5'') xλ= (-1)|λ| sλ′ ( -Xλ1′,…, -Xλn′ ) .$

(3.6) Let $\lambda =\left({\lambda }_{1},{\lambda }_{2},\dots \right)$ be a partition of length $\le n,$ and ${X}_{1},\dots ,{X}_{n}$ elements of a $\lambda \text{-ring}$ $R\text{.}$ Suppose that $i are such that ${\lambda }_{i}={\lambda }_{i+1}=\dots ={\lambda }_{j}$ and

$rk(Xj-Xk)≤ j-kfor i≤k≤ j.$

Then

$sλ/μ (X1,…,Xn)= sλ/μ ( X1,…,Xi-1, Xj,…,Xj, Xj+1,…,Xn ) ,$

that is to say we can replace each ${X}_{k}$ $\left(i\le k\le j\right)$ by ${X}_{j}$ without changing the value of the multi-Schur function.

 Proof. Let $Y={X}_{j}-{X}_{i},$ so that $rk\left(Y\right)\le j-1\text{.}$ For all $m\ge 0$ we have $hm(Xi) = hm(Xj-Y) = ∑k=0j-i (-1)kek (Y)hm-k (Xj).$ It follows that if we replace the ${i}^{\text{th}}$ row of the determinant ${s}_{\lambda /\mu }\left({X}_{1},\dots ,{X}_{i-1},{X}_{j},\dots ,{X}_{j},{X}_{j+1},\dots ,{X}_{n}\right)$ by $∑k=0j-i (-1)kek (Y)rowi+k$ we shall obtain $sλ/μ ( X1,…,Xi-1, Xi,Xj,…,Xj Xj+1,…,Xn )$ with $j-i$ arguments equal to ${X}_{j}\text{.}$ The proof is now completed by induction on $j-i\text{.}$ $\square$

### Duality

Let ${\left({X}_{n}\right)}_{n\in ℤ}$ be a sequence in the $\lambda \text{-ring}$ $R$ such that

$rk(Xn-Xn-1) ≤1$

for all $n\in ℤ\text{.}$

(3.7) Let $I$ be any interval in $ℤ\text{.}$ Then the inverse of the matrix

$H= (hi-j(-Xi)) i,j∈I$

is ${H}^{-1}={\left({h}_{i-j}\left({X}_{j+1}\right)\right)}_{i,j\in I}\text{.}$

 Proof. Let $K$ denote the matrix $\left({h}_{i-j}\left({X}_{j+1}\right)\right)\text{.}$ The $\left(i,k\right)$ element of $HK$ is then $∑jhi-j (-Xi) hj-k (Xj+1)= hi-k (-Xi+Xk+1).$ If $i this is zero; if $i=k$ it is equal to $1\text{;}$ and if $i>k$ it is equal to $(-1)i-k ei-k (Xi-Xk+1)$ which is zero because $rk\left({X}_{i}-{X}_{k+1}\right)\le i-\left(k+1\right) $\square$

(3.8) (Duality Theorem, ${1}^{\text{st}}$ version) Let $\lambda \supset \mu$ be partitions of length $\le n,$ such that $\ell \left(\lambda \prime \right)\le m\text{.}$ Then

$sλ/μ ( -Xλ1-1,…, -Xλn-n ) =(-1)|λ-μ| sλ′/μ′ ( X1-λ1′,…, Xm-λm′ ) .$

 Proof. Let $ξi=λi-i, ηi=μi-i (1≤i≤n), ξj′= λj′-j, ηj′= μj′-j (1≤j≤n),$ Then the integers ${\xi }_{i}$ $\left(1\le i\le n\right)$ and $-{\xi }_{j}^{\prime }-1$ $\left(1\le j\le m\right)$ fill up the interval $\left[-m,n-1\right],$ and so do the ${\eta }_{i}$ and the $-{\eta }_{j}^{\prime }-1\text{.}$ The $\left({\xi }_{1},\dots ,{\xi }_{n};{\eta }_{1},\dots ,{\eta }_{n}\right)$ minor of the matrix $H$ is $det ( hηi-ηj (-Xξi) ) =sλ/μ ( -Xξ1,…, -Xξn ) .$ The complementary cofactor of $\left({H}^{-1}\right)\prime ={\left({h}_{j-i}\left({X}_{i+1}\right)\right)}_{-m\le i,j\le n-1}$ has row indices $-{\xi }_{i}^{\prime }-1$ $\left(1\le i\le m\right)$ and column indices $-{\eta }_{j}^{\prime }-\prime$ $\left(1\le j\le m\right)\text{.}$ Hence it is $(-1)|λ-μ| sλ′/μ′ ( -Xξ1′,…, -Xξm′ ) .$ Since each minor of $H$ is equal to the complementary cofactor of $\left({H}^{-1}\right)\prime$ (because $\text{det} H=1\text{)}$ the result follows. $\square$

Remark. Observe that

$rk ( Xλi-i- Xλi+1-i-1 ) ≤(λi-i)- (λi+1-i-1)= λi-λi+1+1.$

Hence (3.8) gives a duality theorem for the multi-Schur function ${s}_{\lambda /\mu }\left({Y}_{1},\dots ,{Y}_{n}\right)$ provided that $rk\left({Y}_{i+1}-{Y}_{i}\right)\le {\lambda }_{i}-{\lambda }_{i+1}+1$ for $1\le i\le n-1\text{.}$

At first sight the formula (3.8) is disconcerting, because the arguments $-{X}_{{\lambda }_{i}-i}$ on the left are not in general the negatives of the arguments ${X}_{i-{\lambda }_{i}^{\prime }}$ on the right. However, we can use (3.6) to rewrite (3.8), as follows. As in Chapter I, let us write the partition $\lambda$ in the form

$λ= ( p1m1, p2m2,…, pkmk )$

where ${p}_{1}>{p}_{2}>\dots >{p}_{k}>0$ and each ${m}_{i}\ge 1\text{.}$ Then in

$sλ/μ ( -Xλ1-1,…, -Xλn-n )$

the first ${m}_{1}$ arguments are

$-Xp1-1,…, -Xp1-m1$

which by (3.6) may all be replaced by $-{X}_{{c}_{1}},$ where ${c}_{1}={p}_{1}-{m}_{1}\text{.}$ The next ${m}_{2}$ arguments are

$-Xp2-m1-1,…, -Xp2-m1-m2$

which by (3.6) may all be replaced by $-{X}_{{c}_{2}},$ where ${c}_{2}={p}_{2}-{m}_{1}-{m}_{2}\text{.}$ In general, for each $i=1,2,\dots ,k$ the ${i}^{\text{th}}$ group of ${m}_{i}$ arguments may all be replaced by $-{X}_{{c}_{i}},$ where ${c}_{i}={p}_{i}-\left({m}_{1}+\dots +{m}_{i}\right)\text{.}$ Now if

$λ′= ( q1n1, q2n2,…, qknk )$

is the conjugate partition, we have ${m}_{1}+\dots +{m}_{i}={q}_{k+1-i},$ and ${c}_{i}=\pi -{q}_{k+1-i}$ is the content of the square ${s}_{i}=\left({q}_{k+1-i},{p}_{i}\right)$ in the diagram of $\lambda \text{.}$ The squares ${s}_{1},\dots ,{s}_{k}$ are the "salients" of the border of $\lambda ,$ read in sequence from north-east to south-west. Hence the duality theorem (3.8) now takes the form

(3.8') (Duality Theorem, ${2}^{\text{nd}}$ version). With the above notation, we have

$sλ/μ ( (-Xc1)m1,…, (-Xck)mk ) =(-1)|λ-μ| sλ′/μ′ ( (Xck)n1,…, (Xc1)nk ) .$

Finally, if we set ${Z}_{i}=-{X}_{{c}_{i}}$ $\left(1\le i\le k\right)$ we have

$(3.8'') sλ/μ ( Z1m1,…, Zkmk ) =(-1)|λ-μ| sλ′/μ′ ( (-Zk)n1,…, (-Z1)nk )$

provided

$rk(Zi+1-Zi) ≤mi+1+ nk+1-i (1≤i≤k-1).$

Let now ${x}_{1},{x}_{2},\dots$ be independent indeterminates over $ℤ\text{.}$ We may regard $ℤ\left[{x}_{1},{x}_{2},\dots \right]$ as a $\lambda \text{-ring}$ by requiring that each ${x}_{i}$ has rank $1\text{.}$ Let ${X}_{i}={x}_{1}+\dots +{x}_{i}$ for each $i\ge 1\text{.}$ Then we have

$(3.9) ∂ihr(Xi) = hr-1(Xi+1), ∂ier(Xi) = er-1(Xi-1), πihr(Xi) = hr(Xi+1).$

 Proof. Consider the generating functions: ${\partial }_{i}{h}_{r}\left({X}_{i}\right)$ is the coefficient of ${t}^{r}$ in $∂i ( ∑r≥0 hr(Xi) tr ) = ∂i∏j=1i (1-xjt)-1 = ∏j=1i-1 (1-xjt)-1 ·∂i (11-xit),$ and $∂i(11-xit) = 1xi-xi+1 ( 11-xit- 11-xi+1t ) = t (1-xit) (1-xi+1t)$ so that $∂i ( ∑rhr (Xi)tr ) =t∏j=1i+1 (1-xjt)-1= ∑shs (Xi+1) ts+1$ in which the coefficient of ${t}^{r}$ is ${h}_{r-1}\left({X}_{i+1}\right)\text{.}$ The other two relations are proved similarly. $\square$

(3.10) Let $\alpha \in {ℤ}^{n}$ and let ${r}_{1},\dots ,{r}_{n}\ge 0\text{.}$ If $i$ is such that ${r}_{i}\ne {r}_{j}$ for all $j\ne i$ then

$∂risα (Xr1,…,Xrn) = sα-εi ( Xr1,…, Xr+1,…, Xrn ) , ∂risα ( -Xr1,…, -Xrn ) = -sα-εi ( -Xr1,…, -Xri-1,…, -Xrn ) , πrisα (Xr1,…,Xrn) = sα ( Xr1,…, Xr+1,…, Xrn ) ,$

where ${\epsilon }_{i}$ has ${i}^{\text{th}}$ coordinate equal to $1,$ and all other coordinates zero.

 Proof. By definition, we have ${s}_{\alpha }=\text{det}\left({h}_{{\alpha }_{i}-i+j}\left({X}_{{r}_{i}}\right)\right)$ and ${\partial }_{{r}_{i}}$ acts only on the ${i}^{\text{th}}$ row of the determinant, the entries in the other rows being symmetrical in ${x}_{{r}_{i}}$ and ${x}_{{r}_{i}+1}$ (because of the condition ${r}_{j}\ne {r}_{i}$ if $j\ne i\text{).}$ Hence the first of the relations (3.10) follows from the first of the relations (3.9), and the other two are proved similarly. $\square$

Remark. We can use the relations (3.10) to give another proof of duality (3.8) in the form

(3.8''') Let $\lambda$ be a partition such that ${\lambda }_{1}\le m$ and ${\lambda }_{1}^{\prime }\le n\text{.}$ Then

$(*) sλ ( Xm+1-λ1,…, Xm+n-λn ) =(-1)|λ| sλ′ ( -Xm+λ1′-1 ,…, -Xλm′ ) .$

Let $\left(i,j\right)$ be a corner square of the diagram of $\lambda ,$ so that $j={\lambda }_{i}$ and $i={\lambda }_{j}^{\prime }\text{.}$ Let $\mu$ be the partition obtained from $\lambda$ by removing the square $\left(i,j\right)\text{.}$ By operating on either side of $\left(*\right)$ with ${\partial }_{m+i-j}$ we obtain the same relation with $\mu$ replacing $\lambda \text{.}$ Hence it is enough to show that $\left(*\right)$ is true when $\lambda =\left({m}^{n}\right),$ but in that case both sides are equal to ${\left({X}_{1}\dots {X}_{m}\right)}^{n},$ by (3.5'), (3.5'') and (3.6).

(3.11) Let ${w}_{0}$ be the longest element of ${S}_{n}\text{.}$ Then for any $\alpha \in {ℤ}^{n}$ we have

$∂w0sα ( X1+Z1,…, Xn+Zn ) = sα-δ ( Xn+Z1,…, Xn+Zn ) , πw0sα ( X1+Z1,…, Xn+Zn ) = sα ( Xn+Z1,…, Xn+Zn ) ,$

where ${X}_{i}={x}_{1}+\dots +{x}_{i}$ $\left(1\le i\le n\right)$ and the ${Z}_{i}$ are independent of ${x}_{1},\dots ,{x}_{n}\text{.}$

 Proof. The sequence $( n-1, n-2, n-1,…,2,3,…, n-1,1,2,3,…,n -1 )$ is a reduced word for ${w}_{0},$ so that $πw0=πn-1 (πn-2πn-1) …(π2π3…πn-1) (π1π2…πn-1)$ and likewise for ${\partial }_{{w}_{0}}\text{.}$ By (3.10), ${\pi }_{1}{\pi }_{2}\dots {\pi }_{n-1}$ applied to ${s}_{\alpha }\left({X}_{1}+{Z}_{1},\dots ,{X}_{n}+{Z}_{n}\right)$ will produce $sα ( X2+Z1, X3+Z2, …, Xn+Zn-1, Xn+Zn ) .$ We have next to operate on this with ${\pi }_{2}{\pi }_{3}\dots {\pi }_{n-1},$ which will produce $sα ( X3+Z1, X4+Z2,…, Xn+Zn-2, Xn+Zn-1, Xn+Zn ) .$ By repeating this process we shall obtain the formula for ${\pi }_{{w}_{0}}{s}_{\alpha }\text{.}$ That for ${\partial }_{{w}_{0}}{s}_{\alpha }$ is proved similarly. $\square$

Remark. Let $\alpha \in {ℕ}^{n}$ and ${Z}_{1}=\dots ={Z}_{n}=0$ in (3.11). Then by (3.5') we have

$∂w0(xα) = ∂w0sα (X1,…,Xn) = sα-δ(Xn), πw0(xα) = πw0sα (X1,…,Xn) = sα(Xn).$

Thus we have independent proofs of (2.11) and (2.16') and hence (by linearity) of (2.10) and (2.16).

### Sergeev's formula

Let ${x}_{1},\dots ,{x}_{m},{y}_{1},\dots ,{y}_{n}$ be independent variables and let

$Xi=x1+…+xi, Yi=y1+…+yi$

for all $i\ge 1,$ with the understanding that ${x}_{j}=0$ if $j>m$ and ${y}_{j}=0$ if $j>n\text{.}$

(3.12) (Sergeev) For all partitions $\lambda$ we have

$sλ(Xm-Yn)= ∑w∈Sm×Snw ( fλ(x,y)/ D(x)D(y) )$

where

$fλ(x,y)= ∏(i,j)∈λ (xi-yj), D(x)=∏1≤i

 Proof. Let ${w}_{0}^{\left(m\right)}$ (resp. ${w}_{0}^{\left(n\right)}\text{)}$ be the longest element of ${S}_{m}$ (resp. ${S}_{n}\text{)}$ and let ${\pi }_{x}$ (resp. ${\pi }_{y}\text{)}$ denote ${\pi }_{{w}_{0}^{\left(m\right)}}$ acting on the $x\text{'s}$ (resp. ${\pi }_{{w}_{0}^{\left(n\right)}}$ acting on the $y\text{'s).}$ From (3.5) we have, if $r=\ell \left(\lambda \right),$ $(1) fλ(x,y)=sλ ( X1-Yλ1,…, Xr-Yλr )$ and in view of (2.16) Sergeev's formula may be restated in the form $(3.12') sλ(Xm-Yn) =πyπxfλ (x,y).$ From (3.11) and (1) above we have $(2) πxfλ(x,y) =sλ ( Xm-Yλ1,…, Xm-Yλr ) .$ If $\lambda =\left({p}_{1}^{{m}_{1}},\dots ,{p}_{k}^{{m}_{k}}\right),$ (2) can be rewritten in the form $(3) πxfλ(x,y)= sλ(Z1m1,…,Zkmk)$ where ${Z}_{i}={X}_{m}-{Y}_{{p}_{i}}\text{.}$ Since $rk(Zi+1-Zi)= rk(Ypi-Ypi+1) =pi-pi+1,$ the duality theorem (3.8'') applies, and gives $(4) sλ (Z1m1,…,Zkmk) = (-1)|λ| sλ′ ( (-Zk)n1,…, (-Z1)nk ) = (-1)|λ| sλ′ ( (Ypk-Xm)n1,…, (Yp1-Xm)nk ) = (-1)|λ| sλ′ ( Y1-Xm, Y2-Xm,…, Ys-Xm )$ where $s={n}_{1}+\dots +{n}_{k}=\ell \left(\lambda \prime \right)\text{.}$ We can now apply (3.11) again and obtain from (3) and (4) $πyπxfλ = (-1)|λ| πysλ′ ( Y1-Xm,…, Ys-Xm ) = (-1)|λ| sλ′ ( Yn-Xm ) = sλ(Xm-Yn).$ $\square$

## Notes and References

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