Notes on Schubert Polynomials
Chapter 3

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

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 ${\sigma }^r\left(X\right)$ $(={(-1)}^r{\lambda }^r(-X))$ 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

$$s_{\lambda /\mu }\left(X\right)=\text{det}{\left(h_{{\lambda }_i-{\mu }_j-i+j}\left(X\right)\right)}_{1\le i,j\le n}$$

where $n\ge \text{max}(\ell \left(\lambda \right),\ell \left(\mu \right))\text{.}$ 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 $s_{\lambda /\mu }(X_1,\dots ,X_n)$ is defined by

$$\begin{array}{cc}\text{(3.1)}& s_{\lambda /\mu }(X_1,\dots ,X_n)=\text{det}{\left(h_{{\lambda }_i-{\mu }_j-i+j}\left(X\right)\right)}_{1\le i,j\le n}\text{.}\end{array}$$

We also define

$$\begin{array}{cc}\text{(3.1')}& s_{\alpha }(X_1,\dots ,X_n)=\text{det}{\left(h_{{\alpha }_i-i+j}\left(X\right)\right)}_{1\le i,j\le n}\end{array}$$

for any sequence $\alpha =({\alpha }_1,\dots ,{\alpha }_n)$ 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

$${\hat{s}}_{\lambda /\mu }(X_1,\dots ,X_n)=\text{det}{\left(h_{{\lambda }_i-{\mu }_j-i+j}\left(X\right)\right)}_{1\le i,j\le n}$$

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

$${\hat{\lambda }}_i=N-{\lambda }_{n+1-i},{\hat{\mu }}_i=N-{\mu }_{n+1-i}(1\le i\le n)$$

where $N\ge \text{max}({\lambda }_1,{\mu }_1)$ (so that $\hat{\lambda }$ and $\hat{\mu }$ are the respective complements of $\lambda$ and $\mu$ in the rectangle $\left(N^n\right)\text{),}$ then we have

$${\hat{s}}_{\lambda /\mu }(X_1,\dots ,X_n)=s_{\hat{\mu }/\hat{\lambda }}(X_n,\dots ,X_1)$$

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

(3.2) We have

$$s_{\lambda /\mu }(X_1,\dots ,X_n)=0$$

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

		Proof.
	If $\lambda \not\supset \mu$ then ${\lambda }_r<{\mu }_r$ for some $r\le n,$ and hence $${\lambda }_i-{\mu }_j-i+j\le {\lambda }_r-{\mu }_r<0$$ 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 $(n-r+1)\times r$ block of zeros in the south-west corner, and hence its determinant vanishes. $\square$

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

$$s_{\lambda /\mu }(X_1,\dots ,X_n)=s_{\lambda /\mu }(X_1,\dots ,X_r)$$

		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

$$e_r(X+Y)=\sum _{p+q=r}{e}_p\left(X\right)e_q\left(Y\right)$$

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

$$rk(X+Y)\le rk\left(X\right)+rk\left(Y\right)\text{.}$$

(3.4) Let $X_1,\dots ,X_n,Y_1,\dots ,Y_n\in R$ with $rk\left(Y_j\right)\le j-1$ $(1\le j\le n)$ (so that $Y_1=0\text{).}$ Then for all $\alpha \in {\mathbb{Z}}^n,$

$$s_{\alpha }(X_1,\dots ,X_n)=\text{det}\left(h_{{\alpha }_i-i+j}(X_i-Y_j)\right)\text{.}$$

		Proof.
	We have $$h_{{\alpha }_i-i+j}(X_i-Y_j)=\sum _{k=1}^j{h}_{{\alpha }_i-i+k}\left(X_i\right)h_{j-k}(-Y_j),$$ since $h_r(-Y_j)={(-1)}^r{e}_r\left(Y_j\right)=0$ if $r\ge j\text{.}$ Hence the matrix $${\left(h_{{\alpha }_i-i+j}(X_i-Y_j)\right)}_{1\le i,j\le n}$$ is the product of the matrix $${\left(h_{{\alpha }_i-i+j}\left(X_i\right)\right)}_{1\le i,j\le n}$$ and the matrix $${\left(h_{i-j}(-Y_j)\right)}_{1\le i,j\le n}$$ 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 $s_{\alpha }(X_1,\dots ,X_n)$ is mainly of interest when $X_1,\dots ,X_n$ is an increasing sequence in $R,$ in the sense that $rk(X_{i+1}-X_i)<\infty$ for $1\le i\le n-1\text{.}$

(3.5) Let $x_i,y_i$ $(i\ge 1)$ be elements of $R,$ each of rank $\le 1,$ and let

$$X_i=x_1+\dots +x_i,\hspace{0.17em}{Y}_i=y_1+\dots +y_i$$

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

$$s_{\alpha }(X_1-Y_{{\alpha }_1},\dots ,X_n-Y_{{\alpha }_n})=\prod _{i=1}^n\prod _{j=1}^{{\alpha }_i}(x_i-y_j)\text{.}$$

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

$$s_{\lambda }(X_1-Y_{{\lambda }_1},\dots ,X_n-Y_{{\lambda }_n})=\prod _{(i,j)\in \lambda }(x_i-y_j)\text{.}$$

		Proof.
	From (3.4) we have $$\begin{array}{cc}(*)& s_{\alpha }(X_1-Y_{{\alpha }_1},\dots ,X_n-Y_{{\alpha }_n})=\text{det}{\left(h_{{\alpha }_i-i+j}(X_i-Y_{{\alpha }_i}-X_{j-1})\right)}_{1\le i,j\le n}\text{.}\end{array}$$ If $j>i,$ then $$h_{{\alpha }_i-i+j}(X_i-Y_{{\alpha }_i}-X_{j-1})=\pm e_{{\alpha }_i-i+j}(Y_{{\alpha }_i}+X_{j-1}-X_i)$$ which is $0$ because $$rk(Y_{{\alpha }_i}+X_{j-1}-X_i)\le {\alpha }_i+(j-1)-i<{\alpha }_i-i+j\text{.}$$ Hence the determinant at $(*)$ is triangular, with diagonal elements $$\begin{array}{ccc}{h}_{{\alpha }_i}(X_i-Y_{{\alpha }_i}-X_{i-1})& =& h_{{\alpha }_i}(x_i-Y_{{\alpha }_i})\\ & =& \sum _{r\ge 0}{h}_r(-Y_{{\alpha }_i})h_{{\alpha }_i-r}\left(x_i\right)\\ & =& \sum _{r\ge 0}{(-1)}^r{e}_r\left(Y_{{\alpha }_i}\right)x_i^{{\alpha }_i-r}\\ & =& \prod _{j=1}^{{\alpha }_i}(x_i-y_j)\text{.}\end{array}$$ The formula (3.5) now follows. $\square$

In particular, when all the $y_i$ are zero we have

$$\begin{array}{cc}\text{(3.5')}& s_{\alpha }(X_1,\dots ,X_n)=\prod _{i=1}^n{x}_i^{{\alpha }_i}=x^{\alpha }\end{array}$$

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

$$\begin{array}{ccc}{s}_{\lambda }(-Y_{{\lambda }_1},\dots ,-Y_{{\lambda }_n})& =& {(-1)}^{\left|\lambda \right|}\prod _{(i,j)\in \lambda }{y}_j\\ & =& {(-1)}^{\left|\lambda \right|}{y}^{\lambda \prime }\text{.}\end{array}$$

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

$$\begin{array}{cc}\text{(3.5'')}& x^{\lambda }={(-1)}^{\left|\lambda \right|}{s}_{\lambda \prime }(-X_{{\lambda }_1^{\prime }},\dots ,-X_{{\lambda }_n^{\prime }})\text{.}\end{array}$$

(3.6) Let $\lambda =({\lambda }_1,{\lambda }_2,\dots )$ be a partition of length $\le n,$ and $X_1,\dots ,X_n$ elements of a $\lambda \text{-ring}$ $R\text{.}$ Suppose that $i<j$ are such that ${\lambda }_i={\lambda }_{i+1}=\dots ={\lambda }_j$ and

$$rk(X_j-X_k)\le j-k\text{for}\hspace{0.17em}i\le k\le j\text{.}$$

Then

$$s_{\lambda /\mu }(X_1,\dots ,X_n)=s_{\lambda /\mu }(X_1,\dots ,X_{i-1},X_j,\dots ,X_j,X_{j+1},\dots ,X_n),$$

that is to say we can replace each $X_k$ $(i\le k\le j)$ 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 $$\begin{array}{ccc}{h}_m\left(X_i\right)& =& h_m(X_j-Y)\\ & =& \sum _{k=0}^{j-i}{(-1)}^k{e}_k\left(Y\right)h_{m-k}\left(X_j\right)\text{.}\end{array}$$ It follows that if we replace the $i^{\text{th}}$ row of the determinant $s_{\lambda /\mu }(X_1,\dots ,X_{i-1},X_j,\dots ,X_j,X_{j+1},\dots ,X_n)$ by $$\sum _{k=0}^{j-i}{(-1)}^k{e}_k\left(Y\right){\text{row}}_{i+k}$$ we shall obtain $$s_{\lambda /\mu }(X_1,\dots ,X_{i-1},X_i,X_j,\dots ,X_j{X}_{j+1},\dots ,X_n)$$ with $j-i$ arguments equal to $X_j\text{.}$ The proof is now completed by induction on $j-i\text{.}$ $\square$

Duality

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

$$rk(X_n-X_{n-1})\le 1$$

for all $n\in \mathbb{Z}\text{.}$

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

$$H={\left(h_{i-j}(-X_i)\right)}_{i,j\in I}$$

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 $(i,k)$ element of $HK$ is then $$\sum _j{h}_{i-j}(-X_i)h_{j-k}\left(X_{j+1}\right)=h_{i-k}(-X_i+X_{k+1})\text{.}$$ If $i<k$ this is zero; if $i=k$ it is equal to $1\text{;}$ and if $i>k$ it is equal to $${(-1)}^{i-k}{e}_{i-k}(X_i-X_{k+1})$$ which is zero because $rk(X_i-X_{k+1})\le i-(k+1)<i-k\text{.}$ $\square$

(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

$$s_{\lambda /\mu }(-X_{{\lambda }_1-1},\dots ,-X_{{\lambda }_n-n})={(-1)}^{|\lambda -\mu |}{s}_{\lambda \prime /\mu \prime }(X_{1-{\lambda }_1^{\prime }},\dots ,X_{m-{\lambda }_m^{\prime }})\text{.}$$

		Proof.
	Let $$\begin{array}{c}{\xi }_i={\lambda }_i-i,{\eta }_i={\mu }_i-i(1\le i\le n),\\ {\xi }_j^{\prime }={\lambda }_j^{\prime }-j,{\eta }_j^{\prime }={\mu }_j^{\prime }-j(1\le j\le n),\end{array}$$ Then the integers ${\xi }_i$ $(1\le i\le n)$ and $-{\xi }_j^{\prime }-1$ $(1\le j\le m)$ fill up the interval $[-m,n-1],$ and so do the ${\eta }_i$ and the $-{\eta }_j^{\prime }-1\text{.}$ The $({\xi }_1,\dots ,{\xi }_n;{\eta }_1,\dots ,{\eta }_n)$ minor of the matrix $H$ is $$\text{det}\left(h_{{\eta }_i-{\eta }_j}(-X_{{\xi }_i})\right)=s_{\lambda /\mu }(-X_{{\xi }_1},\dots ,-X_{{\xi }_n})\text{.}$$ 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 $-{\xi }_i^{\prime }-1$ $(1\le i\le m)$ and column indices $-{\eta }_j^{\prime }-\prime$ $(1\le j\le m)\text{.}$ Hence it is $${(-1)}^{|\lambda -\mu |}{s}_{\lambda \prime /\mu \prime }(-X_{{\xi }_1^{\prime }},\dots ,-X_{{\xi }_m^{\prime }})\text{.}$$ Since each minor of $H$ is equal to the complementary cofactor of $\left(H^{-1}\right)\prime$ (because $\text{det}\hspace{0.17em}H=1\text{)}$ the result follows. $\square$

Remark. Observe that

$$rk(X_{{\lambda }_i-i}-X_{{\lambda }_{i+1}-i-1})\le ({\lambda }_i-i)-({\lambda }_{i+1}-i-1)={\lambda }_i-{\lambda }_{i+1}+1\text{.}$$

Hence (3.8) gives a duality theorem for the multi-Schur function $s_{\lambda /\mu }(Y_1,\dots ,Y_n)$ provided that $rk(Y_{i+1}-Y_i)\le {\lambda }_i-{\lambda }_{i+1}+1$ 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

$$\lambda =(p_1^{m_1},p_2^{m_2},\dots ,p_k^{m_k})$$

where $p_1>p_2>\dots >p_k>0$ and each $m_i\ge 1\text{.}$ Then in

$$s_{\lambda /\mu }(-X_{{\lambda }_1-1},\dots ,-X_{{\lambda }_n-n})$$

the first $m_1$ arguments are

$$-X_{p_1-1},\dots ,-X_{p_1-m_1}$$

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

$$-X_{p_2-m_1-1},\dots ,-X_{p_2-m_1-m_2}$$

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 $c_i=p_i-(m_1+\dots +m_i)\text{.}$ Now if

$$\lambda \prime =(q_1^{n_1},q_2^{n_2},\dots ,q_k^{n_k})$$

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 $s_i=(q_{k+1-i},p_i)$ 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

$$s_{\lambda /\mu }({(-X_{c_1})}^{m_1},\dots ,{(-X_{c_k})}^{m_k})={(-1)}^{|\lambda -\mu |}{s}_{\lambda \prime /\mu \prime }({\left(X_{c_k}\right)}^{n_1},\dots ,{\left(X_{c_1}\right)}^{n_k})\text{.}$$

Finally, if we set $Z_i=-X_{c_i}$ $(1\le i\le k)$ we have

$$\begin{array}{cc}\text{(3.8'')}& s_{\lambda /\mu }(Z_1^{m_1},\dots ,Z_k^{m_k})={(-1)}^{|\lambda -\mu |}{s}_{\lambda \prime /\mu \prime }({(-Z_k)}^{n_1},\dots ,{(-Z_1)}^{n_k})\end{array}$$

provided

$$rk(Z_{i+1}-Z_i)\le m_{i+1}+n_{k+1-i}(1\le i\le k-1)\text{.}$$

Let now $x_1,x_2,\dots$ be independent indeterminates over $\mathbb{Z}\text{.}$ We may regard $\mathbb{Z}[x_1,x_2,\dots ]$ 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

$$\begin{array}{cc}\text{(3.9)}& \begin{array}{ccc}{\partial }_i{h}_r\left(X_i\right)& =& h_{r-1}\left(X_{i+1}\right),\\ {\partial }_i{e}_r\left(X_i\right)& =& e_{r-1}\left(X_{i-1}\right),\\ {\pi }_i{h}_r\left(X_i\right)& =& h_r\left(X_{i+1}\right)\text{.}\end{array}\end{array}$$

		Proof.
	Consider the generating functions: ${\partial }_i{h}_r\left(X_i\right)$ is the coefficient of $t^r$ in $$\begin{array}{ccc}{\partial }_i\left(\sum _{r\ge 0}{h}_r\left(X_i\right)t^r\right)& =& {\partial }_i\prod _{j=1}^i{(1-x_{j}t)}^{-1}\\ & =& \prod _{j=1}^{i-1}{(1-x_{j}t)}^{-1}·{\partial }_i\left(\frac{1}{1-x_{i}t}\right),\end{array}$$ and $$\begin{array}{ccc}{\partial }_i\left(\frac{1}{1-x_{i}t}\right)& =& \frac{1}{x_i-x_{i+1}}(\frac{1}{1-x_{i}t}-\frac{1}{1-x_{i+1}t})\\ & =& \frac{t}{(1-x_{i}t)(1-x_{i+1}t)}\end{array}$$ so that $${\partial }_i\left(\sum _r{h}_r\left(X_i\right)t^r\right)=t\prod _{j=1}^{i+1}{(1-x_{j}t)}^{-1}=\sum _s{h}_s\left(X_{i+1}\right)t^{s+1}$$ 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 $\alpha \in {\mathbb{Z}}^n$ 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

$$\begin{array}{ccc}{\partial }_{r_i}{s}_{\alpha }(X_{r_1},\dots ,X_{r_n})& =& s_{\alpha -{\epsilon }_i}(X_{r_1},\dots ,X_{r+1},\dots ,X_{r_n}),\\ {\partial }_{r_i}{s}_{\alpha }(-X_{r_1},\dots ,-X_{r_n})& =& -s_{\alpha -{\epsilon }_i}(-X_{r_1},\dots ,-X_{r_i-1},\dots ,-X_{r_n}),\\ {\pi }_{r_i}{s}_{\alpha }(X_{r_1},\dots ,X_{r_n})& =& s_{\alpha }(X_{r_1},\dots ,X_{r+1},\dots ,X_{r_n}),\end{array}$$

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

$$\begin{array}{cc}(*)& s_{\lambda }(X_{m+1-{\lambda }_1},\dots ,X_{m+n-{\lambda }_n})={(-1)}^{\left|\lambda \right|}{s}_{\lambda \prime }(-X_{m+{\lambda }_1^{\prime }-1},\dots ,-X_{{\lambda }_m^{\prime }})\text{.}\end{array}$$

Let $(i,j)$ 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 $(i,j)\text{.}$ By operating on either side of $(*)$ with ${\partial }_{m+i-j}$ we obtain the same relation with $\mu$ replacing $\lambda \text{.}$ Hence it is enough to show that $(*)$ 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 $\alpha \in {\mathbb{Z}}^n$ we have

$$\begin{array}{ccc}{\partial }_{w_0}{s}_{\alpha }(X_1+Z_1,\dots ,X_n+Z_n)& =& s_{\alpha -\delta }(X_n+Z_1,\dots ,X_n+Z_n),\\ {\pi }_{w_0}{s}_{\alpha }(X_1+Z_1,\dots ,X_n+Z_n)& =& s_{\alpha }(X_n+Z_1,\dots ,X_n+Z_n),\end{array}$$

where $X_i=x_1+\dots +x_i$ $(1\le i\le n)$ and the $Z_i$ are independent of $x_1,\dots ,x_n\text{.}$

		Proof.
	The sequence $$(n-1,n-2,n-1,\dots ,2,3,\dots ,n-1,1,2,3,\dots ,n-1)$$ is a reduced word for $w_0,$ so that $${\pi }_{w_0}={\pi }_{n-1}\left({\pi }_{n-2}{\pi }_{n-1}\right)\dots \left({\pi }_2{\pi }_3\dots {\pi }_{n-1}\right)\left({\pi }_1{\pi }_2\dots {\pi }_{n-1}\right)$$ and likewise for ${\partial }_{w_0}\text{.}$ By (3.10), ${\pi }_1{\pi }_2\dots {\pi }_{n-1}$ applied to $s_{\alpha }(X_1+Z_1,\dots ,X_n+Z_n)$ will produce $$s_{\alpha }(X_2+Z_1,X_3+Z_2,\dots ,X_n+Z_{n-1},X_n+Z_n)\text{.}$$ We have next to operate on this with ${\pi }_2{\pi }_3\dots {\pi }_{n-1},$ which will produce $$s_{\alpha }(X_3+Z_1,X_4+Z_2,\dots ,X_n+Z_{n-2},X_n+Z_{n-1},X_n+Z_n)\text{.}$$ 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 $\alpha \in {\mathbb{N}}^n$ and $Z_1=\dots =Z_n=0$ in (3.11). Then by (3.5') we have

$$\begin{array}{ccccc}{\partial }_{w_0}\left(x^{\alpha }\right)& =& {\partial }_{w_0}{s}_{\alpha }(X_1,\dots ,X_n)& =& s_{\alpha -\delta }\left(X_n\right),\\ {\pi }_{w_0}\left(x^{\alpha }\right)& =& {\pi }_{w_0}{s}_{\alpha }(X_1,\dots ,X_n)& =& s_{\alpha }\left(X_n\right)\text{.}\end{array}$$

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

$$X_i=x_1+\dots +x_i,\hspace{0.17em}{Y}_i=y_1+\dots +y_i$$

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

$$s_{\lambda }(X_m-Y_n)=\sum _{w\in S_m\times S_n}w(f_{\lambda }(x,y)/D\left(x\right)D\left(y\right))$$

where

$$\begin{array}{c}{f}_{\lambda }(x,y)=\prod _{(i,j)\in \lambda }(x_i-y_j),\\ D\left(x\right)=\prod _{1\le i<j\le m}(1-x_i^{-1}{x}_j),D\left(y\right)=\prod _{1\le i<j\le n}(1-y_i^{-1}{y}_j)\text{.}\end{array}$$

		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 $r=\ell \left(\lambda \right),$ $$\begin{array}{cc}\text{(1)}& f_{\lambda }(x,y)=s_{\lambda }(X_1-Y_{{\lambda }_1},\dots ,X_r-Y_{{\lambda }_r})\end{array}$$ and in view of (2.16) Sergeev's formula may be restated in the form $$\begin{array}{cc}\text{(3.12')}& s_{\lambda }(X_m-Y_n)={\pi }_y{\pi }_x{f}_{\lambda }(x,y)\text{.}\end{array}$$ From (3.11) and (1) above we have $$\begin{array}{cc}\text{(2)}& {\pi }_x{f}_{\lambda }(x,y)=s_{\lambda }(X_m-Y_{{\lambda }_1},\dots ,X_m-Y_{{\lambda }_r})\text{.}\end{array}$$ If $\lambda =(p_1^{m_1},\dots ,p_k^{m_k}),$ (2) can be rewritten in the form $$\begin{array}{cc}\text{(3)}& {\pi }_x{f}_{\lambda }(x,y)=s_{\lambda }(Z_1^{m_1},\dots ,Z_k^{m_k})\end{array}$$ where $Z_i=X_m-Y_{p_i}\text{.}$ Since $$rk(Z_{i+1}-Z_i)=rk(Y_{p_i}-Y_{p_{i+1}})=p_i-p_{i+1},$$ the duality theorem (3.8'') applies, and gives $$\begin{array}{cc}\text{(4)}& \begin{array}{ccc}{s}_{\lambda }(Z_1^{m_1},\dots ,Z_k^{m_k})& =& {(-1)}^{\left|\lambda \right|}{s}_{\lambda \prime }({(-Z_k)}^{n_1},\dots ,{(-Z_1)}^{n_k})\\ & =& {(-1)}^{\left|\lambda \right|}{s}_{\lambda \prime }({(Y_{p_k}-X_m)}^{n_1},\dots ,{(Y_{p_1}-X_{\mathrm{m}})}^{n_k})\\ & =& {(-1)}^{\left|\lambda \right|}{s}_{\lambda \prime }(Y_1-X_m,Y_2-X_m,\dots ,Y_s-X_m)\end{array}\end{array}$$ 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) $$\begin{array}{ccc}{\pi }_y{\pi }_x{f}_{\lambda }& =& {(-1)}^{\left|\lambda \right|}{\pi }_y{s}_{\lambda \prime }(Y_1-X_m,\dots ,Y_s-X_m)\\ & =& {(-1)}^{\left|\lambda \right|}{s}_{\lambda \prime }(Y_n-X_m)\\ & =& s_{\lambda }(X_m-Y_n)\text{.}\end{array}$$ $\square$

Notes and References

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

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