Notes on Schubert Polynomials
Chapter 5

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

Last update: 2 July 2013

Orthogonality

Recall that

$$\begin{array}{ccc}{P}_n& =& \mathbb{Z}[x_1,\dots ,x_n],\\ {\Lambda }_n& =& \mathbb{Z}{[x_1,\dots ,x_n]}^{S_n}\end{array}$$

where $x_1,\dots ,x_n$ are independent indeterminates.

(5.1) $P_n$ is a free ${\Lambda }_n\text{-module}$ of rank $n!$ with basis

$$B_n=\{x^{\alpha }:0\le {\alpha }_i\le i-1,1\le i\le n\}\text{.}$$

		Proof.
	by induction on $n\text{.}$ The result is trivially true when $n=1,$ so assume that $n>1$ and that $P_{n-1}$ is a free ${\Lambda }_{n-1}\text{-module}$ with basis $B_{n-1}\text{.}$ Since $P_n=P_{n-1}\left[x_n\right],$ it follows that $P_n$ is a free ${\Lambda }_{n-1}\left[x_n\right]\text{-module}$ with basis $B_{n-1}\text{.}$ Now $${\Lambda }_{n-1}\left[x_n\right]={\Lambda }_n\left[x_n\right],$$ because the identities $$e_r(x_1,\dots ,x_n)=\sum _{x=0}^r{(-x_n)}^s{e}_{r-s}(x_1,\dots ,x_n)$$ show that ${\Lambda }_{n-1}\subset {\Lambda }_n\left[x_n\right],$ and on the other hand it is clear that ${\Lambda }_n\subset {\Lambda }_{n-1}\left[x_n\right]\text{.}$ Hence $P_n$ is a free ${\Lambda }_n\left[x_n\right]\text{-module}$ with basis $B_{n-1}\text{.}$ To complete the proof it remains to show that ${\Lambda }_n\left[x_n\right]$ is a free ${\Lambda }_n\text{-module}$ with basis $1,x_n,\dots ,x_n^{n-1}\text{.}$ Since $\prod _{i=1}^n(x_n-x_i)=0,$ we have $$x_n^n=e_1{x}_n^{n-1}-e_2{x}_n^{n-2}+\dots +{(-1)}^{n-1}{e}_n,$$ from which it follows that the $x_n^{n-i}$ $(1\le i\le n)$ generate ${\Lambda }_n\left[x_n\right]$ as a ${\Lambda }_n\text{-module.}$ On the other hand, if we have a relation of linear dependence $$\sum _{i=1}^n{f}_i{x}_n^{n-i}=0$$ with coefficients $f_i\in {\Lambda }_n,$ then we have also $$\sum _{i=1}^n{f}_i{x}_j^{n-i}=0$$ for $j=1,2,\dots ,n,$ and since $$\text{det}\left(x_j^{n-i}\right)=\prod _{i<j}(x_i-x_j)\ne 0,$$ it follows that $f_1=\dots =f_n=0\text{.}$ $\square$

As before, let $\delta =(n-1,n-2,\dots ,1,0)\text{.}$ By reversing the order of $x_1,\dots ,x_n$ in (5.1) it follows that

(5.1') The monomials $x^{\alpha },\alpha \subset \delta$ (i.e., $0\le {\alpha }_i\le n-i$ for $1\le i\le n\text{)}$ form a ${\Lambda }_n\text{-basis}$ of $P_n\text{.}$

We define a scalar product on $P_n,$ with values in ${\Lambda }_n,$ by the rule

$$\begin{array}{cc}\text{(5.2)}& ⟨f,g⟩={\partial }_{w_0}\left(fg\right)(f,g\in P_n)\end{array}$$

where $w_0$ is the longest element of $S_n\text{.}$ Since ${\partial }_{w_0}$ is ${\Lambda }_n\text{-linear,}$ so is the scalar product.

(5.3) Let $w\in S_n$ and $f,g\in P_n\text{.}$ Then

(i)	$⟨{\partial }_{w}f,g⟩=⟨f,{\partial }_{w^{-1}}g⟩$
(ii)	$⟨wf,g⟩=\epsilon \left(w\right)⟨f,w^{-1}g⟩\text{.}$

where $\epsilon \left(w\right)={(-1)}^{\ell \left(w\right)}$ is the sign of $w\text{.}$

		Proof.
	(i) It is enough to show that $⟨{\partial }_{i}f,g⟩=⟨f,{\partial }_{i}g⟩$ for $i\le i\le n-1\text{.}$ We have $$\begin{array}{ccc}⟨{\partial }_{i}f,g⟩& =& {\partial }_{w_0}\left(\left({\partial }_{i}f\right)g\right)={\partial }_{w_0{s}_i}{\partial }_i\left(\left({\partial }_{i}f\right)g\right)\\ & =& {\partial }_{w_0{s}_i}\left(\left({\partial }_{i}f\right)\left({\partial }_{i}g\right)\right)\end{array}$$ because ${\partial }_{i}f$ is symmetrical in $x_i$ and $x_{i+1}\text{.}$ The last expression is symmetrical in $f$ and $g,$ hence $⟨{\partial }_{i}f,g⟩=⟨{\partial }_{i}g,f⟩=⟨f,{\partial }_{i}g⟩$ as required. (ii) Again it is enough to show that $⟨s_{i}f,g⟩=-⟨f,s_{i}g⟩\text{.}$ We have $$⟨s_{i}f,g⟩={\partial }_{w_0}\left(\left(s_{i}f\right)g\right)={\partial }_{w_0{s}_i}{\partial }_i{s}_i\left(f\left(s_{i}g\right)\right)$$ and since ${\partial }_i{s}_i=-{\partial }_i$ this is equal to $$-{\partial }_{w_0{s}_i}{\partial }_i\left(f\left(s_{i}g\right)\right)=-{\partial }_{w_0}\left(f\left(s_{i}g\right)\right)=-⟨f,s_{i}g⟩\text{.}$$ $\square$

(5.4) Let $u,v\in S_n$ be such that $\ell \left(u\right)+\ell \left(v\right)=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\text{.}$ Then

$$⟨{𝔖}_u,{𝔖}_v⟩=\{\begin{array}{cc}1& \text{if}\hspace{0.17em}v=w_{0}u,\\ 0& \text{otherwise.}\end{array}$$

		Proof.
	We have $$\begin{array}{ccc}⟨{𝔖}_u,{𝔖}_v⟩& =& ⟨{\partial }_{u^{-1}{w}_0}{x}^{\delta },{𝔖}_v⟩\\ & =& ⟨x^{\delta },{\partial }_{w_{0}u}{𝔖}_v⟩\end{array}$$ by (5.3). Also $\ell \left(w_{0}u\right)=\ell \left(w_0\right)-\ell \left(u\right)=\ell \left(v\right),$ hence $${\partial }_{w_{0}u}{𝔖}_v=\{\begin{array}{cc}1& \text{if}\hspace{0.17em}v=w_{0}u,\\ 0& \text{otherwise.}\end{array}$$ It follows that $$⟨{𝔖}_u,{𝔖}_v⟩=\{\begin{array}{cc}0& \text{if}\hspace{0.17em}v\ne w_{0}u,\\ ⟨x^{\delta },1⟩={\partial }_{w_0}\left(x^{\delta }\right)=1& \text{if}\hspace{0.17em}v=w_{0}u\text{.}\end{array}$$ $\square$

(5.5) Let $u,v\in S_n\text{.}$ Then

$$⟨w_0{𝔖}_u,{𝔖}_{v{w}_0}⟩=\epsilon \left(v\right){\delta }_{uv}\text{.}$$

		Proof.
	We have $$\begin{array}{ccc}⟨w_0{𝔖}_u,{𝔖}_{v{w}_0}⟩& =& ⟨w_0{𝔖}_u,{\partial }_{w_0{v}^{-1}{w}_0}{x}^{\delta }⟩\\ & =& ⟨{\partial }_{w_{0}v{w}_0}\left(w_0{𝔖}_u\right),x^{\delta }⟩\\ & =& \epsilon \left(v\right)⟨w_0{\partial }_v{𝔖}_u,x^{\delta }⟩\end{array}$$ by (5.3) and (2.12). By (4.2) the scalar product is therefore zero unless $\ell \left(u\right)-\ell \left(v\right)=\ell \left(u{v}^{-1}\right),$ and then it is equal to $\epsilon \left(v\right)⟨w_0{𝔖}_{u{v}^{-1}},x^{\delta }⟩\text{.}$ Now ${𝔖}_{u{v}^{-1}}$ is a linear combination of monomials $x^{\alpha }$ such that $\alpha \subset \delta$ and $\left|\alpha \right|=\ell \left(u\right)-\ell \left(v\right)\text{.}$ Hence $w_0\left({𝔖}_{u{v}^{-1}}\right)x^{\delta }$ is a sum of monomials $x^{\beta }$ where $$\beta =w_0\alpha +\delta \subset w_0\delta +\delta =(n-1,\dots ,n-1)\text{.}$$ Now ${\partial }_{w_0}{x}^{\beta }=0$ unless all the components ${\beta }_i$ of $\beta$ are distinct; since $0\le {\beta }_i\le n-1$ for each $i,$ it follows that ${\partial }_{w_0}{x}^{\beta }=0$ unless $\beta =w\delta$ for some $w\in S_n,$ and in that case $$w_0\alpha =\beta -\delta =w\delta -\delta$$ must have all its components $\ge 0\text{.}$ So the only possibility that gives a nonzero scalar product is $w=1,\alpha =0,u=v,$ and in that case $$\begin{array}{ccc}⟨w_0{𝔖}_u,{𝔖}_{v{w}_0}⟩& =& \epsilon \left(v\right)⟨1,x^{\delta }⟩\\ & =& \epsilon \left(v\right){\partial }_{w_0}\left(x^{\delta }\right)=\epsilon \left(v\right)\text{.}\end{array}$$ $\square$

(5.6) The Schubert polynomials ${𝔖}_w,w\in S_n,$ form a ${\Lambda }_n\text{-basis}$ of $P_n\text{.}$

		Proof.
	Let $u,v\in S_n$ and let $$\begin{array}{cc}\text{(1)}& w_0{𝔖}_u=\sum _{\alpha \subset \delta }{a}_{u\alpha }{x}^{\alpha },\end{array}$$ $$\begin{array}{cc}\text{(2)}& \epsilon \left(v\right){𝔖}_{v{w}_0}=\sum _{\beta \subset \delta }{b}_{v\beta }{x}^{\beta },\end{array}$$ with coefficients $a_{u\alpha },b_{v\beta }\in {\Lambda }_n\text{.}$ Let $c_{\alpha \beta }=⟨x^{\alpha },x^{\beta }⟩\text{.}$ Then from (5.5) we have $$\sum _{\alpha ,\beta }{a}_{u\alpha }{c}_{\alpha \beta }{b}_{v\beta }={\delta }_{uv},$$ or in matrix terms $$\begin{array}{cc}\text{(3)}& AC{B}^t=1\end{array}$$ where $A=\left(a_{u\alpha }\right),$ $B=\left(b_{v\beta }\right)$ and $C=\left(c_{\alpha \beta }\right)$ are square matrices of size $n!,$ with coefficients in ${\Lambda }_n\text{.}$ From (3) it follows that each of $A,B,C$ has determinant $\pm 1\text{;}$ hence the equations (2) can be solved for $x^{\beta },\beta \subset \delta ,$ as ${\Lambda }_n\text{-linear}$ combinations of the Schubert polynomials ${𝔖}_w,w\in S_n\text{.}$ Since by (5.1') the $x^{\beta }$ from a ${\Lambda }_n\text{-basis}$ of $P_n,$ so also do the ${𝔖}_w\text{.}$ $\square$

We have

$$\begin{array}{cc}\text{(5.7)}& ⟨f,g⟩=\sum _{w\in S_n}\epsilon \left(w\right){\partial }_w\left(w_{0}f\right){\partial }_{w{w}_0}\left(g\right)\end{array}$$

for all $f,g\in P_n\text{.}$

		Proof.
	Let $\Phi (f,g)$ denote the right-hand side of (5.7). We claim first that $$\begin{array}{cc}\text{(1)}& \Phi (f,g)\in {\Lambda }_n\text{.}\end{array}$$ For this it is enough to show that ${\partial }_i\Phi =0$ for $1\le i\le n-1\text{.}$ Let $$A_i=\{w\in S_n:\ell \left(s_{i}w\right)>\ell \left(w\right)\},$$ then $S_n$ is the disjoint union of $A$ and $s_{i}A,$ and $s_{i}A=A{w}_0\text{.}$ Hence $$\Phi (f,g)=\sum _{w\in A_i}\epsilon \left(w\right)\{{\partial }_w\left(w_{0}f\right){\partial }_i\left({\partial }_{s_{i}w{w}_0}g\right)-{\partial }_i{\partial }_w\left(w_{0}f\right)\left({\partial }_{s_{i}w{w}_0}g\right)\}\text{.}$$ Since for all $\varphi ,\psi \in P_n$ we have $${\partial }_i(\varphi {\partial }_i\psi -\left({\partial }_i\varphi \right)\psi )=\left({\partial }_i\varphi \right)\left({\partial }_i\psi \right)-\left({\partial }_i\varphi \right)\left({\partial }_i\psi \right)=0,$$ it follows that ${\partial }_i\Phi (f,g)=0$ for all $i$ as required. Next, since each operator ${\partial }_w$ is ${\Lambda }_n\text{-linear,}$ it follows that $\Phi (f,g)$ is ${\Lambda }_n\text{-linear}$ in each argument. By (5.6) it is therefore enough to verify (5.7) when $f=w_0{𝔖}_u$ and $g={𝔖}_{v{w}_0},$ where $u,v\in S_n\text{.}$ We have then $$\Phi (w_0{𝔖}_u,{𝔖}_{v{w}_0})=\sum _{w\in S_n}\epsilon \left(w\right){\partial }_{w^{-1}}\left({𝔖}_u\right){\partial }_{w^{-1}{w}_0}\left({𝔖}_{v{w}_0}\right)$$ which by (4.2) is equal to $$\begin{array}{cc}\text{(2)}& \sum _w\epsilon \left(w\right){𝔖}_{uw}{𝔖}_{vw}\end{array}$$ summed over $w\in S_n$ such that $$\ell \left(uw\right)=\ell \left(u\right)-\ell \left(w^{-1}\right)=\ell \left(u\right)-\ell \left(w\right)$$ and $$\ell \left(vw\right)=\ell \left(v{w}_0\right)-\ell \left(w^{-1}{w}_0\right)=\ell \left(w\right)-\ell \left(v\right)\text{.}$$ Hence the polynomial (2) is (i) symmetric in $x_1,\dots ,x_n$ (by (1) above), (ii) independent of $x_n,$ (iii) homogeneous of degree $\ell \left(u\right)-\ell \left(v\right)\text{.}$ Hence it vanishes unless $\ell \left(u\right)=\ell \left(v\right)$ and $u=w^{-1}=v,$ in which case it is equal to $\epsilon \left(w\right)=\epsilon \left(v\right)\text{.}$ Hence $$\Phi (w_0{𝔖}_u,{𝔖}_{v{w}_0})=\epsilon \left(v\right){\delta }_{uv}=⟨w_0{𝔖}_u,{𝔖}_{v{w}_0}⟩$$ by (5.5). This completes the proof of (5.7). $\square$

Now let $x=(x_1,\dots ,x_n)$ and $y=(y_1,\dots ,y_n)$ be two sequences of independent variables, and let

$$\begin{array}{cc}\text{(5.8)}& \Delta =\Delta (x,y)=\prod _{i+j\le n}(x_i-y_j)\end{array}$$

(the "semiresultant"). We have

$$\begin{array}{cc}\text{(5.9)}& \Delta (wx,x)=\{\begin{array}{cc}0& \text{if}\hspace{0.17em}w\ne w_0,\\ \epsilon \left(w_0\right)a_{\delta }\left(x\right)& \text{if}\hspace{0.17em}w=w_0\text{.}\end{array}\end{array}$$

For

$$\Delta (wx,x)=\prod _{i+j\le n}(x_{w\left(i\right)}-x_j)$$

is non-zero if and only if $w\left(i\right)\ne j$ whenever $i+j\le n,$ that is to say if and only if $w\ne w_0\text{;}$ and

$$\begin{array}{ccc}\Delta (w_{0}x,x)& =& \prod _{i+j\le n}(x_{n+1-i}-x_j)\\ & =& \prod _{j;tk}(x_k-x_j)=\epsilon \left(w_0\right)a_{\delta }\left(x\right)\text{.}\end{array}$$

The polynomial $\Delta (x,y)$ is a linear combination of the monomials $x^{\alpha },\alpha \subset \delta ,$ with coefficients in $\mathbb{Z}[y_1,\dots ,y_n]=P_n\left(y\right),$ hence by (4.11) can be written uniquely in the form

$$\Delta (x,y)=\sum _{w\in S_n}{𝔖}_w\left(x\right)T_w\left(y\right)$$

with $T_w\left(y\right)\in P_n\left(y\right)\text{.}$ By (5.5) we have

$$T_w\left(y\right)={⟨\Delta (x,y),w_0{𝔖}_{w{w}_0}(-x)⟩}_x$$

where the suffix $x$ means that the scalar product is taken in the $x$ variables. Hence

$$\begin{array}{cc}\text{(1)}& \begin{array}{ccc}{T}_w\left(y\right)& =& {\partial }_{w_0}\left(\Delta (x,y)w_0\left({𝔖}_{w{w}_0}(-x)\right)\right)\\ & =& a_{\delta }{\left(x\right)}^{-1}\sum _{v\in S_n}\epsilon \left(v\right)\Delta (vx,y)v{w}_0\left({𝔖}_{w{w}_0}(-x)\right)\end{array}\end{array}$$

by (2.10), where $v\in S_n$ acts by permuting the $x_i\text{.}$

Now this expression (1) must be independent of $x_1,\dots ,x_n\text{.}$ Hence we may set $x_i=y_i$ $(1\le i\le n)\text{.}$ But then (5.9) shows that the only non-zero term in the sum (1) is that corresponding to $v=w_0,$ and we obtain

$$T_w\left(y\right)={𝔖}_{w{w}_0}(-y)\text{.}$$

Hence we have proved

(5.10) ("Cauchy formula")

$$\Delta (x,y)=\sum _{w\in S_n}{𝔖}_w\left(x\right){𝔖}_{w{w}_0}(-y)\text{.}$$

Remark. Let $n=r+s$ where $r,s\ge 1,$ and regard $S_r\times S_s$ as a subgroup of $S_n,$ with $S_r$ permuting $1,2,\dots ,r$ and $S_s$ permuting $r+1,\dots ,r+s\text{.}$ Let $w_0^{\left(r\right)},w_0^{\left(s\right)}$ be the longest elements of $S_r,S_s$ respectively, and let $u=w_0^{\left(r\right)}\times w_0^{\left(s\right)}\text{.}$ If $w\in S_n,$ we have ${\partial }_u{𝔖}_w={𝔖}_{wu}$ if $\ell \left(wu\right)=\ell \left(w\right)-\ell \left(u\right),$ that is to say if $wu$ is Grassmannian (with its only descent at $r\text{),}$ and ${\partial }_u{𝔖}_w=0$ otherwise. Hence by applying ${\partial }_u$ to the $x\text{-variables}$ in (5.10) we obtain

$${\partial }_u\Delta (x,y)=\sum _{v\in G_{r,s}}{𝔖}_v\left(x\right){𝔖}_{vu{w}_0}(-y)$$

where $G_{r,s}\subset S_n$ is the set of Grassmannian permutations $v$ with descent at $r$ (i.e. $v\left(i\right)<v(i+i)$ if $i\ne r\text{).}$ On the other hand, it is easily verified that

$${\partial }_u\Delta (x,y)=\prod _{i=1}^r\prod _{j=1}^s(x_i-y_j)$$

and that $v\prime =vu{w}_0$ is the permutation

$$(v(r+1),\dots ,v(r+s),v\left(1\right),\dots ,v\left(r\right))$$

hence is also Grassmannian, with descent at $s\text{.}$

The shape of $v$ is

$$\lambda =\lambda \left(v\right)=(v\left(r\right)-r,\dots ,v\left(2\right)-2,v\left(1\right)-1)$$

and the shape of $v\prime$ is say

$$\mu \prime =\lambda \left(v\prime \right)=(v(r+s)-s,\dots ,v(r+2)-2,v(r+1)-1)\text{.}$$

The relation between these two partitions is

$${\mu }_i=s-{\lambda }_{r+1-i}(1\le i\le r)$$

that is to say $\lambda$ is the complement, say $\hat{\mu },$ of $\mu$ in the rectangle $\left(s^r\right)$ with $r$ rows and $s$ columns. Hence, replacing each $y_j$ by $-y_j,$ we obtain from (5.10) by operating with ${\partial }_u$ on both sides and using (4.8)

$$\begin{array}{cc}\text{(5.11)}& \prod _{i=1}^r\prod _{j=1}^s(x_i+y_j)=\sum s_{\hat{\mu }}\left(x\right)s_{\mu \prime }\left(y\right)\end{array}$$

summed over all $\mu \subset \left(s^r\right),$ where $\hat{\mu }$ is the complement of $\mu$ in $\left(s^r\right)\text{.}$ This is one version of the usual Cauchy identity [Mac1979, Chapter I, (4.3)'].

Let ${\left({𝔖}_w\right)}_{w\in S_n}$ be the ${\Lambda }_n\text{-basis}$ of $P_n$ dual to the basis $\left({𝔖}_w\right)$ relative to the scalar product (5.2). By (5.3) and (5.5) we have

$$⟨{𝔖}_u,w_0{𝔖}_{v{w}_0}⟩=\epsilon \left(v{w}_0\right){\delta }_{uv}$$

or equivalently

$$⟨{𝔖}_u\left(x\right),w_0{𝔖}_{v{w}_0}(-x)⟩={\delta }_{uv}$$

which shows that

$$\begin{array}{cc}\text{(5.12)}& {𝔖}^w\left(x\right)=w_0{𝔖}_{w{w}_0}(-x)\end{array}$$

for all $w\in S_n\text{.}$ From (5.10) it follows that

$$\Delta (x,y)=\sum _{w\in S_n}{𝔖}_w\left(x\right)w_0{𝔖}^w\left(y\right)$$

or equivalently

$$\begin{array}{cc}\text{(5.13)}& \prod _{1\le i<j\le n}(x_i-y_j)=\sum _{w\in S_n}{𝔖}_w\left(x\right){𝔖}^w\left(y\right)\text{.}\end{array}$$

Let ${\left(x_{\beta }\right)}_{\beta \subset \delta }$ be the basis dual to ${\left(x^{\alpha }\right)}_{\alpha \subset \delta }\text{.}$ If

$$\begin{array}{ccc}{𝔖}_u& =& \sum a_{u\alpha }{x}^{\alpha },\\ {𝔖}^v& =& \sum b_{v\beta }{x}_{\beta },\end{array}$$

then by taking scalar products we have

$$\sum _{\alpha }{a}_{u\alpha }{b}_{v\beta }={\delta }_{uv}$$

and therefore also

$$\sum _w{a}_{w\alpha }{b}_{w\beta }={\delta }_{\alpha \beta },$$

so that

$$\begin{array}{ccc}\sum _{w\in S_n}{𝔖}_w\left(x\right)S^w\left(y\right)& =& \sum _{\alpha ,\beta }\left(\sum _w{a}_{w\alpha }{b}_{w\beta }\right)x^{\alpha }{y}_{\beta }\\ & =& \sum _{\alpha }{x}^{\alpha }{y}_{\alpha }\text{.}\end{array}$$

From (5.13) it follows that $y_{\alpha }$ is the coefficient of $x^{\alpha }$ in ${\prod }_{i<j}(x_i-y_j),$ and hence we find

$$\begin{array}{cc}\text{(5.14)}& x_{\alpha }={(-1)}^{\left|\beta \right|}\prod _{i=1}^{n-1}{e}_{{\beta }_i}(x_{i+1},\dots ,x_n)\end{array}$$

where $\beta =\delta -\alpha \text{.}$

Let

$$C(x,y)=\epsilon \left(w_0\right)\Delta (w_{0}x,y)=\prod _{i<j}(y_i-x_j)\text{.}$$

If $f\left(x\right)\in H_n$ (4.11), let $f\left(y\right)$ denote the polynomial in $y_1,\dots ,y_n$ obtained by replacing each $x_i$ by $y_i\text{.}$ Then we have

$$\begin{array}{cc}\text{(5.15)}& {⟨f\left(x\right),C(x,y)⟩}_x=f\left(y\right),\end{array}$$

where as before the suffix $x$ means that the scalar product is taken in the $x$ variables. In other words, $C(x,y)$ is a "reproducing kernel" for the scalar product.

		Proof.
	From (5.13) we have $$C(x,y)=\sum _{w\in S_n}\epsilon \left(w_0\right){𝔖}_w\left(w_{0}x\right){𝔖}_{w{w}_0}(-y)\text{.}$$ Hence by (5.5) $$\begin{array}{ccc}{⟨C(x,y),{𝔖}_{w{w}_0}\left(x\right)⟩}_x& =& \epsilon \left(w{w}_0\right){𝔖}_{w{w}_0}(-y)\\ & =& {𝔖}_{w{w}_0}\left(y\right)\text{.}\end{array}$$ Hence (5.15) is true for all Schubert polynomials ${𝔖}_u,u\in S_n\text{.}$ Since the scalar product is ${\Lambda }_n\text{-linear}$ it follows from (5.6) that (5.15) is true for all $f\in H_n\text{.}$ $\square$

Let ${\theta }_{yx}$ be the homomorphism that replaces each $y_i$ by $x_i\text{.}$ Then (5.15) can be restated in the form

$$\begin{array}{cc}\text{(5.15')}& {\theta }_{yx}{⟨f\left(x\right),C(x,y)⟩}_x=f\left(x\right)\end{array}$$

for all $f\in H_n\text{.}$

Now let $z=[z_1,\dots ,z_n]$ be a third set of variables and consider

$$\begin{array}{cc}\text{(1)}& {⟨C(x,y),{\partial }_u{v}^{-1}C(x,z)⟩}_x\end{array}$$

for $u,v\in S_n,$ where ${\partial }_u$ and $v^{-1}$ act on the $x$ variables. By (5.3) this is equal to

$$\begin{array}{cc}\text{(2)}& \epsilon \left(v\right){⟨C(x,z),v{\partial }_{u^{-1}}C(x,y)⟩}_x\end{array}$$

and by (5.15') we have

$$\begin{array}{cc}\text{(3)}& {\theta }_{yx}{⟨C(x,y),{\partial }_u{v}^{-1}C(x,z)⟩}_x={\partial }_u{v}^{-1}C(x,z),\\ \text{(4)}& {\theta }_{zx}{⟨C(x,z),v{\partial }_{u^{-1}}C(x,y)⟩}_x=v{\partial }_{u^{-1}}C(x,y)\text{.}\end{array}$$

Since ${\theta }_{yx}$ and ${\theta }_{zx}$ commute, it follows from (1)-(4) that

$$\begin{array}{ccc}{\theta }_{yx}v{\partial }_{u^{-1}}C(x,y)& =& \epsilon \left(v\right){\theta }_{zx}{\partial }_u{v}^{-1}C(x,z)\\ & =& \epsilon \left(v\right){\theta }_{yx}{\partial }_u{v}^{-1}C(x,y)\text{.}\end{array}$$

Hence we have

$$\begin{array}{cc}\text{(5.16)}& \theta \left(v{\partial }_{u^{-1}}{w}_0\Delta \right)=\epsilon \left(v\right)\theta \left({\partial }_u{v}^{-1}{w}_0\Delta \right)\end{array}$$

for all $u,v\in S_n,$ where $\Delta =\Delta (x,y)$ and $\theta ={\theta }_{yx}\text{.}$

Let $E_n$ denote the algebra of operators $\varphi$ of the form

$$\varphi =\sum _{w\in S_n}{\varphi }_{w}w,$$

with coefficients ${\varphi }_w\in {\mathbb{Q}}_n=\mathbb{Q}(x_1,\dots ,x_n)\text{.}$ For such a $\varphi$ we have

$$\begin{array}{cc}\text{(5.17)}& {\varphi }_w=\epsilon \left(w_0\right)a_{\delta }^{-1}\theta \left(\varphi \left(w^{-1}{w}_0\Delta \right)\right)\end{array}$$

for all $w\in S_n,$ where $\varphi$ and $w^{-1}{w}_0$ act on the $x$ variables in $\Delta \text{.}$

For $\theta \left(\varphi \left(w^{-1}{w}_0\Delta \right)\right)={\sum }_{u\in S_n}{\varphi }_u\theta \left(u{w}^{-1}{w}_0\Delta \right),$ and by (5.8) $\theta \left(u{w}^{-1}{w}_0\Delta \right)=\Delta (u{w}^{-1}{w}_{0}x,x)$ is zero if $u\ne w,$ and is equal to $\epsilon \left(w_0\right)a_{\delta }$ if $u=w\text{.}$

Let $u\in S_n,$ and let $(a_1,\dots ,a_p)$ be a reduced word for $u,$ so that ${\partial }_u={\partial }_{a_1}\dots {\partial }_{a_p}\text{.}$ Since ${\partial }_a={(x_a-x_{a+1})}^{-1}(1-s_a)$ for each $a\ge 1,$ it follows that we may write

$$\begin{array}{cc}\text{(5.18)}& {\partial }_u=\epsilon \left(w_0\right)a_{\delta }^{-1}\sum _{v\le u}{\alpha }_{uv}v,\end{array}$$

where $v\le u$ means that $v$ is of the form $s_{b_1}\dots s_{b_q},$ where $(b_1,\dots ,b_q)$ is a subword of $(a_1,\dots ,a_p)\text{.}$

The coefficients ${\alpha }_{uv}$ in (5.18) are polynomials, for it follows from (5.16) and (5.17) that

$$\begin{array}{cc}\text{(5.19)}& \begin{array}{ccc}{\alpha }_{uv}& =& \theta \left({\partial }_u\left(v^{-1}{w}_0\Delta \right)\right)\\ & =& \epsilon \left(v\right)\theta \left(v{\partial }_{u^{-1}}{w}_0\Delta \right)\text{.}\end{array}\end{array}$$

(5.20) For all $f\in P_n$ we have

$$\theta \left({\partial }_u\left(\Delta f\right)\right)=\{\begin{array}{cc}{w}_{0}f& \text{if}\hspace{0.17em}u=w_0,\\ 0& \text{otherwise.}\end{array}$$

		Proof.
	From (5.18) we have $$\theta \left({\partial }_u\left(\Delta f\right)\right)=a_{\delta }^{-1}\sum _{v\le u}{\alpha }_{uv}v\left(f\right)\theta \left(v\Delta \right)\text{.}$$ By (5.9) this is zero if $u\ne w_0,$ and if $u=w_0$ then by (2.10) $$\begin{array}{ccc}\theta \left({\partial }_{w_0}\left(\Delta f\right)\right)& =& a_{\delta }^{-1}\sum _{w\in S_n}\epsilon \left(w\right)w\left(f\right)\theta \left(w\Delta \right)\\ & =& a_{\delta }^{-1}\epsilon \left(w_0\right)w_0\left(f\right)\epsilon \left(w_0\right)a_{\delta }=w_0\left(f\right)\end{array}$$ by (5.9) again. $\square$

The matrix of coefficients $\left({\alpha }_{uv}\right)$ in (5.18) is triangular with respect to the ordering $\le ,$ and one sees easily that the diagonal entries ${\alpha }_{uu}$ are non-zero (they are products in which each factor is of the form $x_i-x_j\text{).}$ Hence we may invert the equations (5.18), say

$$\begin{array}{cc}\text{(5.21)}& u=\sum _{v\le u}{\beta }_{uv}{\partial }_v\end{array}$$

and thus we can express any $\varphi \in E_n$ as a linear combination of the operators ${\partial }_w\text{.}$ Explicitly, we have

$$\begin{array}{cc}\text{(5.22)}& \varphi =\sum _{w\in S_n}\theta \left(\varphi \left({\partial }_{w^{-1}{w}_0}\Delta \right)\right){\partial }_w\text{.}\end{array}$$

		Proof.
	By linearity we may assume that $\varphi =f{\partial }_u$ with $f\in Q_n\text{.}$ Then $$\theta \left(\varphi \left({\partial }_{w^{-1}{w}_0}\Delta \right)\right)=f\theta \left({\partial }_u{\partial }_{w^{-1}{w}_0}\Delta \right)\text{.}$$ Now by (4.2) ${\partial }_u{\partial }_{w^{-1}{w}_0}$ is either zero or equal to ${\partial }_{u{w}^{-1}{w}_0},$ and by (5.20) $\theta \left({\partial }_{u{w}^{-1}{w}_0}\Delta \right)$ is zero if $w\ne u,$ and is equal to $1$ if $w=u\text{.}$ Hence the right-hand side of (5.22) is equal to $f{\partial }_u=\varphi ,$ as required. $\square$

In particular, it follows from (5.22) and (5.21) that

$$\begin{array}{cc}\text{(5.23)}& {\beta }_{uv}=\theta \left(u{\partial }_{v^{-1}{w}_0}\Delta \right),\end{array}$$

hence is a polynomial.

The coefficients ${\alpha }_{uv},{\beta }_{uv}$ in (5.18) and (5.23) satisfy the following relations:

(5.24)

(i) ${\beta }_{uv}=\epsilon \left(uv\right){\alpha }_{v{w}_0,u{w}_0},$ (ii) ${\alpha }_{u^{-1}{v}^{-1}}=v^{-1}\left({\alpha }_{uv}\right),$ (iii) ${\alpha }_{\bar{u},\bar{v}}=\epsilon \left(u{w}_0\right)w_0\left({\alpha }_{uv}\right),$

for all $u,v\in S_n,$ where $\bar{u}=w_{0}u{w}_0,$ $\bar{v}=w_{0}v{w}_0\text{.}$

		Proof.
	(i) By (5.23) and (2.12) we have $$\begin{array}{ccc}{\beta }_{uv}& =& \epsilon \left(v^{-1}{w}_0\right)\theta \left(u{w}_0{\partial }_{w_0{v}^{-1}}{w}_0\Delta \right)\\ & =& \epsilon \left(v^{-1}{w}_0\right)\epsilon \left(u{w}_0\right)\theta \left({\partial }_{v{w}_0}{w}_0{u}^{-1}{w}_0\Delta \right)& \text{by (5.16)}\\ & =& \epsilon \left(uv\right){\alpha }_{v{w}_0,u{w}_0}\text{.}& \text{by (5.19).}\end{array}$$ (ii) From (5.18) we have $$\begin{array}{ccc}\theta \left(v{\partial }_{u^{-1}}{w}_0\Delta \right)& =& \epsilon \left(w_0\right)v\left(a_{\delta }^{-1}\right)\sum _{w}v\left({\alpha }_{u^{-1},w^{-1}}\right)\theta \left(v{w}^{-1}{w}_0\Delta \right)\\ & =& \epsilon \left(v\right)v\left({\alpha }_{u^{-1},v^{-1}}\right)& \text{by (5.9),}\end{array}$$ and likewise $$\begin{array}{ccc}\theta \left({\partial }_u{v}^{-1}{w}_0\Delta \right)& =& \frac{\epsilon \left(w_0\right)}{a_{\delta }}\sum _w{\alpha }_{uw}\theta \left(w{v}^{-1}{w}_0\Delta \right)\\ & =& {\alpha }_{uv}\end{array}$$ again by (5.9). Hence (ii) follows from (5.16). (iii) Since ${\partial }_{\bar{u}}=\epsilon \left(u\right)w_0{\partial }_u{w}_0$ (2.12) we have $$\begin{array}{ccc}\sum _v{\alpha }_{\overline{uv}}\bar{v}& =& \epsilon \left(u{w}_0\right)w_0\left(\sum _v{\alpha }_{uv}v\right)w_0\\ & =& \epsilon \left(u{w}_0\right)\sum _v{w}_0\left({\alpha }_{uv}\right)\bar{v}\end{array}$$ and hence ${\alpha }_{\overline{uv}}=\epsilon \left(u{w}_0\right)w_0\left({\alpha }_{uv}\right)\text{.}$ $\square$

(5.25) Let $E_n^{\prime }$ be the subalgebra of operators $\varphi \in E_n$ such that $\varphi \left(P_n\right)\subset P_n\text{.}$ Then $E_n^{\prime }$ a free $P_n\text{-module}$ with basis ${\left({\partial }_w\right)}_{w\in S_n}\text{.}$

		Proof.
	If $\varphi ={\sum }_{w\in S_n}{\varphi }_w{\partial }_w\in E_n^{\prime },$ then by (5.22) $${\varphi }_w=theat\left(\varphi \left({\partial }_{w^{-1}{w}_0}\Delta \right)\right)\in P_n\text{.}$$ On the other hand, the ${\partial }_w$ are a $Q_n\text{-basis}$ of $E_n,$ and hence are linearly independent over $P_n\text{.}$ $\square$

Notes and References

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

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