## Notes on Schubert PolynomialsChapter 2

Last update: 27 June 2013

## Divided differences

If $f$ is a function of $x$ and $y$ (and possibly other variables), let

$∂xyf= f(x,y)-f(y,x) x-y$

("divided difference"). Equivalently

$∂xyf= (x-y)-1 (1-sxy)$

where ${s}_{xy}$ interchanges $x$ and $y\text{.}$ The operator ${\partial }_{xy}$ takes polynomials to polynomials, and has degree $-1$ (i.e., if $f$ is homogeneous of degree $d,$ then ${\partial }_{xy}f$ is homogeneous of degree $d-1\text{).}$ Explicitly, if $f={x}^{r}{y}^{s}$ we have

$(2.1) ∂xy(xrys) = xrys-xsyr x-y = σ(r-s)∑ xpyq$

where the sum is over $\left(p,q\right)$ such that $p+q=r+s-1$ and $\text{max}\left(p,q\right)<\text{max}\left(r,s\right),$ and $\sigma \left(r-s\right)$ is $+1,0$ or $-1$ according as $r-s$ is positive, zero or negative.

On a product $fg,{\partial }_{xy}$ acts according to the rule

$(2.2) ∂xy(fg)= (∂xyf)g+ (sxyf) (∂xyg).$

In particular we have

$(2.2') ∂xy(fg)= f∂xyg$

if $f\left(x,y\right)=f\left(y,x\right)\text{.}$

(2.3)

 (i) ${\partial }_{xy}{s}_{xy}=-{\partial }_{xy},\phantom{\rule{2em}{0ex}}{s}_{xy}{\partial }_{xy}={\partial }_{xy},$ (ii) ${\partial }_{xy}^{2}=0,$ (iii) ${\partial }_{xy}{\partial }_{yz}{\partial }_{xy}={\partial }_{yz}{\partial }_{xy}{\partial }_{yz}\text{.}$

 Proof. (i) and (ii) are immediate from the definitions, and (iii) is verified by direct calculation: each side is equal to $(x-y)-1 (x-z)-1 (y-z)-1 ∑w∈S3 ε(w)w,$ where the symmetric group ${S}_{3}$ permutes $x,y$ and $z,$ and $\epsilon \left(w\right)$ is the sign of the permutation $w\text{.}$ $\square$

Let ${x}_{1},{x}_{2},\dots ,{x}_{n},\dots$ be independent variables, and let

$Pn=ℤ [ x1,x2,…, xn ]$

for each $n\ge 1,$ and

$P∞ = ℤ[x1,x2,…] = ⋃n=1∞Pn.$

For each $i\ge 1$ let

$∂i=∂xi,xi+1.$

Each ${\partial }_{i}$ is a linear operator on ${P}_{\infty }$ (and on ${P}_{n}$ for $n>i\text{)}$ of degree $-1\text{.}$ From (2.3) we have (compare with (1.1))

$(2.4) { ∂i2=0, ∂i∂j= ∂j∂i if |i-j|>1, ∂i∂i+1 ∂i=∂i+1 ∂i∂i+1$

For any sequence $a=\left({a}_{1},\dots ,{a}_{p}\right)$ of positive integers, we define

$∂a=∂a1 …∂ap.$

Recall that if $w$ is any permutation, $R\left(w\right)$ denotes the set of reduced words for $w,$ i.e. sequences $\left({a}_{1},\dots ,{a}_{p}\right)$ such that $w={s}_{{a}_{1}}\dots {s}_{{a}_{p}}$ and $p=\ell \left(w\right)\text{.}$

(2.5) If $a,b\in R\left(w\right)$ then ${\partial }_{a}={\partial }_{b}\text{.}$

 Proof. We proceed by induction on $p=\ell \left(w\right)\text{.}$ Let us write $a\equiv b$ to mean that ${\partial }_{a}={\partial }_{b}\text{.}$ The inductive hypothesis then implies that $(*) a≡b if either a1=b1 or ap= bp.$ By the exchange lemma (1.8) we have $ci= ( b1,a1,…, aˆi,…,ap ) ∈R(w)$ for some $i=1,\dots ,p\text{.}$ If $i\ne p$ then $b\equiv {c}_{i}\equiv a$ by virtue of $\left(*\right),$ so that $a\equiv b\text{.}$ If $i=p$ and $|{b}_{1}-{a}_{1}|>1$ then by (2.4) and (1.1) $cp′= ( a1,b1,a2, …,ap-1 ) ∈R(w)$ and $a\equiv {c}_{p}^{\prime }\equiv {c}_{p}\equiv b,$ so that again $a\equiv b\text{.}$ Finally, if $i=p$ and $|{b}_{1}-{a}_{1}|=1,$ we apply the exchange lemma again, this time to ${c}_{p}$ and $a\text{;}$ this shows that $di= ( a1,b1,a1,…, aˆi,…,ap-1 ) ∈R(w)$ for some $i=2,\dots ,p-1\text{.}$ But then by (2.4) and (1.1) we have $di′= ( b1,a1,b1, a2,…,aˆi,… ,ap-1 ) ∈R(w)$ and $a\equiv {d}_{i}\equiv {d}_{i}^{\prime }\equiv b\text{.}$ Hence $a\equiv b$ in all cases. $\square$

Remark. For any permutation $w,$ let $GR\left(w\right)$ denote the graph whose vertices are the reduced words for $w,$ and in which a reduced word $a$ is joined by an edge to each of the words obtained from $a$ by either interchanging two consecutive terms $i,j$ such that $|i-j|>1,$ or by replacing three consecutive terms $i,j,i$ such that $|i-j|=1$ by $j,i,j\text{.}$ Then the proof of (2.5) shows that

(2.5') The graph $GR\left(w\right)$ is connected.

From (2.5) it follows that we may define

$∂w=∂a$

unambiguously, where $a$ is any reduced word for w. By (2.2'), the operators ${\partial }_{w}$ for $w\in {S}_{n}$ are ${\Lambda }_{n}$ linear, where

$Λn=ℤ [x1,…,xn] Sn ⊂Pn$

is the ring of symmetric polynomials in ${x}_{1},\dots ,{x}_{n}\text{.}$

A sequence $a=\left({a}_{1},\dots ,{a}_{p}\right)$ will be said to be reduced if $a\in R\left(w\right)$ for some permutation $w\text{.}$

(2.6) If $a=\left({a}_{1},\dots ,{a}_{p}\right)$ is not reduced, then ${\partial }_{a}=0\text{.}$

 Proof. By induction on $p\text{.}$ If $a\prime =\left({a}_{1},\dots ,{a}_{p-1}\right)$ is not reduced, then ${\partial }_{a\prime }=0$ and hence ${\partial }_{a}={\partial }_{a\prime }{\partial }_{{a}_{p}}=0\text{.}$ So we may assume that $a\prime$ is reduced. Let $v={s}_{{a}_{1}}\dots {s}_{{a}_{p-1}},$ $w={s}_{{a}_{1}}\dots {s}_{{a}_{p}}\text{.}$ We have $\ell \left(v\right)=p-1$ and $\ell \left(w\right)\le p-1,$ hence by (1.3) $\ell \left(w\right)=p-2,$ so that $\ell \left(v\right)=\ell \left(w{s}_{{a}_{p}}\right)=\ell \left(w\right)+1\text{.}$ Consequently ${\partial }_{v}={\partial }_{w}{\partial }_{{a}_{p}}$ and therefore ${\partial }_{a}={\partial }_{v}{\partial }_{{a}_{p}}={\partial }_{w}{\partial }_{{a}_{p}}^{2}=0\text{.}$ $\square$

(2.7) Let $u,v$ be permutations. Then

$∂u∂v= { ∂uv if ℓ(uv)= ℓ(u)+ℓ(v) , 0 otherwise.$

 Proof. (2.5), (2.6). $\square$

(2.8) Let $w$ be a permutation, $i\ge 1\text{.}$ Then

$si∂w=∂w⟺ℓ (siw)=ℓ(w) -1.$

 Proof. We have ${s}_{i}{\partial }_{w}={\partial }_{w}⟺{\partial }_{i}{\partial }_{w}=0,$ hence the result follows from (2.7). $\square$

As before let ${w}_{0}=\left(n,n-1,\dots ,2,1\right)$ be the longest element of ${S}_{n}\text{.}$ One element of $R\left({w}_{0}\right)$ is the sequence

$(2.9) ( 1,2,…,n-1,1, 2,…,n-2,…,1, 2,3,1,2,1 ) .$

(2.10) We have

$∂w0=aδ-1 ∑w∈Snε(w) w$

where ${a}_{\partial }=\prod _{1\le i and $\epsilon \left(w\right)=±1$ is the sign of $w\text{.}$

 Proof. From the definition it follows that ${\partial }_{{w}_{0}}$ is of the form $(1) ∂w0=∑w∈Sn cww$ with coefficients ${c}_{w}$ rational functions of ${x}_{1},\dots ,{x}_{n}\text{.}$ By (2.8) we have ${s}_{i}{\partial }_{{w}_{0}}={\partial }_{{w}_{0}}$ for $1\le i\le n-1,$ so that $v{\partial }_{{w}_{0}}={\partial }_{{w}_{0}}$ for all $v\in {S}_{n},$ and therefore $(2) ∂w0=∑w∈Sn v(cw)vw.$ Comparison of (1) and (2) shows that $(3) cvw=v(cw) (v,w∈Sn).$ Hence all the coefficients ${c}_{w}$ are determined by one of them, say ${c}_{{w}_{0}}\text{.}$ From the sequence (2.9) for ${w}_{0}$ it is easily checked that the coefficient of ${w}_{0}$ in ${\partial }_{0}$ is $cw0=ε (w0)aδ-1.$ Hence from (3) we have $cw=ww0 (cw0)=ε (w)aδ-1$ which proves (2.10). $\square$

From (2.10) it follows that, for any $\alpha =\left({\alpha }_{1},\dots ,{\alpha }_{n}\right)\in {ℕ}^{n},$

$(2.11) ∂w0xα= sα-δ (x1,…,xn)$

where ${x}^{\alpha }$ means ${x}_{1}^{{\alpha }_{1}}\dots {x}_{n}^{{\alpha }_{n}},$ $\delta =\left(n-1,n-2,\dots ,1,0\right)$ and ${s}_{\alpha -\delta }$ is the Schur function indexed by $\alpha -\delta \text{.}$ Thus ${\partial }_{{w}_{0}}$ is a ${\Lambda }_{n}\text{-linear}$ mapping of ${P}_{n}$ onto ${\Lambda }_{n}\text{.}$

For $w\in {S}_{n},$ let $\stackrel{‾}{w}={w}_{0}w{w}_{0}\text{.}$ Then

$(2.12) ∂w‾=ε (w)w0∂w w0.$

 Proof. From the definition of ${\partial }_{i}$ we have $w0∂iw0=- ∂n-i$ from which (2.12) follows easily, since ${w}_{0}^{2}=1\text{.}$ $\square$

If $f$ and $g$ are polynomials in ${x}_{1},{x}_{2},\dots ,$ the expression of ${\partial }_{w}\left(fg\right)$ as a sum of polynomials ${\partial }_{u}f·{\partial }_{v}g$ (i.e. the "Leibnitz formula" for ${\partial }_{w}\text{)}$ is in general rather complicated. However, there is one case in which it is reasonably simple, namely when one of the factors $f,g$ is linear:

(2.13) If $f=\sum {\alpha }_{i}{x}_{i}$ then

$∂w(fg)=w(f) ∂wg+∑(αi-αj) ∂wtijg$

summed over all pairs $i such that $\ell \left(w{t}_{ij}\right)=\ell \left(w\right)-1,$ where ${t}_{ij}$ is the transposition that interchanges $i$ and $j\text{.}$

 Proof. Let $\left({a}_{1},\dots ,{a}_{p}\right)$ be a reduced word for $w\text{.}$ Since $f$ is linear it follows from (2.2) that $∂w(fg) = ∂a1…∂ap (fg) = sa1…sap (f)∂a1… ∂apg+∑r=1p sa1…∂ar… sap(f)∂a1 …∂ˆar… ∂apg.$ Now ${\partial }_{{a}_{1}}\dots {\stackrel{ˆ}{\partial }}_{{a}_{r}}\dots {\partial }_{{a}_{p}}=0$ unless $\left({a}_{1},\dots ,{\stackrel{ˆ}{a}}_{r},\dots ,{a}_{p}\right)$ is reduced, and then by (1.11) it is equal to ${\partial }_{wt},$ where $wt={s}_{{a}_{p}}\dots {\stackrel{^}{s}}_{{a}_{r}}\dots {s}_{{a}_{p}}$ has length $p-1=\ell \left(w\right)-1,$ and $t={s}_{{a}_{p}}\dots {s}_{{a}_{r}}\dots {s}_{{a}_{p}}={t}_{ij}$ where $\left(i,j\right)={s}_{{a}_{p}}\dots {s}_{{a}_{r+1}}\left({a}_{r},{a}_{r+1}\right),$ so that $sa1… sar-1 ∂arsar+1 …sap(f)= αi-αj.$ $\square$

We also introduce the operators ${\pi }_{i}\left(i\ge 1\right)$ defined by

$πif=∂i(xif).$

In place of (2.4) we have

$(2.14) { πi2=πi, πiπj= πjπi if |i-j|>1, πiπi+1 πi=πi+1 πiπi+1.$

If we define ${\pi }_{a}$ to be ${\pi }_{{a}_{1}}\dots {\pi }_{{a}_{p}}$ for any sequence $a=\left({a}_{1},\dots ,{a}_{p}\right)$ of positive integers, then corresponding to (2.5) we have

(2.15) If $a,b\in R\left(w\right)$ then ${\pi }_{a}={\pi }_{b}\text{.}$

The proof is the same as that of (2.5), and rests only on the second and third of the relations (2.14). From (2.15) it follows that we may define

$πw=πa$

unambiguously, where $a$ is any reduced word for $w\text{.}$

In place of (2.10) we have

(2.16) For any $f\in {P}_{n},$

$πw0f=aδ-1 ∑w∈Snε(w) w(xδf)=∂w0 (xδf).$

In particular, if $\alpha \in {ℕ}^{n},$

$(2.16') πw0xα=sα (x1,…,xn).$

 Proof. We have $π1f=∂1 (x1f), π1π2f= ∂1 (x1∂2(x2f)) =∂1∂2(x1x2f)$ and generally $π1…πrf=∂1 …∂r(x1…xrf)$ for each $r\ge 1\text{.}$ From this and (2.10) it follows easily that ${\pi }_{{w}_{0}}f={\partial }_{{w}_{0}}\left({x}^{\delta }f\right)\text{.}$ $\square$

Let $\left({a}_{1},\dots ,{a}_{p}\right)$ be a reduced word for $w\text{.}$ Then

$∂w = ∂a1…∂ap = (xa1-xa1+1)-1 (1-sa1) (xa2-xa2+1)-1 (1-sa2)…$

which shows on expansion that ${\partial }_{w}$ is of the form

$∂w=∑v≤w fvwv$

where ${f}_{vw}$ are rational functions of ${x}_{1},{x}_{2},\dots ,$ and in particular (by (1.7))

$fww=(-1)p ∏(i,j)∈I(w-1) (xi-xj)-1$

and thus is $\ne 0\text{.}$ It follows that the ${\partial }_{w}$ are linearly independent over the field of rational functions ${ℚ}_{\infty }=ℚ\left({x}_{1},{x}_{2},\dots \right)\text{.}$

Now from (2.2) we have

$∂a(fg)= (∂af)g+ (saf) (∂ag)$

or equivalently, if $\mu :{P}_{\infty }\otimes {P}_{\infty }\to {P}_{\infty }$ is the multiplication map,

$∂a∘μ=μ∘ ( ∂a⊗1+sa⊗∂a ) .$

From this it follows that

$∂w∘μ=μ∘ ( ∂a1⊗1+sa1⊗ ∂a1 ) ∘…∘ ( ∂ap⊗1+sap ⊗∂ap )$

On expansion this is a sum over subsequences $b$ of $a=\left({a}_{1},\dots ,{a}_{p}\right),$ say

$(1) ∂w∘μ=μ∘ ∑b⊂a ϕ(a,b)⊗ ∂b$

where

$ϕ(a,b)= ϕ1(a,b) ∘…∘ϕp (a,b)$

and

$ϕi(a,b)= { sa1 if ai∈b, ∂a1 if ai∉b.$

Since ${\partial }_{b}=0$ if $b$ is not reduced (2.6), the sum is over reduced subsequences $b$ of $a,$ and by (1.17) we can write

$(2) ∂w∘μ=μ∘ ∑v≤wv∂w/v ⊗∂v$

where for $v\le w$

$(3) ∂w/v=v-1 ∑ϕ(a,b)$

summed over subsequences $b\subset a$ such that $b$ is a reduced word for $v\text{.}$

So for each pair of permutations $w,v$ such that $w\ge v$ we have a well-defined operator ${\partial }_{w/v}$ on ${P}_{\infty },$ defined by (3). Since the ${\partial }_{v}$ are linearly independent, the definition (3) is independent of the reduced word $a\in R\left(w\right)\text{.}$

(2.17) For each pair $w,v\in {S}_{\infty }$ such that $w\ge v$ there is a linear operator ${\partial }_{w/v}$ on ${P}_{\infty }$ such that

$∂w(fg)=∑v≤w v(∂w/vf)· ∂vg.$

${\partial }_{w/v}$ has degree $-\ell \left(w\right)+\ell \left(v\right)\text{.}$

## Examples

1.

Let $v=w,$ then

$∂w/w=w-1 ϕ(a,a)= w-1sa1… sap=1.$
2.

Let $v=1,$ then

$∂w/1=ϕ (a,∅)=∂a1 …∂ap=∂w.$
3.

Suppose that $v\to w,$ so that $v={s}_{{a}_{1}}\dots {ŝ}_{{a}_{r}}\dots {s}_{{a}_{p}}$ for an unique $r\in \left[1,p\right]\text{.}$ Then $b=\left({a}_{1},\dots ,{\stackrel{ˆ}{a}}_{r},\dots ,{a}_{p}\right)$ and

$∂w/v = v-1ϕ (a,b) = v-1sa1… sar-1 ∂ar sar+1… sap = sap…sar+1 ∂arsar+1… sap$

Now $w=vt$ where $t$ is the transposition

$t=tij=sap… sar…sap (i

so that $\left(i,j\right)={s}_{{a}_{p}}\dots {s}_{{a}_{r+1}}\left({a}_{r},{a}_{r}+1\right)$ and therefore

$∂w/v = sap… sar+1 (xar-xar+1)-1 (1-sar) sar+1… sap = (xi-xj)-1 (1-tij)$

is the divided difference operator ${\partial }_{{x}_{i},{x}_{j}}\text{.}$

The product formula for ${\partial }_{w/u}$ is

$(2.18) ∂w/u(fg)= ∑u≤v≤w u-1v (∂w/vf) ∂w/ug.$

 Proof. We have $∂w(fgh)= ∑u≤wu ∂w/u(fg) ∂uh (1)$ and on the other hand $∂w(fgh) = ∑u≤wv ∂w/u(f) ∂v(gh) = ∑u≤v≤wv ∂w/v(f) ·u∂v/u(g) ·∂uh.$ Comparison of (1) and (2) gives $u∂w/u(fg)= ∑u≤v≤wv ∂w/u(f)· u∂v/u(g)$ which gives the result. $\square$

When $u=1,$ this reduces to (2.17).

## Notes and References

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