As the title says, my question is on a specific argument in [Kirillov - Skew divided difference operators and Schubert polynomials](https://arxiv.org/abs/0705.4546) ([journal](http://www.emis.de/journals/SIGMA/2007/072), [MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=2322799)) on positivity of divided difference operators. I recall the relevant passages of the [paper](https://arxiv.org/abs/0705.4546) in the following. 


Let $m\in\mathbb{Z}\_{>0}$ and $n=m-1$. Denote by $\mathcal{S}\_m$ the symmetric group on $m$ letters. For $1\leq\mu\leq n$, let
$$
\mathcal{S}\_m^{(\mu)}=\{w\in\mathcal{S}\_m\mid w(\mu+1)<w(\mu+2)<\cdots<w(m)\}\,.
$$
For $w\in\mathcal{S}\_m$, denote by $\mathfrak{S}\_w\in\mathbb{Z}[x\_1,\ldots,x\_n]$ the Schubert polynomial associated to $w$. For $1\leq a<b\leq m$, denote by $\partial\_{ab}$ the divided difference operator with respect to the variables $x\_a$ and $x\_b$.


The author wants to show that $\partial\_{ab}\mathfrak{S}\_w\in\mathbb{Z}\_{\geq 0}[x\_1,\ldots,x\_m]$. We clearly can assume that $b\leq n$ and $b-a>1$ (since $\mathfrak{S}\_w$ depends only on $x\_1,\ldots,x\_n$ and since $\partial\_{i,i+1}\mathfrak{S}\_w$ is either zero or itself a Schubert polynomial with positive coefficients where $1\leq i\leq n$).


Let $\bar x=(x\_1,\ldots,x\_n),\bar x\_1=(x\_1,\ldots,x\_{a-1}),\bar x\_2=(x\_a,\ldots,x\_b),\bar x\_3=(x\_{b+1},\ldots,x\_n)$. By a theorem in [Macdonald's notes on Schubert polynomials](http://www.ms.unimelb.edu.au/~ram/Resources/NOSPContents.html) (namley Theorem (4.19) in Chapter IV which is restated as Lemma 1 in [Kirillov's paper](https://arxiv.org/abs/0705.4546)), we can write
$$
\mathfrak{S}\_w(\bar x)=\sum\_{u\_1,u\_2,u\_3}d\_{u\_1u\_2u\_3}^w\mathfrak{S}\_{u\_1}(\bar x\_1)\mathfrak{S}\_{u\_2}(\bar x\_2)\mathfrak{S}\_{u\_3}(\bar x\_3)
$$
where the sums runs over $u\_1\in\mathcal{S}\_m^{(a-1)},u\_2\in\mathcal{S}\_m^{(b-a+1)},u\_3\in\mathcal{S}\_m^{(n-b)}$ and where $d\_{u\_1u\_2u\_3}^w\in\mathbb{Z}\_{\geq 0}$.


**Note.** Neither Kirillov nor Macdonald actually show that the elements $u\_1,u\_2,u\_3$ are contained in $\mathcal{S}\_m$ if $w\in\mathcal{S}\_m$. But I think this part is "obvious" because of Theorem (4.11) in [Macdonald's notes](http://www.ms.unimelb.edu.au/~ram/Resources/NOSPContents.html).


If we now apply $\partial\_{ab}$ to this expression, we find that
$$
\partial\_{ab}\mathfrak{S}\_w(\bar x)=\sum\_{u\_1,u\_2,u\_3}d\_{u\_1u\_2u\_3}^w\mathfrak{S}\_{u\_1}(\bar x\_1)(\partial\_{ab}\mathfrak{S}\_{u\_2}(\bar x\_2))\mathfrak{S}\_{u\_3}(\bar x\_3)\,.
$$
Hence, we can assume that $a=1$ and that $w\in\mathcal{S}\_m^{(b)}$.


I come now to the question which concerns page 10 of [Kirillov's paper](https://arxiv.org/abs/0705.4546): He says that we can even assume $b=n$. My question is why we can assume this.


**Thoughts.** (1) If $w\in\mathcal{S}\_m^{(b)}$, it does not mean that $w\in\mathcal{S}\_b$. It is very easy to give examples for this.


(2) We can also assume that $w$ has last descent at $b$. Otherwise the positivity is obvious.


(3) In terms of the Schubert element $\mathfrak{S}^{(m)}(\bar x)$ in the nilCoxeter algebra (cf. Fomin and Stanley's paper), it suffices to prove that
$$
\partial\_{1b}\mathfrak{S}^{(m)}(x\_1,\ldots,x\_b,0,\ldots,0)e\_{b+1}\cdots e\_n e\_{b+1}\cdots e\_{n-1}\cdots e\_{b+1}e\_{b+2}e\_{b+1}
$$
has coefficients in $\mathbb{Z}\_{\geq 0}[\bar x]$. Here, $e\_1,\ldots,e\_n$ denote the generators of the nilCoxeter algebra.


(4) For me, the case $b<n$ looks essentially different from the case $b=n$. Is it possible that there is a mistake in [Kirillov's paper](https://arxiv.org/abs/0705.4546)?


I tried very much to understand the reduction or to give a new proof which clearly distinguishes the case $b<n$ and $b=n$. I had not much success. Every idea or comment is welcome!



Question on a reduction in Kirillov's paper on positivity of divided difference operators
I had the same concern about Kirillov's paper. For $b=n$, $x\_b$ can have power at most $1$, whereas for $b<n$ it could have a larger power. I have an alternative proof that uses different methods than Kirillov. It remains combinatorial, but it uses combinatorial results that appear not to be published (I'm trying to find out; see [Pulling out a variable from a Schubert polynomial](https://mathoverflow.net/questions/448607/pulling-out-a-variable-from-a-schubert-polynomial)). For $\partial\_{ab}$, there is a positive formula for an expression as a product of a Schubert polynomial in $x\_1,\ldots,x\_a,x\_b,\ldots,x\_n$ and some polynomial in $x\_{a+1},\ldots,x\_{b-1}$, with nonnegative coefficients, in which case you can apply a divided difference to the $a$ index, which affects $x\_a$ and $x\_b$, yielding a positive result.


The positive formula actually fits in a MathOverflow answer. It involves double Schubert polynomials $\mathfrak{S}\_{w\_0}(x;y)$ and the Cauchy formula, as well as Sottile's Pieri formula. Define the factorial elementary symmetric polynomial
$$E\_p(x;y\_i)=\prod\_{j=1}^p(x\_j-y\_i)$$
This has an alternative expansion as
$$E\_p(x;y\_i)=\sum\_{j=0}^p (-y\_i)^{p-j}e\_{j,p}(x)$$
Where $e\_{j,p}(x)=e\_j(x\_1,\ldots,x\_p)$ is the elementary symmetric polynomial of degree $j$ in $p$ variables. Then
$$\mathfrak{S}\_{w\_0}(x;y)=\prod\_{i=1}^n E\_{n+1-i}(x;y\_i)$$
Let's say we want to pull out the variable at index $j$. Then there is a permutation $\mu$ such that
$$\mathfrak{S}\_{w\_0}(x;y)=\mathfrak{S}\_\mu(x;y\_1,\ldots,y\_{j-1},y\_{j+1},\ldots,y\_n)E\_{n+1-j}(x;y\_j)$$
Apply the Cauchy formula, and expand the factorial elementary symmetric polynomial, obtaining
$$\mathfrak{S}\_{w\_0}(x;y)=\sum\_{u,q}\mathfrak{S}\_{u\mu}(x)\mathfrak{S}\_u(-y\_1,\ldots,-y\_{j-1},-y\_{j+1},\ldots,-y\_n)(-y\_j)^{n+1-j-q}e\_{q,n+1-j}(x)$$
Use Sottile's Pieri formula to multiply by the elementary symmetric polynomial to get
$$\mathfrak{S}\_{w\_0}(x;y)=\sum\_{u,q}\sum\_{u\mu\xrightarrow[q]{n+1-j} ww\_0}\mathfrak{S}\_{ww\_0}(x)\mathfrak{S}\_u(-y\_1,\ldots,-y\_{j-1},-y\_{j+1},\ldots,-y\_n)(-y\_j)^{n+1-j-q}$$
Then apply the Cauchy formula
$$\mathfrak{S}\_{w\_0}(x;y)=\sum\_w \mathfrak{S}\_{ww\_0}(x)\mathfrak{S}\_w(-y)$$
to get
$$\mathfrak{S}\_w(x)=\sum\_{u\mu\xrightarrow[q]{n+1-j} ww\_0}x\_j^{n+1-j-q}\mathfrak{S}\_u(x\_1,\ldots,x\_{j-1},x\_{j+1},\ldots,x\_n)$$
Apply this repeatedly to get the full result.


