# Question on a reduction in Kirillov's paper on positivity of divided difference operators

As the title says, my question is on a specific argument in Kirillov - Skew divided difference operators and Schubert polynomials (journal, MSN) on positivity of divided difference operators. I recall the relevant passages of the paper 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 (namley Theorem (4.19) in Chapter IV which is restated as Lemma 1 in Kirillov's paper), 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.

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: 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?

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!