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

Question

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!

Answer 1

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). 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.