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!

  • $\begingroup$ Glad you were able to benefit from the answer, even though it came almost 6 years late. I already knew this proof at the time so I'm sorry I didn't see your question. $\endgroup$ Jun 14 at 20:32
  • $\begingroup$ Thank you for your answer! Of course, I will benefit from it, although I am not able to grasp the details right now and to be honest: I haven't been into mathematics for a long time anymore... - but not only me, but the public who runs into the same question. @MattSamuel $\endgroup$ Jun 15 at 13:51
  • $\begingroup$ I have a paper waiting to be published and another two papers in the works with new combinatorial formulas for positivity of applying certain skew divided difference operators to certain Schubert polynomials, with a characterization of what the operation means and why it's likely to always be positive. Someone else thinks they can prove the conjecture geometrically. This is 16 years after Kirillov's paper, but math is slow so that's not too bad. $\endgroup$ Jun 15 at 16:20
  • $\begingroup$ I'm happy to hear from progress in this area! I was myself working on the positivity conjecture of skew divided difference operators applied to Schubert polynomials - but without much success. Thank you for letting me know! :) $\endgroup$ Jun 16 at 6:11

1 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.


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