11
$\begingroup$

Let $e_i^j$ be the elementary symmetric polynomial in $x_1,x_2,\ldots,x_j$. Then the ordinary Schubert polynomial has an expansion of the form $$S_u(x)=\sum_{i_1,i_2,\ldots,i_n}{a^{i_1,i_2,\ldots,i_n}_ue_{i_1}^1\cdots e_{i_n}^n}$$ where the $a_u^{i_1,\ldots,i_n}$ are integers. We can interpret $i_1,\ldots,i_n$ as the code of a unique permutation, so we can rewrite this as $$S_u(x)=\sum_{v\in S_{\infty}}{a_{u}^ve_v}$$ Taking the inverse of the matrix $(a_u^v)$ gives us a matrix $(c_u^v)$ with nonnegative entries, and $$S_{uw_0}(x)=\sum_{v}{c_u^vx_v}$$ The coefficients $c_u^v$ are well understood (they can be expressed in terms of Bruhat paths, RC graphs, etc.). Somehow, though, the $a_u^v$ seem far more important. The universal Schubert polynomials and quantum Schubert polynomials use these coefficients in their definition, replacing the elementary monomials with something else. Have the $a_u^v$ been studied in and of themselves?

$\endgroup$
3
  • 2
    $\begingroup$ The following paper seems relevant: ac.els-cdn.com/S0001870898917303/… $\endgroup$
    – Suvrit
    Nov 16, 2015 at 4:08
  • $\begingroup$ @Suvrit That it most certainly is, thanks. $\endgroup$ Nov 17, 2015 at 4:05
  • $\begingroup$ I found a "positive" (non-cancellative) formula for $a_u^v$ when $u$ is a simple cycle (more generally, it's a "Pieri formula" for the product $x_i^kS_w(x)$). $\endgroup$ Nov 19, 2015 at 4:24

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.