What is the current state of the problem of finding a combinatorial rule for multiplying two Schubert polynomials? Is the problem still open?
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4$\begingroup$ Yes it is still open. $\endgroup$– MishaCommented Feb 21, 2014 at 19:22
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1$\begingroup$ The literature on this topic is vast, but one place to start would be papers by N. Bergeron and Sottile. They solved the Schubert times Schur problem in terms of chains in $k$-Bruhat order. The $k$-Bruhat order is generated by covering relations in ordinary Bruhat order $u \prec v = u \cdot (a,b)$, $\ell(v) = \ell(u) + 1$, with $a \leq k < b$. $\endgroup$– Michael JoyceCommented Feb 22, 2014 at 2:07
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1$\begingroup$ The most recent results I know of are those of Buch-Kresch-Purbhoo-Tamvakis, proving my conjecture for 2-step flag manifolds. Bergeron and Sottile did not solve the Schubert times Schur problem in their paper, but have more recently announced a solution with S. Assaf. $\endgroup$– Allen KnutsonCommented Feb 22, 2014 at 17:34
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1$\begingroup$ @MichaelJoyce: Very minor thing, but you're probably confusing Sottile's solo paper, "Pieri's Rule for Flag Manifolds and Schubert Polynomials"--which is a proof of a Schur-times-Schub version of Pieri's rule--and N. Bergeron and Sottile's later joint work, "Schubert Polynomials, The Bruhat Order, and the Geometry of Flag Manifolds", which restates Sottile's earlier characterization. $\endgroup$– Joshua P. SwansonCommented Jul 24, 2014 at 6:40
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1$\begingroup$ Also, Sottile now has a preprint of his paper with N. Bergeron and S. Assaf on his web site/on the arXiv, "A combinatorial proof that Schubert vs. Schur coefficients are nonnegative". Finally, Knutson's descent cycling paper is very readable and brief, and includes a proof of Monk's rule, so it might be another good place to start. (Same Knutson as above.) $\endgroup$– Joshua P. SwansonCommented Jul 24, 2014 at 6:42
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