Let $S_u(x)$ be a Schubert polynomial and let $S_v(x;y)$ be a double Schubert polynomial. Then their product can be expressed in terms of the double Schubert polynomials as $$S_u(x)S_v(x;y)=\sum_w{c_{uv}^w(y)S_w(x;y)}$$ where the $c_{uv}^w(y)$ are polynomials in the $y$ variables.

If $u,v$ are Grassmannian permutations for the same parabolic subgroup, then the Molev–Sagan rule for factorial Schur functions implies that $c_{uv}^w(y)$ has nonnegative integer coefficients. Is it known whether $c_{uv}^w(y)$ has nonnegative coefficients for general permutations?

As mentioned in the comment below, I've tested this up to $n=6$. Since proving positivity of these coefficients subsumes proving positivity of the structure constants in ordinary cohomology of flag varieties, any proof with current tools would likely be geometric. I think the key would be the double/triple Schubert calculus mentioned in Knutson and Tao's paper "Puzzles and (equivariant) cohomology of Grassmannians," so I'm basically wondering if anyone has studied those concepts enough to have a geometric proof of positivity for the complete flag variety.

This question is almost 9 years old, but in 2015 I mentioned I had a Pieri formula for this. It was super hard to prove, but it is now to appear in JPAA, along with a separated descents formula: A Molev–Sagan type formula for double Schubert polynomials.

I have more unpublished results in this direction. Still looking for a geometric proof.

  • $\begingroup$ Have you tried some computer experiments on this? Of course, proving this is probably very hard unless there is some representation-theoretical way to do it. $\endgroup$ Aug 1, 2015 at 5:12
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    $\begingroup$ @PerAlexandersson I've tested it up to $n=6$. $\endgroup$ Aug 1, 2015 at 11:49
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    $\begingroup$ I believe I have a Pieri formula for this now. $\endgroup$ Aug 25, 2015 at 18:00
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    $\begingroup$ Though it doesn't explicitly answer the question, you might find the end of this note relevant: people.math.osu.edu/anderson.2804/eilenberg/lecture11.pdf $\endgroup$ Jun 5, 2016 at 20:38
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    $\begingroup$ Also, I think the method from this note might adapt to your situation: arxiv.org/abs/0711.0983 $\endgroup$ Jun 5, 2016 at 20:42


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