It is known that the coefficients of the monomials appearing in a Schubert polynomial are always positive. My question is: Is it always true that at least one such coefficient must be $1$? If that is not always true, then is it true that the gcd of the coefficients of the monomials appearing in a Schubert polynomial must always be $1$?
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1$\begingroup$ I suspect that for each permutation $w \in S_n$, each monomial appearing in the Schubert polynomial $\mathfrak{S}_w$ is lexicographically smaller-or-equal to the monomial $\mathbf{x}^{L\left(w\right)} := \prod\limits_{i=1}^n x_i^{\ell_i\left(w\right)}$, where $\ell_i\left(w\right)$ denotes the number of all $j > i$ satisfying $w\left(i\right) > w\left(j\right)$. Moreover, I suspect that the coefficient of this monomial $\mathbf{x}^{L\left(w\right)}$ is $1$. $\endgroup$– darij grinbergCommented Jun 22, 2018 at 15:40
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$\begingroup$ @darij grinberg : can you please explicitly state what lexicographic order you are talking about here ? $\endgroup$– user111492Commented Jun 22, 2018 at 15:42
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1$\begingroup$ Probably it should not be too hard to see this from the "pipe dreams" definition of Schubert polynomials. $\endgroup$– Sam HopkinsCommented Jun 22, 2018 at 15:53
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3$\begingroup$ Look at Corollary 3.9 of "RC-Graphs and Schubert Polynomials" by Bergeron and Billey: projecteuclid.org/download/pdf_1/euclid.em/1048516036. It describes the (unique) leading term of a Schubert polynomial. $\endgroup$– Sam HopkinsCommented Jun 22, 2018 at 17:11
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1$\begingroup$ Darij's suspicion is correct: The leading coefficient is $x^{L(w)}$. This is transparent to see from the Kohnert formula for Schubert polynomials (e.g. arxiv.org/abs/1703.00088). (Some people don't believe some proofs of this formula, but I think everyone believes the formula!) $\endgroup$– OliverCommented Jun 23, 2018 at 13:10
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1 Answer
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It is known that the Schubert polynomials is an integral basis for the polynomials (if it was not a basis, then the problem of expanding a product of Schubert polynomials into Schubert polynomials would be kind of strange). That the basis is integral means that monomials expands into Schubert with integer coefficients.
As mentioned in the comments, there is an ordering such that the leading coefficient in this ordering is $1$ - this property implies that the basis is integral (is it perhaps equivalent to this property)?
answered Jun 23, 2018 at 7:44
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$\begingroup$ The integral basis property is precisely Corollary 3.9 of the paper of Bergeron and Billey I referenced above. $\endgroup$ Commented Jun 23, 2018 at 17:22