In my endless fiddling with formulas I discovered one that fills in the blanks in a generic formula I saw in a paper, but I'm wondering if maybe it's already known and the paper was just mentioning the form of it casually. The formula I saw, which has undetermined constants, expresses a Schubert polynomial in $n$ variables as a sum of products of Schubert polynomials in a smaller number of variables with nonnegative coefficients. It looks like $$S_w(x_1,x_2,\ldots,x_n)=\sum_{u,v}{d_{u,v}^wS_u(x_1,x_2,\ldots,x_k)S_v(x_{k+1},x_{k+2},\ldots,x_n)}$$ where $d_{u,v}^w$ is mentioned to be nonnegative. I discovered that if you let $w_0$ be the longest element of $S_{n+1}$ and let $w_0'$ be the longest element of $S_{n+1-k}$ (identified with the parabolic subgroup of $S_{n+1}$ corresponding to the elements in the first $n+1-k$ positions) then in fact $$S_w(x_1,x_2,\ldots,x_n)=\sum_{a,b}{c_{a,b}^{ww_0}S_{aw_0'w_0}(x_1,x_2,\ldots,x_k)S_{bw_0'}(x_{k+1},x_{k+2},\ldots,x_n)}$$ where $\ell(aw_0'w_0)+\ell(a)=\ell(w_0'w_0)$, $\ell(bw_0')+\ell(b)=\ell(w_0')$, and $c_{a,b}^{ww_0}$ is the corresponding structure constant for multiplying Schubert polynomials, which of course is known to be nonnegative. Is the formula known to this degree of specificity?
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2$\begingroup$ Some of the "Cauchy Identities" in link.springer.com/content/pdf/10.1023/… seem relevant (for double quantum Schubert polynomials) --- though this is purely a web search result.. $\endgroup$– SuvritCommented Oct 8, 2015 at 0:14
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4$\begingroup$ Here's the arXiv link: arxiv.org/abs/q-alg/9703047 $\endgroup$– SuvritCommented Oct 8, 2015 at 0:24
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3$\begingroup$ @Suvrit Thanks. After looking at this I remembered what is called a Cauchy formula and one way to prove the identity in my question is using the Cauchy formula for ordinary double Schubert polynomials. The proof is pretty easy, but the proof of the formula in my dissertation was pretty easy too and nevertheless no one saw it before, so, you know. $\endgroup$– Matt SamuelCommented Oct 8, 2015 at 0:31
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2$\begingroup$ @Suvrit Well, my advisor (Anders Buch) has a paper expressing Schubert polynomials in terms of Schur polynomials, so the determinantal representations of those apply, though I'm not sure that really counts because it's not just one determinant. I haven't heard of a true determinantal formula for Schubert polynomials. $\endgroup$– Matt SamuelCommented Oct 8, 2015 at 0:48
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3$\begingroup$ I think this is just the statement that the "coproduct LR coefficients" are (in type A) the same as the "product LR coefficients" (i.e. the usual ones). Geometrically it has something to do with the duality between cohomology and homology, and combinatorially it has something to do with the Hopf algebra structure on quasi-symmetric functions. You might try looking through Nantel Bergeron's papers. $\endgroup$– Alexander WooCommented Oct 8, 2015 at 23:37
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1 Answer
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This is essentially Theorem 4.6.1 in (http://arxiv.org/pdf/alg-geom/9703001v1.pdf) Schubert polynomials, the Bruhat order, and the geometry of flag manifolds by Bergeron and Sottile.