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This is the culmination of about 11 years of research but after I discovered it I found a proof that was extremely trivial, so I'm wondering if it's already known.

Let $(a,b)$ with $a < b$ positive integers denote the transpositions in the symmetric group. Turn them into a vector space with the relation $$(a,b)+(b,c)=(a,c)$$ (This is the root system thinly disguised.) Let $F$ be the tensor algebra on this vector space. For an element $w\in S_{\infty}$ define $$T_w = \sum_{\substack{(a_1,b_1)\cdots (a_n,b_n) = w\\\ell((a_1,b_1)\cdots (a_i,b_i))=i}} (a_1,b_1)\cdots (a_n,b_n)$$ Then the elements $T_w$ are linearly independent. Define $$T_{w/u} = \sum_{\substack{u(a_1,b_1)\cdots (a_m,b_m) = w\\\ell(u(a_1,b_1)\cdots (a_i,b_i))=\ell(u)+i}} (a_1,b_1)\cdots (a_m,b_m)$$ Then $$T_{w/u}=\sum_{v}{c_{u,v}^wT_v}$$ where $c_{u,v}^w$ is the structure constant in the cohomology ring of the complete flag variety.

So the question is: is this already known?

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  • $\begingroup$ Looks like the "Monk's/Chevalley's rule" way of defining Schubert polynomials... $\endgroup$
    – Sam Hopkins
    Oct 27, 2018 at 21:39
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    $\begingroup$ @Sam Do you have a reference? I'm aware these are dual to Schubert polynomials in a nice way, and you can get a formula for Schubert polynomials in terms of the degree 1 Schubert polynomials by inverting their coefficient matrix. $\endgroup$
    – Matt Samuel
    Oct 27, 2018 at 21:58
  • $\begingroup$ In general that there are two approaches to defining Schubert polynomials: a "top-down" approach where you start by defining $\mathfrak{S}_{w_0} = x_1^{n-1}x_2^{n-2}\cdots x_{n-1}$ and use divided difference operators to get smaller $w$'s from bigger ones going down along edges of the weak order; and a "bottom-up" approach where you start by defining $\mathfrak{S}_{\mathrm{id}}=1$ and use Monk's rule (or maybe more precisely Lascoux's transition formula, see Theorem 3.4 of pi.math.cornell.edu/~allenk/schubnotes.pdf) to get bigger $w$'s from smaller along edges of the strong order. $\endgroup$
    – Sam Hopkins
    Oct 27, 2018 at 23:33
  • $\begingroup$ You are summing over paths from the identity to $w$ in the strong order, so it looks like the bottom-up approach to me. But can't say more precisely than that. $\endgroup$
    – Sam Hopkins
    Oct 27, 2018 at 23:33
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    $\begingroup$ Your result could be the same as Theorem 2 of arxiv.org/pdf/math/0202090.pdf. $\endgroup$
    – Richard Stanley
    Oct 28, 2018 at 17:36

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