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Instead of thinking of the double Schubert polynomial $S_u$ as representing $[\overline{B_- uB}/B] \in H^*_T(GL_n/B)$, equivalently think of it as representing $[\overline{B_- uB}] \in H^*_{T\times B} …

First break $p$ into its homogeneous components, since Schubert polynomials are homogeneous. Now find the lex-last term $c \prod_i x_i^{L_i}$ of $p$ -- look for the largest $n$ such that $x_n$ occurs …

I would say there are three basic reasons for / proofs of positivity. Geometry. [Kleiman 1973] proves that the number one's trying to compute is the number of points in a transverse intersection of …

First, one can resolve the Schubert varieties using Bott-Samelson manifolds, and discover that any two resolutions give the same class upon pushforward. (This good situation ends with K-theory, i.e. i …

In general for a Grassmannian $Gr_k(\mathbb C^n)$, the Schubert polynomials are Schur polynomials in $k$ variables, one for each partition $\lambda$ in a $k\times (n-k)$ rectangle. In this case $S_\la …