Everyone (that is, everyone who cares) knows that double Schubert polynomials represent Schubert classes in equivariant cohomology in type $A$. We also know that we can restrict Schubert classes to fixed points of the torus action to obtain a realization of the equivariant cohomology ring as a subring of a direct product of polynomial rings. Positive formulas for the restrictions in terms of the negative roots are well known.

There's another formula for the restrictions, however, that is not positive, that I figured out but don't remember seeing anywhere. Namely, the restriction of the Schubert class $[X_u]^T$ to the fixed point $t_w$ is given by $$S_u(x_{w(1)},x_{w(2)},\ldots,x_{w(n)};x_1,x_2,\ldots,x_n)$$ where $S_u$ is the corresponding double Schubert polynomial. This can be proved with Macdonald's skew divided difference operators. This formula is pretty logical considering the equivariant cohomology ring as the dual of the nil-Hecke ring, but it's not immediately obvious that the relationship between the rings is so literal that you can just plug things into the polynomials to get the answer.

Any references/insights/comments are welcome.


1 Answer 1


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 $[\overline{B_- uB}] \in H^*_{T\times B}(GL_n) \cong H^*_{T\times T}(GL_n)$. Then setting $y_i \mapsto x_{w(i)}$ corresponds to restricting the $T\times T$ action to the $w$-twisted diagonal $T$, i.e. the map $H^*_{T\times T}(GL_n) \twoheadrightarrow H^*_{w\cdot T_\Delta}(GL_n)$.

The reason to restrict is that $T\times T$ has no fixed points on $GL_n$, i.e. the $T$-fixed points on $GL_n/B$ are not images of $T$-fixed points upstairs. Rather, $wB/B$ is the image of many $w\cdot T_\Delta$-fixed points $p$. Now we can restrict from $H^*_{w\cdot T_\Delta}(GL_n) \to H^*_{w\cdot T_\Delta}(p) \cong H^*_{w\cdot T_\Delta}$ and get the polynomial you wanted.


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