Let $X=G/B$ be a homogeneous space and consider the quantization map $$ S_W\otimes\mathbb{C}[q]\to(S(\mathfrak{h})\otimes\mathbb{C}[q])/I_W^q\,, $$ where

• $S_W$ is the coinvariant algebra of the Weyl group $W$,
• $\mathbb{C}[q]=\mathbb{C}[H_2(X)]$ is the "algebra" of $H_2(X)=$ free $\mathbb{Z}$-lattice spanned by the simple coroots, with parameter $q$,
• $S(\mathfrak{h})$ is the symmetric algebra of the Cartan subalgebra $\mathfrak{h}$ of $G$,
• $I_W^q$ is the ideal of the quantum cohomology of $X$.

Let $\mathfrak{S}_{w_o}$ the image of $\frac{1}{|W|}\prod_{\alpha\in R^+}\alpha$ in $S_W\otimes\mathbb{C}[q]$ (Schubert polynomial of a point), where $R^+$ are the positive roots. Then, we clearly have $w\mathfrak{S}_{w_o}=(-1)^{\ell(w)}\mathfrak{S}_{w_o}$ for all $w\in W$, where $\ell$ is the length function, and where the action of $W$ on $S_W\otimes\mathbb{C}[q]$ is given by the reflection representation of $W$. Let $\mathfrak{S}_{w_o}^q$ be the image of $\mathfrak{S}_{w_o}$ under the above displayed isomorphism (quantum Schubert polynomial of a point).

Question. Is the symmetry of $\mathfrak{S}_{w_o}$, described by the equation "$w\mathfrak{S}_{w_o}=(-1)^{\ell(w)}\mathfrak{S}_{w_o}$", somehow, in any way, reflected in $\mathfrak{S}_{w_o}^q$, or in a slightly modified version of $\mathfrak{S}_{w_o}^q$? In the end, I would like to have $(\mathfrak{S}_{w_o}^q)^2$ invariant under some suitable action of $W$. Does anyone know any work in this diection?

Thanks!

PS. You might take the induced action under the above isomorphism, ok, but I don't know how to describe this action in explicit terms. Such a description is part of the question.

EDIT. More broadly, one can ask the question as follows: For a simple root $\beta$ such that $ws_\beta>w$ where $w\in W$, we know that $s_\beta\mathfrak{S}_w=\mathfrak{S}_w$. How can this symmetry described on the level of the quantized Schubert polynomials $\mathfrak{S}_w^q$? What does the induced action under the above isomorphism do to $\mathfrak{S}_w^q$? And how is $\mathfrak{S}_w^q$ invariant under this action? What does this mean explicitly?