polynomial. This polynomial is an element $\widetilde{R}_n\in\mathbb{C}[x_1^{\pm}, \dots, x_n^{\pm}][z]$ and is formed of variations of the usual symmetric square polynomial as follows: $$\begin{aligned}
    \widetilde{R}_n(x_1, \dots, x_n; z) = \widetilde{R}_n(z) :&= \prod_{\delta_i\in\{\pm 1\}}(1-p^{\frac{n(n+1)}{2}}x_1^{\delta_1}\cdots x_n^{\delta_n}z).\end{aligned}$$ This has degree $2^n$ and has an immediate decomposition of the form $$\widetilde{R}_n(z) = \sum_{m=0}^{2^n}(-1)^m\widetilde{T}_mz^m,$$ where the coefficients $\widetilde{T}_m = \widetilde{T}_m(x_1, \dots, x_n)\in\mathbb{C}[x_1^{\pm}, \dots, x_n^{\pm}]$. These coefficients are clearly invariant under the group of Weyl transformations so, by the Satake isomorphism, there exists a polynomial $$\begin{aligned}
    R_n(z) = \sum_{m=0}^{2^n}(-1)^mT_mz^m \label{RTfact}\end{aligned}$$ whose coefficients $T_m\in\mathcal{R}_p$ satisfy $\widetilde{T}_m = \omega_p(T_m)$. If $n=1$ then notice, from [@Shimurahalf (5.4a)], that $\omega_p(T_1) = \omega_p(T(p))$, so $T_1 = T(p)$ as operators on $f$ in this case.

The polynomial $R(z)$ has a factorisation involving the Frobenius element $U_p$ – this part is similar to the methods found in [@Andrianov pp. 90–91] and [@Panchishkin pp. 42–50].

\[usum\] With $T_m$ and $U_p$ defined as above $$\sum_{m=0}^{2^n}(-1)^mT_mU_p^{2^n-m} = 0.$$

Denote the sum on the left-hand side by $Y$, this belongs to $\mathcal{R}_{0p}$. It is easy to check that $\widetilde{R}_n(z) = (p^{\frac{n(n+1)}{2}}z)^{2^n}\widetilde{R}_n((p^{n(n+1)}z)^{-1})$ so, immediately from \[RTfact\], we have $$\begin{aligned}
    T_m = p^{(n+1)(m-2^{n-1})}T_{2^n-m}. \label{Trel}\end{aligned}$$ By \[Trel\] above and the same argument of Proposition 2.1.2 in [@Andrianov pp. 88–89] (in this 