for all $\beta\in{\mathcal{B}}$. Let ${\varphi}_2\colon X\to{\mathbb{C}}$ be the map for which ${\varphi}_2({\epsilon}_{i_1}+{\epsilon}_{i_1+2})=-{\varphi}_1({\epsilon}_{i_1}+{\epsilon}_{i_1+2})$ and ${\varphi}_2(\beta)={\varphi}_1(\beta)$ for all other $\beta\in{\mathcal{B}}$. One can check that $\xi_{\beta}(\mu_{{\varphi}_1})=\xi_{\beta}(\mu_{{\varphi}_2})$ for all $\beta\in{\mathcal{B}}$. It follows from [@Panov2 Theorem 1.4] that the coadjoint orbits of $\mu_{{\varphi}_1}$ and $\mu_{{\varphi}_1}$ are disjoint, so $J(\mu_{{\varphi}_1})\neq J(\mu_{{\varphi}_2})$. On the other hand, both $J(\mu_{{\varphi}_1})$ and $J(\mu_{{\varphi}_2})$ contain $J=J_c$, and this contradicts the maximality of $J$. The equivalence of (i), (ii), (iii) is now proved. The fact that the map $\xi$ is reconstructed by $J$ via formula (\[formula\_xi\_via\_c\_fd\_An\_Cn\]) follows from the proof of the implication $\mathrm{(iii)}\Longrightarrow\mathrm{(i)}$.$\square$

In the rest of this subsection, given a primitive ideal $J$ of $U({\mathfrak{n}})$, we will construct an explicit set of generators of the quotient algebra $U({\mathfrak{n}})$ satisfying the defining relations of the Weyl algebra of corresponding rank. Pick $\alpha\in\Phi^+$ and put $$\begin{split}
A(\alpha)&=\begin{cases}
{\mathcal{R}}_j\cup\bigcup\nolimits_{i<k<j}\{{\epsilon}_k+{\epsilon}_j\}\cup\{{\epsilon}_{i-1}+{\epsilon}_j,~{\epsilon}_{i-1}-{\epsilon}_i\}&\text{for }\alpha={\epsilon}_i-{\epsilon}_j,~i<j,~i\text{ even},\\
{\mathcal{R}}_j\cup\bigcup\nolimits_{i<k<n}\{{\epsilon}_k+{\epsilon}_j\}&\text{for }\alpha={\epsilon}_i-{\epsilon}_j,~i<j,~i\text{ odd},\\
\bigcup\nolimits_{i<k<j}\{{\epsilon}_k-{\epsilon}_j\}\cup\{{\epsilon}_{i-1}-{\epsilon}_j,~{\epsilon}_{i-1}-{\epsilon}_i\}&\text{for }\alpha={\epsilon}_i+{\epsilon}_j,~i<j,~i\text{ even},\\
\bigcup\nolimits_{i<k<j}\{{\epsilon}_k-{\epsilon}_j\}\cup\{{\epsilon}_{i-1}-{\epsilon}_j\}&\text{for }\alpha={\epsilon}_i+{\epsilon}_j,~i<j,~i\text{ odd},\\
\bigcup\nolimits_{i<k\leq n}\{{\epsilon}_k\}&\text{for }\alpha={\epsilon}_k,
\end{cases}\\
{\mathcal{B}}(\alpha)&=\{\gamma\in{\mathcal{B}}\setminus\Delta\mid{\mathrm{row}}(\gamma)\leq{\mathrm{row}}(\alpha)\},~R(\alpha)=\{{\mathrm{row}}(\gamma)\label{formula:A_B_R_C_Bn_Dn_fd},~-{\mathrm{col}}(\gamma),~\gamma\in{\mathcal{B}}(\alpha)\},\\
C(\alpha)&=\begin{cases}
-R(\alpha)\cup\{{\mathrm{col}}(\alpha)\}\setminus\{-{\mathrm{row}}(\alpha)+1\},&\text{if }{\mathrm{row}}(\alpha)\text{ is even},\\
-R(\alpha)\cup\{{\mathrm{col}}(\alpha)\}\setminus\{-{\mathrm{row}}(\alpha)-1\},&\text{if }{\mathrm{row}}(\alpha)\text{ is odd}.\\
\end{cases}
\end{split}$$ Define the matrix ${\mathcal{U}}$ by the following rule.

  $\Phi$   Size of ${\mathcal{U}}$   ${\mathcal{U}}$
  -------- ------------------------- -----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
  $B_n$    $(2n+1)\times(2n+1)$      ${\mathcal{U}}_{i,j}=-{\mathcal{U}}_{-j,-i}=e_{{\epsilon}_i-{\epsilon}_j}$, ${\mathcal{U}}_{i,-j}=-{\mathcal{U}}_{j, -i}=e_{{\epsilon}_i+{\epsilon}_j}$, $1\leq i<j\leq n$,
                                     ${\mathcal{U}}_{i,0}=-{\mathcal{U}}_{0,i}=e_{{\epsilon}_i}$, $1\leq i\leq n$, ${\mathcal{U}}_{i,j}=0$ otherwise
  $D_n$    $2n\times2n$              ${\mathcal{U}}_{i,j}=-{\mathcal{U}}_{-j,-i}=e_{{\epsilon}_i-{\epsilon}_j}$, ${\mathcal{U}}_{i,-j}=-{\mathcal{U}}_{j, -i}=e_{{\epsilon}_i+{\epsilon}_j}$, $1\leq i<j\leq 