ed}
v_{T}^{(m\,h\,0)} &= c_1\,, &
v_{\Phi}^{(m\,h\,0)} &= c_2\,, &
v_{R}^{(m\,h\,0)} &= c_3\,, &\end{aligned}$$ and $$\begin{aligned}
v_{T}^{(m\,h\,1)} &= -2 [c_3+c_1 (h R T+i m)]\,, \\ {\nonumber}v_{\Phi}^{(m\,h\,1)} &= -2 c_2 (h R T+i m)\,, \\ {\nonumber}v_{R}^{(m\,h\,1)} &= -2  [c_3 (h R T+i m)+c_1-c_2]\,.\end{aligned}$$ Notice that there are three unknown coefficients $c_1$, $c_2$, and $c_3$. They endow us the freedom of choosing a 3-dimensional basis for covectors. In particular, we introduce a specific set of covector bases labeled by $B$ where $B\in\{T,\Phi,R\}$. They are defined by $$\begin{aligned}
\mathbf{V}_T^{(m\,h\,k)}&=\mathbf{V}^{(m\,h\,k)}|_{c_2=c_3=0}\,, \\ {\nonumber}\mathbf{V}_{\Phi}^{(m\,h\,k)}&=\mathbf{V}^{(m\,h\,k)}|_{c_1=c_3=0}\,,  \\ {\nonumber}\mathbf{V}_R^{(m\,h\,k)}&=\mathbf{V}^{(m\,h\,k)}|_{c_1=c_2=0}\,.
\label{eq:specific-choice-of-vector}\end{aligned}$$

### Symmetric tensor bases {#app:tensor-basis-Poincare}

The symmetric tensor bases in Poincaré coordinates can be decomposed using the dual basis one-forms $\{\text{d}T,\text{d}\Phi,\text{d}R\}$ via $$\mathbf{W}^{(m\,h\,k)} = W^{(m\,h\,k)}_{ij}\,\text{d}x^i\otimes \text{d}x^j,\quad x\in\{T,\Phi,R\}\,.$$ The tensor components are given by $$\begin{aligned}
W_{ij}^{(m\,h\,k)} \propto 
  \left[
  \begin{array}{ccc}
   R^{+2}w^{(m\,h\,k)}_{TT}
   & R^{+1} w^{(m\,h\,k)}_{T\Phi}
   & R^{+0} w^{(m\,h\,k)}_{TR}      \\
 * & R^{+0} w^{(m\,h\,k)}_{\Phi\Phi}
   & R^{-1} w^{(m\,h\,k)}_{R\Phi}     \\
 * &
 * & R^{-2} w^{(m\,h\,k)}_{RR}
 \end{array}
 \right] \times {\nonumber}\\
  \times R^{h-k} e^{im\Phi}\,,
 \end{aligned}$$ where $$\begin{aligned}
w_{TT}^{(m\,h\,0)} &=c_1\,, &
w_{\Phi\Phi}^{(m\,h\,0)} &= c_2\,,&
w_{RR}^{(m\,h\,0)} &= c_3\,,\\ {\nonumber}w_{T\Phi}^{(m\,h\,0)} &= c_4\,, &
w_{\Phi R}^{(m\,h\,0)} &= c_5\,,&
w_{RT}^{(m\,h\,0)} &= c_6\,.\end{aligned}$$ Notice that there are six unknown $c$-coefficients. They endow us the freedom of choosing the six tensor bases. In particular, we introduce a specific set of highest-weight tensor bas