\varphi \partial_r W - \frac{2}{r^2} \sin \varphi \cos \varphi \partial_{\varphi} W
        \biggr] {b^{\dagger}}_y b_y \biggr\}  
\end{split}  \label{eq:polar-bis}$$ and $$[r \partial_r, D] = - 2D, \qquad [r \partial_r, \Gamma] = - 2\Gamma, \qquad r \partial_r W = C.
  \label{eq:conditions-bis}$$

Conditions (\[eq:conditions-bis\]) are readily satisfied if we assume that $W$ takes the form $$W = C \ln r + F(\varphi),$$ where $F(\varphi)$ may be any (physically acceptable) function of $\varphi$.

In order to be relevant to the TTW Hamiltonians $H_k$, the realization (\[eq:polar\]), (\[eq:polar-bis\]) should be such that the bosonic part $4\omega K_{0,{\rm B}}$ of ${\cal H}^s$, defined in (\[eq:SUSY\]), reduces to (\[eq:TTW\]), which means that $D$ should be given by $$D = \frac{1}{4\omega} \left\{- \partial_r^2 - \frac{1}{r} \partial_r - \frac{1}{r^2} \partial_{\varphi}^2 +
  \frac{k^2}{r^2} [a(a-1) \sec^2 k\varphi + b(b-1) \csc^2 k\varphi]\right\}.$$ This will be so provided the function $F(\varphi)$ and the constant $C$ satisfy the Riccati equation $$- F'' + F^{\prime 2} + C^2 = k^2 [a(a-1) \sec^2 k\varphi + b(b-1) \csc^2 k\varphi].  \label{eq:riccati}$$ A solution is easily found to be given by $$F(\varphi) = - a \ln \cos k\varphi - b \ln \sin k\varphi, \qquad C = - k(a+b).  \label{eq:F-C}$$ It is worth observing here that to choose (\[eq:F-C\]) among all 