\begin{split}
\gradtsm & = \eta_{m, \, 0}^{-1} \comp \gradtm: \CaSpB^0
\rightarrow \CySpT^{m, \, 1}~, \\
\gradtsm\parT{\ut} & = 
\begin{bmatrix}
\dfrac{1}{m} \parT{\ut + {\rho}\,\pd{\rho}{\ut}}\\
\dfrac{\rho}{m}\pd{z}{\ut}\\
-\ut
\end{bmatrix}~, 
\end{split} \label{eq:gradstar}\\[1.5em]
\begin{split}
\curltsm & = \eta_{m, \, 1}^{-1} \comp \curltm: \CaSpB^1
\times \CaSpB^0 \rightarrow \CySpT^{m, \, 2}~, \\ 
\curltsm \parT{\uvt} & = \begin{bmatrix}
-\ut_z - \pd{z}{\ut_\theta}\\
\ut_\rho + \pd{\rho}{\ut_\theta}\\
\dfrac{1}{m}\parT{\pd{z}\parT{\rho\,\ut_\rho} -
    \pd{\rho}\parT{\rho\,\ut_z} - \pd{z}{\ut_\theta}} 
\end{bmatrix}
\end{split}~,\label{eq:curlstar}\\[1.5em]
\begin{split}
\divergtsm & = \eta_{m, \, 2}^{-1} \comp \divergtm:
\CaSpB^{1*} \times \CaSpB^2 \rightarrow \CySpT^{m, \, 3}~, \\ 
\divergtsm \parT{\uvt} & = \pd{\rho}{\ut_\rho} - \ut_\theta
+ \pd{z}{\ut_z}~.
\end{split}\label{eq:divstar}\end{aligned}$$ Note that, due to the choice of the spaces in Cartesian coordinates, all the differential operators – are well-defined. Moreover, since the discrete spaces in Cartesian coordinates are conforming, we have the following result:

The discrete spaces  are conforming in the spaces defined in .

We have to show that $\CySpT_h^{m, \, k} \subset \CySpT^{m, \,
        k}$, $k=1, \, \ldots, \, 3$. Due to the boundness of $\srfdomflat$, there exists $R > 0$ such that $0 < \rho < R, ~\parT{\rho, \, z} \in \srfdomflat$. Considering $\ut_h \in \CaSpB_h^0$, we have that $\ut_h \in
    \CaSpB^0$ because of the conforming discretization of the Cartesian complexes –. So $u_h \in \CySpT^{m, \, 