\theta}^T}, ~ \parT{\rho, \, z}^T \in \srfdomflat, ~\theta \in [0,
    \, 2\pi)} \cup \Gamma_0~.$$ We have that $\partial \voldomaxis = \Gamma \times [0, \, 2\pi)$ (see Figure \[fig:axis\_scheme\]).

The differential operators involved in the standard de Rham complex in Cartesian coordinates [@buffa:2011] $$\label{eq:3d_cart_seq}
\begin{split}
\tikzsetnextfilename{de_rham/de_rham_3d_cart}
\begin{tikzpicture}[commutative diagrams/every diagram,
arr/.style={commutative diagrams/.cd,
    every arrow,
    every label}]
\def\minW{25mm}
\def\minH{20mm}
\def\hShift{10mm}
\def\vShiftA{20mm}
\def\vShiftB{4mm}
\def\auxA{1mm}
\def\auxB{4mm}
%
% Nodes
\node (X0) [] at (0, 0) {$\Hsp^1\parTs{\voldomaxis}$};
\node (X1) [xshift=1.2*\minW] at (X0) {$\Hsp\parTs{\curlt;\,\voldomaxis}$};
\node (X2) [xshift=1.3*\minW] at (X1) {$\Hsp\parTs{\divergt;\,\voldomaxis}$};
\node (X3) [xshift=1*\minW] at (X2) {$\Lsp^2\parTs{\voldomaxis}$};
%
\node (R) [xshift=-0.6*\minW] at (X0) {$\R$};
\node (Z) [xshift=0.6*\minW] at (X3) {$0$};
% Arrows
\path[draw, arr] (X0) -- (X1) node [midway] {$\gradt$};
\path[draw, arr] (X1) -- (X2) node [midway] {$\curlt$};
\path[draw, arr] (X2) -- (X3) node [midway] {$\divergt$};
%
\path[draw, arr] (R) -- (X0);
\path[draw, arr] (X3) -- (Z);
\end{tikzpicture}
%%\includegraphics{img/de_rham/de_rham_2D_grad_rot_cont}\fsp,
\end{split}$$ correspond to the following ones in cylindrical coordinates: $$\label{eq:cyl_grad}
\renewcommand\arraystretch{1.5}
\gradc u = \begin{bmatrix}
\pd{\rho}{u} \\
\pd{z}{u} \\
\dfrac{1}{\rho}\pd{\theta}{u}
\end{bmatrix}~,$$ $$\label{eq:cyl_curl}
\renewcommand\arraystretch{1.5}
\curlc \uv = \begin{bmatrix}
\dfrac{1}{\rho}\pd{\theta}{u_z} - \pd{z}{u_\theta}\\
\dfrac{1}{\rho}\parT{\pd{\rho}\parT{\rho \,u_\theta} -
    \pd{\theta}{u_\rho}}\\ 
\pd{z}{u_\rho} - \pd{\rho}{u_z}
\end{bmatrix}~,$$ $$\label{eq:cyl_div}
\divergc \uv = \dfrac{1}{\rho}\, \pd{\rho}\parT{\rho \,
    u_\rho} + \dfrac{1}{\rho}\,\pd{\theta}{u_\theta} +
\pd{z}{u_z}~.$$ Since the domain is axisymmetric, we can exploit the 