I wonder if there is a connection between the divergence on the hyperbolic plane
and the divergence in 3D expressed in cylindrical coordinates. According to Divergence operator on the hyperbolic plane $\mathbb{H}^2$ the divergence on the hyperbolic plane (all points in $\mathbb{R}^2$ for which $x>0$ and $y\in \mathbb{R}$) is given by
\begin{align}
\text{div}_{\mathbb{H}^2}(F)=-\frac{-2F_x}{x}+\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}.
\end{align}
Let $F$ be now a vector field on $\mathbb{R}^3$. For the divergence in $3D$ it is
\begin{align}
\text{div}_{\mathbb{R}^3}(F)=\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}=\frac{\partial F_1}{\partial x_1}+\frac{F_1}{x_1}+\frac{1}{x_1}\frac{\partial F_2}{\partial \theta}+\frac{\partial F_3}{\partial x_3}
\end{align}
where $(x_1,\theta,x_3)$ are cylindrical coordinates, $F_1$ is the radial component, $F_2$ the angle component and $F_3$ the height component of $F$. If there is no dependence on the angle this should exactly be the hyperbolic plane. The divergence on the hyperbolic plane and the divergence in cylindrical coordinates look quite alike except for the right constant for the middle term.
Question
Why is the result for the divergence not the same? Is my view on this problem wrong? The result suggests that this is not the hyperbolic plane but why?
Best Answer
The question is interesting and made me go into some detail.
Voss-Weyl Formula
This thread explains well the caveats one has to keep in mind when using the Voss-Weyl formula for the divergence of a vector field $V$ of any dimension: $$\tag{1} \operatorname{div}_gV=\frac{\partial_\mu(\sqrt{\operatorname{det}g}\,V^\mu)}{\sqrt{\operatorname{det}g}}\, $$ where $g_{\mu\nu}$ is a metric tensor.
In short: The Voss-Weyl formula assumes that $V$ is expressed in a coordinate basis, that is, $$\tag{2} V=V^\mu\,\partial_\mu\,. $$
Cylindrical Coordinates
In cylindrical coordinates $(r,\varphi,z)$ one usually works with a different basis, namely, $$\tag{3} \partial_r\,,\;\textstyle\color{red}{\frac{1}{r}}\partial_\varphi\,,\;\partial_z $$ which is orthonormal w.r.t. the well-known cylindrical metric $$\tag{4} g=\begin{pmatrix}1&0&0\\0&r^2&0\\0&0&1 \end{pmatrix}\,. $$ In the orthonormal basis the vector field $V$ from (2) is $$\tag{5} V=V^r_{on}\,\partial_r+\textstyle\frac{1}{r}V^\varphi_{on}\,\partial_\varphi+V^z_{on}\partial_z\,. $$ The most notable difference is $V^\varphi=\frac{1}{r}V^\varphi_{on}\,.$
Since $\sqrt{\operatorname{det}g}=r$ we find from (1) \begin{align} \operatorname{div}_gV&=\frac{\partial_r(rV^r)+\partial_\varphi (r V^\varphi)+\partial_z (rV^z)}{r}\\ &=\frac{\partial(rV^r)}{r}+\partial_\varphi V^\varphi+\partial_z V^z\,.\tag{6} \end{align} To get the well-known divergence formula we have to switch to the orthonormal basis: \begin{align} \operatorname{div}_gV &=\frac{\partial_r(r V^r_{on})}{r}+\frac{1}{r}\partial_\varphi V^\varphi_{on}+\partial_z V^z_{on}\\ &=\partial_rV^r_{on}+\frac{V^r_{on}}{r}+\frac{1}{r}\partial_\varphi V^\varphi_{on}+\partial_z V^z_{on}\,.\tag{7} \end{align}
Hyperbolic Plane
It sounds like, by hyperbolic plane you are referring to the Poincare half-plane model $$\tag{8} \mathbb H=\{x,y\in\mathbb R:y>0\}\,,\quad ds^2=\frac{dx^2+dy^2}{y^2}\,. $$ Its metric tensor is $$\tag{9} g=\begin{pmatrix}y^{-2}&0\\0&y^{-2}\end{pmatrix}\,. $$ Since, $\sqrt{\operatorname{det}g}=y^{-2}$ the divergence of a vector field $$\tag{10} V=V^x\partial_x+V^y\partial_y $$ in the coordinate basis is \begin{align} \operatorname{div}_gV&=\frac{\partial_x\,(y^{-2}\,V^x)+\partial_y(y^{-2} V^y)}{y^{-2}}=\frac{y^{-2}\,\partial_xV^x-2y^{-3}\,V^y+y^{-2}\partial_y\,V^y}{y^{-2}}\\ &=\partial_x\,V^x+\partial_y\,V^y-\frac{2V^y}{y}\,,\tag{11} \end{align} which confirms your calculation.
It looks tempting to seek for a relationship between (11) and (7) when $V$ does not depend on $\varphi$ as $r$ looks very much like $y$ and $x$ looks very much like $z\,.$
However note that the Euclidean space is flat and the half plane $\mathbb H$ is not. Also: I have highlighted the difference of the bases we express the vectorfields in: cylindrical uses orthonormal basis, $\mathbb H$ uses coordinate basis.