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I want to figure out the difference between the divergence operator on $\mathbb{R}^2$ which I denote by $\text{div}_{\mathbb{R}^2}$ and the divergence operator $\text{div}_{\mathbb{H}^2}$ on the hyperbolic plane in two dimensions \begin{align} \mathbb{H}^2=\{(x,y)\in \mathbb{R}_+\times \mathbb{R}\}. \end{align} So what I am interested if possible is to get a representation of $\text{div}_{\mathbb{H}^2}$ in terms of $\partial_x$ and $\partial_y$, i.e. \begin{align} \text{div}_{\mathbb{H}^2}=\text{div}_{\mathbb{R}^2}+...? \end{align} Unfortunately I'm really far from being an expert in differential geometry. If this is possible it would be really kind to exemplary demonstrate how one can derive it if possible :)

user99432
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    In a general Riemannian manifold $(M,g)$, the divergence of a vector field is given by $\text{div}g(X)=\frac{1}{\sqrt{|\det g|}}\frac{\partial}{\partial x^i}\left(\sqrt{|\det g|}X^i\right)$ (Voss-Weyl formula). So, can you compute what this metric determinant is? If that’s still complicated, start off by writing down the metric tensor $g{\Bbb{H}^2}$ first. – peek-a-boo Jan 27 '23 at 19:16
  • @peek-a-boo let me try: – user99432 Jan 28 '23 at 14:26
  • \begin{align}(g_{ij})=\frac{1}{(x^1)^2}\begin{pmatrix}1&0\0&1.\end{pmatrix}\end{align} Then the formula you gave we have \begin{align}\text{div}{g{\mathbb{H^2}}}(X)=\frac{1}{\sqrt{|\text{det}(g)|}}\frac{\partial}{\partial x^{i}}\left(\frac{1}{\sqrt{|\text{det}(g)|}}\right)X^{i}+\frac{\partial}{\partial x^{i}}X^{i}=\frac{1}{x^1}\frac{1}{\partial x^{i}}\left(\frac{1}{x^1}\right)X^{i}+\text{div}{g{\mathbb{R}^2}}(X)=\frac{1}{x^1}\frac{1}{\partial x^{1}}\left(\frac{1}{x^1}\right)X^{1}+\text{div}{g{\mathbb{R}^2}}(X)=-\frac{1}{(x^1)^3}X^1+\text{div}{g{\mathbb{R}^2}}(X)\end{align} – user99432 Jan 28 '23 at 14:26
  • No your divergence formula is wrong inside the partials. You mixed up the numerator and denominator. Look at my comment again. Also your determinant is wrong, it should be squared. – peek-a-boo Jan 28 '23 at 14:52
  • @peek-a-boo You're right I mixed the numerator and denominator but I think what I derived is true, isn't it? Because \begin{align} \sqrt{|\text{det}(g)|}=\sqrt{\frac{1}{(x^1)^2}}=\frac{1}{x^1}\end{align} or am I wrong again? – user99432 Jan 28 '23 at 14:58
  • Determinant of your matrix is $1/(x^1)^4$ (determinant of a constant times identity is constant to the power of the size of the matrix... It is multilinear, not linear). – peek-a-boo Jan 28 '23 at 15:00
  • @peek-a-boo This is embarassing... So it should be \begin{align}\text{div}{g{\mathbb{H}^2}}(X)=-\frac{4}{(x^1)^5}X^1+\text{div}{g{\mathbb{R}^2}}.\end{align} If we now choose \begin{align}\frac{1}{(x^1)^{\frac{1}{2}}}\begin{pmatrix}1&0\0&1\end{pmatrix}\end{align} it is still a metric on $\mathbb{H}^2$ so I could do the same but how does it change the "geometry"? – user99432 Jan 28 '23 at 15:24
  • Still wrong. Power 4 is the determinant, like I said. You have to square root it so $\sqrt{|\det g|}= 1/(x^1)^2 $. It changes the geometry because the lengths of tangent vectors (hence curves) change. So areas also change. What doesn't change (in this case) is the angles between tangent vectors (hence the angles of intersection of curves in the manifold remains the same). – peek-a-boo Jan 28 '23 at 15:35
  • @peek-a-boo Let me one last time state the correct answer now: \begin{align}\operatorname{div}{g{\mathbb{H}^2}}(X)=-\frac{2}{\left(x^1\right)^3} X^1+\operatorname{div}{g{\mathbb{R}^2}}.\end{align} So is it unnatural to choose the metric to be \begin{align}\frac{1}{\left(x^1\right)^{\frac{1}{2}}}\left(\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}\right)?\end{align} Or in other terms would it be "boring" to study the hyperbolic plane with this geometry? – user99432 Jan 28 '23 at 15:43
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    Still wrong divergence. Right partial derivative, but now you forgot to divide by outer determinant. The extra term is $\frac{-2X^1}{x^1}$, which can be written abstractly as $-2, d(\log x^1)(X)$. Anyway, you can certainly study that metric, we just wouldn't call it the hyperbolic plane anymore (it is only conformal to the hyperbolic plane, which itself is only conformal to the Euclidean half-plane). So maybe you want to read up how conformality affects things. – peek-a-boo Jan 28 '23 at 15:49
  • Thanks for your help! – user99432 Jan 28 '23 at 16:02

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