computing Laplacian of hyperbolic spaces Ok, as a beginner to Riemannian manifolds, let me ask the following embarrassing question. Consider the hyperbolic  manifold $(\mathbb H^n,g)$ with hyperbolic metric $g_{ij}=\delta_{ij}\frac{1}{x^2_n}$. Why is that the Laplace-Beltrami operator equal to $$\Delta_g=-x_n^2\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}?$$
Even for  $\mathbb H^2$ my computations does not confirm the above expression (which should be $\Delta_g=-y^2(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})$ ). Or am I missing something in here?
 A: There are several problems with your post. 


*

*Your wrote the hyperbolic metric wrong. The metric should be
$$ g_{ij} = \frac{1}{(x_n)^2} \delta_{ij} $$
in the upper-half-space model. You are missing a square. 

*In $\mathbb{H}^2$ the expression immediately follows from the conformal covariance of the Laplace-Beltrami operator. But in higher dimensions the Laplace-Beltrami operator is not conformally covariant, and there is necessarily a lower-order term. So your claim is in fact entirely false for $\mathbb{H}^n$. See this Wikipedia page for the general formula for the Laplace-Beltrami operator under conformal changes. 

*To show the computations explicitly, using the correct metric that I listed in part 1:
$$ \Delta_g = - \frac{1}{\sqrt{|g|}} \sum_{ij} \partial_i \sqrt{|g|} g^{ij} \partial_j $$
The metric determinant is
$$ \sqrt{|g|} = (x_n)^{-n} $$
and the inverse metric is
$$ g^{ij} = (x_n)^2 \delta^{ij} $$
so we simplify to 
$$ \Delta_g = - x_n^{n} \sum_{ij} \partial_i (x_n)^{2- n} \delta^{ij} \partial_j $$
which gives
$$ \Delta_g = - x_n^2 \sum_i (\partial_i)^2 - (2-n) x_n \partial_n $$
