Orthonormal frame for the Poincaré half plane

Question

I'm trying to find an orthonormal basis for the Poincaré half plane $\mathbb{H}^2$ given the metric $$ds^2 = \frac{dx \otimes dx + dy \otimes dy}{y^2}$$ and I've managed to calculate the matrix $[g_{ij}]$ as $$\begin{pmatrix}\frac{1}{y^2}&0\\ 0&\frac{1}{y^2}\end{pmatrix}$$ but I do not understand the process on how to go from here to an orthonormal frame. What I think I need is that $g(e_i,e_j) =\delta_{ij}$ but how does this work?

Answer 1

Consider the orthonormal basis for the usual metric in $\mathbb{R}^2$, that is $$\frac{\partial}{\partial x}, \, \frac{\partial}{\partial y},$$ which we Will identifie with $e_1$ and $e_2$. We have that the matrix of the metric verifies $g_{ij}=\langle e_i, e_j\rangle$, with the inner product in $\mathbb{H}$. In general you could apply the Gram Schmidt procces to obtain an orthonormal basis, but is straightforward to see that $$y\frac{\partial}{\partial x}, \, y\frac{\partial}{\partial y}, $$ is an orthonormal basis, since the Canonical basis is orthogonal.

Answer 2

In general the standard way is Gramm-Schmidt process on canonical basis for $\mathbb R^n$. Here one can use a simple observation. Poincaré half plane metric can be written as $$ \langle u,v \rangle_p=\frac{u\cdot v}{y^2} $$ Let $e_1$ and $e_2$ to be the canonical frame of $\mathbb R^2$ and try to find $a$ such that $\langle ae_1,ae_1 \rangle_p = 1$.