I'm working on Problem 3-7 of Lee's Introduction to Riemannian Manifolds:
Let $\mathbb{U}^2$ denote the upper half-plane model of the hyperbolic plane (of radius 1), with the metric $\breve{g} = (dx^2 + dy^2)/y^2$. Let $\mathrm{SL}(2, \mathbb{R})$ denote the group of $2 \times 2$ matrices of determinant 1. Regard $\mathbb{U}^2$ as a subset of the complex plane with coordinate $z = x + iy$, and let $$ A \cdot z = \frac{az+b}{cz+d}, \quad A = \left(\begin{array}{cc}a & b \\ c & d\end{array}\right) \in \mathrm{SL}(2, \mathbb{R}). $$ Show that this defines a smooth, transitive, orientation-preserving, and isometric action of $\mathrm{SL}(2, \mathbb{R})$ on $(\mathbb{U}^2, \breve{g})$. Is the induced action transitive on $\mathrm{O}(\mathbb{U}^2)$?
Here, $\mathrm{O}(\mathbb{U}^2)$ means the set of all orthonormal bases for all tangent spaces of $\mathbb{U}^2$, and the induced action of a group on this set is given by $\phi \cdot (b_1, \dots, b_n) = (d\phi_p(b_1), \dots, d\phi_p(b_n))$.
I've actually been able to completely solve this problem, but showing transitivity on $\mathrm{O}(\mathbb{U}^2)$ was...annoying. My idea was to identify the stabilizer group of $i$ (which turned out to be $\mathrm{SO}(2, \mathbb{R})$) and show that its induced action can take any orthonormal basis at $i$ to the standard basis, which is equivalent to transitivity on the bases of $T_i\mathbb{U}^2$. Then I was able to combine this with transitivity on $\mathbb{U}^2$ to take any orthonormal basis $(e_1, e_2)$ at $z$ to the standard basis at $i$ and then to another basis $(f_1, f_2)$ at $w$.
Unfortunately, the only way I could think of to show this was to take some $A = \left[\begin{array}{cc}a & -b \\ b & a\end{array}\right] \in \mathrm{SO}(2, \mathbb{R})$, fully compute the differential $D(\phi_A)_i$ of the action $\phi_A(z) = A \cdot z$ in coordinates, and use the matrix equation $$ D(\phi_A)_i \left[\begin{array}{cc}| & | \\ e_1 & e_2 \\ | & |\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] $$ to solve for $a$ and $b$. This obviously took a lot of messy algebra and felt very clunky. Is there a more elegant way I could have gotten there?
This must be the fastest I've ever gotten an answer on stack exchange. Thanks to Moishe Kohan's comment, I realized that the induced action is in fact not transitive on $\mathrm{O}(\mathbb{U}^2)$, simply because it preserves orientations! That is, it only takes positively oriented bases to other positively oriented bases, and likewise for negatively oriented bases. Therefore, it cannot possibly be transitive. I believe my mistake was not recognizing that the "arbitrary" basis $(e_1, e_2)$ that I calculated with was in fact positively oriented, so of course I was able to transform it to the standard basis! If I had instead tried to transform it to $\left[\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right]$, I would have found no solution. Hopefully this will teach me to think twice before wading hip deep into calculations next time. Thank you again, Moishe Kohan!