I was inspired by R. W. Sharpe's book on doing differential geometry through Cartan connections. Unfortunately, the book is fairly thin in terms of specific examples in Riemannian geometry, so I decided to try a few on my own.

My first thought was to try the hyperbolic plane (modeled on $\mathbb{R}^2$ instead of just the half-plane), since the orthonormal frame bundle is still, essentially, $\mathbb{R}^2\rtimes O(2)$. This actually worked out pretty well. The metric is just $\mathrm{g}=e^{-2y}\mathrm{d}x\otimes\mathrm{d}x+\mathrm{d}y\otimes\mathrm{d}y$, and since we want to work in an orthonormal frame, I made $$\begin{bmatrix}1 \\ 0\end{bmatrix}=e^y\partial_x\text{ and }\begin{bmatrix}0 \\ 1\end{bmatrix}=\partial_y,$$ so the associated covariant derivative is given by* $${\huge\nabla}_{\begin{bmatrix}1 \\ 0\end{bmatrix}}\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}-b \\ a\end{bmatrix} \text{ and } {\huge\nabla}_{\begin{bmatrix}0 \\ 1\end{bmatrix}}\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}0 \\ 0\end{bmatrix}.$$

A few calculations give the Cartan connection as $$\omega\left(\begin{bmatrix}\cos\theta & -\sin\theta & x \\ \sin\theta & \cos\theta & y \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}0 & -t & v^1 \\ t & 0 & v^2 \\ 0 & 0 & 0\end{bmatrix}\right)=\begin{bmatrix}0 & -t-v^1 & v^1 \\ t+v^1 & 0 & v^2 \\ 0 & 0 & 0\end{bmatrix}.$$

The problem came when I tried to find geodesics. Clearly, $\gamma:t\mapsto (0,t)$ is a geodesic in the traditional sense, but it lifts to $$\widehat\gamma:t\mapsto\begin{bmatrix}\cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & t \\ 0 & 0 & 1\end{bmatrix},$$ which has tangent vectors $$\dot{\widehat\gamma}(t)=\begin{bmatrix}\cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & t \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}0 & 0 & \sin\theta \\ 0 & 0 & \cos\theta \\ 0 & 0 & 0\end{bmatrix},$$ so $$\omega(\dot{\widehat\gamma}(t))=\begin{bmatrix}0 & -\sin\theta & \sin\theta \\ \sin\theta & 0 & \cos\theta \\ 0 & 0 & 0\end{bmatrix}.$$ I am fairly confident that this won't develop into a curve whose image projects to a straight line.

On the other hand, the curve $$\sigma:t\mapsto\begin{bmatrix}\cos t & \sin t & \sin t \\ -\sin t & \cos t & \cos t \\ 0 & 0 & 0\end{bmatrix}$$ has tangent vectors $$\dot\sigma(t)=\begin{bmatrix}\cos t & \sin t & \sin t \\ -\sin t & \cos t & \cos t \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}0 & 1 & 1 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},$$ so $$\omega(\dot\sigma(t))=\begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ and it develops into $t\mapsto\exp\left(t\begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\right)$, though $\sigma$ does not project to (the image of) a geodesic.

In short, as the title asks, aren't geodesics of the Riemannian geometry also geodesics of the Cartan geometry?

My ideal answer here includes:

- A "yes" or "no" to the above question.
- If "yes," then where did I go wrong?
- In the unlikely event of "no," why aren't they?
- Possibly a reference to a large collection of worked-out examples of Cartan connections for Riemannian geometry that deals with geodesics

Though, any one of the above will probably work for me.

*I apologize for the poor formatting.