# Aren't Riemannian geodesics also geodesics of the associated Cartan geometry?

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?

• 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.

• For any Riemannian (or pseudo-Riemannian) manifold, or manifold with an affine connection, the geodesics of the Cartan geometry on which the stabilizer part vanishes and the translation part doesn't will project to geodesics. – Ben McKay Aug 4 '17 at 14:26

The geodesics of Euclidean geometry in the plane are straight lines, i.e. curves that do not change direction. Their Cartan geometry is $$\omega=\begin{pmatrix}\gamma&\xi\\0&0\end{pmatrix}$$ on the oriented orthonormal frame bundle, in $$x,y,\theta$$ coordinates, valued in the Lie algebra of rigid motions of the plane, where $$\xi=e^{i\theta}(dx+i \, dy)$$, $$\gamma=i \, d\theta$$. So $$\gamma=0$$ says that the direction does not change: a straight line, i.e. a geodesic.
Now consider the hyperbolic plane. (When you look at the Cartan connection in your coordinates, you spot that it must involve exponential factors $$e^{\pm y}$$, since these arise in coordinates for your metric. You avoid exponentials by working in the moving frame, but then your Cartan connection is not expressed in $$x,y,\theta$$ coordinates anymore, but in the coframing. When you write out the velocity of $$\sigma(t)$$ in the coframing, it doesn't turn out to be as straightforward.)
In a complex notation, the Cartan connection is simpler. On the plane, let $$\underline{\xi}=e^{-y}dx+idy$$ and $$\underline{\gamma}=ie^{-y}dx$$. We see this because we want metric $$\xi_1^2+\xi_2^2$$. Add a variable $$\theta$$ and let $$\xi=e^{i\theta}\underline{\xi}$$, and $$\gamma=\underline{\gamma}+id\theta$$. The Cartan connection is the matrix $$\omega=\begin{pmatrix}\gamma&\xi\\0&0\end{pmatrix}$$ on the oriented orthonormal frame bundle, in $$x,y,\theta$$ coordinates, valued in the Lie algebra of rigid motions of the plane. This differs from your expression, because there are still $$e^{-y}$$ factors. The 1-form $$\gamma$$ does not vanish on your curve $$\sigma(t)$$. So is not a geodesic. Your first curve $$(x(t),y(t),\theta(t))=(0,t,0)$$ is, since $$\gamma=0$$ on it.
Expanding out in these coordinates, the equation of a geodesic is $$\xi=c \, ds, \gamma=0$$, for a complex constant $$c$$. We can assume that $$c\ne 0$$. The 1-forms $$\xi$$ and $$\gamma$$ transform when we rotate $$\theta$$, say the action $$r_{\theta_0} : \theta\mapsto \theta+\theta_0$$, as $$r_{\theta_0}^*\xi= e^{i\theta_0}\xi$$, $$r_{\theta_0}^*\gamma=\gamma$$. So we change the equations by rotation of $$c$$, and we can arrange that $$c>0$$. Rescale the $$s$$ variable to get $$c=1$$. So our equations of geodesics expand out to $$\cos \theta e^{-y} \frac{dx}{ds} - \sin \theta \frac{dy}{ds}=1,$$ $$\sin \theta e^{-y} \frac{dx}{ds} + \cos \theta \frac{dy}{ds}=0,$$ $$\frac{d\theta}{ds}+e^{-y}\frac{dx}{ds}=0,$$ if I have calculated correctly.
• Hi! I did all the computations and it indeed works with $\xi = e^{-i\theta}\underline{\xi}$. However in order to prove straight lines develop into geodesics I had to make an extra choice. Suppose that $(u,v,\phi)$ are coordinates in the Euclidean group. Then the development condition gives me a system of three equations $\cos\theta e^{-y}\dot{x} + \sin\theta \dot{y} = \dot u,$ $\sin\theta e^{-y}\dot{x} - \cos\theta \dot{y} = \dot v,$ $e^{-y}\dot x + \dot \theta = \dot \phi$. Now assume that I want to develop Euclidean geodesics. They are defined by equations $\ddot u =0$, $\ddot v = 0$ – Ivan Feb 11 at 15:42
• Differentiating twice those equations and after eliminating $\theta$ I find the following equations $\ddot x + \dot y (e^y \dot \phi - 2\dot x) =0$ and $\ddot y + e^{-2y}\dot x (\dot x-e^{y}\dot\phi)=0$ which correspond to the geodesic equations only if $\dot\phi = 0$. In Sharpe's book (proposition 4.13, page 209) it is stated that development should not depend on the lift, but here it seems that it does :( – Ivan Feb 11 at 15:56
• Yes, that's exactly what I don't understand! Why $\gamma = 0$? I feel like the answer is because $\gamma$ is the Levi-Cevita connection, but if $\omega$ is the Cartan connection and $\omega_G$ is the corresponding Maurer-Cartan form I thought that the definition of the development of a curve $\eta_{G/H}$ into $\eta_M$ is that for the corresponding lifts one should have $\hat \eta_{G/H}^* \omega_G = \hat\eta_M^*\omega$. I don't see why it would force $\gamma_M^*eta_M = 0$. Is it because it depends in which direction I develop? Should I say that the geodesics are curves which develop into lines? – Ivan Feb 11 at 16:11
• Yes, the geodesics are the curves that develop into lines. It is surprisingly complicated what the other constant vector field curves develop into, but they exist. There is nothing forcing $\gamma=0$ unless you demand that the curve develop into a line. – Ben McKay Mar 12 at 20:11