# Verify that some circles are geodesics on the $2$-dimensional torus

Let $\Bbb T^2$ be the two-dimensional torus. We have the following immersion:

$$\psi:\Bbb T^2\to\Bbb R^3:(e^{i\theta},e^{i\phi})\mapsto((2+\cos\theta)\cos\phi,(2+\cos\theta)\sin\phi,\sin\theta)$$

The metric induced on $\Bbb T^2$ by the euclidean metric on $\Bbb R^3$ is given by

$$g_{(e^{i\theta},e^{i\psi})}=(d\theta)^2+(2+\cos\theta)^2(d \phi)^2$$

I am told to check that $\phi=c$ defines a geodesic for any constant $c\in\Bbb R$.

My first approach was brutal and I determined Christöffel symbols, etc. then checking that the curve $\gamma:t\mapsto (e^{it},e^{ic})$ whose derivative $\dot\gamma(t)$ has coordinates $(1,0)$ in the tangent space at $\gamma(t)$ (where the basis is the tangent vectors associated with local coordinates $\theta,\phi$ where $\theta=x\circ\Phi$, with $\Phi(e^{i\theta},e^{i\psi})=(\theta,\psi)$ and similarly for $\phi$). In other words, $\dot\gamma$ has constant coordinates in the tangent space at $\gamma(t)$ and thus we only need to check that the Christöffel symbols $\Gamma^{1}_{11}$ and $\Gamma^{2}_{11}$ are $0$ to conclude it is a geodesic as we have

$$\frac{D\dot\gamma}{dt}(t)=0\iff\ddot\gamma^{i}(t)+\sum_{k,l=1}^{2}\dot\gamma^{k}(t)\dot\gamma^{l}(t)\Gamma^{i}_{kl}=0,\quad (i=1,2)$$

and many of these terms are $0$, so that it is equivalent to

\begin{align*} (\dot\gamma^{1})^{2}\Gamma^{1}_{11}&=0\\ (\dot\gamma^{1})^{2}\Gamma^{2}_{11}&=0 \end{align*}

However, this involves a lot of computations. Instead of this somehow brutal method, I was told to use an isometry between $\Bbb R^{2}$ endowed with the euclidean metric $g^{2}$ and $\psi(\Bbb T^{2})\subset \Bbb R^3$. But I don't get it, how is that supposed to help me? I understand that the goal is to look at the images of lines (i.e. geodesics of $\Bbb R^2$) to conclude these images are geodesics by the properties of isometries. But which isometry should I take in order to recover the euclidean metric on $\Bbb R^2$ by pull-back?

Hint Each of the circles $C_c := \{\phi = c\}$ are (components of) the intersection of torus $\phi(\Bbb T)$ and a plane of symmetry thereof. The fact that $C_c$ is an (unparameterized) geodesic then follows from the uniqueness of geodesics with prescribed starting point and initial tangent vector.
• Thanks for the answer. I am not sure to understand. The fact that they are in a plane of symmetry, I agree with that; and the map $\Bbb T^2\to\Bbb T^2$ corresponding to a rotation (edit: not rotation, of axis this plane) in $\Bbb R^3$ leaves the metric on $\Bbb T^2$ invariant. But why would this imply that they are geodesics? Jun 21, 2017 at 22:46
• Another way to think of it is this: Pick a point on $C_c$ and an initial direction tangent to $C_c$, and let $\gamma$ be the geodesic corresponding to these data. By symmetry its reflection $R \circ \gamma$ under the reflection $R$ through the plane $\Pi_c$ containing $C_c$ is also a geodesic with the same initial data, and so be uniqueness $R \circ \gamma = g$. Thus, $\gamma$ is contained in the fixed point set of $R$, namely the plane $\Pi_c$. Jun 21, 2017 at 23:06