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?


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Jun 21, 2017 at 22:46
  • $\begingroup$ Ok maybe I understand: the plane leaves the intersection invariant under orthogonal symmetry and the tangent vectors at points of this intersection as well if they are parallel to the plane. Hence, by uniqueness of geodesics, a geodesic with such starting point and initial velocity must be fixed under these transformations. The fixed curves are precisely the circles. Would that be correct? $\endgroup$ Jun 21, 2017 at 22:55
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    $\begingroup$ 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 $\Pi_c$. $\endgroup$ Jun 21, 2017 at 23:06
  • $\begingroup$ Yes this is what I understood from your hint and detailed in my previous comment (but I am not a native speaker so maybe "orthogonal symmetry" is not the appropriate wording for "reflection through a plane"). Thanks a lot! $\endgroup$ Jun 21, 2017 at 23:08
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    $\begingroup$ Okay, great, and I'm glad you found it helpful! $\endgroup$ Jun 21, 2017 at 23:09

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