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I apologize if my post is "silly" because I don't know much about riemannian geometry.

I know that $M = (0,1)$ (the open unit interval) can be seen as a one-dimensional manifold. Since $M$ is an open subset of an euclidean space, a geodesic in $M$ is a curve $\gamma \, : \, I \subset \mathbb{R} \, \longrightarrow \, M$ such that $\ddot{\gamma}(t) \in \big( T_{\gamma(t)}M \big)^{\perp}$ for all $t$. So, the geodesics are the curves $\gamma$ such that $\ddot{\gamma} = 0$, which means that $\gamma$ is a straight line.

Is it possible to change the metric on $M$ such that the geodesics are not straight lines of the form $t \, \longmapsto \, at+b$ ?

I have the feeling that the answer is no because $T_{\gamma(t)}M$ is isomorphic to $\mathbb{R}$ and there are not many possible inner products on $\mathbb{R}$ but I am probably mistaken.

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    $\begingroup$ I think that there's no "silly" question as long as it is serious and there's an effort search in it ;-) Which is obviously the case here $\endgroup$ Jan 29, 2014 at 15:56

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$$ \newcommand{\vvec}{{\bf{v}}} \newcommand{\wvec}{{\bf{w}}} \newcommand{\Mmat}{{\bf{M}}} $$ You can quite easily change the metric on the unit interval. The metric at a point $t$ of the manifold in this case is specified by a $1 \times 1$ symmetric positive definite matrix $\Mmat(t)$, with the inner product of two vectors $\vvec$ and $\wvec$ in the tangent space at $t$ being given by $\vvec^t \Mmat(t) \wvec$. Since $\vvec$ is just a number, and so is $\wvec$, this turns into a product of three numbers $vmw$, where $m$ is a positive number.

As an example, we can try something like $\Mmat(t) = 1/{t^2}$. Now the vector $[1]$ at $t = \frac{1}{4}$ has squared length $4$, but at $t = 1$, it has squared length $[1][1][1] = 1$, hence length $1$ as well. (For this to make sense, I'd need to extend your interval to, say, $(0, 3/2)$, so that $1$ is in the manifold...sorry about that!)

What's a geodesic look like here? It'll still be a curve traversing the unit interval, but in the usual coordinates on $\mathbb R$, it'll have to speed up as it goes from $0$ to $1$. In fact, $\gamma(t) = e^t$ is, I believe, a geodesic in this metric. Let's check that.

The speed of $\gamma$ is the length of $\dot{\gamma}$. Let's look at a particular $t$, say $t_0 = 1/4$. There, we have the following data: \begin{align} \gamma(t_0) &= e^{0.25}\\ \dot{\gamma}(t_0) &= [e^{0.25}]\\ \Mmat(\gamma(t_0)) &= [\frac{1}{\gamma(t_0)^2}] = [\frac{1}{e^{0.5}}]\\ \|\dot{\gamma}(t_0) \|^2 &= \dot{\gamma}(t_0)^t \Mmat(\gamma(t_0)) \dot{\gamma}(t_0) \\ &= [e^{0.25}]^t [\frac{1}{e^{0.5}}] [e^{0.25}]^t \\ &= [1] \end{align} Sure enough, it's a unit-speed curve at that point (and at other points, as I'm sure you can verify).

Great question!

To continue, answering a followup from the comment, we can work through the computation of the Christoffel symbols (there's only 1) at the point $t$: \begin{align} \Gamma_{11}^1 (t) &= \frac{1}{2} g^{11} (\frac{\partial g_{11}}{\partial t} + \frac{\partial g_{11}}{\partial t} - \frac{\partial g_{11}}{\partial t}) \\ &= \frac{1}{2} g^{11} (\frac{\partial g_{11}}{\partial t}) \\ &= \frac{1}{2} t^2 (\frac{-2}{t^3}) \\ &= \frac{1}{t}. \end{align} Note that I've written $\Gamma_{11}^1 (t)$ to indicate that this is a function that varies with the coordinate of the point you're considering. (I've also computed $g^{11} = t^2$, the sole entry of the inverse of the matrix $g_{11} = \frac{1}{t^2}$.)

The equation for geodesics is $$ \frac{d^2 t}{d s^2}(s) + \Gamma_{11}^1 (t(s)) \frac{dt}{ds}(s)\frac{dt}{ds}(s) = 0 $$ where the curve is given by $$ s \mapsto t(s). $$ In our case, the curve is $s \mapsto e^s$. So the equation becomes $$ e^s + \frac{1}{e^s} \left( e^s e^s\right) = 0 $$ which turns out to be true for all $s$. Hence our curve is a geodesic.

There are many alternative characterizations of geodesics, like "locally shortest paths that are unit speed," which is how I knew we were done when I had "unit speed", since in a 1-manifold there's only one direction to go (up to sign). But actually checking the geodesics equation is probably a Good Thing to do as well, just for practice. (I confess, I think I got it wrong 3 times before writing this down.)

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  • $\begingroup$ How would you prove that $\gamma \, : \, t \, \longmapsto \, e^{t}$ is a geodesic in this metric ? I'm not sure this curve has constant speed for this metric. If I was to determine the speed of this curve, I would compute $\left\langle \dot{\gamma}(t), \dot{\gamma}(t) \right\rangle_{\gamma(t)}$, right ? But if $M(t) = e^{-2t}$, then : $\left\langle v,w \right\rangle_{t} = v e^{-2t} w$. So, the speed of $\gamma$ wouldn't be constant. I think it works if $M(t) = \frac{1}{t^{2}}$ because, then, $\left\langle \dot{\gamma}, \dot{\gamma} \right\rangle_{\gamma(t)} = 1$. Am I mistaken ? $\endgroup$
    – pitchounet
    Jan 29, 2014 at 18:43
  • $\begingroup$ I believe that you are correct; I've edited to fix that. $\endgroup$ Jan 29, 2014 at 18:55
  • $\begingroup$ Thanks for the update of your post. Now we're sure $\gamma \, : \, \longmapsto \, e^{t}$ is a unit speed curve. I am sorry to bother you but I still do not see why it would be a geodesic.. How would you prove it ? My try would be : the metric has only one coefficient $g_{11}$ so we could compute the Christoffel symbol $\Gamma_{11}^{1}$ and write the geodesic equations to see if $\gamma$ is a solution of it. What do you think ? $\endgroup$
    – pitchounet
    Jan 30, 2014 at 10:42
  • $\begingroup$ Added computation of Christoffel symbol and proof that $s \mapsto e^s$ is in fact a geodesic at the end of my original answer. $\endgroup$ Jan 30, 2014 at 11:47
  • $\begingroup$ Ah thanks ! I had almost the same thing, except there was a mistake in my geodesic equation because I wrote $\Gamma_{1,1}^{1}(s)$ instead of $\Gamma_{1,1}^{1}(t(s))$... Thanks for everything ! Thanks to you, I understand things better ! $\endgroup$
    – pitchounet
    Jan 30, 2014 at 12:32

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