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I am slowly working through a text on ordinary differential equations and I don't understand what this particular exercise is even asking of me.

The exercise says to determine the geodesics in $\mathbb{R}^3$ of the cylinder with unit radius with respect to the Riemannian metric obtained by restricting the usual dot product on $\mathbb{R}^3$.

My problem is that I do not know what it means to be "respect to the Riemannian metric obtained by restricting the usual dot product on $\mathbb{R}^3$.

I have tried to read about the Riemannian metric assuming that there is just a simple distance function I would need to write down my Langrangian but I am mostly finding a lot of results about Riemannian manifolds using notation far more complex than anything I've been previously introduced to. Otherwise I've tried looking through several books on calculus of variations, but I cannot find anyone else using this kind of language.

I get the impression that all this language is in differential geometry, which I have never studied before so I am a little confused why this is part of a short section in the middle of a differential equations book.

Michael Hardy
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JessicaK
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  • What is the book? – Suzu Hirose Dec 06 '14 at 02:35
  • C. Chicone. Ordinary Differential Equations with Applications. Springer-Verlag, New York 1999.

    I was originally using a book by Gerald Teschl, but it wasn't as detailed as I would have liked.

     – JessicaK Dec 06 '14 at 02:38
  • Are Chicone's explanations on page 71 and 72 not clear? – Zavosh Dec 06 '14 at 03:15
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    It just means that the notion of distance on that particular cylinder is the same as the ambient space it sits in. So if you have a curve on the cylinder and you want to measure its length, you just measure it as a curve in $\mathbb{R}^3$. – Zavosh Dec 06 '14 at 03:17
  • Geodesics are solutions to PDEs, and the PDE itself depends on the choice of metric. Actually, the term 'metric' in Riemannian metric is shorthand for metric tensor. – Passing By Dec 06 '14 at 03:46
  • @Passing by: geodesics are solutions of ODEs of the form ${\ddot x}^i + \Gamma_{jk}^i \dot x^j \dot x^k= 0$, where the $\Gamma_{jk}^i$ depend on the metric tensor. Cheers! – Robert Lewis Dec 06 '14 at 05:42
  • @RobertLewis: ODE, PDE, potato, potatoe ;). My bad, yes ODEs, not PDEs, sorry. – Passing By Dec 06 '14 at 06:13
  • @PassingBy: tomato, tomatoe ;). Enjoyed the "poetry" on your user page. Watch out for the riff-raff! Cheers! – Robert Lewis Dec 06 '14 at 06:24
  • @prometheus Thank you, that explanation helps a lot. I reread page 71-72 several times, but from the word tangent bundle forward, I really struggle with both the definitions and the intuition. (I have very little in the way of completed courses, but I read a lot of books and random stackexchange posts) – JessicaK Dec 06 '14 at 09:58

1 Answers1

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Let $C \subset \mathbb R^3$ denote the cylinder with unit radius and infinite length. We choose the parametrization

$$f: [0, 2 \pi] \times \mathbb R \longrightarrow \mathbb R^3, f(\phi,z) = \begin {array}{1} (cos \ \phi,sin \ \phi, z) \end {array}.$$

The Euclidean metric $< , >$ on $\mathbb R^3$ induces a metric on $C$. Referring to the given parametrization its metric tensor is

$$g =\begin{pmatrix} <f_\phi, f_\phi> <f_\phi, f_z>\\ <f_z,f_{\phi}> <f_z,f_z> \end{pmatrix} =\begin{pmatrix}1 \ 0\\0 \ 1 \end{pmatrix}.$$

Because the metric tensor is constant all Christoffel symbols $\Gamma^i_{j,k}$ vanish. The two differential equations for the parameters of a geodesic reduce to

$$\frac {d^2 \phi}{dt^2} = 0, \frac {d^2 z}{dt^2} = 0$$

with the solutions $\phi(t) = \phi_0 + \phi_1 t, z(t) = z_0+z_1t.$

Note. When rolling the cylinder $C$ onto the plane $\mathbb R^2$ the geodesics become straight lines in the plane.

Jo Wehler
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