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.
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.