Every generalised cylinder is locally isometric to a plane

Question

I'm trying to show that every generalised cylinder is locally isometric to a plane. I understand that they have the same first fundamental form. I know I can parametrise the cylinder as $\sigma(u,v)=\gamma(u)+v\mathbf c$ where $\gamma$ is a curve along the cylinder and the plane can be parametrised as $\pi(u,v)=\mathbf a+u\mathbf p+v\mathbf q$. But I'm not really sure how to come up with a map between curves in each surface that preserves length. Any help is appreciated.

Answer 1

Hint When you parameterize the generalized cylinder $C$, parameterize the (planar) curve $\gamma$ by length, so that (by definition) $\gamma$ is an isometry of an interval onto its image.

Pick any point on the cylinder: The parameterization $\sigma(u, v) := \gamma(u) + v {\bf c}$ can be viewed as directions for getting from an arbitrary reference point on $C$ to any other: By translating coordinates, we can take the reference point to be $\sigma(0, 0) = \gamma(0)$. Since $\gamma$ is parameterized by arc length, $\gamma(u) + v {\bf c}$ is the point in $C$ we reach when traveling from $\gamma(0)$ a distance $u$ along $\gamma$, making a quarter turn to the direction of the vector $\bf c$ orthogonal to the plane that contains $\gamma$ (we may as well normalize $\bf c$ to be orthogonal), and then traveling a distance $v$.

Additional hint Now, $s \mapsto \gamma(s)$ and $t \mapsto \gamma(u) + t {\bf c}$ are geodesics, hence so are their preimages under any local isometry from $\Bbb R^2$ to $C$. But the geodesics of $\Bbb R^2$ are straight lines.