TITLE: Must a bending of the cylinder leave the bases planar?
QUESTION [8 upvotes]: Set $M=\{(\cos(\theta),\sin(\theta),z):\theta\in[0,2\pi],z\in[0,1]\}$. A bending of $M$ is a smooth map $\Gamma:M\times [0,1]\rightarrow \mathbb{R}^3$ such that
1) $\Gamma[M\times\{t\}]$ is a submanifold with boundary of $\mathbb{R}^3$ (I call it $M_t$)
2) For every $m\in M$, we have $\Gamma(m,0)=m$ 
3) For every $t\in[0,1]$, the map $m\rightarrow \Gamma(m,t)$ is an isometry between $M$ and $M_t$
Question: Must any bending of $M$ leave the bases planar?
The only smooth bendings I can think of so far are those which change the shape of the circular bases to elipses
Thank you

REPLY [3 votes]: Take a geodesic $C$ on a smooth surface $S$. Through each point of $C$ draw a tangent to $S$ perpendicularly to $C$. The union of these tangents will be a developable surface, and $C$ is a geodesic on it. In particular, if $C$ is a closed geodesic, the union of tangent segments of length $\epsilon$ centered at the points of $C$ will be isometric to a cylinder. Now, if $C$ is non-planar, this cylinder will have non-planar bases.
In order to construct a continuous deformation of a straight cylinder into a cylinder with non-planar bases, we need a family $(S(t), C(t))$ with, say $S(t)$ a sphere and $C(t)$ its equator (and $C(t)$ non-planar for some $t$). I am quite sure such a family exists, but have no concrete example.