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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

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  • $\begingroup$ @PiotrHajlasz For question 1: Why circle ? M is a cylinder and the M_t(s) are supposed to be 2 dimesnional manifolds with boundaries that are diffeomorphic to a cylinder. For question 2: I mean Isometry of 2 dimesnional riemannian manifolds . I actually see my question clear as it is already. Thank you for your comment though $\endgroup$ – Amr Mar 20 '18 at 21:52
  • $\begingroup$ Sorry, I misunderstood your questions. I will remove my first comment, because that was clearly my misunderstanding. $\endgroup$ – Piotr Hajlasz Mar 20 '18 at 21:54
  • $\begingroup$ @PiotrHajlasz It's OK. Thanks for your time again :) $\endgroup$ – Amr Mar 20 '18 at 21:56
  • $\begingroup$ Why only to ellipses? You can change the base curves to any simple closed planar curves of the same length. Just make a paper model of it and bend. $\endgroup$ – Alexandre Eremenko Mar 20 '18 at 22:07
  • $\begingroup$ @AlexandreEremenko Sure... My wording was bad $\endgroup$ – Amr Mar 20 '18 at 22:08
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There's a nice paper by Halpern and Weaver: Inverting a cylinder through isometric immersions and isometric embeddings which shows (among other things) that:

[a right circular cylinder] can be turned inside out through embeddings which preserve its flat Riemannian metric provided its diameter is greater than $(\pi+2)/\pi$ times its height.

(from the abstract)

So in fact, your cylinder can be fully inverted by a bending (since the ratio of diameter to height is $2>(\pi+2)/\pi\approx 1.637$).

Halpern and Weaver's inversion is described in Section 10 of their paper and involves folding portions of the cylinder like an "accordion" and then twisting those portions around axes in the $xy$-plane.

These figures from the paper show a top view of the cylinder and the initial accordion folding.

Figures 10.1a and 10.1b from Halpern and Weaver

This figure represents the folded cylinder after twisting the accordion around the axes.

Figure 10.1c from Halpern and Weaver

Here's another construction they offer, along the lines of Alexandre Eremenko's answer. Note that their cylinder is the product of the unit circle $S^1\subset\mathbb{R}^2$ with the interval $[-h,h]$:

Figures 10.2 and 10.3 from Halpern and Weaver

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It seems easy to obtain a negative answer by simple experiment. Make this cylinder of paper. Then squeeze one base with your fingers (leaving another base free). Both bases cannot remain planar: indeed the lines $\theta=\mathrm{const}, z\in[0,1]$ in the original non-squeezed position were all of the same length $1$. And they were perpendicular to both base planes. In the squeezed position, their lengths must be still $1$ but they are not all perpendicular to any plane anymore, so both bases cannot remain planar.

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  • $\begingroup$ When I make this experiment, the free base seems to be planar to me $\endgroup$ – Amr Mar 20 '18 at 23:02
  • $\begingroup$ It also seems planar to me, but this implies that the non-free base cannot stay planar. About free base I am not 100% sure. Anyway, they cannot BOTH stay planar, this seems to be clear. $\endgroup$ – Alexandre Eremenko Mar 20 '18 at 23:26
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    $\begingroup$ Just because the ruling lines are not perpendicular to the plane of one base, it does not follow that the other base cannot be planar. Indeed as I explained in the note at the end of my reply, one might glue a pair of equilateral triangles to the bases of the cylinder, where one triangle is "rotated" with respect to the other, to obtain an antiprism. Here both bases will remain planar. You also seem to be assuming that the lines of ruling remain straight lines during a bending, which is not the case, again as shown by the antiprism. $\endgroup$ – Mohammad Ghomi Mar 22 '18 at 20:12
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Here is a general way to construct a large family of bendings of the cylinder $M$, with nonplanar boundaries, via Alexandrov's isometric embedding theorem. All these examples will be convex.

First note that the geodesic curvature of the boundary components of $M$ is zero. Let $D_1$ and $D_2$ be any pairs of convex planar disks with smooth positively curved boundaries $\partial D_1$, $\partial D_2$ which have the same length as the boundary components of $M$. Gluing $D_i$ along the boundaries of $M$ yields a a closed surface $\overline M$ with a metric that has everywhere nonnegative curvature, because geodesic curvatures of $\partial D_i$ are nonnegative. Thus, by Alexandrov's isometric embedding theorem, $\overline M$ admits an isometric embedding into $R^3$ as a convex surface $\overline M'$. Let $M'$ and $D_i'$ be the images of $M$ and $D_i$ in $\overline M'$.

I claim that $M'$ cannot have planar boundary components when $D_1$ and $D_2$ are non-congruent. Indeed, the only time when $M'$ has planar boundaries is when $D_1$ and $D_2$ are congruent, and the corresponding points on their boundaries are glued to the end points of the same line of ruling of $M$. To see this note that if $D_1'$ is planar then the principal normals of $\partial D_1'$ (which are non-vanishing by the positive curvature assumption on $\partial D_i$) must lie in the plane of $D_1'$. But the principal normals of $\partial D_1'$ must be orthogonal to $M'$ since boundary components of $M'$ are geodesics (being a geodesic is an isometric invariant). So $M'$ has to be orthogonal to the plane of $D_1'$.

Thus, if $D_1'$ is planar, then all tangent planes of $M'$ along $\partial D_1'$ are orthogonal to the plane of $D_1'$. But since $M'$ is convex, its tangent planes are support planes. So $M'$ has to lie inside a right cylinder, say $S_1$ generated by $D_1'$. Similarly, if $D_2'$ is planar, then $M'$ has to lie inside a right cylinder $S_2$ generated by $D_2'$. It follows then that, if $D_1'$ and $D_2'$ are both planar, then $S_1=S_2\supset M'$. In particular $D_1'$ and $D_2'$ are congruent, lie in parallel planes, and are directly "above" each other (i.e., the orthogonal projection of $D_1'$ into the plane of $D_2'$ coincides with $D_2'$).

Note: It is important in the above construction that $\partial D_i$ be smooth. Indeed when $D_i$ are triangles, then $M'$ will form the sides of an antiprism, with bases $D_i'$.

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  • $\begingroup$ Why are $M'$ and $D_i'$ smooth surfaces? $\endgroup$ – Ivan Izmestiev May 19 '18 at 8:45
  • $\begingroup$ @Ivan: I don't think that smoothness is needed here, but the phrase "tangent plane" can probably be replaced with "support plane" and things should work out fine. $\endgroup$ – Mohammad Ghomi May 19 '18 at 13:00
  • $\begingroup$ But the question was about a smooth deformation of the cylinder making the bases non-planar. I like your argument and wonder whether it is true/proved that the isometric embedding of two disks with smooth convex boundaries glued isometrically along the boundaries is smooth away from the curve of gluing. $\endgroup$ – Ivan Izmestiev May 22 '18 at 12:24
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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.

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    $\begingroup$ I think one might be able to use a family of oblate ellipsoids for $S(t)$, see figures 11 and 12 and the discussion in this wikipedia page en.wikipedia.org/wiki/… . The idea is that as $b/a$ increases from $(1/2-\epsilon)$ to $1/2$, the nonplanar closed geodesics merge onto the equator; one then reverses this to get $C(t)$. Do you know what conditions must be imposed on the family $(S(t),C(t))$ beyond e.g. requiring the lengths of $C(t)$ to be constant? I guess this must depend on $\epsilon$. $\endgroup$ – j.c. May 22 '18 at 14:11
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    $\begingroup$ @j.c. Yes, I think you are right. And no additional conditions are needed. If we take a long cylinder and deform it too much, then singularities will appear, but an arbitrarily long cylinder can be deformed for a short time, and a short cylinder can be deformed long enough. I guess one can even turn it inside out in this way (probably need to consider geodesics on a family of quadrics). $\endgroup$ – Ivan Izmestiev May 22 '18 at 17:03

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