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I have a question about Geodesics on Cylinders and think I have the right answer but am unsure. The question reads:

Let $C_r:=[(x,y,z)\in\mathbb{R}^3: x^2+y^2=r]$ be the infinite cylinder of radius $r$. Show that $C_{r_1}$ is isometric to $C_{r_2}$ iff $r_1=r_2$.

Now I understand the logic behind this question I think. An isometry preserves geodesics, and because if you intersect a plane parallel to the axis of the cylinder with this cylinder, you get a curve $C$, which is just a circle, that is a geodesic. Now, if the radius between the two cylinders are different, the smaller circle would lie inside of the bigger cylinder, thus not lying on the surface and definitely not a geodesic.

Is this okay to write? Or do I have to explain it mathematically?

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  • $\begingroup$ Do write it down mathematically. $\endgroup$ Commented Apr 17, 2017 at 22:01
  • $\begingroup$ How do I go about doing that? $\endgroup$ Commented Apr 17, 2017 at 22:06
  • $\begingroup$ While the cylinders may both be embedded in $\Bbb R^3$ for convenience, I think it's a mistake to imagine them inside one another. At the very least imagine two separate copies of $\Bbb R^3$ next to one another, each with one cylinder. Ultimately, the space around them is irrelevant. There is no $\Bbb R^3$, there are only the two cylinders (although this is exceptionally hard to visualise; I doubt anyone is truly capable of visualising a cylinder without implicitly embedding it in $\Bbb R^3$ or something similar). $\endgroup$
    – Arthur
    Commented Apr 17, 2017 at 22:09
  • $\begingroup$ Since isometries preserve geodesics, I imagine taking an isometry from the smaller cylinder to the bigger cylinder will give a curve $C$ on the bigger cylinder that will not self-intersect, which is a contradiction to a geodesic no? $\endgroup$ Commented Apr 17, 2017 at 22:17
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    $\begingroup$ @Felicio: Do you know (rigorously) what the geodesics are on a cylinder? (If so, it's easy to show that through any particular point, there exists a shortest closed geodesic; the length of such a curve is preserved by an isometry.) $\endgroup$ Commented Apr 17, 2017 at 22:43

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Let $S_r$ be the circle of radius $r$ in $\mathbb R^2$. Let $C$ be a curve inside this $S_r$ and have length equals that of $S_r$. Then $S_r\times \mathbb R$ is isometric to $C\times \mathbb R$: Let $i: S_r \to C$ be the unit length parametrization of $C$, then $$ \phi : S_r\times \mathbb R \to C\times \mathbb R, \ \ \ \phi(s, t) = (i(s), t)$$ is an isometry. Thus your argument is not rigorous, that one surface is "inside" the other one does not mean that they are not isometric.

However, your idea is definitely a good one. Mathematically, you need to know that if $$\phi: C_{r_1} \to C_{r_2}$$ is an isometry, then $r_1=r_2$. Using your observation, consider the geodesic $ S_{r_1} \times \{0\}\subset C_{r_1}$. The image of this geodesic under $\phi$ is also a closed geodesic in $C_{r_2}$. Can you show that this geodesic is also of the form $S_{r_2} \times \{t\}$ for some $t$? If yes, then as isometry preserves length, one has $$ 2\pi r_1 = 2\pi r_2 \Rightarrow r_1 = r_2.$$

So it really spoils down to this question:

Are all closed geodesics in $C_r$ of the form $S_r \times \{t\}$ for some $t$?

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  • $\begingroup$ I believe so, as every closed geodesic on a cylinder is a circle no? $\endgroup$ Commented Apr 17, 2017 at 22:50
  • $\begingroup$ Yes, it's true. But again you need a proof of this fact. @FelicioGrande $\endgroup$
    – user99914
    Commented Apr 18, 2017 at 2:26
  • $\begingroup$ How would I go about constructing a proof for this? I'm kinda confused by the notation $S_r\times{t}$ $\endgroup$ Commented Apr 18, 2017 at 4:37
  • $\begingroup$ Does $t$ denote the thickness of the circle? $\endgroup$ Commented Apr 18, 2017 at 4:38
  • $\begingroup$ $S_r \times \{t\}$ is the usual set theoteric notation. So $S_r \times \{t\} = \{ (x, y, t) : x^2 + y^2 = r^2\}$. @FelicioGrande $\endgroup$
    – user99914
    Commented Apr 18, 2017 at 6:18
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Another possible way to proceed.

Hint: Isometries preserve the length of geodesics. What is the minimal length of a closed, simple geodesic in a cylinder of radius $r>0$?

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The intersection of a cylinder and plane is never a geodesic. (except there are 2 inflections points) The development is a sine curve which is not a straight line.

To visualize and model a cylinder containing a bent helical line,

take a rectangular $xz$ plane between $ x=\pm a, z=\pm b $ containing a transverse line. Imagine plane, portion of interest to be a thin flexible plastic sheet. Bend the sheet/plane to make bent edge to be a circle segment of a cylinder tangent to the plane. Somewhat as you imagined, the $ r$ in $(x-r)^2 +y^2 =r^2$ is a variable in bending.The series of circular segments which are boundary of bent cylinder are bent circumferentials. The bent circular sheet/segment contains your helix line whose original geodesic length is always preserved.

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