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?
Best Answer
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: