Let $S$ be the surface

$S = \{(x,y,z) : x^2 + y^2/4 = 1\}. $

Minding theorem says that $S $ and $C $ the cylinder $x^2 + y^2 = 1$ are locally isometric, but can anyone build an local isometry between $S $ and $C $?


I am sorry but they are not globally isometric. We can distinguish them using some invariant.

Take a point $p=(p_1,p_2,p_3)$ in $C$, by symmetry (that is, composing with an isometry of $C$) we can assume that $p=(1,0,0)$, and consider the two principal directions $Z_p = (0,0,1)$ and $Y_p=(0,1,0)$ at $p$ (unique up to sign). The unique maximum geodesic starting at $p$ with initial velocity $Z_p$ is $\sigma_p: t\in \mathbb{R}\mapsto p+(0,0,t)\in C$ and the unique maximum geodesic starting at $p$ with initial velocity $Y_p$ is $\tau_p : t \in \mathbb{R} \mapsto (\cos(t),\sin(t),0) \in C$.

On the other hand, take a point $q=(q_1,q_2,q_3)$ in $S$ and again we can assume that $q=(1,0,0)$. In this case, the principal directions are $Z_q=(0,0,1)$ and $Y_q=(0,1,0)$ and the induced geodesics are $\sigma_p:t\in\mathbb{R} \mapsto q+(0,0,t)\in S$ and $\tau_q$, which is the reparameterization of $\tilde{\tau}_q:t\in\mathbb{R}\mapsto (\cos(t),2\sin(t),0)\in S$ by arc-length.

Now, let us suppose that there exists an isometry $\varphi: C\to S$. By symmetry, we can assume that $\varphi (p)=q$. Now $z$-direction is the unique direction in which the maximum geodesic is inyective so we have that $d_p\varphi(Z_p)$ is in $\{Z_q,-Z_q\}$ and $d_p\varphi(Y_p)$ is in $\{Y_q,-Y_q\}$. And again, by symmetry, we can assume that they are in $\{Z_q, Y_q\}$. So $\varphi\circ \sigma_p=\sigma_q$ and $\varphi\circ\tau_p=\tau_q$.

Now, let us focus on $\varphi\circ\tau_p=\tau_q$. Both $\tau_p$ and $\tau_q$ are periodic and the length of the curve over a period is invariant under isometries and under reparameterizations so the lengths $L(\tau_p|_{[0,2\pi]})$ and $L(\tilde{\tau}_q|_{[0,2\pi]})$ are equal. But we actually have that $L(\tau_p|_{[0,2\pi]})=2\pi$ and $$L(\tilde{\tau}_q|_{[0,2\pi]})= \int_0^{2\pi}\sqrt{1+3\cos^2(t)}dt>\int_0^{2\pi}\sqrt{1}dt=2\pi.$$ A contradiction. Note that we can equivalently argue that $\tau_p$ and $\tau_q$ have different period.

However, $C$ and $S$ are locally isometric. For example if $p=q=(1,0,0)$ as above, the formula $$\sigma_p(z)+\tau_p(t)\mapsto \sigma_q(z)+\tau_q(t)$$ defines an isometry between neighborhoods of $p$ and $q$. In fact, the functions $(t,z)\in \mathbb{R}^2\mapsto \sigma_p(z)+\tau_p(t) \in C$ and $(t,z)\in \mathbb{R}^2\mapsto \sigma_q(z)+\tau_q(t)\in S$ are riemannian coverings and so in particular local isometries. Do not worry if you do not know what a covering is, you can check that they are local isometries computing the metric coefficients.

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    $\begingroup$ Sorryyyy, I meant LOCALLY isometric $\endgroup$ – hal97 Feb 15 at 20:21
  • $\begingroup$ Hahaha, please do not change the question! Do not worry but next time check before posting. I am going to edit my answer showing a local isometry. $\endgroup$ – Dante Grevino Feb 15 at 20:53
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    $\begingroup$ thank you very much! $\endgroup$ – hal97 Feb 16 at 0:14
  • $\begingroup$ Done. I think it is a good time to start studying covering maps! $\endgroup$ – Dante Grevino Feb 16 at 5:24

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