Isometry between surfaces

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.

• Sorryyyy, I meant LOCALLY isometric – hal97 Feb 15 at 20:21
• 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. – Dante Grevino Feb 15 at 20:53
• thank you very much! – hal97 Feb 16 at 0:14
• Done. I think it is a good time to start studying covering maps! – Dante Grevino Feb 16 at 5:24