Compact surface with Gaussian curvature 0 everywhere

Question

I'm trying to give an example of a compact surface $S\subset\mathbb{R}^3$ with constant zero Gaussian curvature for all $p\in S$. I know that a plane would have constant zero Gaussian curvature, but it is not compact, so I wonder if part of the plane, for example, $\{(x,y,0):x^2+y^2\leq1\}$, the unit closed disk on the $xy$ plane in $\mathbb{R}^3$. This disk is a subset of $\mathbb{R}^n$ and it's closed and bounded, so by Heine-Borel it should be compact? On the other hand, there is the theorem stating that for a compact oriented surface $S\subset\mathbb{R}^n$, $\exists p\in S$ such that the second fundamental form at $p$ is definite, but would this indicate that the Gaussian curvature must be nonzero at at least one point on a compact surface?

Answer 1

If $M\subset \mathbb{R}^2$ is a compact surface, consider a family of Spheres $S_r(p)$ for some $p\in \mathbb{R}^2$. There will be a smallest radius $r_0>0$ such that $M$ is contained completely in the ball bounded by $S_{r_0}$ - for that radius $r_0$, the sphere $S_{r_0}$ will touch the surface $M$ in (at least) one point with $M$ being 'inside' of that sphere. It is an elementary exercise in manipulation of the second fundamental form that in this case the Gauss curvature of $M$ has to be as large or larger than the Gauss curvature of the sphere (which is $\frac{1}{r_0}>0$).

So your suspicion is right, such a surface $M$ cannot exist as a submanifold of $\mathbb{R}^2$

This does not mean it cannot exist as an abstract manifold, though. But for that case there is a constraint implied by the Theorem of Gauss-Bonnnet, which, for a compact surface withoug boundary says that $$\int_M K \, dA = 2 \pi \chi(M)$$ where $\chi(M)$ is the Euler characteristic of $M$. This means that if $M$ has zero Gauss curvature everywhere, the integral and consequently the Euler characteristic of $M$ has to vanish. This is only possible if $M$ is homeomorphic to a torus.

Edit in response to a comment: an example of such a torus actually does exist. If you look at a square in $\mathbb{R^2}$ and pairwise identify the opposite edges you obtain a manifold diffeomorphic to a torus. If you take the induced metric from $\mathbb{R}^2$ then this torus will be flat, i.e. the Gauss curvature will be $=0$ everywhere.