Are there any compact embedded 2-dimensional surfaces in $\mathbb R^3$ that are also flat?

Solution 1:

Imagine encompassing your surface with a large sphere. Now, slowly shrink the sphere until it's as small as possible while still encompassing your surface. Once it's as small as possible, it must be touching your surface at some point $p$. It must be tangent there, for otherwise the sphere would cross the surface.

I claim that at that point, the surface must have positive curvature. Roughly, it's because if the surface has $0$ or negative curvature, following a geodesic in the direction witnessing this fact for a short time will lead you outside the sphere, contradicting the fact the sphere entirely encompasses the surface.

Edit

Here's how I would make the last sentence precise. By translating, rotating, and scaling (the last of which isn't an isometry, but will still preserve the argument), we may as well assume the following:

The point we care about is $(0,1,0)$. The principal curvature we care about is obtained by intersecting the $xy$ plane with our surface. Finally, the circle of best fit, whose radius is what we're computing, is just the cirlce $x^2 + y^2 = 1$.

In fact, with these assumptions, we can draw the whole picture in $\mathbb{R}^2$ where we're just computing the usual curvature of a curve.

Locally, our surface is given as $y = f(x)$ for some $f$. The curvature of our surface at the point $(0,1)$ is $$\kappa = -\dfrac{f''(x)}{(1+f'(x))^{3/2}}$$ which, by assumption, is less than or equal to $0$. (The choice of sign is to ensure that the circle of radius 1 has curvature $+1$). It follows, of course, that $f''(0)\geq 0$ and we already have $f'(0) = 0$ and $f(0)=1$ due to all the rotating and translating we did. The curvature of our circle is $\frac{1}{\text{radius}} = 1$.

By continuity of $f''$, there is a $\delta > 0$ so that if $|x|<\delta$, then $f''(x) > -\frac{1}{2}$. Now, look at what happens on the interval $(0,\delta)$.

Integrating the condition on $f''(x)$ from $0$ to $x$, we get $f'(x) - f'(0) \geq -\frac{1}{2} x$ so $f'(x) \geq -\frac{1}{2} x$. Integrating again from $0$ to $x$ gives $f(x) - 1 \geq -\frac{1}{4} x^2$.

So, $f(x) \geq 1 - \frac{1}{4}x^2$. Finally, we just need to argue that for all sufficiently small $x$ that this implies $(x,f(x))$ is outside of the circle.

Well, $x^2 + f(x)^2 \geq x^2 + (1-\frac{1}{4}x^2)^2 = x^2 + 1-\frac{1}{2}x^2 + x^4 = 1 + \frac{1}{2}x^2 + x^4$, which is a sum of positive numbers, hence is strictly bigger than $1$ so long as $x\neq 0$. That is, when $x\neq 0$, $x^2 + f(x)^2 > 1$ for any $x$ in $(0,\delta)$, hence these points are outside the circle.

Solution 2:

If I have understood your notation and the question correctly, the answer is Yes under some assumptions on smoothness and No under other assumptions. Jason's No-proof assumes smoothness. But if one loosens the smoothness to $C^1$, then the answer is Yes. See my MO question entitled, $C^1$ isometric embedding of flat torus into $\mathbb{R}^3$, and the references and answers supplied there. See in particular my citing of Zalgaller's proof that the flat torus may be isometrically embedded in $\mathbb{R}^3$, an amazing result that seems little known.