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Let $\overline{g}$ be the flat metric on $\mathbb{R}^3$.

I would like to know if there is any compact embedded 2-dimensional surface $M$ in $\mathbb{R}^3$ (without boundary) such that $\iota^*\overline{g}$ is flat, where $\iota: M \hookrightarrow \mathbb{R}^3$ is the inclusion.

It appears that the answer is "no", but I am having a hard time coming up a with a rigorous proof, which does not use any results about the existence or non-existence of isometric embeddings of abstract surfaces into $\mathbb{R}^3$. Any suggestions would be appreciated.

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I imagine you forbid manifolds with boundaries ? –  Glougloubarbaki Dec 6 '11 at 23:29
    
Also I think this kind of questions are often non trivial –  Glougloubarbaki Dec 6 '11 at 23:37
    
If you allow boundary then of course there are flat surfaces -- the disc, the cylinder, Moebius band, etc. –  Ryan Budney Dec 7 '11 at 0:06
    
@Glougloubarbaki: You're right -- the surface is required to have no boundary. –  Kaloyan Marinov Dec 7 '11 at 0:57
    
Kaloyan, I put a comment at your other question. I edited my answer to that one within an hour of answering, essentially to match what Jason answers below. You were right to be skeptical about what I first wrote. –  Will Jagy Dec 7 '11 at 2:39

2 Answers 2

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.

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Thank you for the feedback. I am really struggling to make the statement in your last sentence precise; do you think you could help me out with that? –  Kaloyan Marinov Dec 7 '11 at 8:36
    
I don't see why your argument would not work in $\mathbb{R}^4$, for which it is false : the torus is a compact flat surface embedded in $\mathbb{R}^4$. –  Glougloubarbaki Dec 7 '11 at 10:03
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@Glougloubarbaki: A 2-sphere in $\mathbb R^4$ does not divide $\mathbb R^4$ into two components. –  Alexander Thumm Dec 7 '11 at 14:35
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@Glougloubarbaki: Try to apply my argument to a circle in $\mathbb{R}^3$ and see where it fails. As a hint: codimension 1 is essential to have principle curvatures. –  Jason DeVito Dec 7 '11 at 15:42
    
@t.b.: I'm embarrassed that you had to edit that for me. Thanks! –  Jason DeVito Dec 7 '11 at 18:27

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.

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