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Problem statement: Let $M \subseteq \mathbb{R}^3$ be a compact, embedded, 2-dimensional Riemannian submanifold. Show that $M$ cannot have $K \leq 0$ everywhere, where $K$ stands for the Gauss curvature of $M$.

I have attempted an approaches (described below), but have not been able to finish it off.

I came across a hint suggesting that I consider the square-of-distance function $f: \mathbb{R}^3 \rightarrow [0,\infty)$ by $f(x)=|x|^2$. As $M$ is compact, there is some $q_0 \in M$ such that \begin{equation} \forall q \in M: \quad f(q) \leq f(q_0). \end{equation} We also know that \begin{equation} \forall v \in T_{q_0} M: \quad df_{q_0}(v) = \frac{d}{dt}\bigg|_{t=0} (f \circ \gamma_v)(t) = 0\quad \text{and} \quad (f \circ \gamma_v)''(0) \leq 0. \end{equation} The problem now is that I do not know how to relate the above statements to a statement about the curvature $K(q_0)$.

I would appreciate any suggestions on either of the above approaches.

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2 Answers 2

This is not a suggestion on your approach, but rather a different approach.

As $M$ is compact, there exists a smallest sphere $S$ containing $M$, and it must touch $M$ at some point $p$. Hence the Gaussian curvature of $M$ at $p$ is the same as the Gaussian curvature of $p$ at $S$, which is positive.

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I had actually given the same answer to a different question: math.stackexchange.com/questions/89061/…. The OP asked me to expand my answer into a full proof and I did, but I'm wondering if you know of a much slicker way of doing it? –  Jason DeVito Jan 31 '13 at 16:07
    
@JasonDeVito: I'm sorry, my answer is backed up only by handwaving, I must admit. Thank you for the link, that's most certainly a more satisfying justification! –  Bruno Stonek Jan 31 '13 at 16:19
    
@JasonDeVito: Also, see this related recent question of mine. –  Bruno Stonek Jan 31 '13 at 16:20
    
I had actually already seen (and upvoted) that other question of yours. I actually found little satisfaction in the justification since it seemed to destroy the beauty of the simple idea. I was hoping to save the beauty with a simple justification ;-). –  Jason DeVito Jan 31 '13 at 16:24

That's very nice. Approach 2 works. You are calling the farthest point $q_0.$ There is a method called Lagrange multipliers. The tangent plane to your surface is orthogonal to the vector from the origin to $q_0.$ Meanwhile, your surface is contained in the closed ball around the origin that passes through $q_0.$ As a result, the principal curvatures are nonzero and have the same $\pm$ sign as the sphere containing $M.$ (If not, there would be points arbitrarily close to $q_0,$ along a principal direction, outside the closed ball, because they would be in the tangent plane at $q_0,$ or between the tangent plane and the tangent sphere, or on the wrong side of the tangent plane). So, the two principal curvatures are nonzero and have the same $\pm$ sign, so their product $K$ is strictly positive.

See http://en.wikipedia.org/wiki/Principal_curvature

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So, if we set $R= |q_0| > 0$ and $S_R(q_0) = \{p \in \mathbb{R}^3: |p| = R\}$, then $T_{q_0} M = T_{q_0} S_R(q_0)$, when considered as affine subspaces of $\mathbb{R}^3$. Your comments in parentheses definitely sound like they are true, but I am having a hard time justifying them rigorously. I read the wikipedia article you suggested, but I'm trying to avoid using normal planes -- what I'm after is a more ``intrinsic'' argument. Do you think you could elaborate on how I might make your parenthetical remarks rigorous? –  Kaloyan Marinov Dec 6 '11 at 7:55
    
If anyone has any suggestions on Approach 1 (even though it uses more advanced machinery), I'd be happy to hear those too. –  Kaloyan Marinov Dec 6 '11 at 7:57
    
There is no completely intrinsic argument. The compact orientable surfaces of higher genus have metrics with constant curvature $K = -1.$ By the Nash embedding theorem, these may be isometrically embedded in some $\mathbb R^n$ for sufficiently high $n.$ So you do need to take advantage of $n=3.$ Furthermore, the torus has an embedding with $K=0$ in $\mathbb R^4,$ the Clifford torus. The fact that $\mathbb R \mathbb P^2$ and the Klein bottle cannot be embedded at all in $\mathbb R^3,$ just immersed, is a whole new ball of wax. –  Will Jagy Dec 6 '11 at 18:26
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Thank you for the comments. I am still having a hard time seeing why the vanishing of some principal curvature $k_i(q_0) = 0$ guarantees that there are points in $M \cap T_{q_0} M \subseteq \mathbb{R}^3$ arbitrarily close to $q_0$. –  Kaloyan Marinov Dec 6 '11 at 20:26
    
Kaloyan, I edited that sentence within an hour of posting. The correct statement is simply that, with a principal curvature 0, there is a point of the surface outside the sphere. You are correct, it need not lie on or on the wrong side of the tangent plane. But this milder condition is still a contradiction to hypothesis, which was the assumption that $q_0$ was the farthest point of the surface from the origin. –  Will Jagy Dec 7 '11 at 2:36

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