Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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

share|improve this answer
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
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

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 at least as big as the Gaussian curvature of $p$ at $S$, which is positive, hence it is itself positive.

share|improve this answer
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! –  lentic catachresis Jan 31 '13 at 16:19
@JasonDeVito: Also, see this related recent question of mine. –  lentic catachresis 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
I think it's true that the curvatures of $M$ and $S$ have the same sign at $p$, but are they actually the same? For example, we could have a smaller sphere sitting inside and tangent to a larger sphere. –  Dtseng Jun 23 at 16:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.