Take the 2-minute tour ×

Let $M$ be an $n$-dimensional hypersurface in $\mathbb R^{n+1}$, such that principal curvatures are bounded from below by a constant $\delta$. Is there any lower bound on the curvature of the curves on $M$? Curves should be intersection of a two plane and the manifold.

share|improve this question
This seems more appropriate for math.stackexchange.com to me. –  Deane Yang Jan 10 '11 at 23:42

2 Answers 2

up vote 4 down vote accepted

Let $P$ be a 2-plane in $\mathbb R^{n+1}$ and $\gamma=P\cap M$. Choose a parametrisation of $\gamma$ by arc-length. Then the curvature of $\gamma$ is $K=||\nabla_{\dot\gamma}\dot\gamma||$, where $\nabla$ is the flat covariant derivative in $P$ (or in $\mathbb R^{n+1}$ as $P$ is totally geodesic). Now, the Levi-Civita covariant derivative $\nabla^g$ of $(M,g)$ is related to $\nabla$ by $$\nabla_XY=\nabla^g_XY+h(X,Y)\xi,$$ where $h$ is the second fundamental form and $\xi$ is a unit normal vector field. Using your assumption $h(X,X)\ge \delta^2 g(X,X)$ for every $X$ tangent to $M$, one gets $$K=||\nabla_{\dot\gamma}\dot\gamma||=\sqrt{||\nabla^g_{\dot\gamma}\dot\gamma||^2+h(\dot\gamma,\dot\gamma)}\ge\sqrt{h(\dot\gamma,\dot\gamma)}\ge \delta.$$ This proves your claim.

share|improve this answer
Yeah, what he said! –  drbobmeister Jan 12 '11 at 0:55

No. For example, on a 2-sphere in $\mathbb{R}^3$, you can draw a circle as small as you like. (Answer withdrawn: this of course answers the question with the bounding in the other direction)

share|improve this answer
The circle is the intersection of a plane and the sphere, so the answer still holds under the assumption. –  John Bentin Jan 10 '11 at 22:57
I am completely confused. Don't small circles have very large curvature? In what sense are they a counterexample to the question? –  Deane Yang Jan 10 '11 at 23:40
@Deane: I don't think you are quite as confused as the OP. –  Igor Rivin Jan 11 '11 at 0:42
@drbobmeister: Correct. The answer is, in fact "Yes" (if $\delta > 0,$ otherwise it is "No") and this looks like a homework problem. –  Igor Rivin Jan 11 '11 at 1:44
I agree that it looks like homework. –  Deane Yang Jan 11 '11 at 2:33

Your Answer


