Lower bound on the curvature of the curves on $M$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T22:37:19Z http://mathoverflow.net/feeds/question/51704 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51704/lower-bound-on-the-curvature-of-the-curves-on-m Lower bound on the curvature of the curves on $M$ MINI 2011-01-10T21:38:49Z 2011-01-11T11:06:22Z <p>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.</p> http://mathoverflow.net/questions/51704/lower-bound-on-the-curvature-of-the-curves-on-m/51707#51707 Answer by John Bentin for Lower bound on the curvature of the curves on $M$ John Bentin 2011-01-10T22:37:54Z 2011-01-11T11:06:22Z <p>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 (Answer withdrawn: this of course answers the question with the bounding in the other direction)</p> http://mathoverflow.net/questions/51704/lower-bound-on-the-curvature-of-the-curves-on-m/51748#51748 Answer by Andrei Moroianu for Lower bound on the curvature of the curves on $M$ Andrei Moroianu 2011-01-11T10:30:19Z 2011-01-11T10:30:19Z <p>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.</p>