2
$\begingroup$

Let $M$ be a smooth surface and let $x,y \in M$. Let $d_{M}(\cdot, \cdot)$ be the geodesic distance metric on $M$, that is the length of the shortest geodesic curve on $M$. Let $\kappa$ be the maximum principle curvature of all points on $M$ along a minimizing geodesic connecting $x$ and $y$. Let $n_{x}$ and $n_{y}$ be the normal vectors at $x$ and $y$. It is well known result in differential geometry that $\angle(n_x, n_y) \leq \kappa d_{M}(x,y)$.

My question is does an analogous statement hold for higher dimensional manifolds, in particular in the case where the codimension is greater than 1. In the case where codimension is greater than 1, I'm asking if the angle between the normal spaces of two points $x$ and $y$ can be bounded similarly.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

The maximum principle curvature is the upper bound for the Lipschitz constant of the Gauss map at the point. Integrating, you get your estimate.

The same can be done in higher dimensions. If $M$ is $m$-dimensional submanifold in $\mathbb R^n$, you have Gauss map $\nu\colon M\to \mathrm{Gr}(m,n)$, where $\mathrm{Gr}(m,n)$ is the Grassmannian of $m$-subspaces of $\mathbb R^n$. The Lipschitz constant of $\nu$ at given point $p$ can be calculated in terms of second fundamental form* $s$: $$\mathrm{lip}_p\nu=\sup_{u\in T_p, |u|=1}\left\{\sqrt{\sum_i|s(e_i,u)|^2}\right\},$$ where $e_i$ is an orthonormal basis in in the tangent space $T_p$.

(*)The second fundamental form $s$ is a quadratic form on the tangent bundle with values in the normal bundle defined as $$s(v,w)=(\nabla_v w)^\bot,$$ where $(\nabla_v w)^\bot$ denotes the orthogonal projection of covariant derivative $\nabla_v w$ onto the normal bundle.

Improve this answer
$\endgroup$
7
  • $\begingroup$ This is great, thank you for your answer! An additional question. For 2-manifolds with codimension > 1 is the Lipschitz constant still bounded by the maximum principle curvature? And then for m-manifolds in $\mathbb{R}^n$ is the Lipschitz constant bounded by some function of the sectional curvature? $\endgroup$
    – user62013
    Nov 23, 2014 at 20:30
  • $\begingroup$ @user62013 Well, the "principle curvatures" are not defined if codimension $>1$. $\endgroup$
    – Anton Petrunin
    Nov 24, 2014 at 4:21
  • $\begingroup$ The Gaussian curvature is still defined. Is there a general bound on the Lipschitz constant in terms of any type of curvature regardless of the dimension and codimension? $\endgroup$
    – user62013
    Nov 24, 2014 at 5:09
  • $\begingroup$ While the principle curvatures may not be defined, it's not immediately obvious that one could not define something similar in the general case. If I consider a surface $M$ in $\mathbb{R}^3$ then principal curvature is defined at a point in terms the intersection with a plane which gives a curve $\gamma$. If I embed $M$ in a higher dimension it still has the same curvature at that point and still has the same $\gamma$ intersecting at $p$. Perhaps one can't intersect it with a plane but perhaps a hyperplane. $\endgroup$
    – user62013
    Nov 24, 2014 at 5:15
  • $\begingroup$ OK, you want to bound the Lipschitz constant of the Gauss map in terms of normal curvatures. This is a question in linear algebra. Set $q(x,y,z,w)=\langle s(x,y),s(z,w)\rangle$ then you get all the values $q(x,x,x,x)$ and you need to estimate $\sum q(u,e_i,u,e_i)$, but I am too lazy. $\endgroup$
    – Anton Petrunin
    Nov 24, 2014 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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