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Let $K\subset \mathbb{R}^n$ be a compact convex set of full dimension. Assume that $0\in \partial K$.

Question 1. Is it true that there exists $\varepsilon_0>0$ such that for any $0<\varepsilon <\varepsilon_0$ the intersection $K\cap \varepsilon S^{n-1}$ is contractible? Here $\varepsilon S^{n-1}$ is the unit sphere centered at 0 of radius $\varepsilon$.

If Question 1 has a positive answer I would like to generalize it a little bit. Under the above assumptions, assume in addition that a sequence $\{K_i\}$ of compact convex sets converges in the Hausdorff metric to $K$.

Question 2. Is it true that there exists $\varepsilon_0>0$ such that for any $0<\varepsilon <\varepsilon_0$ the intersection $K_i\cap \varepsilon S^{n-1}$ is contractible for $i>i(\varepsilon)$?

A reference would be helpful.

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  • $\begingroup$ Wlog you can assume that $K$ lies in the upper half plane. Isn't it clear that $\partial K\cap \epsilon S^{n-1}$ is the graph of a convex function for $\epsilon$ small enough, and doesn't this imply that you can contract along radial lines? $\endgroup$ – Dirk Jan 4 '16 at 10:09
  • $\begingroup$ In Question 2 is $0\in\partial K_i$ still assumed ? $\endgroup$ – Pietro Majer Jan 4 '16 at 14:13
  • $\begingroup$ @PietroMajer: No, only $0\in \partial K$. $\endgroup$ – MKO Jan 4 '16 at 14:26
  • $\begingroup$ Q2 seems true, but delicate. If $0$ belongs to a sharp edge of $K$ (say in 3D) , any small ball $\epsilon B$ around $0$ can be cut by a small translation of $K$, making non-contractible the intersection with $\partial (\epsilon B) $. So for sure $i(\epsilon)$ really depends on $\epsilon$. $\endgroup$ – Pietro Majer Jan 4 '16 at 15:09
  • $\begingroup$ Indeed, the 2nd q. is delicate. $\endgroup$ – Włodzimierz Holsztyński Jan 4 '16 at 23:11
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The answer to Question 1 is yes, which is precisely Lemma 3.6 in the paper:

Boundary torsion and convex caps of locally convex surfaces, J. Differential Geom., 105 (2017), 427-486.

Although the lemma is stated and proved for $R^3$, the same proof works in $R^n$. The proof also indicates how to give a positive answer to Question 2.

The proof is elementary and proceeds as follows: a convex surface can be represented locally as the graph of a convex function over a convex domain in a support plane, which we may identify with $R^{n-1}$. Then the upper half of the sphere corresponds to the graph of a concave function over the same convex domain, assuming that the radius is sufficiently small and after we readjust the domain. Now the portion of the surface cut off by the sphere is the set of points where the convex graph lies below the concave graph. It is easy to check, using the standard inequalities for the convex and concave functions, that this portion projects onto a convex domain. Hence it is a disk.

In the setting of Question 2, we have a sequence of convex functions converging to the convex function mentioned in the previous paragraph. So eventually they will be defined over the same convex domain and lie below the concave function corresponding to the hemisphere, whence again we may conclude that the upper hemisphere cuts out a disk from their graphs. So the answer to Question 2 is yes as well.

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Given a convex set $K$ in Euclidean space and a point $O\in\partial K$, there exists an $\epsilon>0$ such that for all $r<\epsilon$ the intersection $S(O,r)\cap K$ is connected.

Thus, let $O\in \partial K$. Call $P\in \partial K$ a critical point if $\langle O-P, X-P\rangle \geq 0$ for all $X\in K$. Note that if $P_1$ and $P_2$ are critical points with $\angle P_1 O P_2 \leq \frac{\pi}{3}$ then by the Pythagorean theorem $\frac{|OP_1|}{|OP_2|}\leq 2$. Therefore by a simple packing argument $$\epsilon=\inf_{P \; critical} |OP|>0.$$ For every $C>0$ smaller than this infimum, we show that the sphere $S_C$ of radius $C$ centered at $O$ has the property that $S_C\cap K$ is connected. Indeed, if points $X,Y$ were in different connected components of $S_C\cap K$, we would connect them by a path in $K$ and then "push out" the path away from $O$ using a flow in $K$, using that fact that the flow can only get stuck at a saddle point of $\partial K$ for the distance function, and a saddle point is necessarily a critical point. The construction of the flow in the absence of such Grove-Shiohama critical points was described in http://link.springer.com/article/10.1007/BF02187719

For example, for an acute triangle in the plane, the optimal $\epsilon$ for a point on one of the sides will be the smaller of the two distances from $O$ to the remaining two sides. Every circle of radius smaller than $\epsilon$ will meet the triangle in a connected arc.

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  • 1
    $\begingroup$ I'd like to see an explicit formulation of your result at the very beginning of your post (otherwise it is hard for me to start concentrating on your answer). Moreover, most likely you have obtained more than just an answer to @sva's Question(s). Thus an explicit formulation of your result would be extra nice. $\endgroup$ – Włodzimierz Holsztyński Jan 10 '16 at 9:08
  • $\begingroup$ Agree with @WłodzimierzHolsztyński. Are you proving question 1 or also 2? $\endgroup$ – MKO Jan 11 '16 at 5:21
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    $\begingroup$ @sva, I provided an affirmative answer to question 1. $\endgroup$ – Mikhail Katz Jan 11 '16 at 8:35

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