Is the intersection of boundaries of convex bodies a topological sphere? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:32:35Z http://mathoverflow.net/feeds/question/99637 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99637/is-the-intersection-of-boundaries-of-convex-bodies-a-topological-sphere Is the intersection of boundaries of convex bodies a topological sphere? Alfredo Hubard 2012-06-14T19:13:45Z 2012-06-16T11:26:04Z <p>Let $K_1, K_2, \ldots K_n$ be convex bodies in $R^d$. Assume that for any index set $I$, $\cap_{i \in I} K_i$ is not empty and is not properly contained in any body $K_i$ for $i\in I$.</p> <p>Is it true that $\cap_{i \in I} \partial K_i$ is a disjoint union of topological (or even better PL) spheres?</p> http://mathoverflow.net/questions/99637/is-the-intersection-of-boundaries-of-convex-bodies-a-topological-sphere/99645#99645 Answer by Will Sawin for Is the intersection of boundaries of convex bodies a topological sphere? Will Sawin 2012-06-14T20:11:31Z 2012-06-14T20:11:31Z <p>A counterexample is as follows:</p> <p>Let <code>$K_1=\{x,y,z,w|x^2+y^2\leq 1\}$</code> and <code>$K_2=\{x,y,z,w|z^2+w^2\leq 1\}$</code>. The boundaries are just what you get when you replace the inequality with an equality, so their intersection is a torus, $x^2+y^2=1$, $z^2+w^2=1$.</p> <p>Then the interiors intersect, for instance at the origin, fulfilling Patricia Hersh's condition. The boundaries are topologically transverse, fulfilling Richard Kent's condition. I cannot think of any additional reasonable condition that would disallow this counterexample.</p> http://mathoverflow.net/questions/99637/is-the-intersection-of-boundaries-of-convex-bodies-a-topological-sphere/99647#99647 Answer by AndrĂ© Henriques for Is the intersection of boundaries of convex bodies a topological sphere? AndrĂ© Henriques 2012-06-14T20:31:19Z 2012-06-14T20:39:48Z <p>Here's a way of producing <em>any</em> compact subset $X$ of $S^n$ as the intersection of the boundaries of two convex bodies in $\mathbb R^{n+1}$. The first convex body is simply the unit ball $B^{n+1}$, with boundary $\partial B^{n+1}=S^n$.</p> <p>Te second convex body, call it $K$, is constructed as follows.<br> Consider a function $f:S^n\to \mathbb R$ such that $f^{-1}(0)=X$. Such functions exist in great abundance, see e.g. the answers to <a href="http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function/24041" rel="nofollow">this question</a>. By carefully selecting $f$ (i.e. by taking it to be small, and with small first and second derivatives), we can make sure that $$K:=\{x\in\mathbb R^{n+1}:\|x\|\le 1+f(x/\|x\|)\}$$ is convex.</p> <p>It is then clear, by construction, that $\partial B^{n+1}\cap \partial K = X$.</p> <p><hr> If $f$ admits both positive and negative values (which can be arranged iff the complement of $X$ in $S^n$ is disconnected), then neither of $B^{n+1}$ or $K$ is contained in the other one.</p> http://mathoverflow.net/questions/99637/is-the-intersection-of-boundaries-of-convex-bodies-a-topological-sphere/99671#99671 Answer by Tom Goodwillie for Is the intersection of boundaries of convex bodies a topological sphere? Tom Goodwillie 2012-06-15T03:03:40Z 2012-06-15T03:41:25Z <p><a href="http://en.wikipedia.org/wiki/Star_of_David" rel="nofollow">Here</a> is a well known counterexample.</p> <p>Edit: Oh, sorry, a disjoint union of spheres is allowed. I misread the question.</p> http://mathoverflow.net/questions/99637/is-the-intersection-of-boundaries-of-convex-bodies-a-topological-sphere/99702#99702 Answer by Anton Petrunin for Is the intersection of boundaries of convex bodies a topological sphere? Anton Petrunin 2012-06-15T12:52:27Z 2012-06-16T11:26:04Z <p>To get a "YES" answer, you have to assume that at any point of $p\in\partial K_i\cap \partial K_j$ any two supporting hyperplanes to $K_i$ and $K_j$ have angle $> \tfrac{\pi}2$.</p> <p>The proof is by induction on $n$. WLOG we may assume that all $\partial K_i$ are smooth. Assume $S_{n-1}=\partial K_1\cap \partial K_2\cap \dots\cap K_{n-1}$ is a sphere. Note that $f=\mathop{\rm dist}_{\partial K_n}$ is a concave function on $S_{n-1}\cap K_n$, Perturb $f$ so it become smooth. The function has one maximum point and by Morse Lemma the level set $S_n=f^{-1}(0)=\partial K_1\cap \partial K_2\cap \dots\cap K_{n}$ is a sphere.</p>