Is the intersection of boundaries of convex bodies a topological sphere? - MathOverflow

Is the intersection of boundaries of convex bodies a topological sphere?

2012-06-14T19:13:45Z

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 $i\in I$.

Is it true that $\cap_{i \in I} \partial K_i$ is a disjoint union of topological (or even better PL) spheres?

Answer by Will Sawin for Is the intersection of boundaries of convex bodies a topological sphere?

2012-06-14T20:11:31Z

A counterexample is as follows:

Let $K_1=\{x,y,z,w|x^2+y^2\leq 1\}$ and $K_2=\{x,y,z,w|z^2+w^2\leq 1\}$ . The boundaries are just what you get when you replace the inequality with an equality, so their intersection is a torus, $x^2+y^2=1$, $z^2+w^2=1$.

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.

Answer by André Henriques for Is the intersection of boundaries of convex bodies a topological sphere?

2012-06-14T20:31:19Z

Here's a way of producing any 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 $\partial B^{n+1}=S^n$.

Te second convex body, call it $K$, is constructed as follows.
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 this question. 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.

It is then clear, by construction, that $\partial B^{n+1}\cap \partial K = X$.

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.

Answer by Tom Goodwillie for Is the intersection of boundaries of convex bodies a topological sphere?

2012-06-15T03:03:40Z

Here is a well known counterexample.

Edit: Oh, sorry, a disjoint union of spheres is allowed. I misread the question.

Answer by Anton Petrunin for Is the intersection of boundaries of convex bodies a topological sphere?

2012-06-15T12:52:27Z

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 $> \tfrac{\pi}2$.

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.