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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?

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The intersection of the bodies is convex, and the intersection of the boundaries is the boundary of the intersection of the (closed) bodies. This is a sphere of whatever dimension it has. –  Robert Israel Jun 14 '12 at 19:36
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The generality condition you give holds for two cubes intersecting in a single edge. (@Robert: The intersection of the boundaries isn't the boundary of the intersection.) –  Richard Kent Jun 14 '12 at 19:44
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Here's another counterexample: Take a big square and a little square that share a corner. Then the intersection of their boundaries will be two adjacent edges of the little square, forming a topological line, not a topological sphere. –  Will Sawin Jun 14 '12 at 19:56
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Or: Take a cube and a sphere with the same center of mass, but the radius of the sphere equal to the distance from the center of the cube to the center of an edge. Then the intersection of the boundary will be a union of 6 circles on the surface of the cube, but the circles will not be disjoint - they will be tangent to each other, forming a fairly complex graph. –  Will Sawin Jun 14 '12 at 20:00
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A topologically transverse counterexample: $x^2+y^2\leq 1$ and $z^2+w^2\leq 1$. Intersection of boundaries is a torus. –  Will Sawin Jun 14 '12 at 20:03
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4 Answers

up vote 7 down vote accepted

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.

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thanks, the real condition that I wanted was that the set of families satisfying the property is an open and dense set in the space of families of convex sets with a some natural topology (say the Hausdorff distance on each body) I think that this counterexample is generic in this sense. –  Alfredo Hubard Jun 14 '12 at 20:34
    
Yes, I believe so. –  Will Sawin Jun 14 '12 at 20:38
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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.

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It is possible to do this example by just taking $K$ to be the convex hull of $X$, that is, the intersection of all closed, convex subsets of $R^{n+1}$ containing $X$. The set $K$ contains no points outside of $B_n$, because $B_n$ is itself a convex set containing $K$. Also $K$ contains no point $p \in S^n - X$, because if $P$ is the tangent hyperplane of $p$, and then if you move $P$ parallel to itself, just a tiny little bit, inward towards the center of the sphere, then one of the half-spaces bounded by $P$ is a convex set containing $X$ and is disjoint from $p$. –  Lee Mosher Jun 15 '12 at 3:20
    
oh wow. I missed that, thanks. –  Alfredo Hubard Jun 15 '12 at 20:39
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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.

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I wonder if this idea still needs some sort of further refinement, taking into account Will's counterexample. –  Patricia Hersh Jun 15 '12 at 13:29
    
Ups, the inequality has to be strict; now it is corrected. –  Anton Petrunin Jun 15 '12 at 13:39
    
that sounds nice, why? –  Alfredo Hubard Jun 15 '12 at 20:40
    
I do not know "why it sounds nice"; the proof is in the answer now. –  Anton Petrunin Jun 16 '12 at 11:27
    
that's nice, thanks. –  Alfredo Hubard Jul 8 '12 at 17:52
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Here is a well known counterexample.

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

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Isn't the intersection of the boundaries of the (2-dimensional) triangles a union of three 0-spheres in this case? –  Kevin Walker Jun 15 '12 at 3:34
    
Too bad MathOverflow doesn't have style points. –  Patricia Hersh Jun 17 '12 at 23:54
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