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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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    $\begingroup$ 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. $\endgroup$
    – Robert Israel
    Commented Jun 14, 2012 at 19:36
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    $\begingroup$ 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.) $\endgroup$
    – Autumn Kent
    Commented Jun 14, 2012 at 19:44
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    $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Commented Jun 14, 2012 at 19:56
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    $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Commented Jun 14, 2012 at 20:00
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    $\begingroup$ A topologically transverse counterexample: $x^2+y^2\leq 1$ and $z^2+w^2\leq 1$. Intersection of boundaries is a torus. $\endgroup$
    – Will Sawin
    Commented Jun 14, 2012 at 20:03

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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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  • $\begingroup$ 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. $\endgroup$
    – Alfredo Hubard
    Commented Jun 14, 2012 at 20:34
  • $\begingroup$ Yes, I believe so. $\endgroup$
    – Will Sawin
    Commented Jun 14, 2012 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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    $\begingroup$ 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 $p$. $\endgroup$
    – Lee Mosher
    Commented Jun 15, 2012 at 3:20
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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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  • $\begingroup$ I wonder if this idea still needs some sort of further refinement, taking into account Will's counterexample. $\endgroup$
    – Patricia Hersh
    Commented Jun 15, 2012 at 13:29
  • $\begingroup$ Ups, the inequality has to be strict; now it is corrected. $\endgroup$
    – Anton Petrunin
    Commented Jun 15, 2012 at 13:39
  • $\begingroup$ that sounds nice, why? $\endgroup$
    – Alfredo Hubard
    Commented Jun 15, 2012 at 20:40
  • $\begingroup$ I do not know "why it sounds nice"; the proof is in the answer now. $\endgroup$
    – Anton Petrunin
    Commented Jun 16, 2012 at 11:27
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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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  • $\begingroup$ Isn't the intersection of the boundaries of the (2-dimensional) triangles a union of three 0-spheres in this case? $\endgroup$
    – Kevin Walker
    Commented Jun 15, 2012 at 3:34
  • $\begingroup$ Too bad MathOverflow doesn't have style points. $\endgroup$
    – Patricia Hersh
    Commented Jun 17, 2012 at 23:54

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