The limit of edge-midpoint convex polyhedra

Question

   

---

Starting with a convex polyhedron $P_1 \subset \mathbb{R}^3$, replace that with $P_2$, the convex hull of the midpoints of the edges of $P_1$. Continuing this process, we obtain a series of polyhedra approaching a smooth body $B$ (or at least, I think it approaches a smooth body). See above for $P_1,\ldots,P_5$—not to the same scale.

Q1. Is $\lim_{n \to \infty} P_n$ $C^2$-smooth, starting with non-degenerate $P_1$? Or only $C^1$-smooth? Or only $C^0$?

Q2. Does $P_n$ approach an ellipsoid as $n \to \infty$, for every (non-degenerate) starting $P_1$? [My guess: No.]

These are, in some sense, less sophisticated versions of my earlier question, “Derived” polyhedra and polytopes, which focussed on face centroids rather than edge midpoints. But here I am asking specific questions on smoothness and the limit shapes. Still, perhaps @GjergjiZaimi's answer there holds.

The limit objects are almost subdivision surfaces, but not quite.

Answer 1

The limit can be within $\varepsilon$ of any convex polytope:

Start with a polytope $P$ then replace each vertex $v$ by three vertices $v_1,v_2,v_3$ forming a triangle with sides of length (less than) $\varepsilon$ with centroid $v$ and containing no point of $P$ other than $v.$ This gives a polytope $P_1 \supset P$ with three times as many vertices but with all points within $\varepsilon$ of (the closest point of) $P.$ Then as the edge midpoint process is applied to $P_1$ one will get a nested series of polytopes all containing $P.$ $P_1$ will have many faces aside from the tiny triangles and the centroids of these faces will be in the limit body, so the limit is strictly larger than $P.$

I'd still like to know the exact limit for a regular tetrahedron (or irregular, it shouldn't matter) and other simple cases.

---

Here are the first nine stages for $P_1$ the cube (not shown) with vertices $(\pm 2,\pm 2,\pm 2).$ The centroids of the $6$ faces, at distance $\sqrt{4}$ from the origin, will be in the limit body. One is $[2,0,0].$

Then $P_2$ is a cuboctahedron ($14$ faces, $12$ vertices) introducing $8$ new centroids all at distance $\sqrt{16/3}$. One is $[4/3 ,4/3,4/3].$

Then $P_3$ has $26$ faces and $24$ vertices. This introduces $12$ centroids all at distance $\sqrt{9/2}.$ One is $[0,3/2,3/2].$

After this is $P_4$ with $50$ faces and $48$ vertices. The $24$ new centroids are all at distance $\sqrt{147/32}.$ One is $[7/8,7/8,7/4].$

Finally in this picture is $P_5$ with $98$ faces and $96$ vertices. There are $48$ new centroids introduced. Half are like $[0,7/8,15/8]$ at distance $\sqrt{137/32}$ and half are like $[7/8,23/16,23/16]$ at distance $\sqrt{627/128}.$

I'm not sure how much this reveals other than that the limit body is not a sphere. The centroids introduced, which will remain centroids at each stage and are extreme points of the limit body are $6$ at distance $2$ then $8$ at distance$\approx 2.3094$ then $12$ at distance $\approx 2.1213$ then $24$ at distances $\approx 2.1433$ and finally $24$ each at distances $\approx 2.0691,2.2132.$

In general it is non-trivial to find the faces and edges for the convex hull of a set of points (even given that all the points are vertices.) However given the information on $P_i$ it is easy to get $P_{i+1}$: Replace each face by the midpoint polygon and for each vertex find the faces it is on, from each one take an edge joining the midpoints of the two appropriate edges and then use these edges to list these points in a valid order.

Here are $P_i$ for $i=6,7,8,9.$ Each is shown twice. Once centered on a triangle and once not. Notice that the even and odd cases look different. In the top view of $P_8$ it may look as if three triangles and the diamonds meet at a degree $6$ point. However that point is actually a tiny triangle and the "triangles" are trapezoids with a very short side.

The "equator" for $P_9$ looks like this

Answer 2

As Wlodzimierz mentioned, if a plane contains a face of $P_1$, it contains a face of every $P_n$. Now consider a finite set of planes $W_i$ (I think $6$ will do) such that no ellipsoid is tangent to all of them but some convex body has faces in all of them. Then the limiting shape can't be an ellipsoid. For example, this should be the case if the initial polyhedron is a prism with the base pictured here:

Answer 3

I don't have a full answer. I hope that my observation provides some insight and a starting point. Let all vertices have weight $\ 1.\ $ Consider an arbitrary plane $\ W\ $ which contains face $\ F\ $ of $\ P_n\ $ (in particular of $\ P_1).\ $ Then plane $\ W\ $ contains the respective faces of all next generation polyhedrons, and the center of vertices of $\ F,\ $ and of all next generation faces contained in $\ W\ $ is the same point belonging to the respective polyhedrons $\ P_{n+d}\ (d=0\ 1\ \ldots)$. Thus all the mentioned centers belong to the topological boundary of intersection $\ \bigcap_{n=1\ 2\ \ldots} P_n$.

Furthermore (an EDIT session), each such center is the only point of the respective plane $\ W.\ $ This is a very strong indication that the intersection $\ \bigcap_{n=1\ 2\ \ldots} P_n\ $ is strictly convex.

Answer 4

A variant of Robert Israel's argument: Take a convex body $K$. Slice it by planes, to make a polyhedron $K_\epsilon\subset K$ whose Hausdorff distance to $K$ is less than $\epsilon^2$ and the diameter of faces is less than $C\epsilon$. Apply the edge-midpoint process. Then the limit body has many points at distance $<\epsilon^2$ to $\partial K$; one per face. Thus the set of limit bodies is dense in the set of convex bodies. In particular, most limit bodies are not ellipsoids.

Alternately, you can form a polyhedron $K^\epsilon$ containing $K$, arbitrarily close to $K$, by taking finitely many tangent planes. Then again apply the edge-midpoint process to obtain a limit body close to $K$.