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This is a followup to this question, but I'll try to include all the currently relevant information here.

Background summary: I'm investigating polyhedra in R3 which are in some sense "area-minimizing", in analogy with smooth surfaces of constant mean curvature. In particular, given a particular combinatorial "type" of polyhedron homeomorphic to a sphere, I would like to find the realization(s) of that polyhedron which minimize its surface area for a given volume, if any exist. Given a parametrization of the realization space (space of embeddings) M, it would be useful to define an analogue of mean curvature as the gradient of area.

So, what does M look like? If a combinatorial type is "trihedral", i.e. three faces meet at every vertex, then M has three dimensions for each face. Likewise, for a triangulated polyhedron M has three dimensions for each vertex. However, overall six of these dimensions correspond to translation and rotation without changing the actual shape. The "actual" dimension of M, once we quotient out translations and rotations, turns out to be the number of edges. This is true whether the polyhedron is trihedral, triangulated or neither. Furthermore, the "boundaries" between combinatorial types are defined naturally in terms of edges contracting to points.

So I'm interested in parametrizing the realization space of a polyhedron by its edges, and then defining an analogue of mean curvature in terms of this parametrization. A convex triangulated polyhedron is already completely defined by its edge lengths; but I want something that works more generally.


Now for the actual question.

It seems I may have happened upon (part of) the formula for mean curvature before finding a useful parametrization. It's been known for a long time that, for any given combinatorial type, a minimizer must have all faces tangent to a sphere. This means that the surface can be partitioned naturally into pairs of congruent triangles, whose common base is an edge of the polyhedron and whose apices are the points of tangency to the sphere. The area of such a pair of triangles is h = L*tan(θ/2), where L is the length of the edge and θ is the (external) dihedral angle. For all the minimizers I've checked, this value is constant across all edges, which makes it a good candidate for my "discrete mean curvature", at least when restricting to the subspace S of M which has all faces tangent to a sphere.

Given a parametrization of M, define the mean curvature flow as simply following the gradient. Then intuitively, under the flow I would expect polyhedra which start out tangent to a sphere to stay tangent to a sphere (albeit one which "grows with the flow"). That is, it feels like S should be closed under the flow. Furthermore, under the reverse flow all realizations should approach a minimizer.

With these assumptions, how might it be possible to construct a parametrization of M in terms of edges, having h as the gradient of area (at least on S)?

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2 Answers 2

Too long for a comment anyway.

Around 1994 I was interested in a simpler question. What does a minimal polyhedral surface look like? In particular.

Is it true that area minimizing polyhedral surface is saddle.

It turned out that if you fix triangulation then answer is "NO", but if you only fix the number of vertexes then answer is "YES". In the later case in addition to moving each vertex a bit, you may also decrease the area by doing a flip: remove two triangles, say $[xpq]$ and $[ypq]$ and glue instead two triangles $[pxy]$ and $[qxy]$.

Later I found a very rough way which proved much stronger statement, it ended up in my paper on metric minimizing surfaces. So the argument described above was left on the sidelines, but I think it might be useful for you.

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This is not a direct answer to your question, just a few references. The notion of discrete curvature, and in particular, discrete mean curvature, has been studied quite a bit. Perhaps one of the best, concise sources is John Sullivan's 2007 paper, "Curvatures of Smooth and Discrete Surfaces" (arXiv link, PDF link). In Section 4.5, p.7, he seems to be directly addressing your situation, with the combinatorial type and the volume fixed:

A discrete surface which minimizes area among surfaces of fixed combinatorial type and fixed volume will have constant discrete mean curvature $H$ in the sense that at every vertex, $\\mathbf{H}_p = H \mathbf{A}_p$, or equivalently $\nabla_p\operatorname{Area} = -H \nabla_p \operatorname{Vol}$. In general, of course, the vectors $\mathbf{H}_p$ and $\mathbf{A}_p$ are not even parallel... [etc.]

For other literature, perhaps Konrad Polthier's 1996 paper, "An Algorithm for Discrete Constant Mean Curvature Surfaces," could give you ideas and leads. Google scholar shows that 17 later papers cite that one, including Konrad's own 100-page 2002 document, "Polyhedral surfaces of constant mean curvature" (PDF link).

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Thanks. As it happens, I've been reading through the work of Sullivan, Polthier and others from (mostly) TU-Berlin, but for my purposes they have certain shortcomings. While these authors define mean curvature on an edge, in the context of variational problems they always use the curvature at a vertex (sometimes defined as the sum over the surrounding edges) and assume vertices can be varied freely – that is, the surface is triangulated. The exception is when “CMC” surfaces are defined as dual to (triangulated) spherical minimal surfaces; but in this case, they are no longer area-minimizing. –  Robin Saunders Oct 24 '11 at 23:33
    
In general it is not possible to vary all vertices freely, hence my interest in variations of edges. In order for area-minimizing polyhedra to have constant mean curvature on edges, the mean curvature on the edges of such polyhedra must reduce to the form L*tan(θ/2). As Sullivan acknowledges, there are multiple ways to define mean curvature in the discrete case, depending on what identities one wishes to hold. All of them will be roughly linear in θ for small θ, so that all of the identities hold in the smooth limit where θ → 0 – but for nonzero θ, they are still distinct. –  Robin Saunders Oct 24 '11 at 23:34

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