Large-scale Pareto-optimal topologies, or: how to describe a hexahedron

I got to meet Karen Caswelch, the CEO of Madison startup SciArtSoft last week. The company is based on tech developed by my colleague Krishnan Suresh. When I looked at one of his papers about this stuff I was happy to find there was a lovely piece of classical solid geometry hidden in it!

Here’s the deal. You want to build some component out of metal, which metal is to be contained in a solid block. So you can think of the problem as: you start with a region V in R^3, and your component is going to be some subregion W in R^3. For each choice of W there’s some measure of “compliance” which you want to minimize; maybe it’s fragility, maybe it’s flexibility, I dunno, depends on the problem. (Sidenote: I think lay English speakers would want “compliance” to refer to something you’d like to maximize, but I’m told this usage is standard in engineering.) (Subsidenote: I looked into this and now I get it — compliance literally refers to flexibility; it is the inverse of stiffness, just like in the lay sense. If you’re a doctor you want your patient to comply to their medication schedule, thus bending to outside pressure, but bending to outside pressure is precisely what you do not want your metal widget to do.)

So you want to minimize compliance, but you also want to minimize the weight of your component, which means you want vol(W) to be as small as possible. These goals are in conflict. Little lacy structures are highly compliant.

It turns out you can estimate compliance by breaking W up into a bunch of little hexahedral regions, computing compliance on each one, and summing. For reasons beyond my knowledge you definitely don’t want to restrict to chopping uniformly into cubes. So a priori you have millions and millions of differently shaped hexahedra. And part of the source of Suresh’s speedup is to gather these into approximate congruence classes so you can do a compliance computation for a whole bunch of nearly congruent hexahedra at once. And here’s where the solid geometry comes in; an old theorem of Cauchy tells you that if you know what a convex polyhedron’s 1-skeleton looks like as a graph, and you know the congruence classes of all the faces, you know the polyhedron up to rigid motion. In partiuclar, you can just triangulate each face of the hexahedron with a diagonal, and record the congruence class by 18 numbers, which you can then record in a hash table. You sort the hashes and then you can instantly see your equivalence classes of hexahedra.

(Related: the edge lengths of a tetrahedron determine its volume but the areas of the faces don’t.)

Smectic crystals, fingerprints, Klein bottles

Amazing colloquium this week by Randall Kamien, who talked about this paper with Chen and Alexander, this one with Liarte, Bierbaum, Mosna, and Sethna, and other stuff besides.

I’ve been thinking about his talk all weekend and I’m just going to write down a bit about what I learned.  In a liquid crystal, the molecules are like little rods; they have an orientation and nearby molecules want to have nearby orientations.  In a nematic crystal, that’s all that’s going on — the state of the crystal in some region B is given by a line field on B.   A smectic crystal has a little more to it — here, the rods are aligned into layers

(image via this handy guide to liquid crystal phases)

separated by — OK, I’m not totally clear on whether they’re separated by a sheet of something else or whether that’s just empty space.  Doesn’t matter.  The point is, this allows you to tell a really interesting topological story.  Let’s focus on a smectic crystal in a simply connected planar region B.   At every point of B, you have, locally, a structure that looks like a family of parallel lines in the plane, each pair of lines separated by a unit distance.  (The unit is the length of the molecule, I think.)

Alternatively, you can think of such a “local smectic structure” as a line in the plane, where we consider two lines equivalent if they are parallel and the distance between them is an integer.  What’s the moduli space M — the “ground state manifold” — of such structures?    Well, the line family has a direction, so you get a map from M to S^1.  The lines in a given direction are parametrized by a line, and the equivalence relation mods out by the action of a lattice, so the fiber of M -> S^1 is a circle; in fact, it’s not hard to see that this surface M is a Klein bottle.

Of course this map might be pretty simple.  If B is the whole plane, you can just choose a family of parallel lines on B, which corresponds to the constant map.  Or you can cover the plane with concentric circles; the common center doesn’t have a smectic structure, and is a defect, but you can map B = R^2 – 0 to M.  Homotopically, this just gives you a path in M, i.e. an element of pi_1(M), which is a semidirect product of Z by Z, with presentation

The concentric circle smectic corresponds the map which sends the generator of pi_1(B) to F.

So already this gives you a nice topological invariant of a plane smectic with k defects; you get a map from pi_1(B), which is a free group on k generators, to pi_1(M).  Note also that there’s a natural notion of equivalence on these maps; you can “stir” the smectic, which is to say, you can apply a diffeomorphism of the punctured surface, which acts by precomposition on pi_1(B).  The action of (the connected components of) Diff(B) on Hom(pi_1(B), pi_1(M)) is my favorite thing; the Hurwitz action of a mapping class group on the space of covers of a Riemann surface!  In particular I think the goal expressed in Chen et al’s paper of “extending our work to the topology of such patterns on surfaces of nontrivial topology (rather than just the plane)” will certainly involve this story.  I think in this case the Hurwitz orbits are pretty big; i.e. if what you know is the local appearance of the defects (i.e. the image in pi_1(M) of the conjugacy class in pi_1(B) corresponding to the puncture) you should almost be able to reconstruct the homotopy type of the map (up to stirring.)  If I understood Randy correctly, those conjugacy classes are precisely what you can actually measure in an experiment.

There’s more, though — a lot more!  You can’t just choose a map from B to M and make a smectic out of it.  The layers won’t line up!  There’s a differential criterion.  This isn’t quite the way they express it, but I think it amounts to the following:  the tangent bundle of M has a natural line bundle L sitting inside it, consisting of those directions of motion that move a line parallel to itself.  I think you want to consider only those maps from B to M such that the induced map on tangent bundles TB -> TM takes image in L.  More concretely, in coordinates, I think this means the following:  if you think of the local smectic structure at p as the preimage of Z under some real-valued function f in the neighborhood of p, then f should satisfy

This restricts your maps a lot, and it accounts for all kinds of remarkable behavior.  For one thing, it forbids certain conjugacy classes in pi_1(M) from appearing as local monodromy; i.e. the set of possible defect types is strictly smaller than the set of conjugacy classes in pi_1(M).  Moreover, it forbids certain kinds of defects from colliding and coalescing — for algebraic geometers, this naturally makes you feel like there’s a question about boundaries of Hurwitz spaces floating around.

Best of all, the differential equation forces the appearance of families of parallel ellipses, involute spirals, and other plane curves of an 18th century flavor.  The cyclides of Dupin put in an appearance.  Not just in the abstract — in actual liquid crystals!  There are pictures!  This is great stuff.  

Update:  Wait a minute — I forgot to say anything about fingerprints!  Maybe because I don’t have anything to say at the moment.  Except that the lines of a fingerprint are formally a lot like the lines of a smectic crystal, the defects can be analyzed in roughly the same way, etc.  Whether the diffeomorphism type of a fingerprint is an interesting forensic invariant I don’t rightly know.  I’ll bet whoever made my iPhone home button knows, though.