On 9/1/09, Michael Kleber <michael.kleber@gmail.com> wrote:
... But unfortunately the "cusp" branch doesn't actually keep working; the signs turn out to be wrong. Those right angles Fred mentions are exactly the reason: at that point the angles around the degree-6 vertex can be partitioned into two parts (2 and 4 angles) that each sum to pi. So while the intersection point happens to satisfy the angular constraint with all + signs, continuity gets you bupkis.
With Fred's exercise out of the way, ...
Well, yes, that certainly is what happens in detail to the polyhedron! But it doesn't actually answer the question I had in mind --- why doesn't the continuity argument work? Assuming g fixed say, both plane curves defined by the vanishing of the individual quartic factors are analytic around the right-angled / crossover point in the (q,h) plane. But the big polynomial (f_3 or f_4) has a double point there: so it's _not_ analytic, and we cannot rely on its continuity there! If we separate out the quartics and argue each one individually, continuity works just fine. In fact, it works so well that I now realise we didn't actually need those great big sin^2 polynomials at all, or indeed any angles --- plain old Pythagoras does the trick! New drastic rewrites loom ... can this saga get any more embarrassing?
what I really wanted to ask was about this other fact:
There are two classes of polytore corner, "U/V" with 5 faces and "W" with 6. ... The fact that f_5 and f_6 both turn out to be 0 at the same time is at first glance not surprising: Gauss-Bonnet says the total angular defect is 0, so if either type of vertex is flat, the other must be too. But the "cusp" branch *isn't* guaranteeing flatness, we now know; it's instead guaranteeing a 0 of the angle sum with some other combination of signs.
So why on earth does this cusp branch show up as a factor of both f_5 and f_6? What else is forcing these to be simultaneously zero?
(Pity it happens, whatever it is -- otherwise we could eliminate stray branches by taking the gcd of f_5 and f_6!)
I've puzzled over this a good deal. By angle chasing through the triangular faces (without assuming isometric embedding) --- in the same way that Gauss- Bonnet is proved --- it's easy to show that a number of signed angle-sums share the property of holding simultaneously at both types of corner, which sort of accounts for the phenomenon. Otherwise, it almost looks as if there might be another angular invariant lurking in there somewhere, analogous to the Gauss-Bonnet / Euler characteristic. Computing these polynomials for some other polyhedra might be interesting --- if somebody can propose any promising candidates! By the way, I didn'tt actually compute f_5 and f_6 at these corners, but used symmetry and right-angles to get away with f_3 and f_4. The bigger polynomials might well have more common factors --- but we will probably never know! Fred Lunnon