Take a point p on the surface, and its normal N(p). Take a plane P containing N(P), and cut the surface with it: you’ll get a curve C. Since every point is umbilic, C is a line of curvature. This means that, all along the curve C, the normals to the surface all belong to the plane P. This implies that any two of them intersect (possibly at infinity). By choosing a different plane P and repeating, you can cover any pair of points on the surface, so that: any pair of normals to the surface intersect (possibly at infinity).

From here it’s easy. Just use the fact that if n lines in projective space intersect pairwise, they either belong to a plane or have a common intersection. This common intersection is the center of our sphere, QED. ]]>

Of course, a huge amount is known about linear representations of Kahler groups and rigidity of the associated equivariant maps to products of symmetric spaces and buildings, by the work of Siu, Corlette, Gromov-Schoen etc. And since you bring up the subject, it is worth mentioning that there are lots of examples of nonlinear Kahler groups. Anyway, I do plan to try to write something about this!

]]>Pierre

]]>Also, to see the point about the osculating sphere being in the interior: take the radius r tubular neighborhood about the umbilic sphere, for r less than the minimal radius of curvature. Then the interior boundary will also be umbilic (since shrinking the osculating sphere diameter by r will preserve the 2nd order contact), and will become singular at the minimal radius of curvature. At this point, the first point of the cut locus will appear at the focal point of the minimal curvature radius, so a sphere centered at this point will be embedded.

]]>If the 4D smooth Poincare conjecture is true, then the Schoenflies conjecture is also true. This follows because if one has a 3-sphere embedded in the 4-sphere which doesn’t bound a standard 4-ball, then cutting along it produces an punctured exotic sphere on either side, since capping off with a ball cannot produce a standard sphere by the observation you made.

On the other hand, if the 4D smooth Poincare conjecture is false, then the set of oriented exotic 4-spheres forms a monoid under connect sum. Then there is an invertible element of this monoid if and only if the Schoenflies conjecture is true. One direction is the essentially the observation above: if the Schoenflies is false, then a non-standard 3-sphere in the 4-sphere exhibits the 4-sphere as a connect sum of inverse exotic 4-spheres. On the other hand, if an exotic 4-sphere is invertible, then puncturing it along a ball embeds in the standard 4-sphere, so is a standard ball by Schoenflies. But this implies that it is glued from two standard balls, so by the Smale conjecture (proved by Hatcher), the manifold is standard.

]]>I think that the following simple fact about the symmetric group makes your use of commutators a bit more transparent. Let g and h be two elements of the symmetric group and let [g,h] be their commutator. When g and h have no common support (that is, they move different elements of the set) it is easy to see that [g,h] = 1. Now suppose that they have a unique element a in their common support. Then their commutator [g,h] is just the three cycle

(a g(a) h(a)). This is as simple an even element as possible (without being

trivial), and gives many of the desired moves on the cube.

Of course, in practice one would discover this at the end, with only two vertices in the top face (apparently) out of place; one could exchange one for the antipodal vertex using the move which permutes three vertex cubelets. So: not *much* harder.

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