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Triangulating knot complements in 4D

Cappell-Shaneson knot complement

Another picture from our wonderful blackboard artist Ryan Budney in Victoria, Canada. He’s taken a 4-dimensional sphere and removed a knotted 2-dimensional sphere (i.e. the surface of a beach ball), and tried to triangulate the resulting space. Triangulation means breaking up a complicated space into simpler pieces – triangles! – and then explaining how these pieces fit together. The triangulation here has precisely one vertex (a 0-dimensional triangle), one edge (a 1-dimensional triangle), 4 triangles (i.e. your usual garden-variety triangles), 5 tetrahedra (3D triangles) and the two pentachora (4-dimensional triangles).  The bottom picture shows how the pieces are glued together. The top picture is completely unrelated.

The knotted 2-dimensional sphere is called a Cappell-Shaneson knot because it is sitting inside a ‘homotopy 4-sphere’. That is, the space the knot is sitting in looks like a normal sphere from a distance but might turn out not to be upon closer inspection. If it isn’t, then it would provide a counterexample to the Poincaré Conjecture, which is still open in 4 dimensions (but was famously solved for 3 dimensions by Grigori Perelman in 2003).

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6 responses

  1. I was under the assumption that Freedman solved the Poincare Conjecture in 4D back in the 80’s? I know he opened another question though, known as the Smooth Poincare Conjecture. Are these considered the same conjecture? I thought not…

    November 21, 2011 at 5:39 pm

    • haggisthesheep

      You’re right on both counts! Freedman solved what is called the ‘topological’ version of the Poincaré conjecture in dimension 4, where the manifolds don’t come with any structure at all. The PL (“piecewise linear”) and the “smooth” cases are still open in dimension 4. These ask whether manifolds which look like the sphere with some extra structure, are actually a sphere. In the topological case the manifolds can wiggle all over the place and display weird behaviour, but in the smooth case they have to look locally smooth. And the kind of moves you’re allowed to do to make it look like a sphere have to preserve this smoothness.

      In fact, the smooth Poincaré conjecture is also unknown in dimensions above 6 (with the exception of dimensions 12 and 61). A nice recent article about the open cases is http://www.maths.ed.ac.uk/~s1142600/books/hedrick2.pdf.

      November 22, 2011 at 11:56 am

  2. haggis, I think perhaps you’re thinking about something else. The answer to the smooth Poincare conjecture is pretty much known in all dimensions other than 4. For example, there’s only one smooth homotopy 5-sphere, the standard sphere. The number of smooth homotopy n-spheres up to orientation-preserving diffeomorphism is the order of the group usually called “theta_n”. This page has a list of the orders: http://en.wikipedia.org/wiki/Exotic_sphere Technically “theta_n” denotes a slightly different notion, but in dimensions 5 and up it agrees with the group of homotopy n-spheres, where the group operation is connect-sum. The page I link to there describes “theta_n” as the group of h-cobordism classes of oriented homotopy n-spheres, which may be a slightly weaker notion in dimension 4 (but this is as of yet unknown).

    November 22, 2011 at 7:20 pm

    • haggisthesheep

      Hmm..I apologise for any wrong information. Clearly I don’t quite understand the subtleties of this! Is it related to the Kervaire invariant, which is still unknown in dim 126? What does this mean for the smooth Poincare conjecture?

      November 24, 2011 at 10:57 am

  3. Hmm, maybe I should clarify. The number of distinct smooth structures (up to diffeomorphism) on the n-sphere is not known exactly in a “closed form” so I imagine that’s the sense you mean the smooth Poincare conjecture is still an open problem in high dimensions. But that’s kind of like saying the homotopy groups of the spheres are not known. It depends on what type of knowledge you demand. The homotopy groups of spheres are algorithmically computable, they’re known to be finite abelian groups, and depending on precisely which one you’re interested in, variable amounts of information is available, some more ready than others. In particular, lots of them are non-trivial, and many are known exactly. The same largely goes for the groups of exotic spheres. In fact, Milnor, Kervaire and Browder largely reduced the study of the groups of exotic spheres to homotopy groups of spheres. So in a similar sense, the groups of exotic spheres are known, but they’re just hard to compute. The Kervaire invariant problem has to do with this connection between groups of exotic spheres and homotopy groups of spheres — the connection says they’re very similar objects but depending on the answer to this problem, the connection might differ by a factor of 2.

    November 30, 2011 at 5:24 am

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