At present the process isn’t that slick. What I’m doing is the analogue of normal surface theory but for triangulated 4-manifolds. So start off with a complicated triangulation of the 4-sphere, enumerate the vertex-normal 3-manifolds (normal meaning “looks linear” in every simplex, like in the Haken algorithm), and apply geometrization to the discovered 3-manifolds. I’d like to find a proper “reason” for these 3-manifolds to embed. At present it’s just an empirical observation. Currently I have a cluster of about 100 processors doing this kind of normal 3-manifold enumeration. It’s taking a while to figure out how to tune the search to discover new 3-manifolds. For the past few weeks nothing new has shown up…

]]>FYI, I’m getting some examples that are clearly beyond the Mazur manifold level of complexity. The union of a trefoil complement and a (3,5)-torus knot complement along their torus boundaries, but glued together in such a way that the homology of the manifold is a direct sum of two copies of Z/2Z. Embeddings of these kinds of manifolds.

I’m finding these manifolds by taking triangulations of the 4-sphere and enumerating “normal” 3-manifolds in those triangulations then applying geometrization to identify the manifold (whenever I can).

I’m still not finding any “big” 3-manifolds in S^4 via this technique — where “big” would mean, say, H_1 being a direct sum of two copies of Z/7Z or something like that. I hope to correct that soon.

I’m looking forward to showing you the results in detail sometime soon.

]]>Other examples arise from cyclic branched covers of doubly slice knots. This means that the knot is the boundary of two disks in the 4-ball, and those disks fit together to give an unknotted 2-sphere in the 4-sphere. If you can arrange (as is often the case) that the two slice disks are isotopic, then you get the symmetric decomposition that you are asking about. Finally, if M^3 is any manifold that is a n-fold cyclic branched cover of a knot K, then M # -M splits S^4 in a symmetric way. To see this, look at the n-twist spin of K; it is a fibered knot with fiber M. So if you take an embedded copy of M x I as a neighborhood of the fiber, the complement in S^4 is also M x I.

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