Low Dimensional Topology

July 12, 2012

Symmetric decompositions of the 4-sphere

Filed under: 3-manifolds,4-manifolds,Knot theory — Ryan Budney @ 4:43 pm

Rob Kusner recently pointed out to me that the 4-sphere has a very natural differential-geometric decomposition as a double mapping cylinder S^3/Q_8 \to \mathbb RP^2. Here Q_8 is the group \{\pm 1, \pm i, \pm j, \pm k\} in the unit quaternions and \mathbb RP^2 is the real projective plane. Another way to say this is take the Voronese projective plane in S^4, a regular neighbourhood of it is a mapping cylinder S^3/Q_8 \to \mathbb RP^2. Moreover, the *complement* of that regular neighbourhood is another such mapping cylinder.

This decomposition comes about via an elementary linear algebra argument. Consider the space of traceless 3\times 3 symmetric real matrices. SO_3 acts on this vector space (by conjugation), and so it acts on the unit sphere in that vector space, which is S^4. The orbit-type decomposition of this action is the above double mapping cylinder construction. It’s a remarkably symmetric decomposition of the 4-sphere.

Some other similarly symmetric-looking decompositions of the 4-sphere have been coming up in some computations I’ve been working on recently. Take a knot in the 3-sphere. The complement is a homology S^1 \times D^2. So if you glue two knot complements together, longitude to meridian, you get a homology sphere. It appears that many of these homology spheres admit smooth embeddings into S^4, separating S^4 in a somewhat (maybe very) symmetric way. I don’t actually know what these decompositions are yet since I have only a moderately constructive embedding of these manifolds into S^4. But it appears to be cutting S^4 in a fairly symmetric way. The 4-manifolds that these manifolds separate S^4 into are quite “small” so it would be difficult for the decomposition not to be symmetric.

Do other people have any symmetric decompositions of the 4-sphere?


  1. Perhaps this is more trivial than you meant, but most (and all known) examples of contractible 4-manifolds with homology sphere boundaries have the property that their doubles are S^4. For example, this holds for Mazur’s original examples, and for the generalizations studied by Akbulut and Kirby. How general this phenomenon is runs into the Andrews-Curtis conjecture; if your contractible manifold is built with 0, 1, and 2-handles, then its double is the boundary of a 5-manifold built out of the same list of handles. If the A-C conjecture holds, then this 5-manifold is the 5-ball, and its boundary obviously the 4-sphere. I think this was the original context for that conjecture, in fact. Many of the Mazur-type examples have interesting diffeomorphisms (eg involutions) on their boundaries, and the twisted doubles (glue via such a diffeomorphism) tends also to be S^4. But this tends to be harder to show.

    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.

    Comment by Daniel Ruberman — July 28, 2012 @ 6:43 am | Reply

  2. Sorry; left out a line in the second paragraph, regarding the doubly-slice knots. The symmetric decomposition arises from taking a cyclic branched cover of that unknotted 2-sphere, which decomposes as the union of two copies of the branched cover of the two disks in their respective 4-balls.

    Comment by Daniel Ruberman — July 28, 2012 @ 6:47 am | Reply

  3. Hi Danny,

    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.

    Comment by Ryan Budney — July 28, 2012 @ 2:03 pm | Reply

  4. Hi Ryan–That sounds interesting and surprising. Do you know how to start with that gluing of knot complements and end up with the symmetric embedding? (Or any embedding for that matter.) I look forward to hearing more.

    Comment by Daniel Ruberman — July 29, 2012 @ 6:38 am | Reply

    • 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…

      Comment by Ryan Budney — September 19, 2012 @ 1:01 am | Reply

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