Let $K \subset S^3$ be a knot which is both strongly invertible and periodic, that is, $K$ is fixed by both a smooth involution $\tau: S^3 \to S^3$ which preserves the orientation of $S^3$ but reverses the orientation of $K$, and $K$ is fixed by a finite order diffeomorphism $\rho: S^3 \to S^3$ which preserves the orientation of both $S^3$ and $K$ and which has a non-empty fixed-point set.

Many such knots exist, consider for example the figure-eight knot:

     figure-eight knot[source]

In this diagram an order 2 period can be seen by $\pi$ rotation around a line perpendicular to the diagram, and a strong inversion can be seen by $\pi$ rotation around a vertical line. $9_{40}$ is a 3-periodic example:


By the resolution of the Smith conjecture, a periodic or strongly invertible symmetry will have an unknotted fixed set. In these diagrams for $4_1$ and $9_{40}$, the fixed set of the period is an axis perpendicular to the diagram and the fixed set of the strong inversion is an axis contained in the plane of the diagram so that these axes intersect at the origin and infinity. Does every knot which is both strongly invertible and periodic have such a diagram?

  • $\begingroup$ Presumably the answer is yes. By geometrization you can realize the symmetries as linear isometries of $S^3$. For your diagram you use stereographic projection at a point on the fixed-point set, using lines parallel to the fixed point set for the "projection direction" of your diagram. You need a little general position argument to say the knot can be equivariantly perturbed to be in "regular diagram position" with respect to this projection direction. $\endgroup$ Feb 2 at 3:20
  • $\begingroup$ I guess I'm worried that the axis of the strong inversion might be disjoint from the axis of the period. Is there some way to guarantee that they intersect? $\endgroup$ Feb 2 at 3:23
  • $\begingroup$ Yes, compute all the dihedral subgroups of $SO_4$. Basically you do diagonalization, and you can see exactly what axis are possible. This was done by Hopf a long time ago. The main ideas are available in Thurston's book on 3-manifolds, in the classification of elliptic manifolds. Does this make some sense? $\endgroup$ Feb 2 at 3:29
  • $\begingroup$ Ah okay I think so, (I will check the details). Just to check that I understand: the idea is that a period and strong inversion are simultaneously conjugate to the generators of a dihedral subgroup of SO(4) by geometrization, and there is a unique conjugacy class in SO(4) of the dihedral group of each order giving the desired diagram (after an appropriate isotopy)? $\endgroup$ Feb 2 at 3:39
  • $\begingroup$ Correct. I suppose the way to think about is to think of the pair $(S^3,G)$ where $G$ is the symmetry group of the knot. That gives you the quotient orbifold $S^3/G$ which you apply geometrization to. That gives you an isotopy from your knot (and its original position) to one where the symmetry group acts linearly. $\endgroup$ Feb 2 at 4:11

Much like your cited solution to the Smith conjecture works, the solution to the Geometrization Conjecture gives you (nearly) the full solution to your problem.

Let $G$ be your dihedral subgroup of $Diff(S^3,k)$. These diffeomorphisms preserve the orientation of $S^3$ by assumption.

So $S^3 / G$ is an orbifold, so orbifold geometrization tells us that this is a geometric (spherical) manifold. This orbifold equivalence tells us there is an isotopy of $G$ that moves our knot from its original position $K$ to a new position $K'$ such that $(S^3, K')$ has a dihedral group of isometries.

Being a dihedral group there is a Hopf link of "axis". We do stereographic projection at a point on one of these axis, this puts our knot into $\mathbb R^3$. You then need an equivariant general position argument to state that if you project in the plane orthogonal to the axis, this is generically a knot diagram.

That's a sketch of how to one might assemble an argument.


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