Consider a $(p,q)$ torus knot $K$ in 3-dimensional Euclidean space $\mathbb R^3$ where $p,q \geq 2$ and $\operatorname{GCD}(p,q)=1$.

Let $\operatorname{Isom}(\mathbb R^3,K)$ be the isometries of $\mathbb R^3$ that preserve $K$.

It's a fairly standard argument using theorems about uniqueness of Seifert fiberings to prove that it's impossible for $\operatorname{Isom}(\mathbb R^3, K)$ to contain subgroups isomorphic to both $\mathbb Z_p$ and $\mathbb Z_q$. Of course, $\operatorname{Isom}(S^3,K)$ can and does for the standard embeddings of torus knots in $S^3$. In some sense the core issue is that when this does happen, the $\mathbb Z_p$ and $\mathbb Z_q$ subgroups of $\operatorname{Isom}(S^3,K)$ have disjoint fixed point sets.

My question: is there a reasonably elementary proof $\operatorname{Isom}(\mathbb R^3, K)$ does not contain subgroups isomorphic to both $\mathbb Z_p$ and $\mathbb Z_q$ that avoid the use of Seifert-fiber space techniques? I'm particularly interested if any "quantum topology" invariants can make this kind of symmetry argument. I thought a little about this, at least I'm not seeing how one could use the Alexander polynomial.

  • $\begingroup$ In counterpoint to your observation, view the torus knot as lying in $S^3$ where $S^3$ is the unit sphere in $\mathbb{C}^2$, with coordinates $(z,w)$ with $|z|^2+|w|^2=1$, and see it as the points on the Clifford torus $|z|=1/sqrt{2}$ so that $z^p=w^q$, and both $\mathbb{Z}_p$ and $\mathbb{Z}_q$ can be realized as isometries of $S^3$ that have the torus knot as an invariant set. Sounds more like a geometry problem than a topology problem. $\endgroup$ – Charlie Frohman Aug 7 '11 at 22:08
  • $\begingroup$ I'm a little confused by your comment -- what are you making a counter-point to? $\endgroup$ – Ryan Budney Aug 7 '11 at 22:27
  • $\begingroup$ By seeing the knot as lying in a slightly different homogenous space, and thinking about isometries there, both the $\mathbb{Z}_p$ and $\mathbb{Z}_q$ groups can be realized as isometries. $\endgroup$ – Charlie Frohman Aug 8 '11 at 0:13
  • $\begingroup$ Notice that my two different families of symmetries generate a copy of $\mathbb{Z}_p\times \mathbb{Z}_q$ in the isometry group of the sphere. However, except for the Klein four group, I don't think such groups can be realized as a subgroup of the isometries of Euclidian space. It your hypotheses were just a little stronger, I'd be done as there is no $(2,2)$ torus knot. $\endgroup$ – Charlie Frohman Aug 8 '11 at 0:40
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    $\begingroup$ I oversimplified, the point is that the $\mathbb{Z}_p\times \mathbb{Z}_q$ I produced has no global fixed point. The classification of finite subgroups of $SO(3)$ can be found in Joe Wolf's "Spaces of Constant Curvature". $\endgroup$ – Charlie Frohman Aug 8 '11 at 2:49

The answer no. Neither quantum invariants nor the Alexander polynomial sees the difference between a knot in the three sphere and a knot in Euclidian three space. In the case of the Alexander polynomial the missing point does not interfere with the first homology of the infinite cyclic cover. In the case of quantum invariants, it is often proved that the coefficients of the vector assigned to the the knot complement by whatever TQFT you are working with are given by evaluations of one of the standard knot polynomials, which are the same for knots in the sphere and knots in Euclidian space.

No matter how you present the torus knot there is a subgroup of the homeomorphism group of the complement of the $(p,q)$-torus knot ( in the sphere) that is isomorphic to $\mathbb{Z}_p\times \mathbb{Z}_q$. So you cannot rule it out with topological invariants that don't distinguish between knots in $\mathbb{R}^3$ and knots in $S^3$.

  • $\begingroup$ In the question I don't explicitly state that the $\mathbb Z_p$ and $\mathbb Z_q$ actions commute. So there's the fussy issue to deal with that perhaps they generate an infinite group. But getting to more the point of my question -- many quantum invariants work with a planar projection, so in some sense are prejudiced towards a "knots in $\mathbb R^3$" perspective, which is why I was hopeful one might be able to extract this kind of symmetry argument. $\endgroup$ – Ryan Budney Aug 8 '11 at 17:57

It seems to me that the isometry group of a $(2,q)$ torus knot can be a dihedral group, thus containing both $\mathbb{Z}_2$ and $\mathbb{Z}_q$. The standard realization (on a torus of revolution, take the curve having the right homotopy class and whose latitude and longitude move at constant speed) should do the trick. It is indeed $q$ symmetric with respect to an axis passing in the hole of the torus, and seems $2$ symmetric (with respect to any axis that is orthogonal to the first one and meets the knot) to me.

Edit: as precised by Ryan, the knot is in fact oriented and we only consider symmetries preserving this orientation.

Further edit: let me give a partial elementary proof under this assumption. Assume that there are a $\mathbb{Z}_p$ and a $\mathbb{Z}_q$ in the symmetry group of the knot. Since they preserve its orientation, they act by translation and therefore their actions on the knot commute. Since the knot is not planar and the symmetries are linear, they must commute globally. Except the ruled out case of $p=q=2$, this implies that the two subgroups of isometries have the same axis. From here the knot can be cut off into $pq$ isometric pieces glued by rotations, and I guess that someone more used to knots than me can produce a contradiction.

  • $\begingroup$ My apologies, I should have mentioned that I want the groups to act on the knot by translations -- preserving the orientations of the knot. $\endgroup$ – Ryan Budney Aug 8 '11 at 17:52
  • $\begingroup$ Your 2nd edit makes Charlie's point about this being a geometry problem much more clear, thanks. I had been thinking of this question rather inefficiently. $\endgroup$ – Ryan Budney Aug 9 '11 at 16:41

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