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I have a question about whether Ryan Budney's question:

Torus knots in Euclidean space -- a symmetry argument

can be extended to links. He asks:

Suppose you have a $(p,q)$ torus knot $K$ in $\mathbb{R}^3$ fixed by a subgroup $G$ of $\operatorname{SO}(3)$. Budney asks (more or less) for a nicer proof that $G$ cannot contain $\mathbb{Z}/p\mathbb{Z}$ and $\mathbb{Z}/q\mathbb{Z}$ as subgroups.

Edit: I think I understand Charlie Frohman's point in the comments to that question now that this follows from knowing the periods of the torus knots AND the structure of the finite subgroups of SO(3), which are very different than the finite subgroups of SO(4).

I'd like to extend this same result to $(p,q)$ torus links: there should be no configuration in $\mathbb{R}^3$ which is both $p$-fold and $q$-fold symmetric. I feel like this sort of thing ``should be known'', but so far my literature search has not turned up anything.

EDIT: I think you could probably do this in a similar way if you knew the periods of the torus links (as opposed to the torus knots). But is \emph{this} known?

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You mean that you want to extend it from $(p,q)$ torus links to arbitrary configurations, not that you want to extend it to torus links, right? How do you plan to rule out $pq$-gon? Or a circle, even, with symmetry group $\mathbb R/\mathbb Z$ containing both $\mathbb Z/p$ and $\mathbb Z/q$ as subgroups? –  Will Sawin Jul 9 '12 at 20:48
    
No, I really mean $(p,q)$ torus links ($(p,q)$ not relatively prime) as opposed to $(p,q)$ torus knots ($(p,q)$ relatively prime). Certainly the circle, for instance, is an example of something with all the $\mathbb{Z}/q\mathbb{Z}$ as subgroups of the symmetry group. For instance, I can find all kinds of references about the periods of torus knots, but nothing (so far) about the periods of torus links. –  Jason Cantarella Jul 10 '12 at 13:22
    
Actually, that's sort of a subquestion, which maybe I should have asked as well: what are the periods of torus links (as opposed to torus knots)? Is it still true that they are the divisors of $p$ and $q$? –  Jason Cantarella Jul 10 '12 at 13:23

1 Answer 1

A pretty nice way to interpet your question is two consider the action of the symmetry group on $\mathbb{S}^3$. This is done for knots in Boileau, Boyer, Cebanu, and Walsh section 3, but we can get enough of it to work to answer your question. The point is that the (p,q) torus link will also admit a symmetry group with an element $a$ of order $m=lcm(p,q)$.

The quotients of $\mathbb{S}^3$ under the action of $a$ are dubbed orbi-lens spaces in the language of the above paper because we can describe them similar to the way that Rolfen describes lens spaces in 9B(III) of Knots and Links. Namely, let $w$ be a primitive $m$-root of unity. We consider $\mathbb{S}^3 =$ { $(z_1,z_2)| z_1,z_2\in \mathbb{C} \mbox{ and } |z_1|^2+|z_2|^2=1$ }. The $a$ can be realized as $a(z_1,z_2)=(w^q z_1,w^p z_2)$.

If $gcd(p,q)=d$, then $a^{p/d}$ will fix the core of a Hopf fibration and $a^{q/d}$ will fix the other core. Also, according the above paper, the quotient has base space a lens space with homology $\mathbb{Z}/d\mathbb{Z}$. We note that the quotient under the action of $a^{m/d}$ will be a lens space and that $p,q \geq 2$ and are not relatively prime so $d\geq 2$ .

Since any finite order element $g$ of $SO(3)$ fixes an axis, the image of $a^{m/d}$ will fix an axis. But the map given by the 1 point compactification of $\mathbb{R}^3$, $f: \mathbb{R}^3 \cup \{\infty\} \rightarrow \mathbb{S}^3$, is 1-1. So we could induce a quotient map on $\mathbb{S}^3$ via the image under $f$ of the points equivalent under $g$ which has base space $\mathbb{S}^3$. That would contradict the above paragraph.

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