Let $M$ be a connected $3$-manifold and let $\alpha$ and $\beta$ be elements in $\pi_1(M)$. Then $\alpha$ and $\beta$ can be represented by two knots $a$ and $b$ in $M$. We may further require that the images of $a$ and $b$ are disjoint. My question is:

Is there any necessary/sufficient condition on the knots $a$ and $b$ (or for the link $a\bigcup b$) for $\alpha$ and $\beta$ to commute in $\pi_1(M)$?

I believe that there are conclusions (what conclusions?) for satellite knots. Any comment or reference will be greatly appreciated.

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    $\begingroup$ I find strange that you can take them to be disjoint since in order to talk about the fundamental group you have to choose a base point where every path starts and ends. $\endgroup$ – Fernando Muro Sep 6 '13 at 4:59
  • $\begingroup$ You are right. I want to look at the link $a\bigcup b$ and as $a$ and $b$ necessarliy intersect at the base point, we can actually have two ways to create the link $a\bigcup b$ if we shift $b$, say, a little bit from the basepoint so that $a$ and $b$ do not intersect. Or perhaps it is better to simply consider $a\bigcup b$ as a virtual knot? $\endgroup$ – Zuriel Sep 6 '13 at 5:08
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    $\begingroup$ The question still makes no sense to me: Every element of $\pi_1(M,x)$ is represented by infinitely many (isotopy classes of) knots. It is even worse for a pair of knots with the common base-point. Lastly, you have to deal with non-uniqueness of resolution of the crossing of $\alpha$ and $\beta$. I suggest you think more about the question you are trying to ask. $\endgroup$ – Misha Sep 6 '13 at 7:49

Since every homotopy class of loops is represented by infinitely many knots, the only way I can interpret the question is:

Is there a characterization of commuting pairs of elements $\alpha, \beta$ of $\pi_1(M)$ in terms of existence of some "nice" knots representing these based homotopy classes?

(I am dropping the disjointness condition since it does not make much sense to me.)

Then the answer is: If and only if $\alpha, \beta$ are represented by powers of torus knots in $M$ lying on the same torus which is either embedded in $M$ or is a "vertical" torus immersed in one of the Seifert components of the JSJ decomposition of $M$.

The proof is quite easy and follows from the JSJ theory, once you know what it is. The point is that either the group generated by $\alpha, \beta$ is cyclic (possibly finite) in which case both loops could be simultaneously (based) homotoped to an embedded solid torus in $M$ or it is free abelian of rank 2, in which case you apply the JSJ theory.


I believe Misha's comment directly addresses some of the problems with you question as stated. However, let me say a few words about the issues here and then mention some observations that might help you refine your ideas of your question.

Ignoring the issues of base points (which are also important) and instead concerning ourselves with only free homotopy classes of curves still leaves some issues. In the picture below, consider the green curve as $a$. The homotopy class of $a$, $\alpha$ is trivial, and so it commutes with everything in $\pi_1(M)$. (This picture highlights one distinction between homotopy classes and isotopy classes.)

(This picture highlights one distinction between homotopy classes and isotopy classes.)

Also, in this case the underlying manifold is $S^1 \times D^2$ so all elements in the fundamental group commute. In fact, we could choose a more complicated manifold, say a Seifert fibered space, and still see elements commuting. For example, Seifert fibered spaces have non-trivial center. Thus, choosing $M$ to be a Seifert fibered space and $\alpha$ to be in the center of $\pi_1(M)$ will give the desired $a$ and $b$. Again, this has nothing to do with the isotopy classes of the knots $a$ and $b$.

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    $\begingroup$ Or just consider knots in the 3-sphere. $\endgroup$ – Misha Sep 7 '13 at 13:20

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