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Suppose I have a knot $K$ in $S^3$. Given a diagram $D$ of $K$ I get the Wirtinger presentation $\langle x_1, \dots, x_a \mid r_1, \dots, r_c\rangle$ of its knot group $\pi(K) = \pi_1(S^3 \setminus K)$. There is a generator $x_i$ for each arc of $D$ and a relation $r_j$ for each crossing of $D$. (Arc means over-arc: an arc starts at an under-crossing and continues through over-crossings until the next under-crossing.) The generators represent meridians of $K$ and the relations all say that the generators of the two under-arcs at a crossing are conjugate via the over-arc.

Suppose $x_1$ and $x_2$ are two arcs related at a crossing as below:

Wirtinger relation at a crossing

and that $x_1$ and $x_2$ correspond to distinct arcs (if I follow the knot from $x_1$ to $x_2$ I travel under at least one crossing). Is it possible for $x_1$ and $x_2$ to commute? More formally, is there a knot $K$ for which the map $$\phi : \pi(K) \to \pi(K)/\langle x_1 x_2 x_1^{-1} x_2^{-1}\rangle$$ is injective?

I'm interested in this question in the context of hyperbolic knots, so a more specific question is if there's a hyperbolic knot $K$ for which $\phi$ is injective.

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    $\begingroup$ They will always commute if it’s the diagram of a trivial knot. But I think this points to the general phenomenon: in a 3-manifold group, two infinite-order elements commute if and only if they either lie in a common cyclic subgroup or they are in the image of a pi_1-injective torus. Since your loops are meridians, thinking about the fact that they both map to the same generator of H_1 implies that they will commute if and only of they are identical elements of pi_1. $\endgroup$
    – Andy Putman
    Commented Apr 26, 2023 at 13:38
  • $\begingroup$ (it is easy to find examples where they are the same — take a diagram and do a bunch of silly Reidemeister moves to complicate it unnecessarily) $\endgroup$
    – Andy Putman
    Commented Apr 26, 2023 at 13:42
  • $\begingroup$ Thanks! That's quite helpful. I thought there was some fact like this about 3-manifold groups but didn't know it precisely. $\endgroup$
    – Calvin McPhail-Snyder
    Commented Apr 26, 2023 at 17:19
  • $\begingroup$ The goal of the distinct arcs condition was to try to guarantee that $x_1$ and $x_2$ are actually different elements of $\pi(K)$ but I think it might be a little harder to ensure this. $\endgroup$
    – Calvin McPhail-Snyder
    Commented Apr 26, 2023 at 17:20
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    $\begingroup$ Yeah, I don't think there is any simple condition on the knot diagram that ensures that two particular generators in the Wirtinger Presentation are distinct. As far as the general condition about 3-manifold groups, what it follows from is the classification of abelian subgroups of 3-manifold groups, which can be found in Hempel's book. $\endgroup$
    – Andy Putman
    Commented Apr 26, 2023 at 17:21

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