TITLE: Thurston universe gates in knots: which invariant is it?
QUESTION [5 upvotes]: Today I discovered this nice video of a lecture by Thurston:
https://youtu.be/daplYX6Oshc
in which he explains how a knot can be turned into a "fabric for universes". For example, the unknot can be thought as a portal to Narnia, and when you pass again you switch back to the Earth. This forms in a sense a $\mathbb{Z}/2\mathbb{Z}$. Then he proceeds to explore what fabric one gets with the treefoil and you get an $S_3$. I am sure there is some real geometry behind but I can't grasp how to translate a portal into something homotopic.
A way to formalize this would be the following. Take a knot $K$ in $\mathbb{R}^3$. Fix a tubular neighborhood $N$ of $K$. For each point $x$ in the knot take the loop $L_x$ obtained as the sphere bundle of $N$ at $x$ (a small circle around $x$ that jumps into the portal). Then there exist a connected 3-manifold $M$ with a (finite?) cover $M \to \mathbb{R}^3 \setminus K$ such that the "monodromy in small circles" around $L_x$ has order two for all $x$. Then we set the "group of universes" as the group of cover automorphisms.
I think this captures the previous idea in the following sense: to each locus of $\mathbb{R}^3 \setminus K$ we have $n$ counterimages that represent the different worlds. Some branch should be chosen to make distinguishing between worlds possible. The constraint on monodromy ensures that if you jump twice through the same portal (at least for the portals very close to the boundary) you get back.
Does such a manifold exist for all knots? Is this construction just some simplification of the fundamental group of the complement?

REPLY [6 votes]: Here is a higher-quality video of the same material.  My answer is a more algebraic version of Thurston's presentation, but I will tie this back to Thurston's "intention" at the end.

Suppose that $L$ is an oriented knot diagram of a knot $K$.  Let $A_i$ enumerate the "over-strands" of the diagram $L$.  There is a presentation of $\pi_1(S^3 - K)$ with the $A_i$ as generators; here $A_i$ represents a meridional loop that goes (using the righthand rule) about the corresponding strand. The crossings give relations; if $A$, $B$, and $C$ are the strands at a crossing of $L$ then we deduce the relation $BA = AC$ or $CA = AB$ (depending on the sign of the crossing). This is called the Wirtinger presentation
Thurston is adding the relation $A_i^2 = 1$ for all $i$.  This gives a quotient of $\pi_1(S^3 - K)$.  The quotient group has a presentation with, again, the $A_i$ as generators and with relations $A_i^2 = 1$ and $BA = AC$ (with $A$, $B$, and $C$ as above). In the second half of the video he builds a Cayley graph of the quotient group in the case of the trefoil.
It is an exercise (using the Reidemeister moves) to show that the quotient group is independent of the given diagram $L$.  Thus the quotient is a knot invariant.

As mme points out, in their comment above, we can also obtain this group by quotienting by the smallest normal subgroup containing the square of any meridional element.  This gives the orbifold fundamental group of the “orbifold” Dehn filling of the knot complement $S^3 - K$ along the slope $(2, 0)$.  This orbifold has a universal cover which is a manifold.  The covering degree is the number of lands in Thurston's discussion.  For "most" hyperbolic knots this Dehn filling is hyperbolic and the covering degree is infinite.  It is a nice exercise to show that the covering degree is finite for all two-bridge knots.