What is the degree of an n-fold branched cover over a trefoil?

The order-2 cyclic branched cover over a trefoil has degree 6, meaning the preimage of any point off the trefoil has cardinality six. (You can find a wonderful video of this here, made by Moritz Sümmermann.) The order-3 cyclic branched cover over a trefoil has degree 24.

What is the formula for the degree of an order-$$n$$ cyclic branched cover over a trefoil?

(I'm unsure of the proper terminology for this.)

I'm pretty sure this is $$|G_n|$$, where $$G_n=\langle x,y\mid xyx=yxy,~x^n=y^n=1\rangle;$$ however, I have no idea how to find the size of this group.

– Karl
Feb 18, 2021 at 17:30
• The algorithm I have used to render youtube.com/watch?v=1TMY2U4_9Qg (and obtain 24) can find the following values: 1, 6, 24, 96, 600, ... (the current implementation does not seem to have enough RAM for the next item, I will try to optimize it) Feb 18, 2021 at 18:23
• I tried running GAP on the order-6 cover group, but it's having trouble too. Feb 18, 2021 at 19:56
• You've defined the "degree" but not the "order" (when reading the first few words of your post I immediately interpreted "order" to mean what you are calling "degree"; imagine my surprise a few words later). Feb 18, 2021 at 21:29

The group $$G_n = \langle x,y \mid xyx=yxy,\, x^n=1,\, y^n=1\rangle$$ is infinite for all $$n\geq 6$$.

To see this, observe first that the third relation $$y^n=1$$ follows from the first two, since by the first relation $$y=(xy)x(xy)^{-1}$$ and hence $$y^n=(xy)x^n(xy)^{-1}=1$$. Thus $$G_n = \langle x,y\mid xyx=yxy,\, x^n=1\rangle.$$

Now, it is well-known that the braid group $$B_3=\langle x,y\mid xyx=yxy\rangle$$ can also be presented as $$\langle a,b\mid a^2=b^3\rangle$$, where $$a=xyx$$ and $$b=xy$$. Since $$x=b^{-1}a$$, it follows that $$G_n = \big\langle a,b \;\bigl|\; a^2=b^3,\,(b^{-1}a)^n = 1\big\rangle$$ Now consider the following quotient of $$G_n$$: $$Q_n = \big\langle a,b \;\bigl|\; a^2=b^3=1,\,(b^{-1}a)^n = 1\big\rangle$$ (It doesn't affect the argument, but $$\langle a,b \mid a^2=b^3=1\rangle$$ is a presentation for the modular group $$\mathrm{PSL}(2,\mathbb{Z})$$, which is a quotient of $$B_3$$.)

What does the Cayley graph of $$Q_n$$ look like? If we treat $$a$$ edges as undirected, then the Cayley graph is the 1‑skeleton of a regular tiling of a simply connected surface by $$2n$$-gons corresponding to $$(b^{-1}a)^n$$ and triangles corresponding to $$b^3$$, with two $$2n$$-gons and one triangle meeting at every vertex.

If we try to make the polygons regular and Euclidean, then each $$2n$$-gon has angles of $$\pi\bigl(1-\frac{1}{n}\bigr)$$ and each triangle has angles of $$\pi/3$$, so the total angle at each vertex is $$2\pi\biggl(1-\frac{1}{n}\biggr) + \frac{\pi}{3} = 2\pi\biggl(\frac{7}{6}-\frac{1}{n}\biggr).$$ For $$n<6$$ this sum is less than $$2\pi$$, so the Cayley graph of $$Q_n$$ is the 1‑skeleton of a tiling of the sphere, and hence $$Q_n$$ is finite. Indeed, the Cayley of $$Q_n$$ for $$n=2$$, $$n=3$$, $$n=4$$, and $$n=5$$, are respectively the 1‑skeleta of a triangular prism, a truncated tetrahedron, a truncated cube, and a truncated dodechedron. Zeno Rogue's computer code shows that $$G_n$$ is finite in these cases as well.

For $$n=6$$, the sum is equal to $$2\pi$$, so the Cayley graph of $$Q_n$$ is the 1‑skeleton of the truncated hexagonal tiling of the Euclidean plane by equilateral triangles and regular dodecagons. For $$n>6$$, the sum is greater than $$2\pi$$, which means that the Cayley graph of $$Q_n$$ is the 1‑skeleton of a tiling of the hyperbolic plane. For example, when $$n=7$$ this is the truncated heptagonal tiling of the hyperbolic plane. In particular, $$Q_n$$ is infinite for all $$n\geq 6$$, and hence $$G_n$$ is as well.

• Ha! Who'da thunk it! Feb 19, 2021 at 3:14
• And it's beautiful how this question and answer brings together topology, group theory, and geometry. Two different visual problems - in knot theory and tiling theory - linked together by algebra. Feb 19, 2021 at 3:19
• I think $a^2=b^3=1$, in the knot theory view, corresponds to a branched cover where going through the center of the trefoil three times doesn't take you anywhere new. (This is already the case in the 2-fold cover. For the 3-fold cover, trusting your "truncated tetrahedron" calculation, this would reduce the number of worlds (or sheets) from 24 to 12.) Feb 19, 2021 at 3:28
• The next question, raised by Zeno in a Twitter thread, is "How many knots are there?" Group theoretically, I believe this is the same as asking for the size of the conjugacy class of $x$ (equiv. $y$). Zeno gave the sequence 1, 3, 4, 6, 12. From your answer, it seems this is the number of $2n$-gons in the tiling, which means it continues as $\infty$ as well. Feb 19, 2021 at 3:55
• My friend Balarka points out that $Q_n$ is exactly the Von Dyck group $D(2,3,n)$, the group of orientation-preserving isometries of tilings of triangles with angles $\pi/2$, $\pi/3$, and $\pi/n$. This tiling is Euclidean or hyperbolic for $n\ge6$. Feb 19, 2021 at 6:08