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.

Sümmermann's KnotPortal program imagines a trefoil-shaped portal connecting six worlds.


1 Answer 1


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.

  • $\begingroup$ Ha! Who'da thunk it! $\endgroup$ Feb 19, 2021 at 3:14
  • $\begingroup$ 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. $\endgroup$ Feb 19, 2021 at 3:19
  • $\begingroup$ 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.) $\endgroup$ Feb 19, 2021 at 3:28
  • $\begingroup$ 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. $\endgroup$ Feb 19, 2021 at 3:55
  • $\begingroup$ 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 $n\ge6$. $\endgroup$ Feb 19, 2021 at 6:08

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