12
$\begingroup$

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.

$\endgroup$
6

1 Answer 1

8
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.