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Suppose we let $G = \mathbb{Z}^{2} = \left<x,y \mid xyx^{-1}y^{-1}\right>$ for $x = (1,0)$, and $y=(0,1)$. Then for a given index $n \in \mathbb{Z}$, we can find all index $n$ subgroups of $G$ by considering all lattices $\left<{(r,0),(p,q)}\right>_{\mathbb{Z}}$ for $r\cdot q = n$, and $0\leq p < r$. This comes from the smith normal form of a finitely generated-submodule of a free module (considering abelian groups as $\mathbb{Z}$-modules).


For example, consider the index 4 subgroups, we get a subgroup for every combination $(r,p,q)$ such that $r,p,q \in \mathbb{Z}_{\geq 0}$, $r\cdot q = n$ and $0 \leq p < r$. This gives the subgroups:

\begin{align*} & \Lambda_{1} = \left<{(4,0),(0,1)}\right>_{\mathbb{Z}} && \Lambda_{2} = \left<{(4,0),(1,1)}\right>_{\mathbb{Z}} && \Lambda_{3} = \left<{(4,0),(2,1)}\right>_{\mathbb{Z}} & \\ & \Lambda_{4} = \left<{(4,0),(3,1)}\right>_{\mathbb{Z}} && \Lambda_{5} = \left<{(2,0),(0,2)}\right>_{\mathbb{Z}} && \Lambda_{6} = \left<{(2,0),(1,2)}\right>_{\mathbb{Z}} \end{align*}

Then since $G$ is abelian, every subgroup of $G$ is normal, and since $G$ is the fundamental group of the space $X$ (i.e the Torus) gained by attaching the 2-cell along the loop $xyx^{-1}y^{-1}$ to $B_{2}$ (the bouquet of two circles), there is a one to one correspondence between the index $n$ subgroups of $G$ and regular $n$-degree coverings of $X$.

A nice way to visualise the coverings corresponding to $\Lambda_{i}$, is to construct a surjective group homomorphisms $\phi_{i} : G \rightarrow H_{i}$, for $H_{i}$ an appropriate group of order 4, such that $\operatorname{ker}(\phi_{i}) = \Lambda_{i}$. In our case it's quite simple to see that the following work for $\Lambda_{i}$, $1\leq i \leq 5$, but I can't seem to find such a map for $\Lambda_{6}$.

\begin{align*} & \phi_{1}: G \rightarrow C_{4} = \left< \xi \mid \xi^{4}=1 \right> && (1,0) \mapsto \xi && (0,1) \mapsto 1 & \\ & \phi_{2}: G \rightarrow C_{4} = \left< \xi \mid \xi^{4}=1 \right> && (1,0) \mapsto \xi && (0,1) \mapsto \xi^{3} & \\ & \phi_{3}: G \rightarrow C_{4} = \left< \xi \mid \xi^{4}=1 \right> && (1,0) \mapsto \xi && (0,1) \mapsto \xi^{2} & \\ & \phi_{4}: G \rightarrow C_{4} = \left< \xi \mid \xi^{4}=1 \right> && (1,0) \mapsto \xi && (0,1) \mapsto \xi & \\ & \phi_{5}: G \rightarrow C_{2}\times C_{2} = \left<\eta \mid \eta^{2}\right> \times \left< \kappa \mid \kappa^{2}\right> && (1,0) \mapsto (\eta,0) && (0,1) \mapsto (0,\kappa) & \\ \end{align*}

Since, from here, it is relatively trivial (since there is a low number of generators of low order) to produce the corresponding coverings of the torus, if I could find the map for $\Lambda_{6}$, I would have classified all 6 (up to covering transformation) regular degree 4 coverings of the torus.

Could anyone help here?

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    $\begingroup$ Probably this is not important in the end, but the fundamental group of a bouquet of circles is the free product of copies of $\mathbb{Z}$, not the direct product. $\endgroup$ – Javi May 28 '18 at 20:45
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    $\begingroup$ Gah! My bad, I meant to mention that I added the 2-cell to $B_{2}$ that adds the relation $xyx^{-1}y^{-1}$. Will fix now $\endgroup$ – Adam Higgins May 28 '18 at 20:46
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You can realise $\Lambda_6$ as the kernel of the homomorphism $$ \phi_6 : \mathbb Z^2 \to C_4=\left< \xi\mid \xi^4 = 1\right>,$$

$$ (0,0) \mapsto 0, \ \ (1,1) \mapsto \xi, \ \ (1, 0) \mapsto \xi^2, \ \ (0, 1) \mapsto \xi^3.$$

I would suggest that you visualise the covering space as the torus $\mathbb R^2 / \Lambda_6 $, with the covering map $\mathbb R^2 / \Lambda_6 \to \mathbb R^2 / \Lambda$ being the map $[(x,y)] \mapsto [(x,y)]$. It should clear that the loops from the original torus $\mathbb R^2 / \Lambda$ that lift to closed loops on covering space $\mathbb R^2 / \Lambda_6$ are precisely those loops that correspond to elements of the subgroup $\Lambda_6 \subset \mathbb Z^2$, and that the group of deck transformations for this covering is the group ${\rm ker}(\phi_6) = \mathbb Z^2 /\Lambda_6\cong C_4$ described above.

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  • $\begingroup$ Thanks so much! This is a perfect answer, and I find your visualisation to be very useful. Thanks again! $\endgroup$ – Adam Higgins May 28 '18 at 21:34

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