In my revision on covering maps, I came across a problem asking me to construct a degree four regular based covering map of the torus, and the corresponding presentation of the subgroup of the fundamental group of the torus . This is the process I went through to it, and I would appreciate any feedback on my the validity and efficiency of my constructions.
Firstly, I shall describe the torus (which I shall call $T$) using its cell-complex construction (which I shall call $\mathscr{T}$), using a single $0$-cell labeled $v$, two oriented $1$-cells labelled $a$ and $b$, and a single $2$-cell attached along the boundary word $aba^{-1}b^{-1}$. Then $ \pi_{1}(T) \cong \pi_{1}(\mathscr{T},v) \cong \left< a, b \mid aba^{-1}b^{-1} \right> = G$, say.
I shall also use $\mathscr{T}^{(1)}$ to denote the $1$-skeleton of $\mathscr{T}$. The diagram below shows the chosen edge orientations for $\mathscr{T}^{(1)}$.
Then, regular coverings of $T$ of finite degree, correspond to regular coverings of $\mathscr{T}^{(1)}$ with some relations, which in turn correspond to normal subgroups of $G$ of finite index. Then if we're looking for a normal subgroup of $G$ of index 4, we should be looking at a surjective homomorphism from $G$ to a group of order $4$. To this end consider the following map:
\begin{align*} & \phi : G \rightarrow C_{4} = \left< \xi \mid \xi^{4} \right> &\phi(a) = \phi(b) = \xi & \\ \end{align*}
Then let $H = \operatorname{ker}(\phi)$, so that if we find the based covering maps of $\mathscr{T}^{(1)}$ corresponding to $H$ (with the appropriate relations) we should be done. Then let:
\begin{align*} p : \left( \tilde{\mathscr{T}}_{H}, \tilde{v}_{H}\right) \rightarrow \left( \mathscr{T}, v \right) \\ \end{align*}
be the regular covering maps corresponding to $H$, and we shall begin by calculating the structure of $\left( \tilde{\mathscr{T}}_{H}, \tilde{v}_{H}\right)$.
Now we know that a loop $\ell$ based at $v$ in $\mathscr{T}^{(1)}$ lifts to a loop in $\tilde{\ell}$ based at $\tilde{v}_{H}$ in $\tilde{\mathscr{T}}_{H}^{(1)}$ if and only if $[\ell] \in H$. So we know that $\left( \tilde{\mathscr{T}}_{H}^{(1)}, \tilde{v}_{H} \right)$ is a regular graph with $4$ vertices, and $4$-edges leaving each vertex, and that $a^{4}, b^{4}, ab^{-1}, a^{-1}b, a^{2}b^{2}, a^{3}b, b^{4}$ are all loops based at $\tilde{v}_{H}$ in $\tilde{\mathscr{T}}_{H}^{(1)}$. We conjecture then, that $\left( \tilde{\mathscr{T}}_{H}, \tilde{v}_{H}\right)$ has the structure of the below diagram, where we have 4 2-cells attached along the boundary word $aba^{-1}b^{-1}$ starting at each vertex. The covering is regular so we can pick $\tilde{v}_{H}$ to be any of the four vertices, but in my diagram below I was considering $v_{1} = \tilde{v}_{H}$.
Then, we calculate a maximal tree (in red), and label each edge with $e_{i}$ (in purple) not in the maximal tree (in cyan) as shown:
Then for an edge $e$ let $\iota(e)$ denote the initial vertex, and $\tau(e)$ denote the vertex edge. Then for $v$ a vertex, let $\theta(v)$ denote the unique path through the maximal tree from $v_{1}$ to $v$. Then, the generating set for $H$ is $\{ g_{i} \}_{i=1}^{5}$ where $g_{i} = \theta(\iota(e_{i})) \cdot e_{i} \cdot \theta(\tau(e_{i})^{-1})$. These elements are tabulated below:
\begin{array}{c | c c c c} i & \theta(\iota(e_{i})) & e_{i} & \theta(\tau(e_{i})) & g_{i} \\ \hline 1 & a & b^{-1} & \varnothing & ab^{-1}\\ 2 & a & b & a^{2} & aba^{-2}\\ 3 & a^{2} & b & a^{-1} & a^{2}ba\\ 4 & a^{-1} & b & \varnothing & a^{-1}b\\ 5 & a^{2} & a & a^{-1} & a^{4}\\ \end{array}
The relations are given by the edges not in the maximal tree that are transversed by loops in $\tilde{\mathscr{T}}_{H}$ traced out by the path $aba^{-1}b^{-1}$ starting at each of the 4 vertices.
That is, if $\alpha_{i}$ is the loop based at $v{i}$ in $\tilde{\mathscr{T}}_{H}$ traced out by the path $aba^{-1}b^{-1}$, then we get a relation whose letters are the edges transversed that are not in the maximal tree. These are tabulated below:
\begin{array}{c | c c } i & \text{edges} & r_{i} \\ \hline 1 & e_{2}e_{1} & g_{2}g_{1} \\ 2 & e_{3}e_{5}^{-1}e_{2}^{-1} & g_{3}g_{5}^{-1}g_{2}^{-1} \\ 3 & e_{5}e_{4}e_{3}^{-1} & g_{5}g_{4}g_{3}^{-1} \\ 4 & e_{1}e_{4}^{-1} & g_{1}g_{4}^{-1} \\ \end{array}
Then $H$ has the group presentation: $\left<\alpha, \beta, \gamma, \delta, \epsilon \mid \beta\alpha, \gamma\epsilon^{-1}\beta^{-1}, \epsilon\delta\gamma^{-1}, \alpha\delta^{-1} \right>$ where
\begin{array}{c | c } \alpha & ab^{-1}\\ \beta & aba^{-2}\\ \gamma & a^{2}ba\\ \delta & a^{-1}b\\ \epsilon & a^{4}\\ \end{array}
Then, I think this is a valid solution to the problem?
I was bored so I decided to try and come up a second distinct solution. And likewise would appreciate feedback. This time I considered the map:
\begin{align*} & \psi : G \rightarrow C_{2}^{2} = \left< \nu, \mu \mid \nu^{2}, \mu^{2}, \nu\mu\nu^{-1}\mu^{-1} \right> &\psi(a) = \nu, \psi(b) = \mu & \end{align*}
Then let $K = \operatorname{ker}(\psi)$, and let
\begin{align*} q : \left( \tilde{\mathscr{T}}_{K}, \tilde{v}_{K}\right) \rightarrow \left( \mathscr{T}, v \right) \\ \end{align*}
be the unique regular based covering corresponding to $K$. Then similarly we know that a loop $\ell$ based at $v$ in $\mathscr{T}^{(1)}$ lifts to a loop in $\tilde{\ell}$ based at $\tilde{v}_{K}$ in $\tilde{\mathscr{T}}_{K}^{(1)}$ if and only if $[\ell] \in K$. So we know that $\left( \tilde{\mathscr{T}}_{K}^{(1)}, \tilde{v}_{K} \right)$ is a regular graph with $4$ vertices, and $4$-edges leaving each vertex, and that $a^{2}, b^{2}$ and are both loops based at $\tilde{v}_{K}$ in $\tilde{\mathscr{T}}_{K}^{(1)}$. We then conjecture that $\tilde{\mathscr{T}}_{K}$ has the structure of this diagram, where similarly we have 4 2-cells attached along the boundary word $aba^{-1}b^{-1}$ starting at each vertex. Similarly, since the covering is regular so we can pick $\tilde{v}_{K}$ to be any of the four vertices, but in my diagram below I was considering $v_{1} = \tilde{v}_{K}$.
Then, similarly, we calculate a maximal tree (in red), and label each edge with $e_{i}$ (in purple) not in the maximal tree (in cyan) as shown:
Then we calculate the generators $\{g_{i}'\}_{i=1}^{5}$ tabulated below:
\begin{array}{c | c c c c} i & \theta(\iota(e_{i})) & e_{i} & \theta(\tau(e_{i})) & g_{i}' \\ \hline 1 & \varnothing & a & a^{-1} & a^{2} \\ 2 & a^{-1} & b & a^{-1}b^{-1} & a^{-1}b^{2}a \\ 3 & a^{-1}b^{-1} & a & b & a^{-1}b^{-1}ab^{-1} \\ 4 & b & b & \varnothing & b^{2} \\ 5 & b & a & a^{-1}b^{-1} & baba \\ \end{array}
Then the relations $\{ r'_{i} \}_{i=1}^{4}$ are tabulated below:
\begin{array}{c | c c } i & \text{edges} & r’_{i} \\ \hline 1 & e_{1}e_{2}e_{5} & g’_{1}g’_{2}g’_{5} \\ 2 & e_{3}e_{2} & g’_{3}g’_{2} \\ 3 & e_{3}e_{4} & g’_{3}g’_{4} \\ 4 & e_{5}e_{1}e_{4} & g’_{5}g’_{2}g’_{4} \\ \end{array}
Giving the group presentation $\left< \alpha,\beta,\gamma,\delta,\epsilon \mid \alpha\beta\epsilon, \gamma\beta, \gamma\delta, \epsilon\beta\delta \right>$ for $K$, where:
\begin{array}{c | c } \alpha & a^{2}\\ \beta & a^{-1}b^{2}a\\ \gamma & a^{-1}b^{-1}ab^{-1} \\ \delta & b^{2} \\ \epsilon & baba \\ \end{array}
Is this second one correct?