Show that the map $(z,w) \mapsto (z^{am} w^{bn}, z^{cm} w^{dn})$ is a covering map.

Question

Show that the map $f : S^1 \times S^1 \longrightarrow S^1 \times S^1$ defined by $(z,w) \mapsto (z^{am} w^{bn}, z^{cm} w^{dn})$ is a covering map of degree $mn$ where $ad - bc = 1$ and $a,b,c,d,m,n \in \mathbb Z.$

Our instructor says that it can be written as a composition of covering maps of finite degree and hence it's a covering map. But I can't see how can it be written as composition of covering maps.

Any help in this regard would be warmly appreciated. Thanks for investing your valuable time on my question.

Answer 1

Define maps $\phi, \psi : S^1 \times S^1 \to S^1 \times S^1$ by $$\phi(z,w) = (z^n,w^m), $$ $$\psi(u,v) = (u^av^b,u^cv^d) .$$ Clearly $f = \psi \circ \phi$.

The maps $\omega_k : S^1 \to S^1, \omega_k(z)= z^k$, are covering maps of degree $p$. We have $\phi = \omega_n \times \omega_m$. We now invoke the following

Theorem. Let $p : \tilde X \to X$ and $q : \tilde Y \to Y$ be covering maps. Them $p \times q$ is a covering map. If $\deg p = n$ and $\deg q = m$, then $\deg p \times q = nm$.

This is well-known and the proof is very easy.

We conclude that $\phi$ is a covering map of degree $mn$.

In order that the claim in your question be true, $\psi$ should be a covering map of degree $1$. But such maps are homeomorphisms. This heuristic consideration shows that we should prove that $\psi$ is a homeomorphism.

Given a $(2 \times 2)$-matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with entries in $\mathbb Z$, let us define $$A^* : S^1 \times S^1 \to S^1 \times S^1, A^*(u,v) = (u^av^b,u^cv^d) .$$

Clearly the unit matrix $E$ gives $$ E^* = id \tag{1} .$$ Let us next show that for $B = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$ one has $$(A \cdot B)^* = A^* \circ B^* \tag{2} .$$

We have $A \cdot B = \begin{pmatrix} a e + b g & a f + b h \\ c e + d g & c f + d h \end{pmatrix}$, thus $$(A \cdot B)^*(u,v) = (u^{a e + b g} v^{a f + b h}, u^{c e + d g }v^{c f + d h}) ,$$

$$A^*(B^*(u,v)) = A^*(u^ev^f,u^gv^h) = ((u^ev^f)^{a}(u^gv^h)^{b}, (u^ev^f)^{c}(u^gv^h)^{d}) \\= (u^{e a} v^{f a} u^{g b} v^{h b},u^{e c} v^{f c} u^{g d} v^{h d}) = (u^{e a + g b}v^{f a + h b},u^{e c + g d}v^{f c + h d}) .$$

This proves $(2)$.

Formulae $(1)$ and $(2)$ show that if $A$ is invertible and $A^{-1}$ has all entries in $\mathbb Z$, then $A^*$ is a homeomorphism.

$\det A = ad - bc \ne 0$ means that $A$ is invertible as a real metrix. But we even have $\det A = 1$ which implies that $A^{-1}$ has all entries in $\mathbb Z$.

Since $\psi = A^*$, we are finished.

Update:

That $A^{-1}$ has integer entries follows from

Lemma. Let $A$ be an $(n \times n)$-matrix with entries in $\mathbb Z$. Then $A$ has an inverse $A^{-1}$ with entries in $\mathbb Z$ if and only if $\det A = \pm 1$.

Proof. 1. For all matrices $A, B$ with entries in $\mathbb R$ we have $\det A \cdot B = \det A \cdot \det B$. If both $A, B$ have entries in $\mathbb Z$, then also $\det A, \det B \in \mathbb Z$.

1. Assume that $A$ has entries in $\mathbb Z$ and has an inverse $A^{-1}$ with entries in $\mathbb Z$. Then $$1 = \det E = \det A \cdot A^{-1} = \det A \cdot \det A^{-1} \tag{3}.$$ Since both $\det A, \det A^{-1} \in \mathbb Z$, $(3)$ can only be satisfied when $\det A = \det A^{-1} = 1$ or $\det A = \det A^{-1} = -1$.

2. Assume that $A$ has entries in $\mathbb Z$ and has $\det A = \pm 1$. Then $A$ is invertible as a real matrix. It inverse is given by $$A^{-1} = \frac{1}{\det A} \left(-1)^{i+j} A_{ij}\right)^T \tag{4}$$ where the $A_{ij}$ are the minors of $A$ and $^T$ denotes transposition. The minors are the determinants of the $((n-1) \times (n-1))$-submatrices of $A$ obtained by deleting the $i$-th row and $j$-th column. If $A$ has entries in $\mathbb Z$ then all $A_{ij} \in \mathbb Z$ and therefore $A^{-1}$ has entries in $\mathbb Z$ since $\det A = \pm 1$.