Let $A$ be a $2\times 2$ matrix with integer coefficients and $$ p_A : S^1 \times S^1 \to S^1 \times S^1 $$ such that $$ (z,w) \to (z^{a_{11}}w^{a_{12}},z^{a_{21}}w^{a_{22}}). $$ Show that $p_A$ is a covering with degree $|\det(A)|$ if $\det(A) \neq 0$.
I have no clue. Could you help me, please?
You have the cover $p:R^2\rightarrow S^1\times S^1$ defined by $p(x,y)=(e^{ix},e^{iy})$. The linear map $A$ of $R^2$ is a diffeomorphism iff $det(A)=0$.
The $f_A:S^1\times S^1\rightarrow S^1\times S^1$ defined by $f_A(e^{ix},e^{iy})=e^{ia_{11}x}e^{ia_{12}y},e^{ia_{21}x}e^{ia_{22}y})$ verifies $p\circ A=f_A\circ p$. This implies that $dp\circ A=df_A\circ dp$. Since $dp_{x,y}:T_{x,y}R^2\rightarrow T_{e^{ix},e^{iy}}S^1\times S^1$ is an isomorphism, we deduce that ${df_A}_{e^{ix},e^{iy}}$ is an isomorphism if and only if $A$ is an isomorphism. This is equivalent to the fact that $f_A$ is a local diffeomorphism defined on a compact manifold i.e a covering iff $A$ is a diffeomorphism.
A proper local diffeomorphism between manifolds is a covering map.
To compute the degree of $f_A$, remark that if $dx\wedge dy$ is the volume form of $R^2$, it is invariant by the translations so it is invariant by the Deck transformations of the covering $p$. You deduce the existstence of a volume form $\omega$ of $S^1\times S^1$ such that $p^*\omega =dx\wedge dy$, you have $det(A)dx\wedge dy= A^*dx\wedge dy =A^*p^*\omega=(p\circ A)^*\omega =(f_A\circ p)^*\omega=p^*(f_A^*\omega)=p^*(det A\omega)$. This implies that $f_A^*\omega =detA\omega$ since $dp_{x,y}$ is an isomorphism.
