$\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\RR}{\mathbb{R}}$Let $S = S_2$ be the genus two surface. In this case, $\ZZ^4$ is the deck group of the desired covering. Consider $\ZZ^4$ inside of $\RR^4$ and add to these points the usual edges labelled $a, b, c, d$, parallel to the four coordinate axes. This gives a Cayley graph for $\ZZ^4$.

To make this graph a surface, we form a two-complex $S'$ by attaching two-cells. At every vertex we attach a two-cell via the attaching map $abcdABCD$ (capital letters denote inverses). This is possible because the boundary word describes a closed loop in the graph. Thus very edge of $S'$ meets a pair of two-cells while every vertex meets eight two-cells. The eight corners give the vertex a disk neighborhood in $S'$.

Thus $S'$ is a surface. Taking the quotient by the action of $\ZZ^4$ gives the original surface $S$. By the Galois correspondence, $S'$ is the desired covering space. Note that $S'$ is quasi-isometric to $\ZZ^4$ so it is one-ended. The loops $abAABa$ and $cdCCDc$, based at the origin, meet in exactly one point. Thus $S'$ has genus and so has infinite genus.

This construction works in general. It even works when $g = 1$ (except for the infinite genus part).

$\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\RR}{\mathbb{R}}$Let $S = \Sigma_2$ be the genus two surface. In this case, $\ZZ^4$ is the deck group of the desired covering. Consider $\ZZ^4$ inside of $\RR^4$ and add to these points the usual edges labelled $a, b, c, d$ parallel to the four coordinate axes. This gives a Cayley graph for $\ZZ^4$.

Next, starting at every vertex of the graph we attach a two-cell via the attaching map $abcdABCD$ (capital letters denote inverses). This is possible because the boundary word describes a closed loop in the graph. Let $S'$ be the resulting two-complex. Every edge of $S'$ meets a pair of two-cells while every vertex meets eight two-cells. The eight corners give the vertex a disk neighborhood in $S'$.

Thus $S'$ is a surface. Taking the quotient by the action of $\ZZ^4$ gives the original surface $S$. By the Galois correspondence, $S'$ is the desired covering space. Note that $S'$ is quasi-isometric to $\ZZ^4$ so it is one-ended. The loops $abAABa$ and $cdCCDc$, based at the origin, meet in exactly one point. Thus $S'$ has genus, and so has infinite genus.

This construction works in any genus. When $g = 1$ the construction produces the universal cover.

$\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\RR}{\mathbb{R}}$Let $S = S_2$ be the genus two surface. In this case, $\ZZ^4$ is the deck group of the desired covering. Consider $\ZZ^4$ inside of $\RR^4$ and add to these points the usual edges labelled $a, b, c, d$, parallel to the four coordinate axes. This gives a Cayley graph for $\ZZ^4$.

To make this graph a surface, we form a two-complex $S'$ by attaching two-cells. At every vertex we attach a two-cell via the attaching map $abcdABCD$ (capital letters denote inverses). This is possible because the boundary word describes a closed loop in the graph. Thus very edge of $S'$ meets a pair of two-cells while every vertex meets eight two-cells. The eight corners give the vertex a disk neighborhood in $S'$.

Thus $S'$ is a surface. Taking the quotient by the action of $\ZZ^4$ gives the original surface $S$. By the Galois correspondence, $S'$ is the desired covering map. Note that $S'$ is quasi-isometric to $\ZZ^4$ so it is one-ended. The loops $abAABa$ and $cdCCDc$, based at the origin, meet in exactly one point. Thus $S'$ has infinite genus.

This construction works in general. It even works when $g = 1$ (except for the infinite genus part).

$\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\RR}{\mathbb{R}}$Let $S = S_2$ be the genus two surface. In this case, $\ZZ^4$ is the deck group of the desired covering. Consider $\ZZ^4$ inside of $\RR^4$ and add to these points the usual edges labelled $a, b, c, d$, parallel to the four coordinate axes. This gives a Cayley graph for $\ZZ^4$.

To make this graph a surface, we form a two-complex $S'$ by attaching two-cells. At every vertex we attach a two-cell via the attaching map $abcdABCD$ (capital letters denote inverses). This is possible because the boundary word describes a closed loop in the graph. Thus very edge of $S'$ meets a pair of two-cells while every vertex meets eight two-cells. The eight corners give the vertex a disk neighborhood in $S'$.

Thus $S'$ is a surface. Taking the quotient by the action of $\ZZ^4$ gives the original surface $S$. By the Galois correspondence, $S'$ is the desired covering space. Note that $S'$ is quasi-isometric to $\ZZ^4$ so it is one-ended. The loops $abAABa$ and $cdCCDc$, based at the origin, meet in exactly one point. Thus $S'$ has genus and so has infinite genus.

This construction works in general. It even works when $g = 1$ (except for the infinite genus part).