Obtaining two-holed torus as a quotient of $\Bbb C$

Question

The torus $T^2$ arises as a quotient space $\Bbb{C}/\Gamma$ for some lattice $\Gamma=\Bbb{Z}\gamma_1+\Bbb{Z}\gamma_2$ for $\gamma_1,\gamma_2\in\Bbb C$.

One could think of this as gluing a $4g$-gon where $g=1$.

Can we find the two holed torus $T^2\# T^2$ as a quotient space $\Bbb{C}/Y$? I can see that we can obtain it from gluing an $8$-gon, but am not sure that we can get it as a quotient space, given there isn't a 'fundamental region' in $\Bbb{C}$ that tessellates.

Do I need to quotient $\Bbb{C}^n$ for some $n$ instead?

Answer 1

Yes, although the action isn't holomorphic or isometric.

$\mathbb{C}$ is diffeomorphic to the hyperbolic / upper half plane $\mathbb{H}$, and by the uniformization theorem any complex structure on the $2$-holed torus $\Sigma_2$ has universal cover the upper half plane, as a Riemann surface. The upper half plane has automorphism group $PSL_2(\mathbb{R})$, and so there is a Fuchsian group $\Gamma \subset PSL_2(\mathbb{R})$, abstractly isomorphic to the fundamental group of $\Sigma_2$, such that

$$\mathbb{H}/\Gamma \cong \Sigma_2.$$

The same is true for the $g$-holed torus $\Sigma_g$, $g \ge 2$. There are some corresponding hyperbolic tesselations of the hyperbolic plane by $4g$-gons; below is a picture of a tesselation of the hyperbolic plane (in the Poincare disk model, I think) by octagons. In general the hyperbolic plane can be tiled by regular $n$-gons for all $n \ge 7$.