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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?

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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 $n \ge 7$.

enter image description here

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  • $\begingroup$ Very interesting. Thank you $\endgroup$
    – user600999
    Commented Oct 10, 2018 at 11:01
  • $\begingroup$ I know the result that any surface is a quotient of the hyperbolic plane by the action of some Fuchsian group — is it always true that this Fuchsian group is (isomorphic to) the fundamental group of the surface? $\endgroup$ Commented Oct 11, 2018 at 10:51
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    $\begingroup$ @Santana: if the action is free, then yes, it follows from covering space theory. $\endgroup$ Commented Oct 11, 2018 at 11:40
  • $\begingroup$ Does any octagon refer to the double torus? Is it possible to label the side consistently? Misha commented that's not possible. Any idea? $\endgroup$
    – draks ...
    Commented Sep 9, 2019 at 14:40

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