I'm trying to build $2:1$ covering space for $2$- genus surface by $3$-genus surface.
I can see that if I take a cut of $3$-genus surface in the middle (along the mid hole) I get $2$ surfaces each one looks like $2$-genus surface which are open in one side so it's clear how to make the projection from these $2$ copies to the $2$-genus surface.
my question is how can I see this process in the polygonal representation of 3-genus surface
( $12$ edges: $a_{1}b_{1}a_{1}^{-1}b_{1}^{-1}……a_{3}b_{3}a_{3}^{-1}b_{3}^{-1})$
. I can't visualize the cut I make in the polygon.
thanks.
[Math] covering space of $2$-genus surface
algebraic-topologycovering-spaces
Related Solutions
Let's focus on the surface $\Sigma_3$ of genus $3$ just to be concrete.
Here's the geometric idea:
The basic idea is that the lattice $\mathbb{Z}^3$ has to act by deck transformations on the covering space. So we have to have three curves on $\Sigma_3$ which "unwind" to the grid pictured on the right in the covering space. From there it's not a stretch to the cut-and-paste operation in the picture.
Formally - without exhibiting explicit formulae - I think of this two ways.
Three disjoint essential simple closed curves on $\Sigma_3$, $\alpha_1,\alpha_2,\alpha_3$, generate a copy of $F_3$, the free group on three generators, in $\pi_1\Sigma_3$. Associated to $F_3$ is a covering $\widehat{\Sigma_3}\to \Sigma_3$, with $F_3$ acting by deck transformations on $\widehat{\Sigma_3}$ so that $\widehat{\Sigma_3}/F_3 = \Sigma_3$. Since $Z(F_3)<F_3$, we have a quotient cover $\widehat{\Sigma_3}/Z(F_3)\to \Sigma_3$, with the deck group $F_3/Z(F_3)\cong \mathbb{Z}^3$. Geometrically, take the "asterisk" fundamental domain (the one you get by cutting along the red curves) and glue it so it looks like the Cayley graph of $F_3$. Then quotient by the action of the commutators of the generators.
The fundamental group $\pi_1(\Sigma_3)$ projects onto the first homology group $H_1(\Sigma_3;\mathbb{Z})$. Associated to $H_1(\Sigma_3;\mathbb{Z})$ is an abelian cover $A(\Sigma_3)\to \Sigma_3$. The images of $\alpha_1,\alpha_2,\alpha_3$ generate a copy of $\mathbb{Z}^3$ in $H_1$, which corresponds to a $\mathbb{Z}^3$-covering space of $\Sigma_3$.
I think there should be a nice picture of a fundamental domain of this cover in $\mathbb{H}^2$ by stringing together right-angled dodecagons, but I don't quite see it.
Imagine a closed chain of tori (that is, torus $i$ is connected-summed to tori $i+1$ and $i-1$ mod $2k.$) That is a (cyclic) $k$-fold cover of a double torus (surface of genus $2.$).
Best Answer
Take the dodecagon at the origin with one pair of edges intersecting the $y$-axis (call them the top and bottom faces) and one pair intersecting the $x$-axis. Cut the polygon along the $x$ axis, and un-identify the left and right faces. This gives two octagons, each with an opposing pair of unmatched edges. Identify these new edges, and you should have what you're looking for.