I need to know the degree of this covering map $R \rightarrow S$:
$$T^{2}\#T^{2}\#T^{2}\#T^{2}\#T^{2}\#T^{2} \rightarrow T^{2}\#T^{2}$$
I have that genus of $R$, $g_{R} = 6$, and genus of $S$, $g_{S} = 2$, then:
So using the Riemann-Hurwitz formula $\chi(R) = N\chi(S)$, where $N$ is the degree of the covering map, then solving for $N$ I have that $N = 5$. Am I right? How would I draw a picture of this covering map?
Thanks.
Let $\Sigma_g$ denote the connected sum of $g$ tori.
You are correct that there is a degree five covering map $\Sigma_6 \to \Sigma_2$. One way to view this covering map is to think of $\Sigma_6$ as in the following (poorly drawn) picture:
The arrows indicate a map which generates the group of deck transformations (which is isomorphic to $\mathbb{Z}_5$). The five regions are fundamental domains for the action of the deck transformations. After quotienting $\Sigma_6$ by this action, we have identified all of these regions with each other (so we can pretend we just have one), and we identify the two boundary components of this region which leaves us with a surface of genus two.
More generally, if $\chi(\Sigma_g) = k\chi(\Sigma_h)$, then $g = 1 + k(h - 1)$, so we can view $\Sigma_g$ with $k$ equally spaced chains of $\Sigma_{h-1}$ attached to a single torus, then the group $\mathbb{Z}_k$ acts by rotations and the quotient is $\Sigma_h$. In particular,
For $k \geq 0$, $\chi(\Sigma_g) = k\chi(\Sigma_h)$ if and only if there is a $k$-sheeted covering $\Sigma_g \to \Sigma_h$.
The coverings described this way all have deck groups $\mathbb{Z}_k$, but there are many other possible deck groups.