Find all double covers $\Sigma_2\to T$ with specified branched points

Question

Given a torus $E$ with two distinct points $O,P\in E$, there are, up to isomorphism, 4 double corves $X\to E$ branched exactly at $O,P$. See the Kodaira-Parshin construction on p.4 here: http://virtualmath1.stanford.edu/~conrad/mordellsem/Notes/L01.pdf
By Riemann-Hurwitz formula, $g(X)=2$. I find one example, here $E=X/G$, $G=\Bbb{Z}/2\Bbb{Z}$. But how to find the other 3 cases?

Answer 1

From your comments to an earlier draft, it seems that you do already know some of the theory of covering spaces and the classification of covering maps. But I'll leave my outline of the approach here anyway, before turning to a description of actual examples.

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First, for any double branched covering map $f : X \to E$ branched exactly over $O,P \in E$, the restricted map $f : X - f^{-1}\{O,P\} \to E-\{O,P\}$ is an ordinary covering map. Furthermore, the degree of the restricted map is equal to the degree of the given map. Knowing this, your goal shifts to classifying the double (unbranched) covers of $E - \{O,P\}$, and then determining which of those determine double covers of $E$ branched exactly over $\{O,P\}$.

Second, you have to know the general classification theorem for covering spaces (that link is pretty terrible, but on the other hand this theorem is covered very well in topology textbooks, such as Munkres "Topology"). Then you have to apply this theorem to classify double covering maps over $E - \{O,P\}$ of degree $d$, which up to covering equivalence are in one-to-one correspondence with conjugacy classes of index $2$ subgroups of $\pi_1(E - \{O,P\},q)$ where I have chosen an arbitrary base point $q \in E - \{O,P\}$.

Third, you have to know that $\pi_1(E - \{O,P\})$ is a free group of rank $3$, you need $3$ loops on $E - \{O,P\})$ which represent a free basis, and you have to know how to write out exactly all its subgroups of index $2$ and collect them into equivalence classes; here your task simplifies somewhat, because index $2$ subgroups are always normal, hence distinct index $2$ subgroups are never conjugate to each other. Furthermore, the quotient group is cyclic of order $2$, so the index $2$ subgroups correspond one-to-one with the surjective homomorphisms $\pi_1(E-\{O,P\}) \to \mathbb Z / 2$.

Finally, for each index $2$ subgroup, you take the corresponding (unbranched) covering map, you re-insert $\{O,P\}$ and $f^{-1}\{O,P\}$ to get a branched covering map, and now you have to determine whether it is branched exactly over $\{O,P\}$. To do that you must examine two loops in $E - \{O,P\}$ encircling $O$ and $P$, and determine for each one whether or not the element of $\pi_1(E-\{O,P\})$ represented by that loop it is or is not in the classifying subgroup of the covering map: the covering is branched over $O$ if and only the corresponding element is not in the subgroup, and similarly for $P$.

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So now let's list all of the degree two covering maps of $E - \{O,P\}$, and the associated branched covers of $E$.

One way to represent $E$ topologically is as the quotient of a regular hexagon in the plane, by gluing opposite sides of that hexagon, using the gluing pattern $a \, b \, c \, \bar a \, \bar c \, \bar c$. The six vertices of the hexagon fall into two vertex cycles of cardinality $3$ representing the two points $O,P$, and as you walk around the boundary of the hexagon you alternate between those two vertex cycles. We can therefore represent $E - \{O,P\}$ topologically as the quotient of the hexagon with its 6 vertices deleted.

Let's use the center point of the hexagon as the base point $q$ of $E - \{O,P\}$. The rank 3 free group $\pi_1(E-\{O,P\})$ has a free basis $A,B,C$ where, for example, $A$ is represented by the loop based at $q$ which first goes from $q$ along the line segment to the midpoint of the $a$ side, which is identified with the midpoint of the $\bar a$ side, and then goes along the line segment back to $q$. Using the description of the vertex cycles given earlier, the elements of $\pi_1(E-\{O,P\})$ representing loops around $O$ and $P$ are $$A \, C \, \bar B \quad\text{and}\quad B \, \bar A \, \bar C $$

Now let's pick a random index 2 subgroup to see what happens. We choose the images of $A,B,C$ in $\mathbb Z/2$, which I shall do by flipping a coin .... hmmm... where's a coin when you need it ..... $$A \mapsto 1, \quad B \mapsto 1, \quad C \mapsto 0 $$ it follows that $A \, C \, \bar B \mapsto 0$ and that $B \, \bar A \, \bar C$ maps to $0$. Thus both of those elements are in the kernel of the homomorphism, hence the associated covering map $X \mapsto E$ is branched over neither $O$ nor $P$.

From this experience, we can see that we get exact branching over both of $O$ and $P$ if and only if exactly one of $A,B,C$ maps to $1$ or all three of $A,B,C$ maps to $1$. That gives exactly four distinct branch coverings of $E$ such that the branch set is exactly $O$ or $P$.

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Finally, how to depict these?

Let me depict just one of them branched coverings, namely the one which is determined by the homomorphism where all three of $A,B,C$ map to $1$. Perhaps by working through this example, one can get enough experience and courage to try the rest oneself.

Starting with our side-labelled original hexagon, which we denote $H_W$ and we picture in the color white, we take another such hexagon depicted by moving the original one off to the right, denoting it $H_B$ and picturing it in the color black. We now do our gluing: the $A$ side of $H_W$ glues to the $\bar A$ side of $H_B$, and the $\bar A$ side of $H_W$ glues to the $A$ side of $H_B$; and similarly for the other side pairs.

So there's the best visualization I can offer: representing the branched covering surface by gluing two hexagons together.