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Let $E:y^2=x(x-1)(x-\lambda)$ be the Legendre form of an elliptic curve $E$ defined over $\mathbb{C}$. The ramified covering $E\to \mathbb{P}_{1}$ defined so that $(x,y)\mapsto x$ has two branches and ramification points $0,1,\lambda,\infty$.

The two branches $y=\pm \sqrt{x(x-1)(x-\lambda)}$ are therefore glued at the ramification points. $E$ would turn out to be hence the glueing of two spheres $\mathbb{P}_{1}$ at the 4 distinct points $0,1,\lambda,\infty$.

Such an object has not genus 1, therefore there has to be a mistake.

In J. Silverman's AEC at the very beginning of chapter 6 it is explained that what one does is actually to glue the spheres along the two segments $\overline{\infty 0}$ and $\overline{1\lambda}$. This gives a torus, but still I don't understand why all the points of the segments has to be removed by the two branches? And why not to choose another couple of disjoint segments like $\overline{01}$ and $\overline{\infty\lambda}$? All integrals of $dx/y$ over paths in the complement of the ramification points are perfectly defined in the same branch once the sign of the square root is chosen, see for example figure 6.5 at page 159, where a closed path $\alpha$ is taken across both segments.

So my question is: why do we remove the entire segments if the junction points of the two branches are only 4?

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    $\begingroup$ It's been a while since I've thought about this picture, but IIRC, any two segments should work. The point is that you want to disallow loops which wind around an odd number of branch points, since then the integral of $dx/y$ around such a loop is only well defined up to $\pm 1$. After all, your observation that two spheres glued along 4 points is not genus 1 is correct. Indeed, if you instead glue the spheres along the complement of two segments, then essentially by stretching the "slits", you can see that each sphere becomes half a donut (chopped "in the way that one might chop a donut") $\endgroup$ – Will Chen Dec 22 '16 at 21:44
  • $\begingroup$ Also note that in this picture, a loop that winds around a puncture once gives rise to a path in the torus which is not closed (ie, not a loop)! Hopefully this helps. $\endgroup$ – Will Chen Dec 22 '16 at 21:46
  • $\begingroup$ In other words, if it would really be two $\mathbb P^1$s glued at four branch points, then the restriction of $(x,y)\mapsto x$ to a small circle centered at a branch point would be a trivial covering, consisting of two disjoint circles (one in each of the glued $\mathbb P^1$s). Whereas in fact it is a nontrivial covering, consisting of a single circle. $\endgroup$ – მამუკა ჯიბლაძე Dec 22 '16 at 22:02
  • $\begingroup$ Sorry but I do not understand. Why a loop around 1 or 3 branch points is lifted to two disjoint paths if the spheres are glued in the 4 points only? Why things are different for loops around 2 points? And what about around 4? If things are also fine then why do not we glue the spheres along the entire area between the 4 points? That would even give a genus 0 object. An object covering another one is the glueing of the branches in the ramification points (am I right?). Of course we expect a torus but it is not. Something is missing or wrong. $\endgroup$ – Hair80 Dec 23 '16 at 8:37
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    $\begingroup$ "once the sign of the square root is chosen" -- you're not choosing a sign, you're choosing a branch, and that's why you need the cuts. You can't define $\sqrt{z}$ on the complex plane continuously, but you can't define it on the complex plane with 0 removed either -- you need to remove a line from 0 to infinity. Now generalise to your situation and paradox resolved. $\endgroup$ – znt Dec 23 '16 at 19:55

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