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Algebraic geometry predicts a degree 2 branched cover from an elliptic curve to the projective line. What does this map look like topologically?

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The Weierstrass pe-function will do. – Anweshi Feb 3 '10 at 21:27
up vote 5 down vote accepted

One example: lay your $g$-holed torus $T$ out flat and draw a line the long way through each hole. It hits the torus in $2g + 2$ points. Consider the 180 degree rotation $w$ through that line. Now consider the space $T/w$ formed by identifying two points $P$ and $Q$ if $P = wQ$ (since $w^2 = 1$ we also have $Q = wP$). I claim that it's not too hard to see that $T/w$ is isomorphic to the projective line, and the $2g+2$ points which hit the line are the ramification points.

edit: This is of course more general than you were asking, but the picture is completely general when you're talking about topology.

edit 2: A picture of this (albeit approached from the perspective of starting on the projective line and cutting slits) can be found in section 20e of Fulton's Algebraic Topology book.

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An elliptic curve is an abelian group. The quotient with respect to the equivalence relation $x\sim -x$ is a genus 0 curve. The branched cover is the projection to the quotient and the singular points are the 4 points of order 2 on the curve.

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This quotient is not a torus? – Anweshi Feb 3 '10 at 19:22
Anweshi -- applying Riemann-Hurwitz we get the Euler characteristic of the quotient: it is (0-4)/2+4=2 (since the images of all singular points are distinct). – algori Feb 3 '10 at 19:25
By the way, the Euler characteristic of the quotient by the hyperelliptic involution described in the stankewicz's answer can also be computed using Riemann-Hurwitz: we get ((2-2g)-(2+2g))/2+2+2g=2. – algori Feb 3 '10 at 19:31

If your elliptic curve is $\{(x,y)~|~y^2=x^3 + ax + b\}$ then take the projection $(x,y) \to x$. This has 4 branch points at the 3 roots of $x^3 + ax + b$ and $\infty$. From this perspective, it's easier to see the resulting $\mathbb{CP}^1$, and harder to see that the elliptic curve is a topological torus.

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You can do a reverse construction: start with a sphere without 4 points; now add two points over each one in such a way that every time you go around one hole the two points get interchanged.

The same Riemann-Hurwitz calculation guarantees that you get a torus. If you have complex structure on you sphere without 4 points you get one on top as well; a beautiful fact is that you get all complex structures on a torus — in other words, all elliptic curves over $\mathbb C$ — that way.

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Cover of Francis' "A topological picturebook". (I've linked to the correct edition - the new edition has a different picture on the cover.)

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