6
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ The Weierstrass pe-function will do. $\endgroup$
    – Anweshi
    Feb 3 '10 at 21:27
5
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ This quotient is not a torus? $\endgroup$
    – Anweshi
    Feb 3 '10 at 19:22
  • 1
    $\begingroup$ 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). $\endgroup$
    – algori
    Feb 3 '10 at 19:25
  • 1
    $\begingroup$ 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. $\endgroup$
    – algori
    Feb 3 '10 at 19:31
5
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

The cover of the first edition of Francis' book: "A topological picturebook".

Birman's favorite involution

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.