Degree 2 branched map from the torus to the sphere - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:42:56Z http://mathoverflow.net/feeds/question/14024 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14024/degree-2-branched-map-from-the-torus-to-the-sphere Degree 2 branched map from the torus to the sphere Randy Brown 2010-02-03T19:06:41Z 2010-02-03T21:11:29Z <p>Algebraic geometry predicts a degree 2 branched cover from an elliptic curve to the projective line. What does this map look like topologically?</p> http://mathoverflow.net/questions/14024/degree-2-branched-map-from-the-torus-to-the-sphere/14026#14026 Answer by algori for Degree 2 branched map from the torus to the sphere algori 2010-02-03T19:14:47Z 2010-02-03T19:14:47Z <p>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.</p> http://mathoverflow.net/questions/14024/degree-2-branched-map-from-the-torus-to-the-sphere/14027#14027 Answer by stankewicz for Degree 2 branched map from the torus to the sphere stankewicz 2010-02-03T19:19:35Z 2010-02-03T19:26:24Z <p>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.</p> <p>edit: This is of course more general than you were asking, but the picture is completely general when you're talking about topology.</p> <p>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.</p> http://mathoverflow.net/questions/14024/degree-2-branched-map-from-the-torus-to-the-sphere/14037#14037 Answer by Douglas Zare for Degree 2 branched map from the torus to the sphere Douglas Zare 2010-02-03T19:46:10Z 2010-02-03T19:46:10Z <p>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. </p> http://mathoverflow.net/questions/14024/degree-2-branched-map-from-the-torus-to-the-sphere/14043#14043 Answer by Ilya Nikokoshev for Degree 2 branched map from the torus to the sphere Ilya Nikokoshev 2010-02-03T20:15:20Z 2010-02-03T20:15:20Z <p>You can do a reverse construction: start with a <strong>sphere without 4 points</strong>; now add two points over each one in such a way that every time you go around one hole the two points get interchanged.</p> <p>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 <strong>all complex structures on a torus</strong> &mdash; in other words, all elliptic curves over $\mathbb C$ &mdash; that way.</p> http://mathoverflow.net/questions/14024/degree-2-branched-map-from-the-torus-to-the-sphere/14047#14047 Answer by Sam Nead for Degree 2 branched map from the torus to the sphere Sam Nead 2010-02-03T20:41:15Z 2010-02-03T20:41:15Z <p><a href="http://www.scribd.com/doc/14009839/George-K-Francis-A-Topological-Picture-Book-1988-213p-0387964266" rel="nofollow">Cover</a> of Francis' "A topological picturebook". (I've linked to the correct edition - the new edition has a different picture on the cover.)</p>