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Can someone explain this in simple terms?

What I don't get is, how can a 1-D curve be topologically equivalent to a 2-D surface? I guess isomorphism cannot hold between the 2 sets, and I don't see the homeomorphism.

Yan King Yin
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  • Is it a complex curve? – Lucien Sep 11 '13 at 11:12
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    Homeo between 2-dim torus and 1-dim curve is impossible due to the invariance of domain theorem. What kind of equivalence are you talking about? What cubic curve? EDIT: I assume it is a real curve, but probably it is not. – savick01 Sep 11 '13 at 11:14
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    The confusion here is with the term "curve" and "surface". A complex curve of genus 1, that is, a torus, is actually a real surface. Or, for example, a complex line (which is locally like $\mathbb{C}$) is really homeomorphic to the Riemann sphere (a real 2-dimensional surface). So when someone says "curve", make sure you know if they're talking about a complex curve or a real curve! – rfauffar Sep 11 '13 at 11:23
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    Mind that the complex line is identified to a real plane (Argand-Gauss). – Andrea Mori Sep 11 '13 at 11:31
  • Thanks, that answered my question: I forgot that a complex curve has 2x real dimensions. – Yan King Yin Sep 11 '13 at 14:51

2 Answers2

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If your curve is given by a polynomial $p(x,y)=0$, you need to consider all complex $x,y$ satisfying the equation (so it's not surprising that you're getting something 2-dimensional), and you also need to add points at infinity to get a closed surface.

If the equation if $y^2=q(x)$, where $q$ is a cubic polynomial: for every $x\in\mathbb C$ s.t. $q(x)\neq0$ you have 2 solutions for $y$, and if $q(x)=0$ then you have just one solution (namely $y=0$). So you can imagine the complex cubic curve as lying over the plane of (complex) $x$-es, in two layers everywhere, except for the 3 roots of $q$. There is just one "point at infinity" to be added, we can call it $x=\infty$, $y=\infty$. Now we see the surface as lying over the (Riemann) sphere of $x$-es.

To understand that the surface is actually a torus requires some work: we slit the $x$-sphere along two cuts, one connecting 2 roots of $q$ and the second connecting the third root with $\infty$. As we remove these arcs, the surface above splits to two spheres with the same cuts. To get the surface back in whole, we need to re-glue the cuts: imagine two spheres with two holes each, when we glue the corresponding holes, we get a torus:

enter image description here

user8268
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  • I am a bit confused. Since the Riemann sphere $S^2$ can be visualised according to what you say making those branch cuts would mean that we would "cut" the piece from $x_1$ to $x_2$ and the piece $x_3$ to $\infty$, as you say as well, but removing the arcs I do not see how one gets $S^2$ to be separated into two $S^2$ and additionally I do not see how when we glue them back together we get $T = S^1 \times S^1$. Would you mind to explain a bit further? – Marion Apr 19 '16 at 20:42
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An elliptic curve is a $1$-dimensional smooth variety of genus $1$ over some field. If the field is $\mathbb{C}$, the complex dimension is $1$, hence the real dimension is $2$. Using Weierstrass's elliptic functions one can prove that the group of $\mathbb{C}$-valued points of a complex elliptic curve is a torus, i.e. isomorphic to $\mathbb{C}/\Lambda$ for some lattice $\Lambda$.

Martin Brandenburg
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