# Hyperelliptic curve of genus 2 over R

I know that the points of an elliptic curve over $\mathbb{Q}$, $\mathbb{R}$ or other field $K$ form a group, particularly the most common example to explain the naive way is with this curve $y^2=x^3-x$, internally what is happening is that line with coefficients over the ground field intersects with $K$-rational points, my questions are:

1. Why a curve $y^2=x^5-x$ which is of genus 2 and "looks like" $y^2=x^3-x$ if it is taken over $\mathbb{R}$ what is going to fail in the group operation? (every line must intersect 3 $\mathbb{R}$-rational points (with counted multiplicity) , what I believe is that 'associativity' is going to fail... but I dont see it, but I know the points doesn't form a group (The jacobian does)

2. Are the points of the last genus 2 curve a set which forms a group under the usual line-tangent rule for some (finite) field?, I think no... but why

I know that the points of this curve does not form a group, but in $\mathbb{R}$ the curve 'looks' the same for having a similar naive definition for the explanation of the group operation,

1. If I 'force' the group operation in this curve calculating $nP$ (generating it) for some point $P\in H(\mathbb{R})$ where $H(x,y)=y^2-x^5+x$ do I get a group? (Again... It shouldnt but why?)

Thanks

• In general a curve $C$ of genus $g>0$ has a Jacobian $J(C)$, which is a group of dimension $g$. If $C$ has a rational point then it embeds into $J(C)$, but not as a subgroup except in the $g=1$ case of an elliptic curve in which case $C$ is its own Jacobian. – Noam D. Elkies Feb 22 '15 at 18:58
• And, aren't all genus 2 curves over any field hyperelliptic? – Qfwfq Feb 22 '15 at 19:46
• @Qfwfq: Yes, they are. – Michael Stoll Feb 22 '15 at 19:58
• My inclination would be take an accurate drawing of the real locus of $y^2=x^5-x$ and see for myself that something (probably associativity) breaks down. – Lubin Mar 17 '15 at 16:15
• Lubin, yes, that's what I'm in fact asking, As @Michael Stoll replied, there are more intersection points, and there is not a way to build a group with the points, but what I ask is that if I take two P,Q $\mathbb{R}$-rational points on $y^2=x^5-x$, the line between them is going to intersect another $\mathbb{R}$-rational point, (can this be assumed? I think thats the main problem) If I take that point naively to be -(P+Q), what I get ?, does associativity fails? What if I fix a $\mathbb{R}$-rational point R not lying in the x axis and I generate $nR$ naively also nR, that should fail also. – Eduardo R. Duarte Oct 9 '15 at 14:04

Also, it is a result of topology that a topological group has to have vanishing Euler characteristic, but the Riemann surface you get here from the complex points has Euler characteristic $-2$, so there is not even a continuous (in the complex topology) group law on it. (Of course, you can take a homeomorphism of the real points with $S^1 \times {\mathbb Z}/2{\mathbb Z}$ and so force a continuous group law on the real points, but this is not what you are asking.)