Group over algebraic curves having genus greater than 1 Elliptic curves have a group structure over the rational points. Why is this impossible for curves having genus greater than 1? I read somewhere that this impossibility is implied by Faltings theorem, which states that the number of rationals is finite. I don't see the implication. The group over the rational points could be finite.
 A: You seem to be interested in group structures on complete (= proper) curves. In general, if you have a complete group variety X of dimension $d$ over $\mathbb{Q}$, then you can prove that the tangent (or cotangent) sheaf is free. So you get an isomorphism $\Omega_X \cong \mathcal{O}_X^{\oplus d}$.
This means that $\dim H^0(X,\Omega_X) = d \cdot \dim H^0(X,\mathcal{O}_X) = d$.
If $X$ is a curve, so $d = 1$, note that $\dim H^0(X,\Omega_X)$ is the genus of $X$. Hence $X$ has genus $1$.
(Fun fact: Since $X$ is proper, one can also show that the group structure must be commutative!)
A: You might consider complex Lie group structures. If you want a holomorphic Lie group structure on a reduced complex space, the space acts transitively on itself, so there are no singular points, so it is a complex manifold. One dimensional connected complex Lie groups are abelian, because the bracket is a vector valued 2-form. So you get the complex line and curves covered by the complex line, i.e. elliptic curves and the punctured complex line.
Sophus Lie classified all connected Lie groups in low dimensions, modulo replacing by covering groups. Lie's proofs work on any analytic manifolds, real or complex, without modification. His proofs need some help for smooth manifolds, but smooth Lie groups are always analytic, via the exponential map. You can see the classification of the Lie algebras in low dimension in Sophus Lie, Gesammelte Abhandlungen. Band 5, Johnson Reprint Corp., New York, 1973, .767–773.
A: Let $G/k$ be a proper smooth connected non-trivial group variety. For $1 \neq x \in G$, the translation by $x$ has no fixed point, so by the Lefschetz fixed point formula for $\ell$-adic cohomology (using $G$ connected), the Euler characteristic of $G$ is $0$. If $G/k$ is $1$-dimensional, this implies that $0 = \chi(G) = 2-2g$, so $g=1$.
A: The following simple proof works as soon as the curve contains infinitely many points (for instance, when it is defined over an algebraically closed field).
If you have an algebraic group $G$, then for every fixed $a \in G$ the translation $x \mapsto x+a$ is an automorphism of the underlying algebraic variety $X_G$, in particular $G$ embeds into $\mathrm{Aut}(X_G)$.
But it is well known that a curve of genus $g \geq 2$ has at most finitely many automorphisms (for instance, in characteristic $0$ there are at most  $84(g-1)$ of them).
A: Curves of $g>1$ corresponds to Riemann surfaces with hyperbolic metric. But a Lie group is an $H$-space, so the first fundamental group must be abelian. And we have a contradiction. For $g=0$ case it follows from a version of hairy ball theorem. If my memory serves, this can also be proved using cohomology of lie algebra. 
I am not entirely sure if the argument can be carried through for curves over a number field, though. I understand your main interest is in the rational points. I suspect the etale fundamental group may be related. 
