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.
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2$\begingroup$ A curve of genus $g\geq 1$has an associated group (abelian variety), its Jacobian variety, of dimension $g$. For $g>1$ the Jacobian is not a curve. $\endgroup$– KConradCommented Jul 19, 2018 at 7:20
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5$\begingroup$ It's not just that there's a group structure on the rational points; e.g. there's a group structure on the points over any field extension $K$ of the ground field, and these group structures are compatible with inclusions of field extensions. More precisely, elliptic curves are what are called group schemes. $\endgroup$– Qiaochu YuanCommented Jul 19, 2018 at 7:20
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$\begingroup$ Does this property not hold for curves with genus greater than 1? In the case, one could have a group over the rationals that is not compatible with group over extended fields. At least in principle. $\endgroup$– AlmCommented Jul 19, 2018 at 7:46
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5$\begingroup$ Every non-empty set can be endowed with a group structure. But that is not (should not be?) what you are after. I do not see why you would be interested in putting a group structure on the rational points of some variety if you don't ask for some compatibility with the geometric structure. (It is like have a set, endowing it with a group structure and a topology, but not caring whether multiplication is continuous...) $\endgroup$– jmcCommented Jul 19, 2018 at 7:53
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3$\begingroup$ By the way: there exist elliptic curves over the rationals that have finitely many rational points. $\endgroup$– jmcCommented Jul 19, 2018 at 7:54
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5 Answers
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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 $g=1$.
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1$\begingroup$ Is there also a Lefschetz fixed point formula for étale cohomology with compact supports? Then one could drop the assumption "proper". $\endgroup$– user19475Commented Jul 19, 2018 at 8:19
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1$\begingroup$ It seems to me that the topological formula $\chi(G)=2g-2$ only works over $\mathbb{Q}$ and in the proper case. $\endgroup$ Commented Jul 19, 2018 at 8:34
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1$\begingroup$ I think you're right: One needs properness since $\mathrm{H}^i_c(\mathbf{G}_a,\mathbf{Q}_\ell)$ is non-trivial only in dimension $i = 2$. $\endgroup$– user19475Commented Jul 19, 2018 at 9:48
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$\begingroup$ Concerning your answer, there is something that I didn’t get. Algebraic groups can be built also on curves with genus 0, whereas you end up with g=1. For example, a line has a natural group where the group law is given by summing each coordinate (provided that line passes through the origin). What did I miss? $\endgroup$– AlmCommented Nov 6, 2018 at 19:44
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1$\begingroup$ @AlbertoMontina: $\mathbf{G}_a$ and $\mathbf{G}_m$ are not proper. $\endgroup$– user19475Commented Nov 7, 2018 at 2:31
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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).
answered Jul 19, 2018 at 8:05
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2$\begingroup$ I think that "automorphism of $G$" must mean only as a variety, not (as it appears to mean) as a group. $\endgroup$– LSpiceCommented Jul 19, 2018 at 15:35
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$\begingroup$ A translation is never a group variety, since it does not fix the origin, so I do not think that any confusion is possible. However, I clarified that part. $\endgroup$ Commented Jul 19, 2018 at 18:33
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$\begingroup$ Does the bound on the number of automorphisms also have a simple proof? $\endgroup$– WojowuCommented Jul 20, 2018 at 11:22
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$\begingroup$ At least over $\mathbb{C}$, the proof is rather elementary. You can find it in R. Miranda's texbook Algebraic curves and Riemann Surfaces (and in many other places). $\endgroup$ Commented Jul 20, 2018 at 11:37
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3$\begingroup$ Over other algebraically closed fields of characteristic zero, this follows from the Lefshetz principle. In characteristic $p$, the specific bound is false. However, the only thing this argument needs is that there are not infinitely many automorphisms, which follows from the observing that the tangent space to the space of automorphisms is the global sections of the tangent bundle. $\endgroup$ Commented Jul 21, 2018 at 7:37
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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!)
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1$\begingroup$ For a group variety $G/k$, one always has $\Omega^1_{G/k} \cong \Omega^1_{G/k}(e) \otimes_k \mathscr{O}_G$, one does not need properness. $\endgroup$– user19475Commented Jul 19, 2018 at 9:05
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$\begingroup$ Right! It's an artifact of me trying to answer a question about complete curves... $\endgroup$– jmcCommented Jul 19, 2018 at 9:08
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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.
answered Jul 19, 2018 at 8:21
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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.
answered Jul 21, 2018 at 7:29
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2$\begingroup$ One can do this for the etale fundamental group. This requires the fact that the curve is projective (or a characteristic zero assumption) to ensure $\pi_1(X \times X) \cong \pi_1(X) \times \pi_1(X)$, it is false for some groups in characteristic $p$. $\endgroup$ Commented Jul 21, 2018 at 7:38
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$\begingroup$ @WillSawin: Thanks! I am really happy to know. $\endgroup$ Commented Jul 21, 2018 at 7:39