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I'm reading Milne's Elliptic Curves and came across this statement: If a nonsingular projective curve has a group structure defined by polynomial maps, then it has genus 1. In this question a similar question was asked, and an answer given there (the one that was not accepted) someone backs this up using machinery/notation that I do not understand. Can someone give or direct me to a proof of this statement? It doesn't have to be very formal; this isn't homework. Also, in the affine case this statement clearly breaks down - the affine line has the group structure of addition, for instance. Is there some sort of correction or does the group structure bear no relation to the genus in this case?

Community
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Julien Clancy
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3 Answers3

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Just to throw another way to formalize this fact, if $X$ is a variety that is a group, then the canonical sheaf $\omega_{X}$ must be trivial. The intuitive reason is that there is a canonical way to identify the tangent space $T_{x}$ at any point $x \in X$ with the tangent space $T_{e}$ at identity. (Namely, the map $y \leadsto y * x^{-1}$).

If $X$ is a curve, then Riemann-Roch implies that $deg(K) = 2g-2$, where $K$ is the divisor corresponding to $\omega_{X}$. Since $\omega_{X}$ is trivial, $deg(K) = 0$. This implies that $g = 1$.

Piotr Pstrągowski
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For complex algebraic curves, this is actually very easy. Suppose we have an algebraic group law on a complex algebraic curve. Then it is necessarily continuous in a classical topology, so we get a topological group. Now, a fundamental group of a topological group has to be abelian (this is an interesting and not that hard exercise). But, the curves with genus > 1 don't have abelian fundamental groups!

xyzzyz
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    Alternatively, the Euler characteristic of a compact Lie group is zero (e.g. because it admits non-vanishing vector fields), and a compact Riemann surface of genus $g$ has Euler characteristic $2 - 2g$. – Qiaochu Yuan Apr 22 '13 at 21:06
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    Can you explain the last assertion, i.e. genus $\neq 1 \Rightarrow$ fundamental group not abelian? – Julien Clancy Apr 22 '13 at 22:15
  • Complex algebraic curve with genus n topologically is a sphere with n handles attached. Usual topological methods of calculating a fundamental group give you desired result. – xyzzyz Apr 23 '13 at 14:06
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I accepted another answer but I'm going to post a more elementary (from my perspective) way to prove this. The Riemann-Hurwitz Formula gives the nice bound $|\text{Aut}(\mathcal{C})| < \infty$ for a $g(\mathcal{C}) > 1$ (we'll assume characteristic zero so all maps are tamely ramified). But each point gives the translation automorphism, and these are all clearly distinct, whence we obtain a contradiction for $g(\mathcal{C}) > 1$. So the question becomes: does the projective line have a group law? Of course not, for then a hyperelliptic curve $E$ ($g(E) > 1$) would have one too. The gist is that if we select a degree-2 map $f \colon E \to \mathbb{P}^1$, totally ramified, then it must be injective, so we can pull back the group law.

Julien Clancy
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  • How do you prove the finiteness of Aut(C) using just Riemann-Roch ? Your proof of $P^1$ having no group structure is incomplete. You can't pull-back a group law by $f$ (moreover $f$ is never injective). –  May 01 '13 at 07:48
  • @QiL'8 It's true that it's incomplete. Can you think of a way to show that $\mathbb{P}^1$ has no group law by morphisms? Also, why would $f$ not be injective if it were totally ramified? I meant Riemann-Hurwitz, not Riemann-Roch. – Julien Clancy May 01 '13 at 19:15
  • if your morphism to $\mathbb P^1$ is injective, over an algebraically closed field of characteristic $0$, it will be an isomorphism because the degree of the morphism will be one. See this question http://math.stackexchange.com/questions/341281/. The easiest proof that $\mathbb P^1$ is not an algebraic group I know is again the answer given by Piotr. –  May 02 '13 at 14:46