Why can projective varieties just have abelian group operations? - MathOverflow

Why can projective varieties just have abelian group operations?

2011-03-21T13:04:24Z

I just started to read Shimura - Automorphic forms and number theory (Lecture notes in mathematics, 54). On page 20 or so, he mentions that every projective variety which is an algebraic group, is necessary abelian.

Why?

Answer by Qfwfq for Why can projective varieties just have abelian group operations?

2011-03-21T13:17:52Z

A very accessible proof (at the beginning of the book for the case over $\mathbb{C}$, and further on in the book for any characteristic) of that statement is present in Mumford's book Abelian varieties.

Answer by Francesco Polizzi for Why can projective varieties just have abelian group operations?

2011-03-21T13:23:22Z

I borrow this proof from [Birkenhake-Lange, Complex Abelian Varieties, Lemma 1.1.1].

Let $X$ be a projective variety having a group structure. I assume that we are working over $\mathbb{C}$.

Consider the commutator map $\Phi(x,y)=xyx^{-1}y^{-1}$, and let $U$ be a coordinate neighborhood of $1 \in X$. By the continuity of $\Phi$, and since $\Phi(x,1) =1 \in U$, for all $x \in X$ we can find open neighborhoods $U_x$ and $W_x$ such that $\Phi(U_x, W_x) \subset U$.

Since $X$ is compact, finitely many $V_x$ cover $X$. Calling $W$ the intersection of the corresponding subsets $W_x$, we get $\Phi(X, W) \subset U$.

Now $\Phi(1, y)=1$ for all $y \in W$. Since holomorphic functions on a compact variety are constant, it follows $\Phi(X, W)\equiv 1$. Being $W$ open and non-empty, this in turn implies $\Phi(X, X) \equiv 1$, which is our claim.

Notice that "projective" is not really necessary, in fact what we actually use in the proof is "compact complex". Indeed, pushing further this argument (by a straightforward use of the exponential map) one can show that any compact complex connected Lie group is a complex torus.

Answer by Emerton for Why can projective varieties just have abelian group operations?

2011-03-21T13:23:54Z

There are several different ways to see this. Here is one:

Let $G$ be our irreducible projective algebraic group variety over the field $k$, with identity element $e$. The group $G$ acts on itself by conjugation, and this action fixes $e$. Thus this induces a $k$-linear action of $G$ on the local ring $\mathcal O_e,$ and hence on the (finite-dimensional) quotients $\mathcal O_e/\mathfrak m_e^n$ for each $n$. Now any morphism from the irreducible projective variety $G$ to the affine variety $End_k(\mathcal O_e/\mathfrak m_e^n)$ must be constant, and so the $G$-action on each $\mathcal O_e/\mathfrak m_e^n$, and hence on $\mathcal O_e$ itself, must be trivial.

Since $G$ is irreducible, it is now easy to see, using the fact that the conjugation action induces a trivial action on $\mathcal O_e$, that the conjugation action is in fact trivial on $G$ itself, and hence that $G$ is commutative.

(This argument breaks down if $G$ is not projective, because then $G$ can have non-trivial morphisms to the matrix rings $End_k(\mathcal O_e/\mathfrak m_e^n)$; it is instructive to think about this in the case $G = GL_n(k)$. It also breaks down if $G$ is not irreducible, e.g. if $G$ is a finite non-abelian group, then we can think of it as a zero-dimensional projective algebraic group. The point in this case is that one can't make the "analytic continuation" argument from the action on $\mathcal O_e$ to all of $G$.)