## Why can projective varieties just have abelian group operations?

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?

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More generally, every abelian scheme is abelian in the honest sense of the word. The standard proof uses the rigidity lemma: Every morphism between abelian schemes fixing $e$ is a group morphism. In particular, $x \mapsto x^{-1}$ is a group morphism, i.e. the group is abelian. – Martin Brandenburg Mar 21 2011 at 16:35

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.

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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 $G$.)

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Since the title of the questions mentions <i>group operations</i>, it may be interesting to observe that this proof shows the more general result: let $G$ be an algebraic group acting faithfully on an algebraic variety, if $G$ has a fixed point, then $G$ is linear. (The same holds if $G$ is an abstract group of finite type, or in the analytic category and gives strong constraints to possible actions of lattices on varieties.) Take $G$ is a projective algebraic group and apply this to the action by conjugation of $G$: the quotient of $G$ by its center is linear and projective, hence trivial. – ACL Mar 23 2011 at 8:14

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.

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See also Milne's notes at jmilne.org. – Chandan Singh Dalawat Mar 21 2011 at 13:33
Emerton's argument actually coincides with the one given by Mumford. – Martin Brandenburg Mar 22 2011 at 9:01