# Which algebraic variety can become a algebraic group?

First, I know the algebraic group must be non-singular and the index of the identity component must be finite.

Now given a algebraic variety (especially for a algebraic curve or a algebraic surface whose picture is beautiful) with these conditions, how to judge whether we can give it a group structure and make it as a algebraic group?

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I don't understand what you mean when you say that "the index of the center must be finite." This is clearly false in any reasonable sense for an algebraic group such as $\text{GL}_n$ (for $n > 1$), and it is also not a condition on varieties but a condition on groups, so I don't understand what you mean by "these conditions" when applied to this condition. – Qiaochu Yuan Dec 22 '12 at 6:07
Oh, you are right. I make a mistake, "the center" should be change to the identity component. I will edit it. – Strongart Dec 24 '12 at 11:11

This homogeneity condition already prevents complete smooth curves of genus $\geq 2$ from being algebraic groups (because they have finite groups of automorphisms).
Over $\mathbb C$ the complete connected algebraic groups have been classified: they are exactly the tori $\mathbb C^g/\Lambda$, where $\Lambda$ is a lattice satisfying the Riemann bilinear conditions: see Theorem (4.2.1) page 73 of Birkenhake-Lange's Complex Abelian Varieties.
The Euler characteristic of an algebraic group should be zero. This follows from the trace formula. In fact, for any non-trivial element $a$, the translation $t_a$ by $a$ has no fixed points. By the trace formula, the trace of $t_a$ on the cohomology of your algebraic group should be zero. The latter equals the euler characteristic. In particular, this shows that in dimension $1$, the genus has to be zero because the Euler characteristic equals $2g-2$. Moreover, note that the Euler characteristic of a torus $\mathbf C^g/Lambda$ is indeed zero thus it all works out. – Harry Dec 22 '12 at 16:04