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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
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up vote 6 down vote accepted

An algebraic group must be smooth (as you mention) and also homogeneous in the sense that given two points on it there is an automorphism of the variety sending the first to the second.
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.

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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
    
The components of an algebraic group are irreducible. Also, although this follows from homogeneity but is likely easier to check, they are isomorphic. It is also a general fact that an affine algebraic group is (algebraically) isomorphic to a matrix group (in other words, it admits a faithful linear algebraic representation). –  tomasz Dec 24 '12 at 11:48
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