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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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    Oh, you are right. I make a mistake, "the center" should be change to the identity component. I will edit it.2012-12-24

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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 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).2012-12-24