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Let $G$ be an abelian connected complex linear algebraic group.

Is it true that $G$ is isomorphic to $(\mathbb{G}_m)^k\times (\mathbb{G}_a)^\ell$, where the nonnegative exponents denote repeated direct products of linear algebraic groups, $\mathbb{G}_m=\mathbb{C}^*$, and $\mathbb{G}_a=(\mathbb{C},+)$?

This seems like a very elementary statement, but I cannot find a good reference for it and it is the foundation for a paper that I'm writing. I've written down a proof using results from the 'Linear Algebraic Groups' books of Humphreys and Springer, but there must be a good citation (by someone who knows more about linear algebraic groups than I!). I can only speculate that finding a reference is difficult because these books wish to include statements that work over other fields, and the answer is much more complicated in these cases. (It would make a good exercise, though.)

My proof goes like this: use the Jordan decomposition so that $G=G_s\times G_u$; since $G$ is abelian and connected, this is the direct product of connected linear algebraic groups. Then $G_s$ is a 'd-group', so (since connected) is a torus. $G_u$ is 'elementary unipotent'. Over $\mathbb{C}$, these are vector groups isomorphic to $(\mathbb{G}_a)^\ell$.

Thanks for answering my question!

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  • $\begingroup$ I voted to close, since this seems like homework for me. I would ask on math.exchange, if I have questions, which I assume are easy if you know the right tools. To the topic, I recall that every connected locally compact abelian group is isomorphic to a bunch of copies of $\mathbb{R}$ and circles. There is a book of Deitmar and Echterhoff "Principles of Harmonic analysis", which adresses this. $\endgroup$
    – Marc Palm
    Commented Jul 20, 2012 at 15:52
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    $\begingroup$ Reference requests aren't homework. $\endgroup$
    – Will Sawin
    Commented Jul 20, 2012 at 16:05
  • $\begingroup$ It ia a reference request, but easy to answer. Tori are uniform in all characteristics. The notes at the end of Section 20 in my book list older sources on commutative unipotent groups, simplest in char 0. For instance, Demazure-Gabriel V.2.4 show that only vector groups occur in char 0, emphasizing the correspondence between algebraic groups and Lie algebras. Serre's *Groupes algebriques et corps de classes" (also in English) in VII.2 emphasizes Ext vanishing. The books by Borel, Springer, and me treat char 0 only in passing, focusing on Chevalley's uniform theory. $\endgroup$
    – Jim Humphreys
    Commented Jul 20, 2012 at 17:39
  • $\begingroup$ @OP and WS: Then I apologize.... $\endgroup$
    – Marc Palm
    Commented Jul 20, 2012 at 20:38
  • $\begingroup$ Welcome to MO, Brian! :D $\endgroup$
    – Chuck Hague
    Commented Jul 23, 2012 at 16:08

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A reference: Lie Groups and Algebraic Groups (Springer Series in Soviet Mathematics) by Arkadij L. Onishchik, Ernest B. Vinberg and Dimitry A. Leites (new printing will appear on Amazon.com on Jul 31, 2012). See Ch. 3, Section 2.5, page 116, Corollary of Theorem 8: Any irreducible commutative (linear) algebraic group is the direct product of a torus and a vector group.

I highly recommend this book for beginners, and not only for beginners. It is mainly a collection of problems (with hints of proofs at the end of each section).

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