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The basic question is whether there is a notion of chief factor of a connected solvable algebraic group that matches my intuition. A few smaller assertions are sprinkled through the explanation, and the implicit question is if these are correct (are unipotent groups nilpotent, are their chief factors all isomorphic to subgroups of Ga, etc.).

In J.S. Milne's course notes on the basic theory of algebraic groups, theorem 14.30:

A connected solvable smooth group over a perfect field has a connected unipotent normal subgroup whose quotient is of multiplicative type.

In other words, the derived subgroup is contained in the unipotent radical.

Now, a connected group acts on a group of multiplicative type trivially, by 13.21.

I did not see it mentioned, but I believe unipotent groups are nilpotent in both the group theoretic sense and whatever fancy definition one might cook up for these functors. I think that the lower central factors should be direct products of subgroups of the additive group Ga.

By the Jordan decomposition or 13.13 or 13.15, I think any action of Gm on (Ga)^n is diagonal.

It looks like a connected solvable affine algebraic group over an algebraically closed field has a chief series consisting of subgroups of Ga and Gm, all of which I would describe as being one-dimensional.

The analogy with finite groups takes the unipotent radical to be O_p(G), the p-core, and so it appears that a finite group of solvable algebraic type always has [G,G] <= O_p(G), so that not only is G supersolvable nilpotent-by-abelian, it is p-closed and its eccentric chief factors are all for the same prime p. In the algebraic case, the central chief factors would be the subgroups of Gm, and the eccentric chief factors would be the subgroups of Ga.

In other words, connected solvable affine algebraic groups over algebraically closed fields are very dissimilar from finite solvable groups in that the representations they define on their own chief factors are all one-dimensional. More briefly, connected solvable affine algebraic groups over algebraically closed fields are supersolvable.

Edit: I think the answer to my question is relatively simple: "supersolvable" is a little tricky to directly generalize, but "nilpotent-by-abelian" is quite easy and true, and still implies that any chief factors will be one-dimensional. In Jim's answer, it appears Borel-Serre-Mostow (at least by the time they are translated into Russian) also considered these groups to be "supersolvable", so the name is reasonable, even if the correct definition is just "nilpotent-by-abelian".

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    $\begingroup$ What's your base field? The answer to "are the chief factors [of a unipotent group] all isomorphic to Ga" is "no" if you'll allow me a wacky base field (characteristic p with inseparable extensions). $\endgroup$
    – Kevin Buzzard
    Commented Mar 31, 2010 at 21:51
  • $\begingroup$ I'm happy with whatever nice field assumptions one wants as long as it allows thinking about finite groups. I assume something like "let the field be the algebraic closure of its prime field" is general enough for me. The only examples I've heard of where the chief factors of a unipotent group are not exactly Ga have the property that they are subgroups of Ga, and so, I think, are still one-dimensional. It seems similar to how groups of multiplicative type don't need to be Gm, but are subgroups of Gm. $\endgroup$
    – Jack Schmidt
    Commented Mar 31, 2010 at 22:49
  • $\begingroup$ Edited to reflect more of the intent (added "over an alebraically closed field" and "subgroups of"). Basically there is an intuition that the action of a solvable group in this fancy sense on its abelian normal subquotients is fairly trivial compared the action of a solvable group in the group theoretic sense. For instance, there is nothing like S4 = AGL(2,2) where the unipotent radical, K4, does not contain the derived subgroup A4. Since it is a natural matrix group, this is surprising to me. However, it exists precisely because it violates Lie–Kolchin's theorem. $\endgroup$
    – Jack Schmidt
    Commented Mar 31, 2010 at 23:06

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Properties of solvable linear algebraic groups have been explored sporadically for over 50 years, with various assumptions on the base field. There seems to be no single viewpoint about how this subject interacts with finite groups or with general algebraic groups. But to get perspective it may be helpful to consult older work originating with Lie groups (Borel, Mostow, Serre) as well as the long paper by Platonov, V. P., The theory of algebraic linear groups and periodic groups. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 30 1966 573--620. This appears in Series 2, Volume 69 of the AMS Translations. For example, in 3.1 a solvable algebraic group is defined to be "supersolvable" if its finite component group is supersolvable. This may or may not be everyone's preferred definition, but given this history one has to be careful about the underlying assumptions when asking new questions about solvable algebraic groups.

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  • $\begingroup$ Thanks! That sounds like the same viewpoint I'm working on. I'll check the paper tomorrow (the AMS Translations are not online, right?). $\endgroup$
    – Jack Schmidt
    Commented Apr 2, 2010 at 1:11
  • $\begingroup$ Probably the best you can do online is to type in a search word like "supersoluble", which should return many but not all pages of Platonov's article: e-math.ams.org/bookstore-getitem/item=TRANS2-69 I've never explored this area of the literature on algebraic groups, which got developed by algebraists in the former Soviet Union influenced by abstract group theory. So I'm not sure what's really there. $\endgroup$
    – Jim Humphreys
    Commented Apr 2, 2010 at 12:39
  • $\begingroup$ Thanks, this finally arrived. Yes it approaches things from the linear groups perspective, which is much easier for me to understand. It doesn't cover chief factors directly, but it covers many other finite ideas in this context and so is a great help in understanding these fancy ideas in more concrete terms. $\endgroup$
    – Jack Schmidt
    Commented Apr 7, 2010 at 3:27
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Edit: According to Wikipedia, $A_4$ is not supersolvable, but $A_4 \cong (\mathbb{G}_a \rtimes \mathbb{G}_m)(\mathbb{F}_4)$. I think this answers your question in the negative.

Original response: What definition of unipotent algebraic group are you using? If you're only looking at groups over an algebraically closed field, then unipotent groups are defined in SGA3 Exp. 17 by the existence of a composition series with quotients given by subgroups of the additive group. The description of the central series is given in loc. cit. Theorem 3.5, and agrees with your conjecture.

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