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Cherlin's "Main Conjecture" from his 1979 paper "Groups of Small Morley Rank" is the following: Every simple $\omega$-stable group is an algebraic group over an algebraically closed field. Zilber was simultaneously doing similar work (as Cherlin notes). The finite Morley rank version of the conjecture is sometimes called the Cherlin-Zilber conjecture or the Algebraicity conjecture.

There is an extensive literature for the finite Morley rank case. I am not asking about the finite rank case. I am interested in the status of the conjecture for the infinite rank case. Cherlin notes that this conjecture would imply that any simple $\omega$-stable group is of finite rank - modulo the finite rank version of the conjecture, this is essentially the content of the infinite rank version of the conjecture.

Further, Cherlin notes that one could formulate a linear version of the conjecture, in which the group acts as a subgroup of the linear transformations of some vector space.

What is the status of the infinite rank version of Cherlin's "Main Conjecture"?

Edit: See the comments made by S. Thomas below. (Thanks for the clarification Simon +1).

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Your question would be clearer and more accurate if you pointed out that you are discussing groups in the pure language of group theory. Of course, the "conjecture" is false for suitably chosen richer languages. For example, consider $PSL(3,K)$ in the language of pairs of algebraically closed fields $F \subset K$. Alternatively, you could take $PSL(3,K)$ for a differentially closed field $(K, \delta)$ of characteristic zero. – Simon Thomas Aug 2 '11 at 11:30
This is a fair criticism, but I think your comment makes it clear. One should note that in the differential context any (quasi) simple differential algebraic group will actually be an algebraic group (perhaps the F-points for a proper definable subfield). So, one won't find counterexamples in either of the contexts which Simon mentions. But, as he says, the question of the language does not matter for the abstract simplicity of the group, so you can start adding equivalence relations, etc., to any simple algebraic group and get something simple of infinite rank. – James Freitag Aug 2 '11 at 14:46
Rumor has it that (if I remember it correctly) Zilber wanted his name to be detached from the conjecture, since he does not believe that it is true anymore. – Jizhan Hong Aug 18 '11 at 2:00
@Jizhan: To be clear, are you referring to the finite rank version of the conjecture or the infinite rank version? This is mostly a curiosity for me - I was reading an old paper of Cherlin's and saw the full conjecture stated, but was surprised by it since I had only heard the conjecture about groups of finite Morley rank. I wouldn't be surprised by either clarification here. I am just wondering. But since we don't here much about the infinite rank version here, I suspect you are referring to the finite Morley rank conjecture. – James Freitag Aug 18 '11 at 16:02
@James: Yes, as you mentioned in your question, Zilber's name is attached to the finite rank version only (at least that's my impression). I heard it during the Oléron meeting. Someone announced it to the audience; I vaguely remember that it was because the conjecture was mentioned by the speaker (I don't remember if Zilber was still there or who the speaker and the announcer were); and another person said that it is impossible once it has been attached (or something like that). And then people just laughed. Maybe someone can "confirm" this "rumor". – Jizhan Hong Aug 18 '11 at 17:19

I think the 'status' might be described as : Pillay has shown using used Selah's work that the free group is not CM-trivial.

All known counterexamples to Zilber's conjecture are CM-trivial. A non-abelian simple group of finite Morley rank is not CM-trivial. We therefore suspect that the current methods based upon Hrushovski's counterexamples cannot produce even an infinite rank counterexample who is a simple group.

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Hi Jeff. Thanks for the answer. I am not precisely sure I understand though. What does the free group have to do with the question, since free groups on two or more generators are stable, but not superstable, right? Or am I mistaken? Or am I right and are the free groups involved for other reasons? I don't really see why though... – James Freitag Oct 4 '11 at 2:39
I meant merely that the free group strongly resembles a counterexample nothing more, i.e. stable, non-CM-trivial, etc. There have actually been attempts to build counterexamples inspired by Selah's work, but none got very far. – Jeff Burdges Oct 4 '11 at 18:13

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