# Countable model theory for $\omega$-stable theories?

This is a bit of a fishing expedition, because I'm not sure what I'm looking for. Very vaguely stated, here's the driving question:

What conditions on an $\omega$-stable theory make the class of countable models "well-behaved?"

Here conditions would mean model-theoretic conditions, e.g. few types, although that's a bad example since of course all $\omega$-stable theories have few types.

"Well-behaved" would mean, in terms of descriptive set theory, that the isomorphism relation for countable models of the theory is not Borel-complete. But I'd be happy even knowing when it's (say) smooth, or even when it has only countably many countable models.

Of course the Ryll-Nardzewski theorem tells us when there's exactly one countable model, but beyond that, I don't know anything in this context.

I know some results in the other direction- the conditions "eni-DOP" and "eni-deep" both imply the isomorphism relation is Borel-complete. This is a recent result of Laskowski and Shelah. But the converse doesn't seem helpful (to me); knowing (say) that the theory has eni-NDOP and is eni-shallow implies a certain technique for proving complexity won't work, but it doesn't seem to imply that models are simple, only that you'd need a new proof.

So I'm just looking for any results in the "positive" direction- some reasonable condition on the theory which implies any kind of "niceness" for the class of countable models.

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Have you thought about "eni-nmd", that is, any eventually nonisolated type is nonorthogonal to the emptyset? – John Goodrick Aug 13 '14 at 19:48
This comment is helpful; it turns out if T is eni-nmd, then isomorphism of countable models is determined by which types they realize (a countable subset of a countable set). This is very low in the hierarchy. Thank you! – Richard Rast Aug 20 '14 at 14:07

Let T be a countable superstable theory of finite rank with $< 2^{\aleph_0}$ many countable models. Then for M a countable model of T there is a finite A ⊂ M and a J ⊂ M such that M is prime over A∪J , J is A−independent and {stp(a/A) : a ∈ J} is finite.
If $T$ has only countably many models, then $T$ is Borel (by a Scott height argument) and therefore smooth. So unfortunately, knowing Vaught's Conjecture holds is enough to sort of trivialize that side of things, without using the fine characterization that Buechler provides. – Richard Rast Aug 20 '14 at 14:13