It's well known that a countable theory is uncountably categorical if and only if it is $\omega$-stable and has no Vaughtian pairs. One of the old definitions of unidimensional theory (all $\omega\_1$-saturated models of the same (sufficiently large) size are isomorphic) is a weak categoricity notion, so it's natural to wonder if an analogous characterization can be made.


Hrushovski showed that unidimensional theories are supserstable. Although I'm sure it's already known, I couldn't find a proof that unidimensional theories do not have any Vaughtian pairs, so I worked out a silly proof:



> 
> Let $T$ be a countable, superstable theory with a Vaughtian pair. By Vaught's two-cardinal theorem, since $T$ has a Vaughtian
>  pair, it has a $(\omega\_1,\omega)$-model. By a result of Shelah,
>  in a stable theory the existence of any two-cardinal model implies the
>  existence of $(\kappa,\lambda)$-models for any
>  $\kappa\geq\lambda\geq\omega$, so we can find a
>  $(\kappa,\omega)$-model, $\mathfrak{M}$, of $T$, with $\kappa \gg 2^\omega$. Then if we take
>  any non-principal ultrafilter $\mathcal{U}$ on $\omega$,
>  $\mathfrak{M}^\omega/\mathcal{U}$ will still be a two-cardinal,
>  $\omega\_1$-saturated model of $T$, contradicting unidimensionality.
> 
> 
> 


Again, I'm sure there's a better, published proof somewhere, but I couldn't find it.


I would guess it's far too much to hope that a countable theory $T$ is unidimensional if and only if it is superstable and has no Vaughtian pairs, but I also can't quite find a counterexample. So my first question is: 



> 
> What is an example of a countable superstable theory with no Vaughtian pairs
>  that fails to be unidimensional, if such a thing exists?
>  EDIT: I believe theory of the following structure is a counterexample to this: $(2^\omega, P\_i, f\_i)$ where $P\_i$ is a sequence of unary predicates such that $P\_i(\alpha)$ is true if and only if $\alpha(i)=1$, and $f\_i$ is a sequence of unary functions such that $f\_i(\alpha)(j)=\alpha(j)$ if $i\neq j$ and $f\_i(\alpha)(i)=1-\alpha(i)$.
> 
> 
> 


Another related characterization of uncountably categorical countable theories is that a countable theory $T$ is uncountably categorical if and only if there is a definable strongly minimal set $\varphi$ such that the entire structure is layered by $\varphi$ (where layering is in the sense of Hodges' "Model Theory"). In Pillay's "Geometric Stability Theory," he shows in chapter 6 proposition 1.1 that if $T$ is a countable, superstable, unidimensional theory that fails to be $\omega$-stable, then there is a weakly minimal group definable over $\text{acl}(\varnothing)$. Since we know that $T$ doesn't have any Vaughtian pairs, by the results on layerings in Hodges we know that the entire theory is layered over this weakly minimal group. 


I suspect that it's also way too much to hope that any weakly minimal group is unidimensional, so my next questions are:



> 
> What is an example of a weakly minimal group that fails to be unidimensional? 
>  Is there a good characterization of which weakly minimal groups are unidimensional?
> 
> 
> 


Aside from those parts, I'm pretty confident that if a theory is layered over a unidimensional set, then it will be unidimensional, so if there is a good characterization of which weakly minimal groups are unidimensional, then I'm expecting there is a statement like: "A theory $T$ is unidimensional if and only if it is layered over a strongly minimal set or a weakly minimal group with property X." Where 'property X' is the missing ingredient to ensure that a weakly minimal group is unidimensional.



Is there a Baldwin-Lachlan style characterization of countable unidimensional theories?
I believe the answer to the second question is that all weakly minimal groups are unidimensional. 


**Proof:** Let $T$ be the theory of a weakly minimal group. It suffices to show that if $p$ and $q$ are non-algebraic $1$-types over a sufficiently saturated model $G$, then $p$ and $q$ are not orthogonal. So fix such $p$ and $q$. By weak minimality, $p$ and $q$ have $U$-rank $1$, and thus are generic. By stability, there is some $g\in G$ such that $q=gp$. So $p$ and $q$ are not orthogonal.


  

Edit: The fact that stable unidimensional theories have no Vaughtian pairs is in Lemma IX.1.10 of Shelah's book. 


