Regular type (stability theory)

A stationary type $$p(x)$$ in a stable theory is said to be regular if it is orthogonal to all its forking extensions. If $$p = \operatorname{tp}(a/C)$$, this means that whenever $$D \supseteq C$$ and $$a \not \downarrow_C D$$ and $$\operatorname{tp}(a/D)$$ is stationary, then $$\operatorname{tp}(a/D)$$ and $$\operatorname{tp}(a/C)$$ are orthogonal. Regularity is a parallelism invariant: if $$p'$$ is a forking extension of $$p$$, then $$p'$$ is regular if and only if $$p$$ is regular.

Types of U-rank 1 are regular, because their non-forking types are algebraic, hence orthogonal to every type. More generally, any type whose U-rank is a power $$\omega$$ is regular.

Proof. This comes from the fact that if $$p$$ and $$q$$ are types of rank $$\omega^\alpha$$ and rank $$\beta < \omega^\alpha$$, then $$p$$ and $$q$$ are orthogonal. Indeed, if $$U(a/C) = \omega^\alpha$$ and $$U(b/C) < \omega^\alpha$$, then by the Lascar inequalities, $$U(a/bC) \oplus U(b/C) \ge U(ab/C) \ge U(a/C) = \omega^\alpha$$. Since $$U(b/C)$$ is less than $$\omega^\alpha$$, this forces $$U(a/bC) \ge \omega^\alpha$$. Then $$U(a/bC) \ge U(a/C)$$, so $$a \downarrow_C b$$. QED

In DCF, the generic type of the home sort (which has rank $$\omega$$) and the generic type of the constant field (which has rank 1) are two examples of regular types.

If $$p$$ is a regular type over a set $$C$$, there is a natural pregeometry structure on the set of realizations of $$p$$. A set $$\{a_1, \ldots, a_n\}$$ is independent if and only if it is independent in the sense of stability theory, i.e., $$a_1 \downarrow_C a_2$$, $$a_1a_2 \downarrow_C a_3$$, and so on. If $$S$$ is a set of realizations of $$p$$, a tuple $$a$$ is in the closure of $$S$$ if and only if $$a \not \downarrow_C S$$.

Regular types have weight 1.

Regular types are prevalent in superstable theories, in some sense…