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…