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, $U(a/bC)\oplus U(b/C)\geq U(ab/C)\geq U(a/C)=\omega ^{\alpha }$ . Since $U(b/C)$ is less than $\omega ^{\alpha }$ , this forces $U(a/bC)\geq \omega ^{\alpha }$ . Then $U(a/bC)\geq U(a/C)$ , 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., $a_{1}\downarrow _{C}a_{2}$ , $a_{1}a_{2}\downarrow _{C}a_{3}$ , 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… ::: ::: :::