## FANDOM

78 Pages

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…