## FANDOM

78 Pages

If $p(x)$ and $q(y)$ are two complete types over a set $C$, $p$ and $q$ are set to be almost orthogonal if there is a unique complete type in the variables $(x,y)$ extending $p(x) \cup q(y)$. That is, if $a, a' \models p$ and $b, b' \models q$, then $ab \equiv_C a'b'$.

If $p(x)$ and $q(y)$ are stationary types in a stable theory, then one can easily check that $p(x)$ and $q(y)$ are almost orthogonal if and only if $a \downarrow_C b$ for all $a$ realizing $p(x)$ and $b$ realizing $q(y)$.

In a stable theory $T$, two stationary types $p$ and $q$ are orthogonal if $p|C$ and $q|C$ are almost orthogonal for every set $C$ containing the bases of $p$ and of $q$. Here $p|C$ and $q|C$ denote the unique non-forking extensions of $p$ and $q$ to $C$. It turns out that if $p|C$ and $q|C$ fail to be almost orthogonal for some $C$, then $p|C'$ and $q|C'$ also fail to be almost orthogonal for all $C' \supseteq C$. Therefore, it suffices to check the orthogonality at sufficiently large sets $C$, and orthogonality depends only on the parallelism class of $p$ and $q$.

Roughly speaking, $p$ and $q$ are orthogonal if there are no interesting relations between realizations of $p$ and realizations of $q$. For example, if $p$ and $q$ are the generic types of two strongly minimal sets $P$ and $Q$, then $p$ and $q$ are orthogonal if and only if there are no finite-to-finite correspondences between $P$ and $Q$, i.e., no definable sets $C \subset P \times Q$ with $C$ projecting onto $P$ and onto $Q$ with finite fibers in both directions.

The relation of non-orthogonality is an equivalence relation on strongly minimal sets, or more generally, on stationary types of U-rank 1. If $p$ and $q$ are two non-orthogonal types of rank 1, then $p$ and $q$ have the same underlying geometry. A theory is uncountably categorical if and only if it is $\omega$-stable and unidimensional (e.g. every pair of stationary non-algebraic types is non-orthogonal).