Lascar rank

In a stable theory, Lascar rank (or U-rank) gives a well-behaved notion of "dimension" or "rank" to complete types and definable sets. This notion is especially well-behaved in superstable theories, which can be described as the theories in which Lascar rank is never $$\infty$$.

Definition
Work in a monster model of a stable theory. For $$\alpha$$ an ordinal, $$a$$ a finite tuple, and $$B$$ a small set, one inductively defines $$U(a/B) \ge \alpha$$ as follows:


 * $$U(a/B) \ge 0$$ always holds.
 * If $$\lambda$$ is a limit ordinal, $$U(a/B) \ge \lambda$$ means that $$U(a/B) \ge \alpha$$ for every $$\alpha < \lambda$$.
 * $$U(a/B) \ge \alpha + 1$$ means that there is some $$C \supset B$$ such that $$a \not \downarrow_B C$$ and $$U(a/C) \ge \alpha$$.

Then the Lascar rank $$U(a/B)$$ of a complete type $$\operatorname{tp}(a/B)$$ is defined to be the largest ordinal $$\alpha$$ such that $$U(a/B) \ge \alpha$$, or the error value of $$\infty$$ if $$U(a/B) \ge \alpha$$ for every ordinal $$\alpha$$.

This essentially means that $$U(a/B)$$ is the length of the longest forking chain beginning with $$\operatorname{tp}(a/B)$$.

Finally, if $$\Sigma(x)$$ is a partial type over some parameters $$B$$, then the Lascar rank of $$\Sigma(x)$$ is defined to be the supremum of $$\operatorname{tp}(a/B)$$ as $$a$$ ranges over realizations of $$\Sigma(x)$$, or -1 if $$\Sigma(x)$$ is inconsistent. The maximum need not be attained. It turns out that the choice of $$B$$ does not matter; the Lascar rank of $$\Sigma(x)$$ depends only on $$\Sigma(x)$$. Also, the Lascar rank of $$\operatorname{tp}(a/B)$$ (thought of as a partial type over $$B$$) agrees with $$U(a/B)$$.

In particular, the Lascar rank of a definable set $$D$$ is the Lascar rank of the (finite) partial type picking out $$D$$.

One can alternatively define $$U(a/B)$$ in terms of the fundamental order: Lascar rank is exactly the foundation rank of the fundamental order.

Basic properties
Lascar rank has the basic properties one expects of a rank:


 * $$U(a/B) = 0$$ if and only if $$a \in \operatorname{acl}(B)$$.
 * If $$a' \in \operatorname{acl}(aB)$$, then $$U(a'/B) \le U(a/B)$$.
 * If $$\Sigma(x)$$ is a partial type, and $$\Sigma'(x)$$ is a bigger partial type (i.e., one picking out a smaller type-definable subset), then $$U(\Sigma') \le U(\Sigma)$$. In particular, $$U(a/B) \ge U(a/BC)$$.

Moreover, Lascar rank is closely related to forking:


 * If $$a \downarrow_B C$$, then $$U(a/B) = U(a/BC)$$.
 * If $$U(a/B) < \infty$$, the converse holds: if $$a \not \downarrow_B C$$, then $$U(a/B) > U(a/BC)$$.

It turns out that superstable theories are exactly the theories in which $$U(a/B) < \infty$$ for all $$a$$ and $$B$$.

Lascar rank is related to Shelah's $$\infty$$-rank and Morley Rank by the following inequalities: $$U(a/B) \le R^\infty(a/B) \le RM(a/B).$$ There is a certain sense in which Lascar rank is the smallest nice rank.

Lascar inequalities
The principal advantage of Lascar rank above other ranks is that one has the following Lascar inequality $$U(a/bC) + U(b/C) \le U(ab/C) \le U(a/bC) \oplus U(b/C)$$ for any small set $$C$$ and tuples $$a$$ and $$b$$. Here $$+$$ denotes usual sum of ordinals, and $$\oplus$$ denotes the so-called "natural sum." In the case where all the ranks are finite, the left side and the right side agree, and one obtains an equality $$U(ab/C) = U(a/bC) + U(b/C)$$

On the level of definable sets, this implies that if $$f : X \to Y$$ is a definable surjection, and every fiber $$f^{-1}(y)$$ has Lascar rank $$d$$, then $$U(X) = U(Y) + d,$$ which is a very intuitive property that one would expect.

One also has a useful auxiliary result: if $$a \downarrow_C b$$, then $$U(ab/C) = U(a/C) \oplus U(b/C)$$, (right?). This implies for definable sets that $$U(X \times Y) = U(X) \oplus U(Y)$$.

The disadvantage
Both Morley rank and $$\infty$$-rank have the continuity property that if a partial type $$\Sigma(x)$$ has rank $$\alpha$$, then some finite subtype has rank $$\alpha$$ (rather than higher rank). This translates into a semi-continuity property of the function from complete types to rank. Lascar rank lacks this property. TODO: example.

Relation to infinity rank
I believe the following is true: if a definable set $$X$$ has $$U(X) = n$$, for $$n < \omega$$, then $$R^\infty(X) = n$$. So on the level of definable sets, Lascar rank and Shelah's infinity rank agree in the finite-rank setting. This can fail in infinit rank settings, however.

Relation to pregeometry ranks
In a strongly minimal set $$X$$, Lascar rank agrees with Morley rank and Shelah's $$\infty$$ rank, as well as with the notion of rank coming from the inherent pregeometry. That is, if $$a$$ is a tuple from $$X$$ and $$B$$ is a set over which $$X$$ is defined, $$U(a/B)$$ equals the size of a maximal $$\operatorname{acl}$$-independent subtuple of $$a$$.

More generally, if $$X$$ is a set of $$\infty$$-rank 1, then it turns out that $$\infty$$-rank and Lascar rank agree in powers of $$X$$. There is a pregeometry in this setting, and the pregeometry rank agrees with Lascar rank.

Even more generally, if $$\Sigma(x)$$ is a partial type of Lascar rank 1, then $$\operatorname{acl}$$ still yields a pregeometry structure on realizations of $$\Sigma(x)$$, and the rank of a tuple is still its Lascar rank.

In the first two cases (strongly minimal sets, and definable sets of $$\infty$$-rank 1), it turns out that Lascar rank is definable in families. That is, if $$\phi(x;y)$$ is a formula, with $$x$$ living in the set of rank 1, then the set of $$b$$ such that $$\phi(x;b)$$ has rank $$n$$ is definable, for every $$b$$.

SU rank
The definition of Lascar rank given above works just as well in simple theories. It is conventional to denote this rank by $$SU(a/B)$$ rather than $$U(a/B)$$, though, for historical reasons. The Lascar inequalities continue to hold. One defines a theory to be supersimple if $$SU(a/B) < \infty$$ for every $$a$$ and $$B$$. Note that in a supersimple theory, one has no analogs of Morley rank or Shelah $$\infty$$-rank.

There is also a version for rosy theories.