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 ∞.
Work in a monster model of a stable theory. For α an ordinal, a a finite tuple, and B a small set, one inductively defines U(a/B) ≥ α as follows:
Then the Lascar rank U(a/B) of a complete type tp(a/B) is defined to be the largest ordinal α such that U(a/B) ≥ α, or the error value of ∞ if U(a/B) ≥ α for every ordinal α.
This essentially means that U(a/B) is the length of the longest forking chain beginning with tp(a/B).
Finally, if Σ(x) is a partial type over some parameters B, then the Lascar rank of Σ(x) is defined to be the supremum of tp(a/B) as a ranges over realizations of Σ(x), or -1 if Σ(x) is inconsistent. The maximum need not be attained. It turns out that the choice of B does not matter; the Lascar rank of Σ(x) depends only on Σ(x). Also, the Lascar rank of 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.
Lascar rank has the basic properties one expects of a rank:
Moreover, Lascar rank is closely related to forking:
It turns out that superstable theories are exactly the theories in which U(a/B) < ∞ for all a and B.
Lascar rank is related to Shelah's ∞-rank and Morley Rank by the following inequalities: U(a/B) ≤ R∞(a/B) ≤ RM(a/B). There is a certain sense in which Lascar rank is the smallest nice rank.
The principal advantage of Lascar rank above other ranks is that one has the following Lascar inequality U(a/bC) + U(b/C) ≤ U(ab/C) ≤ U(a/bC) ⊕ U(b/C) for any small set C and tuples a and b. Here + denotes usual sum of ordinals, and ⊕ 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 → 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↓Cb, then U(ab/C) = U(a/C) ⊕ U(b/C), (right?). This implies for definable sets that U(X × Y) = U(X) ⊕ U(Y).
Both Morley rank and ∞-rank have the continuity property that if a partial type Σ(x) has rank α, then some finite subtype has rank α (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.
I believe the following is true: if a definable set X has U(X) = n, for n < ω, then R∞(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.
In a strongly minimal set X, Lascar rank agrees with Morley rank and Shelah's ∞ 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 acl-independent subtuple of a.
More generally, if X is a set of ∞-rank 1, then it turns out that ∞-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 Σ(x) is a partial type of Lascar rank 1, then acl still yields a pregeometry structure on realizations of Σ(x), and the rank of a tuple is still its Lascar rank.
In the first two cases (strongly minimal sets, and definable sets of ∞-rank 1), it turns out that Lascar rank is definable in families. That is, if ϕ(x; y) is a formula, with x living in the set of rank 1, then the set of b such that ϕ(x; b) has rank n is definable, for every b.
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) < ∞ for every a and B. Note that in a supersimple theory, one has no analogs of Morley rank or Shelah ∞-rank.
There is also a version for rosy theories. ::: ::: :::