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 {\displaystyle \infty }.


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

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

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

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

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

One can alternatively define {\displaystyle 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:

Moreover, Lascar rank is closely related to forking:

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

Lascar rank is related to Shelah's {\displaystyle \infty }-rank and Morley Rank by the following inequalities: {\displaystyle U(a/B)\leq R^{\infty }(a/B)\leq 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 {\displaystyle U(a/bC)+U(b/C)\leq U(ab/C)\leq U(a/bC)\oplus U(b/C)} for any small set {\displaystyle C} and tuples {\displaystyle a} and {\displaystyle b}. Here {\displaystyle +} denotes usual sum of ordinals, and {\displaystyle \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 {\displaystyle U(ab/C)=U(a/bC)+U(b/C)}

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

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

The disadvantage[]

Both Morley rank and {\displaystyle \infty }-rank have the continuity property that if a partial type {\displaystyle \Sigma (x)} has rank {\displaystyle \alpha }, then some finite subtype has rank {\displaystyle \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 {\displaystyle X} has {\displaystyle U(X)=n}, for {\displaystyle n<\omega }, then {\displaystyle 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 {\displaystyle X}, Lascar rank agrees with Morley rank and Shelah's {\displaystyle \infty } rank, as well as with the notion of rank coming from the inherent pregeometry. That is, if {\displaystyle a} is a tuple from {\displaystyle X} and {\displaystyle B} is a set over which {\displaystyle X} is defined, {\displaystyle U(a/B)} equals the size of a maximal {\displaystyle \operatorname {acl} }-independent subtuple of {\displaystyle a}.

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

Even more generally, if {\displaystyle \Sigma (x)} is a partial type of Lascar rank 1, then {\displaystyle \operatorname {acl} } still yields a pregeometry structure on realizations of {\displaystyle \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 {\displaystyle \infty }-rank 1), it turns out that Lascar rank is definable in families. That is, if {\displaystyle \phi (x;y)} is a formula, with {\displaystyle x} living in the set of rank 1, then the set of {\displaystyle b} such that {\displaystyle \phi (x;b)} has rank {\displaystyle n} is definable, for every {\displaystyle 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 {\displaystyle SU(a/B)} rather than {\displaystyle U(a/B)}, though, for historical reasons. The Lascar inequalities continue to hold. One defines a theory to be supersimple if {\displaystyle SU(a/B)<\infty } for every {\displaystyle a} and {\displaystyle B}. Note that in a supersimple theory, one has no analogs of Morley rank or Shelah {\displaystyle \infty }-rank.

There is also a version for rosy theories. ::: ::: :::