FANDOM


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 Edit

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 Edit

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 Edit

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 Edit

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 Edit

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 Edit

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 Edit

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.