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.