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 $U(a/B)\geq \alpha$ as follows:

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

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 $U(a/B)\geq \alpha$ , or the error value of $\infty$ if $U(a/B)\geq \alpha$ 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 $U(a'/B)\leq U(a/B)$ .
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 $U(\Sigma ')\leq U(\Sigma )$ . In particular, $U(a/B)\geq U(a/BC)$ .

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: $U(a/B)\leq R^{\infty }(a/B)\leq RM(a/B).$ 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 $U(a/bC)+U(b/C)\leq U(ab/C)\leq U(a/bC)\oplus U(b/C)$ 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. ::: ::: :::