## FANDOM

78 Pages

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 $U(X \times Y) = U(X) \oplus U(Y)$.

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.