## FANDOM

78 Pages

The Lascar inequalities say that if $a, b$ are finite tuples and $C$ is a set of parameters in a stable theory, then $U(a/bC) + U(b/C) \le U(ab/C) \le U(a/bC) \oplus U(b/C).$ Here $U(x/S)$ denotes the Lascar $U$-rank of $\operatorname{tp}(x/S)$. On the left, $+$ denotes the usual ordinal sum, while on the right, $\oplus$ denotes the natural sum of ordinals, defined by adding the coefficients in Cantor normal form: $\sum_{\alpha} \omega^\alpha \cdot n_\alpha \oplus \sum_{\alpha} \omega^\alpha \cdot m_\alpha := \sum_\alpha \omega^\alpha \cdot (n_\alpha + m_\alpha).$

There is also a related auxiliary result: if $a \downarrow_C b$, then $U(ab/C) = U(a/C) \oplus U(b/C)$.

More generally, the same results hold in simple theories, with $SU$-rank rather than $U$-rank. The same proofs work.

## Preliminary Results Edit

Recall the definition of Lascar rank: $U(a/S) \ge \alpha + 1$ if and only if there is some $S' \supset S$ such that $\operatorname{tp}(a/S')$ is a forking extension of $\operatorname{tp}(a/S)$ and $U(a/S') \ge \alpha$.

We will use the following basic facts.

Fact 1: If $S' \supseteq S$, then $U(a/S') \le U(a/S)$, so Lascar rank is appropriately monotone.

Proof. We prove by induction on $\alpha$ that $U(a/S') \ge \alpha$ implies $U(a/S) \ge \alpha$. The cases where $\alpha = 0$ or $\alpha$ is a limit ordinal are completely trivial, so consider the successor ordinal case. Suppose $U(a/S') \ge \alpha + 1$. Then there is some $S'' \supseteq S'$ such that $U(a/S'') \ge \alpha$ and $a \not \downarrow_{S'} S''$. As $S \subseteq S' \subseteq S''$ and $\operatorname{tp}(a/S'')$ forks over $S'$, it certainly forks over $S$. So $U(a/S'') \ge \alpha$ implies $U(a/S) \ge \alpha + 1$, completing the inductive step. (I guess we only used induction in the limit ordinal case.) QED

Fact 2: If $S' \supseteq S$ and $a \downarrow_S S'$, then $U(a/S') = U(a/S)$. Non-forking extensions have the same rank.

Proof. In light of Fact 1, we only need to show that $U(a/S') \ge U(a/S)$. We prove by induction on $\alpha$ that $U(a/S) \ge \alpha$ implies $U(a/S') \ge \alpha$. As before, the cases where $\alpha$ is zero or a limit ordinal are completely trivial. Consider the successor ordinal case: $U(a/S) \ge \alpha + 1$. Then there exists $S'' \supseteq S$ such that $a \not\downarrow_S S''$ and $U(a/S'') \ge \alpha$. We may move $S''$ by an automorphism over $aS$ so that $S'' \downarrow_{aS} S'$. Since $a \downarrow_S S'$, it follows by left transitivity that $S'' a \downarrow_S S'$, which in turn implies $a \downarrow_{S''} S'$. Since $U(a/S'') \ge \alpha$, the inductive hypothesis implies $U(a/S'S'') \ge \alpha$. If $a \downarrow_{S'} S''$, the fact that $a \downarrow_S S'$ would imply by right-transitivity that $a \downarrow_S S'S''$, contradicting the choice of $S''$. So $a \not\downarrow_{S'} S''$. Therefore $U(a/S'S'') \ge \alpha$ implies $U(a/S') \ge \alpha + 1$, by definition of Lascar rank. This completes the inductive step, and the proof. QED

Fact 3: If $a,b$ are tuples, then $U(a/S) \le U(ab/S)$.

Proof. We show by induction on $\alpha$ that $U(a/S) \ge \alpha$ implies $U(ab/S) \ge \alpha$. As before, the only non-trivial case is the successor ordinal case. Suppose $U(a/S) \ge \alpha + 1$. Then there is $S' \supset S$ with $a \not \downarrow_S S'$ and $U(a/S') \ge \alpha$. By induction, $U(ab/S') \ge \alpha$. By monotonicity of forking, $ab \not \downarrow_S S'$, so $U(ab/S') \ge \alpha$ implies $U(ab/S) \ge \alpha + 1$, completing the inductive step. QED

## Proof of the Lascar Inequalities Edit

For the left-hand side, we prove by induction on $\beta$ that if $U(b/C) \ge \beta$, then $U(a/bC) + \beta \le U(ab/C)$. For the base case $\beta = 0$, note that $U(a/bC) \le U(a/C) \le U(ab/C)$ by Facts 1 and 3 above. For the successor case, suppose that $U(b/C) \ge \beta + 1$. Then by definition of Lascar rank, there is some $C' \supseteq C$ such that $U(b/C') \ge \beta$ and $b \not \downarrow_C C'$. Move $C'$ by an automorphism over $bC$ so that $C' \downarrow_{bC} a$. This implies that $U(a/bC') = U(a/bC)$, by Fact 2.

By the inductive hypothesis, $U(a/bC) + \beta = U(a/bC') + \beta \le U(ab/C').$ Because $b \not \downarrow_C C'$, monotonicity of $\downarrow$ implies that $ab \not \downarrow_C C'$, that is, $\operatorname{tp}(ab/C')$ is a forking extension of $\operatorname{tp}(ab/C)$. Consequently, $U(ab/C') \ge U(ab/C') + 1 \ge U(a/bC) + \beta + 1,$ completing the inductive step, in the case of successor ordinals. Finally in the limit ordinal case, suppose that $\lambda$ is a limit ordinal, and $U(b/C) \ge \lambda$. Then $U(b/C) \ge \beta$ for every $\beta < \lambda$. By the inductive hypothesis, $U(a/bC) + \beta \le U(ab/C)$ for every $\beta < \lambda$. Since ordinal addition is continuous in the right operand, $U(a/bC) + \lambda \le U(ab/C)$ holds, completing the inductive step in the limit ordinal case. We have shown that for every $\beta$, $U(b/C) \ge \beta \implies U(a/bC) + \beta \le U(ab/C).$ Taking $\beta = U(b/C)$ (or taking $\beta$ arbitrarily large when $U(b/C) = \infty$), we conclude that $U(a/bC) + U(b/C) \le U(ab/C),$ the left-hand side of the Lascar inequalities holds.

For the right-hand side, we prove by induction on $\alpha$ that $U(ab/C) \ge \alpha \implies U(a/bC) \oplus U(b/C) \ge \alpha.$ The $\alpha = 0$ case is trivial, since Lascar rank is always at least 0. The limit ordinal case is also completely trivial. So consider the successor ordinal case: suppose that $U(ab/C) \ge \alpha + 1$. Then by definition of Lascar rank, there exists $C' \supseteq C$ such that $ab \not \downarrow_C C'$, and $U(ab/C') \ge \alpha$. By the inductive hypothesis, $U(a/bC') \oplus U(b/C') \ge \alpha.$ If $b \downarrow_C C'$ and $a \downarrow_{bC} C'$, then by left-transitivity of forking independence, $ab \downarrow_C C'$, a contradiction. So $b \not \downarrow_C C'$ or $a \not\downarrow_{bC} C'$. In the first case, $U(b/C) \ge U(b/C') + 1 = U(b/C') \oplus 1.$ Since $U(a/bC) \ge U(a/bC')$ by monotonicity of Lascar-rank (Fact 1), we see that $U(a/bC) \oplus U(b/C) \ge U(a/bC') \oplus U(b/C') \oplus 1 \ge \alpha \oplus 1 = \alpha + 1,$ completing the inductive step. In the second case, $a \not\downarrow_{bC} C'$, so $U(a/bC) \ge U(a/bC') \oplus 1,$ so we similarly have $U(a/bC) \oplus U(b/C) \ge U(a/bC') \oplus 1 \oplus U(b/C') = U(a/bC') \oplus U(b/C') \oplus 1 \ge \alpha + 1.$ Either way, the inductive step holds. This completes the induction on $\alpha$. Now setting $\alpha = U(ab/C)$, we get the right-hand side of the Lascar inequalities.

Finally, we prove that if $a \downarrow_C b$, then $U(ab/C) = U(a/C) \oplus U(b/C)$. We have already seen that $U(ab/C) \le U(a/bC) \oplus U(b/C) = U(a/C) \oplus U(b/C),$ so we only need to show that $U(a/C) \oplus U(b/C) \le U(ab/C)$. We prove by joint induction on $\alpha$ and $\beta$ that $U(a/C) \ge \alpha \wedge U(b/C) \ge \beta \rightarrow U(ab/C) \ge \alpha \oplus \beta.$ If $\alpha$ or $\beta$ is 0, this follows from the fact that $U(a/C) \le U(ab/C)$ and $U(b/C) \le U(ab/C)$. Otherwise, recall that the natural sum $\alpha \oplus \beta$ is the smallest ordinal greater than all ordinals of the form $\alpha' \oplus \beta$ and $\alpha \oplus \beta'$ for $\alpha' < \alpha$ and $\beta' < \beta$. So, we need to show that if $\alpha' < \alpha$ or $\beta' < \beta$, then $U(ab/C) > \alpha' \oplus \beta$ or $U(ab/C) > \alpha \oplus \beta'$, respectively. We handle the first case; the second is completely symmetric. Since $U(a/C) > \alpha'$, it follows that $U(a/C) \ge \alpha' + 1$, so there is some $C' \supseteq C$ with $a \not\downarrow_C C'$ and $U(a/C') \ge \alpha'$. Moving $C'$ over $aC$, we may assume that $C' \downarrow_{aC} b$. But then since $a \downarrow_C b$, it follows by left-transitivity that $C'a \downarrow_C b$, and in particular $C' \downarrow_C b$ and $a \downarrow_{C'} b$. Since $C' \downarrow_C b$, we have $U(b/C') = U(b/C) \ge \beta$ (Fact 2). And since $a \downarrow_{C'} b$, $U(a/C') \ge \alpha'$, and $U(b/C') \ge \beta$, the inductive hypothesis ensures that $U(ab/C') \ge \alpha' \oplus \beta$. Now $ab \not \downarrow_C C'$, since $a$ forks with $C$, so $U(ab/C) \ge U(ab/C') + 1 \ge \alpha' \oplus \beta \oplus 1 > \alpha' \oplus \beta.$ This completes the proof.