Proofs:Lascar Inequalities

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
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
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.