FANDOM


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.