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

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

Preliminary Results[]

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

We will use the following basic facts.

Fact 1: If {\displaystyle S'\supseteq S}, then {\displaystyle U(a/S')\leq U(a/S)}, so Lascar rank is appropriately monotone.

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

Fact 2: If {\displaystyle S'\supseteq S} and {\displaystyle a\downarrow _{S}S'}, then {\displaystyle 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 {\displaystyle U(a/S')\geq U(a/S)}. We prove by induction on {\displaystyle \alpha } that {\displaystyle U(a/S)\geq \alpha } implies {\displaystyle U(a/S')\geq \alpha }. As before, the cases where {\displaystyle \alpha } is zero or a limit ordinal are completely trivial. Consider the successor ordinal case: {\displaystyle U(a/S)\geq \alpha +1}. Then there exists {\displaystyle S''\supseteq S} such that {\displaystyle a\not \downarrow _{S}S''} and {\displaystyle U(a/S'')\geq \alpha }. We may move {\displaystyle S''} by an automorphism over {\displaystyle aS} so that {\displaystyle S''\downarrow _{aS}S'}. Since {\displaystyle a\downarrow _{S}S'}, it follows by left transitivity that {\displaystyle S''a\downarrow _{S}S'}, which in turn implies {\displaystyle a\downarrow _{S''}S'}. Since {\displaystyle U(a/S'')\geq \alpha }, the inductive hypothesis implies {\displaystyle U(a/S'S'')\geq \alpha }. If {\displaystyle a\downarrow _{S'}S''}, the fact that {\displaystyle a\downarrow _{S}S'} would imply by right-transitivity that {\displaystyle a\downarrow _{S}S'S''}, contradicting the choice of {\displaystyle S''}. So {\displaystyle a\not \downarrow _{S'}S''}. Therefore {\displaystyle U(a/S'S'')\geq \alpha } implies {\displaystyle U(a/S')\geq \alpha +1}, by definition of Lascar rank. This completes the inductive step, and the proof. QED

Fact 3: If {\displaystyle a,b} are tuples, then {\displaystyle U(a/S)\leq U(ab/S)}.

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

Proof of the Lascar Inequalities[]

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

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

For the right-hand side, we prove by induction on {\displaystyle \alpha } that {\displaystyle U(ab/C)\geq \alpha \implies U(a/bC)\oplus U(b/C)\geq \alpha .} The {\displaystyle \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 {\displaystyle U(ab/C)\geq \alpha +1}. Then by definition of Lascar rank, there exists {\displaystyle C'\supseteq C} such that {\displaystyle ab\not \downarrow _{C}C'}, and {\displaystyle U(ab/C')\geq \alpha }. By the inductive hypothesis, {\displaystyle U(a/bC')\oplus U(b/C')\geq \alpha .} If {\displaystyle b\downarrow _{C}C'} and {\displaystyle a\downarrow _{bC}C'}, then by left-transitivity of forking independence, {\displaystyle ab\downarrow _{C}C'}, a contradiction. So {\displaystyle b\not \downarrow _{C}C'} or {\displaystyle a\not \downarrow _{bC}C'}. In the first case, {\displaystyle U(b/C)\geq U(b/C')+1=U(b/C')\oplus 1.} Since {\displaystyle U(a/bC)\geq U(a/bC')} by monotonicity of Lascar-rank (Fact 1), we see that {\displaystyle U(a/bC)\oplus U(b/C)\geq U(a/bC')\oplus U(b/C')\oplus 1\geq \alpha \oplus 1=\alpha +1,} completing the inductive step. In the second case, {\displaystyle a\not \downarrow _{bC}C'}, so {\displaystyle U(a/bC)\geq U(a/bC')\oplus 1,} so we similarly have {\displaystyle U(a/bC)\oplus U(b/C)\geq U(a/bC')\oplus 1\oplus U(b/C')=U(a/bC')\oplus U(b/C')\oplus 1\geq \alpha +1.} Either way, the inductive step holds. This completes the induction on {\displaystyle \alpha }. Now setting {\displaystyle \alpha =U(ab/C)}, we get the right-hand side of the Lascar inequalities.

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