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) ≤ U(ab/C) ≤ U(a/bC) ⊕ U(b/C). Here U(x/S) denotes the Lascar U-rank of tp(x/S). On the left, + denotes the usual ordinal sum, while on the right, ⊕ denotes the natural sum of ordinals, defined by adding the coefficients in Cantor normal form: $\sum\limits_{\alpha}\omega^{\alpha} \cdot n_{\alpha} \oplus \sum\limits_{\alpha}\omega^{\alpha} \cdot m_{\alpha}:=\sum\limits_{\alpha}\omega^{\alpha} \cdot (n_{\alpha} + m_{\alpha}).$

There is also a related auxiliary result: if a↓_Cb, then U(ab/C) = U(a/C) ⊕ 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) ≥ α + 1 if and only if there is some S^′ ⊃ S such that tp(a/S^′) is a forking extension of tp(a/S) and U(a/S^′) ≥ α.

We will use the following basic facts.

Fact 1: If S^′ ⊇ S, then U(a/S^′) ≤ U(a/S), so Lascar rank is appropriately monotone.

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

Fact 2: If S^′ ⊇ S and a↓_SS^′, 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^′) ≥ U(a/S). We prove by induction on α that U(a/S) ≥ α implies U(a/S^′) ≥ α. As before, the cases where α is zero or a limit ordinal are completely trivial. Consider the successor ordinal case: U(a/S) ≥ α + 1. Then there exists S^″ ⊇ S such that $a\operatorname{\downarrow\not{}}_{S}S^{''}$ and U(a/S^″) ≥ α. We may move S^″ by an automorphism over aS so that S^″↓_aSS^′. Since a↓_SS^′, it follows by left transitivity that S^″a↓_SS^′, which in turn implies a↓_S^″S^′. Since U(a/S^″) ≥ α, the inductive hypothesis implies U(a/S^′S^″) ≥ α. If a↓_S^′S^″, the fact that a↓_SS^′ would imply by right-transitivity that a↓_SS^′S^″, contradicting the choice of S^″. So $a\operatorname{\downarrow\not{}}_{S^{\prime}}S^{''}$. Therefore U(a/S^′S^″) ≥ α implies U(a/S^′) ≥ α + 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) ≤ U(ab/S).

Proof. We show by induction on α that U(a/S) ≥ α implies U(ab/S) ≥ α. As before, the only non-trivial case is the successor ordinal case. Suppose U(a/S) ≥ α + 1. Then there is S^′ ⊃ S with $a\operatorname{\downarrow\not{}}_{S}S^{\prime}$ and U(a/S^′) ≥ α. By induction, U(ab/S^′) ≥ α. By monotonicity of forking, $ab\operatorname{\downarrow\not{}}_{S}S^{\prime}$, so U(ab/S^′) ≥ α implies U(ab/S) ≥ α + 1, completing the inductive step. QED

Proof of the Lascar Inequalities[]

For the left-hand side, we prove by induction on β that if U(b/C) ≥ β, then U(a/bC) + β ≤ U(ab/C). For the base case β = 0, note that U(a/bC) ≤ U(a/C) ≤ U(ab/C) by Facts 1 and 3 above. For the successor case, suppose that U(b/C) ≥ β + 1. Then by definition of Lascar rank, there is some C^′ ⊇ C such that U(b/C^′) ≥ β and $b\operatorname{\downarrow\not{}}_{C}C^{\prime}$. Move C^′ by an automorphism over bC so that C^′↓_bCa. This implies that U(a/bC^′) = U(a/bC), by Fact 2.

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

For the right-hand side, we prove by induction on α that U(ab/C) ≥ α ⇒ U(a/bC) ⊕ U(b/C) ≥ α. The α = 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) ≥ α + 1. Then by definition of Lascar rank, there exists C^′ ⊇ C such that $ab\operatorname{\downarrow\not{}}_{C}C^{\prime}$, and U(ab/C^′) ≥ α. By the inductive hypothesis, U(a/bC^′) ⊕ U(b/C^′) ≥ α. If b↓_CC^′ and a↓_bCC^′, then by left-transitivity of forking independence, ab↓_CC^′, a contradiction. So $b\operatorname{\downarrow\not{}}_{C}C^{\prime}$ or $a\operatorname{\downarrow\not{}}_{bC}C^{\prime}$. In the first case, U(b/C) ≥ U(b/C^′) + 1 = U(b/C^′) ⊕ 1. Since U(a/bC) ≥ U(a/bC^′) by monotonicity of Lascar-rank (Fact 1), we see that U(a/bC) ⊕ U(b/C) ≥ U(a/bC^′) ⊕ U(b/C^′) ⊕ 1 ≥ α ⊕ 1 = α + 1, completing the inductive step. In the second case, $a\operatorname{\downarrow\not{}}_{bC}C^{\prime}$, so U(a/bC) ≥ U(a/bC^′) ⊕ 1, so we similarly have U(a/bC) ⊕ U(b/C) ≥ U(a/bC^′) ⊕ 1 ⊕ U(b/C^′) = U(a/bC^′) ⊕ U(b/C^′) ⊕ 1 ≥ α + 1. Either way, the inductive step holds. This completes the induction on α. Now setting α = U(ab/C), we get the right-hand side of the Lascar inequalities.

Finally, we prove that if a↓_Cb, then U(ab/C) = U(a/C) ⊕ U(b/C). We have already seen that U(ab/C) ≤ U(a/bC) ⊕ U(b/C) = U(a/C) ⊕ U(b/C), so we only need to show that U(a/C) ⊕ U(b/C) ≤ U(ab/C). We prove by joint induction on α and β that U(a/C) ≥ α ∧ U(b/C) ≥ β → U(ab/C) ≥ α ⊕ β. If α or β is 0, this follows from the fact that U(a/C) ≤ U(ab/C) and U(b/C) ≤ U(ab/C). Otherwise, recall that the natural sum α ⊕ β is the smallest ordinal greater than all ordinals of the form α^′ ⊕ β and α ⊕ β^′ for α^′ < α and β^′ < β. So, we need to show that if α^′ < α or β^′ < β, then U(ab/C) > α^′ ⊕ β or U(ab/C) > α ⊕ β^′, respectively. We handle the first case; the second is completely symmetric. Since U(a/C) > α^′, it follows that U(a/C) ≥ α^′ + 1, so there is some C^′ ⊇ C with $a\operatorname{\downarrow\not{}}_{C}C^{\prime}$ and U(a/C^′) ≥ α^′. Moving C^′ over aC, we may assume that C^′↓_aCb. But then since a↓_Cb, it follows by left-transitivity that C^′a↓_Cb, and in particular C^′↓_Cb and a↓_C^′b. Since C^′↓_Cb, we have U(b/C^′) = U(b/C) ≥ β (Fact 2). And since a↓_C^′b, U(a/C^′) ≥ α^′, and U(b/C^′) ≥ β, the inductive hypothesis ensures that U(ab/C^′) ≥ α^′ ⊕ β. Now $ab\operatorname{\downarrow\not{}}_{C}C^{\prime}$, since a forks with C, so U(ab/C) ≥ U(ab/C^′) + 1 ≥ α^′ ⊕ β ⊕ 1 > α^′ ⊕ β. This completes the proof. ::: ::: :::