Hod up to AD R + Θ is measurable

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1 Hod up to AD R + Θ is measurable Rachid Atmai Department of Mathematics University of North Texas General Academics Building Union Circle # Denton, TX atmai.rachid@gmail.com Grigor Sargsyan Department of Mathematics Rutgers University 110 Frelinghuysen Rd. Piscataway, NJ gs481 grigor@math.rutgers.edu, Abstract Suppose M is a transitive class size model of AD R + Θ is regular. We say M is a minimal model of AD R + Θ is measurable if (i) there is µ M such that M µ is a normal R-complete measure on Θ and (ii) for any transitive class size N M, N there is no R-complete measure on Θ. Continuing Trang s work in [9], we compute HOD of a minimal model of AD R + Θ is measurable. The computation of HOD of models of determinacy has been one of the central themes in descriptive inner model theory. Steel s seminal [6] jumpstarted the project and a later work of Steel and Woodin (for instance see [4] or [5]) established connections with the Mouse Set Conjecture, core model induction and the inner model problem. The analysis of HOD presented in the above papers, however, only compute VΘ HOD. Woodin computed the full HOD of L(R) under AD L(R) (the proof can be found in [7]) Mathematics Subject Classifications: 03E15, 03E45, 03E60. Keywords: Mouse, inner model theory, descriptive set theory, hod mouse. First author s work is partially based upon work supported by the National Science Foundation under Grant No DMS and DMS

2 Trang continued this work in [9]. He presented the exact inner model theoretic structure of HOD of models of determinacy that are contained inside the minimal model of AD R + Θ is regular. We extend Trang s work to minimal models of AD R + Θ is measurable. Assume AD +. We say Θ is measurable if there is an R-complete normal measure µ on Θ. We then say that M is a minimal model of AD R +Θ is measurable if for any N M such that R, Ord N, N Θ is not measurable. It is then not hard to characterize minimal models of AD R + Θ is measurable. Lemma 0.1. Suppose that V is a minimal model of AD R + Θ is measurable. Let M = L(P(R)). Let µ be a normal R-complete measure on Θ. Then V = L(P(R))[µ]. Proof. We have that L(P(R))[µ] AD R + Θ is measurable. Because L(P(R))[µ] V, we have that V = L(P(R))[µ]. Acknowledgments. The authors would like to thank Nam Trang for communicating them the problem considered in this paper. 1. The main theorem Suppose V is a minimal model of AD R + Θ is measurable. Let M = L(P(R)) and working in M, let F = {(P, Σ) : (P, Σ) is a hod pair and Σ has branch condensation and is fullness preserving} Following [4], for (P, Σ), (Q, λ) F, we let (P, Σ) (Q, Λ) if for some α λ Q, (Q(α), Λ Q(α) ) is a tail of (P, Σ), i.e., Q(α) I(P, Σ) and Λ Q(α) = Σ Q(α). It follows from comparison theory of hod pairs (see Chapter 2 of [4]) that is directed. We then let M be the direct limit of the system F under the maps πp,q Σ : P Q(α) where α λ Q is such that Q(α) is a Σ-iterate of P. It is shown in [4] (see Theorem 4.24) that in M, M = VΘ HOD. The following is essentially the generic interpretability result of [4] (see Theorem 3.10 of [4]). Given a hod premouse P and X generic over P, we let Σ P[X] be the interpretation of Σ P onto P[X] according to the procedure described in the proof of Theorem 3.10 of [4]. Lemma 1.1. In M (and hence in V ), M is (Θ, Θ)-iterable via a strategy Σ such that given any T according to Σ, Σ( T ) = b M [ T ] Σ M [ T ] ( T ) = b. We let Σ be the strategy of M described in Lemma 1.1. extending M. Given N M we say N is good if Next, we define a model 2

3 1. N is sound, and 2. whenever π : N N is elementary and N is countable, then N is ω 1 -iterable above π 1 (Θ) as a Σ π - premouse. The next two lemmas are basic lemmas about good mice. Lemma 1.2. Suppose N 0, N 1 are good such that for some η N 0 N 1, η is a cutpoint of both N 0 and N 1, N 0 η = N 1 η and ρ ω (N 0 ), ρ ω (N 1 ) η. Then either N 0 N 1 or N 1 N 0. Proof. We start with clause 1. Suppose that neither N 0 N 1 nor N 1 N 0 holds. Let then π : H L ξ (P(R))[µ] be elementary such that ξ Θ, N i rng(π), Σ rng(π), η rng(π) and H = ω. We let Ni = π 1 (N i ) for i = 0, 1. Because Σ has hull condensation, it follows that Σ H = Σ π. It follows from elementarily that for i = 0, 1, ρ ω ( N i ) = π 1 (η). But now because N i are sound Σ π -mice, we have that N 0 N 1 or a vice versa. Lemma 1.3. Suppose N is good such that ρ ω (N ) Θ. Then ρ ω (N ) = Θ. Proof. Towards a contradiction assume ρ(n ) < Θ. Let π : H L ξ (P(R))[µ] be such that H Θ Θ, cp(π) > ρ(n ) and N rng(π). Let P = π 1 (N ). It follows from the proof of clause 1 that M P is OD. It then follows that P M. This is a contradiction. We then let M = {N : M N, N is good and ρ ω (M) = N }. Notice that we have that M HOD. Let η = o(m ). Let now µ V be a normal R-complete measure on Θ. It follows from Lemma 0.1 that V = L(P(R))[µ]. Working in HOD µ, let π µ be the ultrapower embedding via µ HOD µ. Let λ µ = (Θ + ) πµ(m ). Notice that the ordinal λ µ may depend on µ. Let then M = π µ (M ) λ µ and let E µ be (Θ, η)-extender derived from π µ M 1. More precisely, (a, A) E µ a (o(m )) <ω, A [Θ] a M, and a π µ (A) We now make the following minimality assumption on µ. Minimality Assumption on Measures: We say µ satisfies MAM if λ µ above is the least possible. The following is our main theorem. Theorem 1.4 (Main Theorem). HOD = J [M, E µ ]. 1 Recall that o(x) = X Ord. 3

4 We will present the proof as a sequence of lemmas. Before we go into the proof of the main theorem, we list some of the complications involved with proving it. First we will show that E µ is amenable to M (see Lemma 1.6). It follows from its definition that it coheres M. It then follows that (M, E µ ) is a hod premouse. The next challenge is to show that no level of J [M, E µ ] projects to or below Θ (see Lemma 1.8). A consequence of this is that VΘ HODM = V J [M,Eµ] Θ. This then allows us to show that P(R) can be symmetrically added to J [M, E µ ] (which is done as part of proving Lemma 1.8). We then finish by showing that there is only unique normal R-complete measure over Θ (see Lemma 1.12). It follows from here that µ HOD HOD implying that J [M, E µ ] HOD. Combining the aforementioned results it is then not hard to see that V is a symmetric extension of J [M, E µ ], which then easily implies that in fact HOD J [M, E µ ]. We start by showing that 1.1. Amenability Lemma 1.5. M = π µ (M ) η, M = M (Θ + ) M and whenever M N M is such that ρ(n ) = η, N is good. Proof. We start by proving the first equality. We have that M HOD and therefore, M HOD µ. It is then enough to show that whenever M N M is such that ρ(n ) = Θ, there is f : Θ M such that f HOD µ and π µ (f)(θ) = N. Working in HOD µ, construct a club C and a continuous chain (N κ : κ C) along with embeddings π κ : N κ N such that cp(π κ ) = θ κ = κ. We have that N κ M. It then easily follows that π µ (N κ : κ C)(Θ) = N. This shows that M π µ (M ), and therefore, M = π µ (M ) η. For the second equality, it is enough to show that whenever M N M is such that ρ ω (N ) = Θ then N is good. Let then f : Θ M be such that π µ (f)(θ) = N. It then follows that for a µ-measure one set of κ, there is an elementary embedding π κ : f(κ) N. It then also follows that whenever π : N N is an elementary embedding such that N is countable, there is a µ-measure one set of κ such that there is an embedding σ : N f(κ) with the property that π = π κ σ. It then follows that N is Σ π -good. The proof of the third equality is very similar. In what follows, we will sometimes write M for M. Next, is our amenability result. Lemma 1.6. µ is amenable to M, that is µ M M Proof. It is enough to show that if N M is such that ρ ω (N ) = Θ then µ N M. Fix N and let p N be the standard parameter of N. Construct an elementary chain (N κ : κ S) and elementary embeddings (π κ : κ S) such that S Θ is a club and for κ < λ S 1. π κ : N κ N, 2. cp(π κ ) = κ, κ = θ κ, rng(π κ ) rng(π λ ) and 4

5 3. N κ Θ = κ. Let for κ < λ S, π κ,λ = def π 1 λ π κ : N κ N λ. For κ S let p κ be the standard parameter of N κ. Let µ κ = {A N κ : κ π κ (A)} be the measure derived from π κ. We claim that Claim. {κ : π κ [µ κ ] µ} µ. Proof. Let µ = µ HOD µ. Notice that M HOD and µ M = µ M. Moreover, we can choose S in a way that S HOD µ. Since S is a club, Θ π µ (S). Let π µ (N κ : κ S) = (N κ : κ π µ (S)) π µ (µ κ : κ S) = (ν κ : κ π µ (S)) π µ (π κ : κ S) = (σ κ : κ π µ (S)). Then we have that ν Θ is just the measure derived from the embedding π µ N Θ. It follows that µ N Θ = ν Θ. But because π µ [µ ] π µ (µ ), we have that Θ {κ : π µ [ν κ ] π µ (µ)} as π µ [ν Θ ] π µ (µ ). But then {κ : π κ [µ κ ] µ} µ. Next, let A = {κ : π κ [µ κ ] µ}. Since (1) π κ [µ κ ] µ λ, π κ = π λ π κ,λ and π λ [µ λ ] µ, we have that (2) µ κ is just the measure derived from the embedding π κ,λ, i.e., µ κ = {A N κ : κ π κ,λ (A)}. It follows from condensation that N κ, N λ M. It then follows that π κ,λ HOD M. This is simply because π κ,λ is generated by the embedding that sends p κ to p λ and is the identity on ρ ω (N κ ) = κ where p κ and p λ are the standard parameters of N κ and N λ respectively. More precisely, (3) π κ,λ (x) = τ N λ (s, pλ ) where s κ <ω and τ is such that x = τ Nκ (s, p κ ). It follows from (3) that π κ,λ HOD M. We then have that µ κ HOD M. Let M κ M be the least such that µ κ M κ for κ A. Let π µ (M κ : κ < Θ)(Θ) = Q. Notice that we must have that Q is good and hence, Q M. This follows from the fact that if τ : R Q is a countable hull of Q then for some κ there is τ : R M κ such that Σ τ = Σ τ. We then have that µ N Q because 5

6 N = π µ (N κ : κ S)(Θ) and µ N = π µ (µ κ : κ S)(Θ) A strategy for countable submodels of J [M, E µ ] Suppose ξ is such that E µ is a total extender in J ξ [M, E µ ]. Let π : Q J ξ [M, E µ ] be elementary such that Q is countable. In this section, we show that Q has a π-realizable iteration strategy. Given any such R, we let E R be the preimage of E µ. We also let R = R ((δ R ) + ) R. Recall from [4] that if T is a stack on some model M and R is a node in T then T R is the portion of T after stage R. Given a stack T on Q, it can easily be partitioned into segments by considering when the image of E Q is used. Thus, we say (M 0 α, E α, M 1 α, T α : α < η) are the essential components of T if 1. M 0 0 = Q, 2. E 0, M 1 0 are defined if and only if the first extender used in T is E Q in which case M 1 0 = Ult(M 0 0, E Q ). 3. T 0 is the largest initial segment of T that is based on (M 1 1). 4. If α + 1 < η then M 0 α+1 is the last model of T α. Then E α+1 = E M0 α+1 and M 1 α = Ult(M 0 α+1, E α+1 ). Again, T α+1 is the largest portion of T M 1 α+1 that is based on M 1 α If α < η is limit then M 0 α is the direct limit of (M 0 β : β < α) under the iteration embeddings. The rest of the objects are defined as in the successor case. Suppose now that T is a stack on Q with essential components (M 0 α, E α, M 1 α, T α : α η). Suppose that we also have embeddings (π 0 α, π 1 α : α < η) such that 1. π 0 0 = π. 2. For α < η and i 2, π i α : M i α J ξ [M, E µ ]. 3. For α < β < η and i, j 2, π i α = π j β π T M i α,m j β 4. For α < η, T α is according to π 1 α-pullback of Σ. We would like to define embeddings (π 0 η, π 1 η) such that 1. For i 2, π i η : M 0 η J ξ [M, E µ ]. 2. For α < η and i, j 2, π i α = π j η π T M i α,mj η and π 1 α = π 1 α π T M 0 α,m1 α. and π 1 η = π 1 η π T M 0 η,m 1 η. We say T is π-realizable if there is π = def (π 0 α, π 1 α : α < η) witnessing the above clauses. Suppose first η = α + 1. Let (S, Λ) F be such that 6

7 sup(π 1 α+1[δ M1 α+1 ]) = δ M (S,Λ). The above equality simply says that the direct limit of all Λ-iterates of S reaches the ordinal mentioned on the left side. Since M 1 α+1 is countable, we can also require that π 1 α+1[m 1 α] rng(π Λ S, ). Let then k = def (π Λ S, ) 1 π 1 α+1 : M 1 α S. We then have that T α is according to k-pullback of Λ. It follows then that letting R be the last model of k T α there is an embedding l such that (1) l : M 0 η R and π k T α k = l π T α. We then set π 0 η = π Λ R R, l. Notice now that (2) π Λ S, = πλ R R, π k T α. (1), (2) and our choice of (S, Λ) imply that (3) π 1 α = π 0 η π T M 1 α,m0 η. (3) then implies that πη 0 is as desired. To define πη 1 we use countable completeness of µ. First let ζ = o((m 0 η) ). Thus, ζ is the successor of δ M0 η in M 0 η. For each a ζ <ω, let A a = {πη(a) 0 : (a, A) E M0 η }. Fix now a fiber f for the set {(πη(a), 0 A a : a ζ <ω }. We can now define π 1 η([a, g] E M 0 η ) = π0 η(g)(f(a)). It is a standard argument to show that π 1 η is as desired. It is now easy to show that Lemma 1.7. There is an (ω 1, ω 1 ) iteration strategy Λ for Q such that whenever T is a stack according to Λ then T is π-realizable J [M, E µ ] is a hod premouse In this section we show that no level of J [M, E µ ] projects across or to Θ. Lemma 1.8. N = J (M, E µ ) is a hod premouse such that for every α, ρ ω (N α) > Θ. Proof. We start with the following claim. Later we will show that in fact for every α, ρ ω (J α (M, E µ )) Θ. Claim 1. For every α, ρ ω (J α (M, E µ )) Θ Proof. Suppose not and fix the least α such that ρ ω (J α (M, E)) < Θ. Now fix a β such that β < λ M and θ β ρ ω (J α (M, E)) < θ β+1. Fix a hod pair (Q, Λ ) such that M (Q, Λ ) = M(β + 2) and an elementary hull σ : H L ξ (P(R))[µ] such that: 1. ξ Θ, 7

8 2. H = ℵ 0 3. (Q, Λ ) rng(σ), 4. J α (M, E) rng(σ). Now let Q = σ 1 (J α (M, E))) and let γ = σ 1 (β). Then by elementarity we have that ρ ω ( Q) < θ Q γ+1. Let Q = C Q ω (p, Q(γ + 1)) where p is the standard parameter of Q. Notice that Q is δ Q γ+1-sound. Let l : Q Q be the core embedding. Let Λ be a σ l-realizable strategy of Q (see Lemma 1.7). It follows from the branch condensation of Λ that (1) Λ Q(γ+1) = Λ Q(γ+1). Consider the pointclass generated by (Q, Λ): Since Q is δ Q γ+1-sound (2) Q is OD ΛQ(γ+1). Γ(Q, Λ) = {A : ( U, R) B(Q, Λ)(A < W Code(Λ R, U ))}. This is because Q is the unique Λ Q(γ+1) -hod mouse generating the pointclass Γ(Q, Λ). But since Q(γ + 2) is Λ Q(γ+1) -full, we must have Q Q(γ + 2). But ρ(q) < δ Q γ+1 and this is a contradiction. The following claim is easier and finishes the proof of the lemma. Claim 2. For every α, ρ(j α (M, E))) > Θ Proof. Suppose not. Let α be the least such that ρ(j α (M, E))) = Θ. Then let A Θ be definable over J α (M, E) such that A / J α (M, E). Notice that for every κ < Θ, A κ M. Let S be the set of κ < Θ such that θ κ = κ. Given κ S we let M κ M be the least such that A θ κ M κ. Let j : M Ult(M, µ) be the ultrapower embedding. Let j((m κ : κ S)) = (N κ : κ j(s)). But then A = j(a) Θ N Θ. Notice that N Θ implying that N Θ M. It follows that A M, contradiction! 8

9 1.4. P(R) is symmetrically generic over J [M, E µ ] We start by recalling Vopenka algebra. We work in M = L(P(R)). Let ϕ be a formula and s Θ <ω. Given t Θ <ω, define We write A s φ,t = {a : dom( a) = dom(s), a(i) s(i) ω, L(P(R)) φ[a, t]}. (φ, t) s (ψ, v) if and only if A s φ,t = As ψ,v. We then let [φ, t] s be the s -equivalence class of (φ, t). Next define Q = {(s, [ϕ, t]) : s Θ <ω, ϕ is a formula, t Θ <ω } The ordering on Q is defined as follows: (v, [φ, s]) (u, [ψ, t]) if and only if u v, A u ψ,t A v ϕ,s and A v ϕ,s dom(u) A u ψ,t where A v ϕ,s dom(u) = {a dom(u) : a A v ϕ,s}. Let P = {(s, a) : s Θ <ω, dom(a) = dom(s), i n, a(i) s(i) ω }. We set (s, a) P (t, b) if and only if t s and b a. Notice that if (f, h) is P-generic then f : ω Θ and h : ω (Θ ω ) V are surjections such that for each i ω, f(i) (h(i)) ω. The following lemma is standard and is really due to Vopenka. Lemma 1.9 (Vopenka s lemma). Suppose (f, h) is P-generic. Let G = {(f n, [ϕ, v]) : n < ω, (f n, [ϕ, v]) Q h n A f n φ,v }. Then G is Q/M-generic. Therefore, it is J [M, E µ ]-generic over Q. Proof. We first show that for ξ < Θ, we can bound the ordinal parameters used to define OD M subsets of P(ξ). Claim 1. In M, there is a function F : Θ <ω Θ such that whenever t Θ <ω and a are such that dom(a) = dom(t), a(i) t(i) and a is OD M then a is OD in L(Γ F (t) ) where Γ β = {A R : w(a) < β}. Proof. For A R 2, we say (R, A) codes an OD structure if (R, A) is a well-founded, extensional model of some fragment of ZF C and its transitive collapse is OD. Notice that by a standard Skolem hull argument, in M, if A R 2 and (R, A) codes an ordinal definable structure then for some β < Θ, (R, A) codes an ordinal definable structure in L(Γ β ). Fix now α < Θ. Then we must have that sup{β : A R 2 (L(Γ β ) (R, A) codes an OD structure )} < Θ. This is because otherwise we will have a function G : Γ β Θ which is unbounded. As Θ is regular, this is impossible. Let then 9

10 F (t) = sup{β + ω : A R 2 (A Γ sup(rng(t))+ω L(Γ β ) (R, A) codes an OD structure )}. Let then D M be a dense subset of Q. We want to see that G D. Let E be the set of (s, a) such that dom(s) = dom(a), s Θ <ω, for every i < dom(s), a(i) s(i) ω and for some [φ, v] such that (s, [φ, v]) D, a A s φ,v. We claim that E is dense in P. Fix then (s, a) P. Claim 2. {A t φ,v dom(s) : s t, (t, [φ, v]) D} = {b : dom(b) = dom(s) and for all i < dom(s), b(i) s(i) ω }. Proof. Notice that D M, which makes the claim non-trivial. Consider the set A = {A t φ,v dom(s) : (t, [φ, v]) D}. Each member of A is OD M. We want to see that A itself is OD in M. To see this let N M be such that ρ ω (N ) = Θ and D N. Fix ζ Θ and let π : H L ζ (P(R))[µ] be such that 1. N, F, D rng(π), 2. cp(π) = κ = θ κ. Let ( N, F, D) = π 1 (N, F, D). We have that F = F θ κ and D = {(t, ([φ, v]) L(P(R))H ) : (t, [φ, v]) D}. Let Ā = π 1 (A). Then N M = VΘ HODM. We then have that D OD M. Thus Ā is ODM. Notice next that Ā = A. Indeed, fix B A. We have that B is ODM. It follows that B is OD L(Γ F (s)) and hence, B is OD L(Γ F (s)). Let then (φ, v) be such that v (Θ L(Γ F (s)) ) <ω and B is definable over L(Γ F (s) ) via (φ, v). Since D is dense in H, we must have (t, [ψ, u]) Q (s, [φ, v]) such that (t, [ψ, u]) D. But then B = A t ψ,u Ā. It now follows that A OD M implying the claim. It follows from Claim 2 that there is (t, [φ, v]) D such that a A t φ,v dom(s). Let then b A t φ,v dom(s) be such that b dom(a) = a. Then (t, b) E and (t, b) P (s, a). Since we now have that E is dense, we can fix n < ω such that (f n, h n) E. Let [φ, v] be such that h n A f n φ,v. We then have that (f n, [φ, v]) G D. We leave it to the reader to verify that G is a filter. The following is the main lemma of this section. Lemma V is a symmetric extension of J [M, E µ ]. In fact, V = J [M, E µ ](Θ ω ). Proof. Let (f, h) be P-generic. Let G be as in Lemma 1.9. Then Θ ω J [M, E µ ][G]. It is a consequence of AD + that every set of reals A is (OD t ) M for some t Θ ω. Therefore, we have that P(R) HOD M (Θ ω ) J [M, E µ ](Θ ω ). 10

11 The reader can find more on the above equality is proved by consulting Section 2 of [1]. It then follows that J [M, E µ ](P(R)) = J [M, E µ ](Θ ω ). Because every set A P(Θ) J [M, E µ ](Θ ω ) is added to J [M, E µ ] by a small forcing (in fact by the Vopenka algebra at some θ α < Θ), we have that E µ has a canonical extension E µ + to J [M, E µ ](Θ ω ) (see Theorem 2.4 of [1]). Let ν = {A : (Θ, A) E µ + }. We then have that J [M, E µ ](Θ ω ) ν is a normal R-complete measure on Θ. It then follows that V = J [M, E µ ](Θ ω ) The uniqueness of µ In this section we show that there is a unique normal R-complete measure µ. We start with the following lemma, which is a simple consequence of Lemma 1.10 and the fact that small forcing doesn t create new measures. Lemma Suppose ν is a normal R-complete measure on Θ. Then ν J (M, E µ ) J (M, E µ ). Lemma There is a unique R-complete measure on Θ that satisfies MAM. Proof. The proof is a combination of the comparison argument and the argument that shows that there are no bicephali. Suppose that there is a normal R-complete measure ν such that µ ν and λ ν = λ µ. We then have two normal measures µ 0, ν 0 J [M, E µ ] such that letting µ + 0, ν 0 + be the extensions of µ 0 and ν 0 in J [M, E µ ](Θ ω ), µ = µ + 0 and ν = ν 0 +. It then follows that µ 0 ν 0. Let then F = (Θ, (Θ + ) M ))-extender derived from π ν0 (in J [M, E µ ]). Let E = E µ. We now have that J [M, E, F ] is a bicephali. Let now ξ = (Θ +ω ) J [M,E,F ] and let π : Q J ξ [M, E, F ]. It follows from the construction of Subsection 1.2 that there is an iteration strategy Λ for Q. The rest of the argument is as follows. We compare (Q, Λ) with itself. The goal is to iterate Q to some R such that R πq,r Λ (E) = πλ Q,R (F ). The complete proof of this comparison process is beyond the scope of this paper. It can be proved using Theorem 1.4 of [3] and Theorem 9.2 of [2]. We first show that 1.6. Computation of HOD Lemma VΘ HOD = M. Moreover, if A R 2 is such that (R, A) codes an OD structure then in M, (R, A) codes an structure. Proof. Fix A θ α < Θ such that A is OD over L(P(R))[µ]. It follows from Lemma 1.10 that J (M, E)[Θ ω ] A is OD. Fix ζ > Θ such that J ζ (M, E)[Θ ω ] A is OD. Let M = Hull J ζ(m,e) (Θ). We then have that 2 In fact, ν = µ but we do not need this. M(Θ ω ) A is OD. 11

12 Let Q M be such that M Q. Let π : H L ζ (P(R))[µ] be such that cp(π) = κ = θ κ > θ α, H = κ and M, Q rng(π). Let ξ = o(h), N = π 1 ( M), and P = π 1 (Q). Then it follows from elementarity that N (κ ω ) A is OD and N P. However, P M as ρ ω (P) = θ κ. It then follows that N is OD M implying that A OD M. Hence, A M. The proof of the second clause of the lemma is very similar. To finish we will use the following two lemma due to Trang and Woodin. Lemma 1.14 (Trang-Woodin, Theorem 1.1 of [9]). Suppose Q is the Vopenka algebra of M. Then HOD M = L[Q] = L[M ]. Finall we show that Lemma HOD = J [M, E µ ] Proof. We have that J [M, E µ ] HOD. It is then enough to show that HOD J [M, E µ ]. Because µ is ordinal definable, we have that HOD = L[P ](µ) where P is the Vopenka algebra (see for instance Theorem of [8]). However, notice that it follows from Lemma 1.13 and the proof of Lemma 1.9 that if Q is the Vopenka algebra of M, then in fact L[Q](µ) = HOD. Since we also have that Q M it follows that HOD J [M, E µ ]. References [1] Andres Caicedo, Paul Larson, Grigor Sargsyan, Ralf Schindler, John R. Steel, and Martin Zeman. Square in P max extensions, in preparation. [2] William J. Mitchell and John R. Steel. Fine structure and iteration trees, volume 3 of Lecture Notes in Logic. Springer-Verlag, Berlin, [3] Grigor Sargsyan. The analysis of HOD below the theory AD + + the largest Smuslin cardinal is a member of the Solovay sequence, available at math.rutgers.edu/ gs481. [4] Grigor Sargsyan. Hod mice and the Mouse Set Conjecture. Available at gs481/. [5] Grigor Sargsyan. Descriptive inner model theory. Bull. Symbolic Logic, 19(1):1 55, [6] John R. Steel. HOD L(R) is a core model below Θ. Bull. Symbolic Logic, 1(1):75 84, [7] John R. Steel. An outline of inner model theory. In Handbook of set theory. Vols. 1, 2, 3, pages Springer, Dordrecht, [8] Nam Trang. Generalized Solovay Measures, the HOD Analysis, and the Core Model Iinduction. PhD Thesis, UC Berkeley, [9] Nam Trang. HOD in natural models of AD +. Ann. Pure Appl. Logic, 165(10): ,