2 Revised answer for case mu_1 not in ran(j_01).

This is an excellent and interesting question! You are asking whether the 2-step iteration of a normal measure μ μ on a measurable cardinal κ κ is uniquely factored by the steps of the iteration itself.

(I think I have just been exchanging email with you--orperhaps with one of your colleagues, with a differentname?--about the one-dimensional version of this question.)

My answer is that No, you haven't quite got all the possibilities.

The reason is that actually, there answer is a commuting square ofelementary embeddings for the 2-step iteration, as followsYes.The ultrapower

Let me denote κ0 just by κ and j02 is just the ultrapower of V by the measure j. Sinceμx μ, and it 1 is a standardmodel-theoretic fact that the ultrapower of any structureby a product measure μ x in M1, it has theform j01(m)(κ), where m =(να | α < κ). Since youhave said that μ1 is to first take not inran(j01), we may choose theultrapower by ν, να to be all different, and then take the ultrapower by μ.The lower half of the square differentfrom μ0. In this case, there is a partitionof κ as the iteration you have defineddisjoint union of Xα,with the diagonal being j02Xα in να and nonein μ0. The upper half of the square is Let x = (Xα | α< κ). Note that j02κ is also equal to the compositionj01(Xα) for any α <κ, and similarly κ1 is not inj(Xα). But κ is inj01(x)(β) for some β <κ1, since this is a partition ofκ1. This Apply j12 to conclude thatκ1 is not obviousin j(x)(β) for this β.Thus, but itcan be verified by proving there is some β in the model-theoretic fact interval [κ,κ1) having the form β =j(f)(κ1) for the function f that Imentionedpicks theindex. ThusFrom this, it follows from normality ofμ0 that we can take i write κ =j(g)(κ1) for some function g, since anyβ < κ1 generates κ viaj01:V-->M. In my favored terminology, the seedκ1 generates κ via j and k = in factgenerates all β in [κ,κ1) viaj01restricted to M.

Similarly, suppose that δ is in the interval[κ1,and this will map elementarilyinto j(κ)). We know δ =j12(f)(κ1) for some function fon κ1 in M2 with 1. We also know f =j0201(F)(κ) for some F in V. Thus, δ =k.ij(F)(κ, κ1). But in thiscaseIn Y, we let(α,β) be the smallest pair with δ =j(F)(α,β). It cannot be that both are belowκ1, since this would be insideran(j12) and so the least pair must have N β= Mκ1. Thus, δ generatesκ1, which we already observed generates

To summarize, every ordinal in the interval[κ1,j(κ)) generatesκ1, which generates all the ordinalsβ in [κ,κ1), any of whichgenerate κ and all the other such β.

This is enough to answer your question. The k " N in yourquestion is just j01 " an arbitrary elementary substructure ofM12 containing ran(j), so suppose we have Yelementary in M2 and ran(j) subset Y.

This The case Y= ran(j) is not equal to any one of the sets on your listcases. It isOtherwise, Y has somethingnot j02 " Vin ran(j). Every object in M2 has formj(h)(κ,κ1) for some function h, since it includesby looking at the smallest pair of ordinals to generate agiven object with j(h), we see that there must be ordinalsbelow j01(κ). It is not j12 "Mκ) in Y. If Y contains any ordinal δ inthe interval [κ1 since it does not include κ. And ,j(κ)), then it iscontain both κ and κ1, since weobserved that any such δ generates these ordinals. Inthis case, Y = M2, since it is not transitivethose two ordinalsgenerate everything.

A much So we assume that Y contains no suchδ. In this last case, Y must contain some ordinalβ in the interval [κ,κ1). Sinceany such β generates κ, Y contains all suchordinals. It follows that ran(j12) subset Y andin fact = Y, since if Y contained anything more interesting follow-up it wouldhave to have an additional ordinal δ in[κ1,j(κ)).

So we've seen that your three cases are the onlypossibilities. And like your previous question, there is whether thefactorings arising in noneed to assume that Y or N is somehow internally definable.

By the commuting square are exhaustiveway, and I will give this some was a problem that I had solved many yearsago for my dissertation, although perhaps other people hadalso thought about it. I think it may be thecase was interested in understandingwhich pairs of ordinals (α,β) generate productmeasures via an embedding j, and this question is very muchrelated to thatif you add .

(Click the edit history to see my extra previous answer, which was just about the case when μ1 is in the range of j01, then a case for which the list answer is exhaustive...no.)

1

This is an excellent and interesting question! You are asking whether the 2-step iteration of a normal measure μ on a measurable cardinal κ is uniquely factored by the steps of the iteration itself.

(I think I have just been exchanging email with you--or perhaps with one of your colleagues, with a different name?--about the one-dimensional version of this question.)

My answer is that No, you haven't quite got all the possibilities. The reason is that actually, there is a commuting square of elementary embeddings for the 2-step iteration, as follows. The ultrapower j02 is just the ultrapower of V by the measure μ x μ, and it is a standard model-theoretic fact that the ultrapower of any structure by a product measure μ x ν is to first take the ultrapower by ν, and then take the ultrapower by μ. The lower half of the square is the iteration you have defined, with the diagonal being j02. The upper half of the square is that j02 is also equal to the composition j01.j01. This is not obvious, but it can be verified by proving the model-theoretic fact that I mentioned. Thus, we can take i = j01:V-->M1 and k = j01 restricted to M1, and this will map elementarily into M2 with j02 = k.i. But in this case, we have N = M1 and k " N is just j01 " M1.

This is not equal to any of the sets on your list. It is not j02 " V, since it includes j01(κ). It is not j12 " M1 since it does not include κ. And it is not M2 since it is not transitive.

A much more interesting follow-up question, is whether the factorings arising in the commuting square are exhaustive, and I will give this some thought. I think it may be the case that if you add my extra case, then the list is exhaustive...