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I'm trying to get a deeper understanding of Whitehead's problem.

It is possible to construct a group of cardinality $\aleph_1$ that satisfies Chase's condition, and is not free. This group is Whitehead when we add Martin's axiom to ZFC, it is not Whitehead when we add $Z=L$ to ZFC. Why is it not possible to check if the group is Whitehead within ZFC? I heard someone say that we can't know it in ZFC, because checking if every homomorphism with kernel isomorphic to $\mathbb{Z}$ onto the group splits, would take an uncountable amount of steps, and when a proof can be given in uncountable steps, it can also be given in countable steps, but this is not the case here. This is merely what I recall from what I heard, can someone either explain me or point me to literature on the subject? I am quite familiar with Eklof's essay, people who answer me can refer to it.

I know that when we add Martin's axiom, we can prove that the non-free Chase group is Whitehead, we can do it because Martin's axiom allows us to connect homomorphisms with a countable domain to form a big splitting homomorphism.

1234aaa
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  • I would rather say it is independent in ZFC. – André Nicolas Aug 04 '14 at 04:26
  • Related, though perhaps you know everything mentioned there already: http://mathoverflow.net/questions/23825/shelahs-proof-of-the-independence-of-the-whitehead-problem –  Aug 06 '14 at 15:30
  • @Powerlust: What, exactly, is the "deeper knowledge" you are seeking regarding Whitehead's problem and its undecidability in ZFC? You seem (at first glance) to understand the situaton rather well. – Thomas Benjamin Aug 11 '14 at 15:49
  • " Why is it not possible to check if the group is Whitehead within ZFC? " – 1234aaa Aug 11 '14 at 15:50

2 Answers2

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Because uncountable cardinals are unwieldy.

Note that the countable case is provable, and if you have seen the proof that $\lozenge$ implies the Whitehead conjecture is true, then you would have seen that to some extent the proof is somewhat similar.

The major difference is that when we deal with the countable case we can do it with a simple induction. In the uncountable case we need to go to transfinite induction, which passes through limit cases. Lots and lots of limit cases. And it turns out that these limit cases don't "go down easily".

I don't think it's something you can easily understand without understanding the proofs for both directions of the Whitehead conjecture's independence, as well other independence proofs. In particular, I think, that the independence of the axiom of countable choice is relevant here. In both cases we seem to want to "glue together" provable bits (in the Whitehead case, every countable Whitehead group is free; in the countable choice case, every finite subset has a choice function). But in both cases the induction doesn't quite catch the limit case.

I hope that this can be of some help.

Asaf Karagila
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  • Hey Asaf. Your answer gives me some idea of what it's about. I understand the first two paragraphs, the third one is a bit vague to me. I added a bounty, maybe you want to elaborate on your answer? – 1234aaa Aug 06 '14 at 05:25
  • I'll try to give it some thought. – Asaf Karagila Aug 06 '14 at 05:54
  • By the way, how much do you know about set theory and general independence proofs? – Asaf Karagila Aug 06 '14 at 06:00
  • I followed a course on set theory. This is the only independence proof I am familiar with. – 1234aaa Aug 06 '14 at 06:21
  • @Powerlust There is no way that you have seen the independence proof of the Whitehead problem and have not seen any other independent proof. Surely you must have seen the independence of the continuum hypothesis. – William Aug 06 '14 at 11:27
  • @William: You can take diamonds and Martin's Axiom as blackboxes. – Asaf Karagila Aug 06 '14 at 11:32
  • Hey Asaf, are you planning on elaborating on your answer? If not, I'll ask my set theory professor. – 1234aaa Aug 08 '14 at 08:01
  • @Powerlust: I'm a bit preoccupied these past few days. I'll try to extend my answer, but I'm still not sure how to accommodate your knowledge about independence proofs without writing a few pages worth of answer. In either case, before asking your professor, you should at least let the bounty play out and see if someone else posts an answer (and I'll try to elaborate, or rewrite if need be) in the remaining few days. – Asaf Karagila Aug 08 '14 at 15:30
  • Is it possible to recast Shelah's result regarding Whitehead's problem completely in terms of Category theory using Topoi and Sheaves, making his result completely algebraic, so to speak? Would such a result allow one to come to a 'deeper understanding' (whatever that might mean) of Whitehead's problem? – Thomas Benjamin Aug 10 '14 at 12:53
  • @Thomas: I don't know much about topos theory and category theory. I also think that the independence is coming from set theoretic issues. But Whitehead's Problem, and the independence of it from ZFC are two different things. – Asaf Karagila Aug 10 '14 at 13:41
  • @Thomas: I also don't quite understand why an algebraic formulation would give a "deeper understanding" in any way, to this problem. – Asaf Karagila Aug 10 '14 at 14:48
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In some sense (I think), you have, in your question, enough information to adequately answer it yourself. When you wrote me (in response to my question as to what deeper knowledge you were seeking regarding Whitehead's problem), "why is it not possible to check if the group is Whitehead in ZFC", I am going to infer that "the group" you had in mind is the group you spoke about in your question, the "group of cardinality $\aleph_1$ that satisfies Chase's condition, and is not free" that is "Whitehead when we add Martin's axiom" (and $2^{\aleph_0}$ $\gt$ $\aleph_1$) and "is not Whitehead when we add V=L" to ZFC.

To show that you have, in fact, answered your own question, I would like to present a mathematical 'thought experiment' regarding the aforementioned group. The gist of the experiment is to use the following theorem, due to Solovay and Tennenbaum

Assume GCH [in V, the 'universe' formed by the 'cumulative hierarchy'--my comment] and let $\kappa$ be a regular cardinal greater than $\aleph_1$. There exists a c.c.c. notion of forcing P such that the generic extension V[G] by P satisfies Martin's Axiom and $2^{\aleph_0}$=$\kappa$

to relate the domain of sets in which ZF+V=L holds to the domain of sets in which ZFC+Martin's Axiom+$2^{\aleph_0}$ $\gt$$\aleph_1$ holds. That this theorem does is clear by the fact that ZF+V=L $\vdash$GCH so that the domain of sets in which ZFC+Martin's axiom +$2^{\aleph_0}$ $\gt$$\aleph_1$ is a forcing extension of the domain of sets in which ZF+V=L holds, both domains via the c.c.c. having the same cardinals and cofinalities.

Now according to your question, the group of cardinality $\aleph_1$ which is not free and satisfies Chase's condition, is not Whitehead in the domain of sets in which ZFC+V=L holds. Since this is a domain in which ZFC holds, by constructing this group in ZFC+V=L, you 'checked' that his group is not Whitehead in this domain (it should be noted that in the Eklof paper you mention in your question the construction of the group A of cardinality $\aleph_1$ satisfying Chase's condition but was not free was carried out entirely in ZFC). Now by the theorem of Solovay and Tennenbaum one can use the notion of forcing P (which satisfies the c.c.c.) to construct from the domain of sets which satisfies ZFC+V=L the domain of sets satisfying ZFC+Martin's Axiom +$2^{\aleph_0}$ $\gt$ $\aleph_1$ and since Goedel proved that L is a subdomain of sets in ZFC (actually ZF) in which not only the axioms of ZF hold but AC and GCH hold as well, one might rightfully infer that the domain of sets in which ZFC+Martin's Axiom +$2^{\aleph_0}$ $\gt$ $\aleph_1$ exists as well and reconstruct the group A in this domain so it is Whitehead--by construction you have checked that A is Whitehead within this domain in which ZFC holds. This shows that it is impossible to check whether the group A you speak about in your question is or is not Whitehead in ZFC because ZFC in itself does not contain sufficient information (i.e. the extra axioms, by the 'facts' each of the domains prove about the group A, the axioms are incompatible) to decide the question.

Thomas Benjamin
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  • I don't understand this answer. After extending the universe you add a new group which Whitehead but not free. The groups in $L$ are still free. – Asaf Karagila Aug 15 '14 at 07:32
  • @AsafKaragila: If Powerlust was correct as to the facts in his question, there is a group A of cardinality $\aleph_1$ that satisfies Chase's condition and is not free which is not Whitehead in ZFC+V=L i.e this group can be constructed in ZFC+V=L and shown (by construction) that it is not Whitehead when V=L holds. After extending the 'universe' V=L via the forcing construction in the proof of the Solovay-Tennenbaum theorem I quoted, Martin's Axiom +$2^{\aleph_0}$ $\gt$ $\aleph_1$ now holds in the new 'universe'. The groups in L are still free and the group A that satisfies Chase's condition – Thomas Benjamin Aug 15 '14 at 19:18
  • (cont.) but is not free (in V=L) also exists in the new universe. However, because Martin's Axiom and $2^{\aleph_0}$ $\gt$ $\aleph_1$ hold, the group A can be shown to have a new property it did not have when it existed in V=L--that of being Whitehead. Because both 'universes' are consistant if ZFC is, I infer that it is not possible to check whether the group A is Whitehead or not in ZFC because ZFC by itself does not have the extra, incompatible axioms need to do this check. If I have made any errors in reasoning or my inferences let me know so I can either make the necessary corrections – Thomas Benjamin Aug 15 '14 at 19:36
  • (cont.) or withdraw the question. Thanks. – Thomas Benjamin Aug 15 '14 at 19:37
  • Alright, fine. But look, a theorem of $\sf ZFC$ states that a group satisfying Chase's condition is free if and only if in a continuous sequence of subgroups with such and such properties, only a non-stationary set of the groups do not satisfy some property. When we add new reals and Martin's Axiom, we effectively add both new subgroups and possibly new clubs to $\omega_1$. So the old group has a new decomposition. This is a nontrivial fact, and you seem to have ignored it, or not clarified it properly. Which is what bothers me in general here. The entire answer is very unclear. – Asaf Karagila Aug 15 '14 at 19:39
  • @AsafKaragila: Actually, the theorem you mention is the following in Eklof's A.M.M. paper: The group A [satisfying Chase's condition--my comment] is free if and only if E[=the set of all limit ordinals $\lambda$ $\lt$ $\omega_1$ such that $A_{\lambda}$ is not $\aleph_1$-pure in A – Thomas Benjamin Aug 26 '14 at 22:51
  • --my comment] is not a stationary subset of A. The theorem you want to use (since since the group A you want to use satisfying Chase's condition is not free) is an easy consequence of the aforementioned theorem: The group A is not free if and only if the set E is a stationary subset of $\omega_1$. Note that both the group A in Eklof's paper and the theorem you mention were proved in ZFC. Since this is the case, and since ZFC is consistent with ZFC+V=L and ZFC+MA+$2^{\aleph_0}$ $\gt$ $\aleph_1$, I cannot see how new subgroups are added to A under MA+$2^{\aleph_0}$ $\gt$ $\aleph_1$. – Thomas Benjamin Aug 26 '14 at 23:09
  • (since A is the same group via construction entirely in ZFC). Also, since the notion of forcing P in the Solovay-Tennenbaum theorem satisfies the c.c.c., there is a theorem in Jech's paper on stationary sets that states that if P satisfies the $\kappa$-chain condition, then every club C $\in$ V[G] has a club subset D in the ground model V so that every stationary set S remains stationary in V[G]. Does adding a new club in V[G] mean that the club set D in the ground model is a proper subset of the club set C in V[G]? Also, both Jech (in his book "Multiple Forcing") and Eklof (in his – Thomas Benjamin Aug 26 '14 at 23:35
  • A.M.M. paper) don't mention anything you mention to me in your comment. Instead they both concentrate on using MA to construct (as Powerlust, now known as GlennGulda mentions in his question) the "big splitting homomorphism" that makes the group A satisfying Chase's condition Whitehead. Why? I don't doubt that you are correct but why don't Jech and Eklof mention the points that you bring up since A having a new decomposition in the forcing extension that makes A Whitehead is a very nontrivial fact. – Thomas Benjamin Aug 26 '14 at 23:46
  • Some remarks: (1) The fact that you build something in $\sf ZFC$ doesn't mean that it remains the same object after forcing. You build the real numbers in $\sf ZFC$, but they don't stay the same when forcing Martin's Axiom to hold, because you have to add reals in the process. As for why Eklof and Jech don't mention things, I don't know. I suppose Eklof might want to skip this because it's a "refined set theoretic point" that you can somewhat ignore; and Jech is just well known for being telegraphic (in that book more than usually, too). It seems reasonable to think that [...] – Asaf Karagila Aug 27 '14 at 04:36
  • [...] the group in discussion is not the same group in $L$. It has the same definition, yes. But as remarked the same definition being reinterpreted, does not mean the same set. – Asaf Karagila Aug 27 '14 at 04:37
  • @AsafKaragila: Have you any references in which it is shown that by adding new reals and MA to ZFC you are effectively adding new subgroups and possibly new clubs to $\omega_1$ so that the group A has a new decomposition which allows it to be both not free and Whitehead in the forcing extension? Is this what Shelah shows in his original papers? – Thomas Benjamin Aug 27 '14 at 06:37
  • I don't know how to continue this discussion with you. You stipulate that there is a group in $L$ which is not free and not Whitehead. And without changing almost anything related to it, it is suddenly Whitehead. I don't know what Shelah did in his original papers and with Eklof (or rather don't fully remember what he did, since I have read them a long time ago). I can tell you that "proving something exists" or rather proving a certain formula defines an object with such and such properties, doesn't mean that the same formula defines the same object in larger models. [...] – Asaf Karagila Aug 27 '14 at 06:43
  • [...] Which may or may not be the source of your mistake here. Since otherwise you really seem to stipulate that you had a group which had size $\aleph_1$, was not free, was $\aleph_1$-free, and the set of points where the decomposition had a certain failure was non-stationary; but after forcing without collapsing $\omega_1$, this set is suddenly non-stationary and you did not add sufficiently many new subgroups to allow a new decomposition, or so on. That is just impossible. – Asaf Karagila Aug 27 '14 at 06:46
  • @AsafKaragila: I don't doubt that you are correct--My sources only concern themselves with using MA+$2^{\aleph_0}$ $\gt$ $\aleph_1$ to construct the splitting homomorphism that makes the nonfree group A Whitehead. Can you show that the c.c.c. forcing notion P that can 'construct' from the model L the forcing extension V[G] in which MA+$2^{\aleph_0}$ $\gt$ $\aleph_1$ holds adds new subgroups to A and possibly new clubs to V[G] or know of a source that explicitly shows this? That would help a lot so I can amend my answer accordingly. This would even better show why one cannot check if A – Thomas Benjamin Aug 29 '14 at 23:30
  • is Whitehead in ZFC. – Thomas Benjamin Aug 29 '14 at 23:30