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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJul 13th 2016

I have started a page localization of abelian groups with a bunch of facts and their proofs.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeJul 14th 2016

What to make of that remark 3.2 in Neisendorfer 08, that one would expect local morphisms of abelian groups to be characterized by $Ext^\bullet$ into local groups, but that instead one needs to use just $Hom$ into local groups in order to get the correct concept of $p$-localization as a special case?

This sounds like some piece of a story is missing. What’s going?

1. Yes, I agree that this is not satisfactory. I’d investigate the following line of thought. The notion of localisation discussed here can be formulated in a simplical model category. The correct notion for abelian groups should agree with the one obtained by viewing an abelian group as a simplicial abelian group, and using the model structure on simplicial abelian groups. Is this the same as the definition in Neisendorfer?

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeJul 14th 2016
• (edited Jul 14th 2016)

I’d think so. The $Ext^\bullet$ is the derived hom of abelian groups regarded as connective chain complexes, and by Dold-Kan their homotopy theory is the same as that of the corresponding simplicial groups.

2. But this would say that the definition should be that which Neisendorfer terms ’strong local equivalence’, which indeed seems the correct choice to me. The reason that Neisendorfer chooses the other definition seems to be the example of $Z(p^{\infty})$ in section 8. But I do not follow the reasoning.

Indeed, in this section, it is shown that one can construct a $p$-completion explicitly as a certain Ext group if one takes the definition of $p$-completion to be the one involving Homs rather than Exts. But then one can hardly expect this notion of $p$-completion to give rise to a ’strong local equivalence’: one would need instead to use a ’strong $p$-completion’, if such a thing exists in general.

3. It would not surprise me if it did not exist in general (of course everything is fine in the finitely generated case). In particular, note that the homotopy groups of the topological p-completion do not necessarily coincide with the p-completion as defined by Neisendorfer of the homotopy groups in the non-finitely generated case.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeJul 14th 2016

in this section, it is shown that one can construct a $p$-completion explicitly as a certain $Ext$ group

Yes, but he claims that it satisfies its universal property with respect to $Hom$, not $Ext$. That’s the thing.

• CommentRowNumber8.
• CommentAuthorRichard Williamson
• CommentTimeJul 14th 2016
• (edited Jul 14th 2016)

That’s exactly my point! There is no reason at all that I see to expect it to satisfy the universal property with respect to Ext. So this does not justify re-defining local equivalence for abelian groups to mean the Hom notion, in my opinion: the correct notion is the Ext one; they agree in the finitely generated case, where one can construct a p-completion; and in the non-finitely generated case one might not (I do not know) be able to construct a p-completion (in the Ext sense).

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeJul 14th 2016
• (edited Jul 14th 2016)

This is my point from #2, but I am glad we agree.

The thing is that he shows (on p. 16) that when using $Hom$ Instead of $Ext^\bullet$ to characterize local morphisms, then the functor $Ext^1( \mathbb{Z}(p^\infty,-) )$ agrees on finitely generated groups with $p$-adic completion.

This is curious because, as you are amplifying, too, one would naively expect $Ext^\bullet$ instead of $Hom$. Therefore my question in #2: what’s going on here, how should one think of this fact? (I mean the fact proven on p.16, which is not a matter of opinion or taste, but something that is asking for conceptual explanation).

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeJul 14th 2016

Maybe it’s related to the fact that the $p$-adic completion of an abelian group really wants to be an abelian prop-group, not just an abelian group? I.e. maybe the conceptual mismatch goes away as we pass to the category of abelian pro-groups?

4. I’m not sure I’ve quite conveyed my point yet :-). What I am saying is that the result that Neisendorfer proves illustrates that one can define a p-completion in the Hom sense for an arbitrary abelian group; but in the finitely generated case it illustrates just as well that one can define a p-completion in the Ext sense. The non-finitely generated case could be thought of as a curiosity, not necessarily a p-completion in the ’correct’ sense, especially from the point of view of topology (due to the fact about p-completions of homotopy groups that I mentioned above); in particular, not as something which means that the Ext definition of local equivalence is incorrect.

5. Yes, that thought also occurred to me: I think indeed that one would obtain the nicest correspondence between the topological and abelian group cases in the pro-setting. But still the fact that one can stay in the same category in the finitely generated case is useful.

• CommentRowNumber13.
• CommentAuthorRichard Williamson
• CommentTimeJul 14th 2016
• (edited Jul 14th 2016)

[Edited: removed something that was not of particular significance.]

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeJul 14th 2016
• (edited Jul 14th 2016)

What I am saying is that the result that Neisendorfer proves illustrates that one can define a p-completion in the Hom sense for an arbitrary abelian group; but in the finitely generated case it illustrates just as well that one can define a p-completion in the Ext sense.

Sorry, where does he show this?

Theorem 8.6 states that $\mathbb{Z}[1/p]$-localization in the Hom-sense is $p$-adic completion when applied to finitely generated abelian groups. But where does the author say anything about $\mathbb{Z}[1/p]$-localization in the strong sense?

• CommentRowNumber15.
• CommentAuthorRichard Williamson
• CommentTimeJul 14th 2016
• (edited Jul 14th 2016)

He does not show it explicitly, as far as I see. I would have thought it true, though, that if one defines $p$-completion to be the $p$-adic one, then the $p$-completion morphism should be (in the finitely generated case) a local equivalence in the Ext sense. I also expected that if one went through the proof of 1) and 2) on pages 16 and 17 using the $p$-adic completion, then one would probably find that the morphism $k$ is a strong local equivalence, but I have not checked this, so I should have been more circumspect in the lines that you quoted!

• CommentRowNumber16.
• CommentAuthorRichard Williamson
• CommentTimeJul 15th 2016
• (edited Jul 15th 2016)

Edit: ignore this!

• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeJul 15th 2016

Thanks, Richard. That’s excellent! I’ll try to dig out that reference. Thanks for this.

6. Sorry Urs, this was a red herring. The reference is nice enough (Corollary 33.44 in Modern classical homotopy theory by Strom; I deleted it before seeing Urs’ reply, but I guess it should be visible now!), and maybe offers some complements to Neisendorfer’s treatment with regard to the use of Eilenberg-MacLane spaces, but it does not address the fundamental point we were discussing. I do still think that the result I claimed should be true, though!

• CommentRowNumber19.
• CommentAuthorUrs
• CommentTimeJul 15th 2016

Okay, thanks for your efforts. If you find out anything, I’d be grateful if you could drop me a note.

(I don’t strictly need the statement for what I am doing. But it is irritating not to have it! :-)

• CommentRowNumber20.
• CommentAuthorRichard Williamson
• CommentTimeJul 15th 2016
• (edited Jul 15th 2016)

Indeed, I’d like to resolve this too!

Because $Hom$ and $Ext^1$ commute with finite direct sums (it is the first variable that is relevant here), it suffices to check it for the $p$-completion of the groups $\mathbb{Z}/p^n \mathbb{Z}$ and $\mathbb{Z}$. In the first case, the $p$-completion is just the group itself, so here the statement is obvious.

In the second case, the $p$-completion is the $p$-adic integers. Now, for any abelian group $A$, we have that $Hom(\mathbb{Z},A)$ is just $A$, and the corresponding $Ext$ group is $0$. So the question is: is this also true for the $p$-adic integers instead of the integers, at least for a $p$-complete abelian group? For the Hom part, Neisendorfer’s results show that this is true for a $p$-complete group. So we can just focus on $Ext$ groups with the $p$-adic integers in the first variable. Is there a reason for these to be $0$, at least when the second variable is a $p$-complete abelian group?

Alternatively, could we use Theorem 8.4 in Neisendorfer? The question then is whether the co-kernel of the completion morphism from the integers to the $p$-adic integers is a $Z[1/p]$-module? I guess not, but I could be missing something.