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I am interested in those objects in the ("topological") stable homotopy category $SH$(I call them spectra) whose homology (with integral coefficients; should I call it singular or stable, or $H\mathbb{Z}$-one? how can one denote it?) is zero (in all degrees). My questions are:

1) Is it ok to call these spectra acyclic?

2) Does there exist any "description" of all acyclic spectra?

3) Is it true that the ($H\mathbb{Z}$-)cohomology of any acyclic spectrum vanishes? Possibly, this fact can be deduced from Proposition 16.2 of the book, Margolis H.R., Spectra and the Steenrod Algebra: Modules over the Steenrod Algebra and the Stable Homotopy Category, North-Holland, Amsterdam-New York, 1983; yet I am not sure.

4) Is it possible to localize $SH$ by the full subcategory of acyclic objects (so, do we obtain a category whose morphism classes are sets this way)? If this is possible, then we would obtain a "better $SH$", and this should contradict a result of Schwede (on the Margolis's axiomatisation conjecture); yet I am not sure in this argument (see the Upd. below).

Did anyone consider this localization?

5) Can one describe the left or the right orthogonal to all acyclic spectra, i.e., the objects that are only connected with acyclic spectra by zero morphisms? Note in particular that there are no non-zero morphisms from acyclic spectra to connective ones.

Any hints or references would be very welcome! A related matter: I am interested in texts that treat Atiyah-Hirzebruch spectral sequences for arbitrary spectra.

Upd. So, 3 is fine; thanks! Is the converse implication true (are spectra with vanishing cohomology acyclic)?

About 4: note that $SH$/acyclic spectra contains the category of finite spectra (and the category of connective ones also). So, why does not one consider this localization as a "reasonable" substitute of $SH$?

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    $\begingroup$ I think the term is "acyclic," or maybe "$H \mathbb{Z}$-acyclic." $\endgroup$ – Qiaochu Yuan Sep 12 '15 at 18:14
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    $\begingroup$ Yes to 3. If $R$ is $H\mathbb Z$ or any other unital ring spectrum than any map of spectra $X\to R$ factors through a map $R\wedge X\to R$. $\endgroup$ – Tom Goodwillie Sep 12 '15 at 20:40
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    $\begingroup$ 4 is much older than EKMM. Bousfield showed how to localize with respect to the class of $E$-acyclic spectra for any spectrum $E$. $\endgroup$ – Tom Goodwillie Sep 12 '15 at 21:10
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    $\begingroup$ A good example of an acyclic spectrum that should not be thrown away is mod $p$ periodic $K$-theory, $KU\wedge H\mathbb Z/p$. The integral homology groups of $KU$ are rational vector spaces. $\endgroup$ – Tom Goodwillie Sep 12 '15 at 22:08
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    $\begingroup$ Yes, certainly for connective spectra acyclic implies (weakly) contractible; if the homotopy groups of a spectrum $X$ vanish in degrees less than $n$ then $\pi_n(X)\cong H_n(X)$. $\endgroup$ – Tom Goodwillie Sep 12 '15 at 22:09
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Let me address what hasn't been answered in comments (not in an optimal way, though).

1) is OK and, modulo the meaning of your quotation marks, the answer to 2) is 'no'. I mean, don't expect anything very explicit or much beyond the very definition, it's a very complicated problem. As for 5), the right othogonal is by definition the category of $H\mathbb Z$-local spectra, which is equivalent to $SH/$acyclic spectra by Bousfield localisation. I don't know about the left orthogonal, but Bousfield localisation does not apply since the category of acyclic spectra is localising but not colocalising, because integral homology doesn't preserve infinite products.

The converse of 3) is set theory. By universal coefficients, this is equivalent to ask whether there is a non-trivial abelian group $A$ with $\operatorname{Hom}(A,\mathbb Z)=0=\operatorname{Ext}(A,\mathbb Z)$. The answer 'no' is independent of the usual axioms of set theory by Shelah. More precisely, abelian groups satisfying $\operatorname{Ext}(A,\mathbb Z)=0$ are called Whitehead groups (this name has also other uses) and it is undecidable whether all of them are free. In that case $\operatorname{Hom}(A,\mathbb Z)$ wouldn't vanish unless $A=0$.

What your observation about 5) shows is that the category $SH/$acyclic spectra is not compactly generated, nor the category of acyclic spectra. If so, by Neeman an Thomason $SH/$acyclic spectra would be compactly generated by finite spectra and, since the triangulated category of $H\mathbb Z$-local spectra has a model, this would contradict Schwede's uniqueness theorem, as you remark. Neeman's more general theory of well generated triangulated categories says that $SH/$acyclic spectra is well generated. I dare say it is even $\aleph_1$-well generated, but definitely not $\aleph_0$. Coproducts in $H\mathbb Z$-local spectra are not just ordinary coproducts of spectra since these wouldn't be $H\mathbb Z$-local. It would be interesting to find an explicit example where the homotopy groups of an infinite coproduct of $H\mathbb Z$-local spectra is not the colimit of the homotopy groups of the finite subcoproducts. That would be a very explicit proof of the fact that the sphere spectrum is not compact in $SH/$acyclic spectra.

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  • $\begingroup$ Thank you for your great answer! Yet I am somewhat confused: the category of acyclic spectra is closed with respect to coproducts; doesn't this mean that localizing by them respects coproducts and compact objects? $\endgroup$ – Mikhail Bondarko Sep 13 '15 at 8:58
  • $\begingroup$ @MikhailBondarko I guess that the confusion is in the fact that we have a left adjoint $SH\to SH/$acyclic spectra, but $H\mathbb Z$-local spectra are the image in $SH$ of the right adjoint of the previous functor. I've identified $SH/$acyclic spectra and $H\mathbb Z$-local spectra, as it is usual, but this identification does not preserve coproducts since it involves a right adjoint. Localisation w.r.t. $H\mathbb Z$ is not smashing, in topological terminology. Concerning compact objects, left adjoints do not preserve them since right adjoints do not preserve direct sums. I hope this helps. $\endgroup$ – Fernando Muro Sep 13 '15 at 9:09
  • $\begingroup$ Thank you! So it seems that the localization functor does respect coproducts (by Corollary 3.2.11 of Neeman's "Triangulated categories"); yet it does not respect the compactness of objects (though it does preserve morphism groups with compact targets). $\endgroup$ – Mikhail Bondarko Sep 13 '15 at 18:38
  • $\begingroup$ @MikhailBondarko I think something is going wrong with your argument. If localization functor commutes with arbitrary coproducts then $HZ$-localization is a smashing localization, which is not true as Fernando noticed. A localization functor commutes with finite coproducts always. I suspect that Neeman's corollary is implicitly related to some cardinality issues of the set which indexes how big is your coproduct (how many factors). Unfortunately I don't have the book to check. $\endgroup$ – Ilias A. Sep 13 '15 at 19:50
  • $\begingroup$ Moreover, if the $HZ$-localization commutes with arbitrary coproducts (which is not true), then it will send compact objects to compact objects. $\endgroup$ – Ilias A. Sep 13 '15 at 19:58

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