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I know very little about algebraic topology, and more about $k$-linear stable $\infty$-categories (i.e. homological algebra).

Given an abelian group $A$, there is the Eilenberg-Mac Lane spectrum $HA$, for which: $\pi_{< 0}(\cdot) = 0$, $\pi_0 (\cdot) = A$, $\pi_{>0}(\cdot) = 0$. There is also the Moore spectrum $SA$, for which $\pi_{<0}(\cdot) = 0$, $\pi_{=0} (\cdot)= A$, $H_{> 0} (\cdot , \mathbb{Z}) = 0$.

I more-or-less have some feeling for $HA$ (under Dold-Kan, it is simply $A$, so one simply gets connected deloopings $BA$, $BBA$, etc. making the infinite loop space structure explicit).

I have no feeling what-so-ever about $SA$.

But reading about Bousfield localization, it seems that most the standard examples are with $SA$. How to understand this for a novice? Say, for $p$-localization and $p$-completion, how do I understand that I "want" to use the Moore spectrum and not the Eilenberg-Mac Lane spectrum?

Or, at least, what is the rationale for introducing Moore spectra?

Thank you

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    $\begingroup$ Denis's answer is 100% the reason why the specific object S/p appears in the theory of localization. You could still ask whether you could use HZ/p, and here the answer is somewhat unintuitive: for connective spectra, localization at S/p and at HZ/p agree, and so you cannot tell the difference before passing to nonconnective spectra, which are somewhat exotic objects. The spectrum $KU := \Sigma^\infty_+ \mathbb CP^\infty[\beta^{-1}]$ is a concrete example of a spectrum with $H_*(KU; Z)$ rational, hence $H_*(KU; Z/p)$ and $L_{HZ/p} KU$ null, but $L_{S/p} KU$ nontrivial. $\endgroup$
    – Eric Peterson
    Commented Feb 2, 2019 at 13:26
  • $\begingroup$ @EricPeterson: Thank you! I will have to think about it. $\endgroup$
    – Sasha
    Commented Feb 2, 2019 at 19:40

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I'm not sure I can answer to the general question, but I can explain why the Moore spectrum $\mathbb{S}/p$ shows up in the discussion of Bousfield localizations.

This is just the cofiber of multiplication by $p$ from $\mathbb{S}$ to itself. Hence saying that for a spectrum $X$ we have $X\wedge \mathbb{S}/p=0$ means simply that multiplication by $p$ acts on $X$ as an equivalence. So $X$ is $\mathbb{S}[1/p]$-local iff it is $\mathbb{S}/p$-acyclic. From standard localization arguments (plus the fact that $\mathbb{S}/p$ is a finite spectrum) you get the fracture square relating $\mathbb{S}[1/p]$-localization and $p$-completion (i.e. the "complementary localization" of $\mathbb{S}[1/p]$-localization).

So, essentially, the reason it shows up is because the Moore spectrum is the layer of the diagram $$ \mathbb{S} \xrightarrow{p} \mathbb{S}\xrightarrow{p}\mathbb{S}\xrightarrow{p}\mathbb{S}\xrightarrow{p}\cdots $$ that computes $\mathbb{S}[1/p]$-localization. From this it is clear that $\mathbb{S}/p^n$ is also going to be relevant to the study of $p$-completion (since these are just the "higher" layers of the tower), and from there to $$\mathbb{S}/p^\infty = \mathbb{S}\mathbb{Q}_p/\mathbb{Z}_p = \mathrm{colim}\,\mathbb{S}/p^n$$ and $$\mathbb{S}\mathbb{Q}/\mathbb{Z}=\bigvee_p \mathbb{S}/p^\infty$$ is just a small step.

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  • $\begingroup$ Thank you! This now makes perfect sense. But for p=2, it seems that the cofiber of multiplication by 2 on the sphere spectrum has $\pi_0$ with four elements, so this should differ from the Moore spectrum as defined in my question, no? $\endgroup$
    – Sasha
    Commented Feb 2, 2019 at 19:40
  • $\begingroup$ @Sasha No, $π_0\mathbb{S}/2=H_0\mathbb{S}/2=\mathbb{Z}/2$. Maybe you're confounding it with $[\mathbb{S}/2,\mathbb{S}/2]=\mathbb{Z}/4$? (Yes, it is a bit counterintuitive but this nontrivial extension is secretly coming from $\pi_1\mathbb{S}$) $\endgroup$
    – Denis Nardin
    Commented Feb 2, 2019 at 19:41
  • $\begingroup$ Oops, I am used to cohomological convention, so I simply calculated wrongly the long exact sequence of homotopy groups for the cofiber sequence. Thank you! $\endgroup$
    – Sasha
    Commented Feb 2, 2019 at 19:46
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    $\begingroup$ @Sasha Well, as I'm fond of saying, the cohomological convention is the tool of the Devil. Don't fall for it! ;) $\endgroup$
    – Denis Nardin
    Commented Feb 2, 2019 at 19:49
  • $\begingroup$ I don't quite have a say, since this is so common in homological algebra (representaiton theory etc.) that I can't go against it. But perhaps it makes sense to put $\pi_1$ in negative degree, since it represents "smaller" stuff? (like the cardinality of $BG$ is $|G|^{-1}$ etc.) Anyway, probably one can find justification for each. $\endgroup$
    – Sasha
    Commented Feb 2, 2019 at 19:52

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