23
$\begingroup$

The motivation for this question is that I want "toy examples" of how to prove/disprove the existence of multiplicative structures on examples of spectra. The class of examples I am thinking of is the Moore spectrum. For concreteness this is defined as a spectrum $X$ such that $\pi_n(X) = 0$ for $n <0$ $H_n(X)= 0$ for $n >0$ and $H_0(X) = R$ for some ring $R$.

There are some curious phenomenon that happens:

  • On one extreme, the Mod 2 Moore spectrum has no unital multiplication at all (by simple arguments in, say, Difficulties with the mod 2 Moore Spectrum)
  • The Mod 3 Moore spectrum is not $A_{\infty}$ by Massey product arguments.
  • The comment here on top of page 838: http://www.math.uni-bonn.de/people/schwede/rigid.pdf says that the mod $p$ Moore spectrum for $p \geq 5$ is homotopy associative by folklore (I would like to see an argument for this too!)
  • On another extreme, since we can model the $\mathbb{Z}[q^{-1}]$ by localizing the sphere spectrum they are $E_{\infty}$.

In this light, my questions are:

  • First and foremost, I would love to see a proof of the folklore result above about $p \geq 5$
  • Is there a "general pattern" about multiplicative structures of the Moore spectrum as the ring/abelian group varies
$\endgroup$
2
  • 3
    $\begingroup$ For what it's worth, it seems like what you're getting at is this fundamental difficulty we have in spectra of talking about "modding out by ideals," since you're only looking at localization versus taking quotients. $\endgroup$ Commented Jul 12, 2014 at 16:09
  • $\begingroup$ You should check V. Angeltveit's thesis, or his paper on THH and cohomology of $A_\infty$ ring spectra. $\endgroup$ Commented Mar 29, 2015 at 7:21

5 Answers 5

11
$\begingroup$

There is an unpublished result of Hopkins that none of the Moore spectra (modulo any power of $p$) admit $A_\infty$-structures. Although I do not know the proof, I believe the obstruction is $L_1$-local.

It might also be worth noting that none of the Moore spectra (including generalized ones) can have $E_\infty$-structures. More generally, any $E_\infty$-ring which is a finite spectrum must have nontrivial rational homology or be contractible. This is due to T. Lawson and appears as Remark 4.3 of this paper.

$\endgroup$
5
  • $\begingroup$ wow the last statement is pretty useful - neat proof as well! $\endgroup$ Commented Jul 14, 2014 at 0:12
  • $\begingroup$ What is the general belief regarding the ring structure of generalized Moore spectra? For example, I would like to believe that none of the Moore spectra is even $A_{\infty}$. $\endgroup$
    – Prasit
    Commented Jul 14, 2014 at 0:28
  • 5
    $\begingroup$ I just need to state that, in reality, all I did was point out that this result was a consequence of the theorems in the linked paper of Mathew-Naumann-Noel. $\endgroup$ Commented Jul 14, 2014 at 3:48
  • $\begingroup$ @Prasit: It is conjectural that generalized Moore spectra of sufficiently high powers admit actions of any given finitely presented operad (the powers depend on the operad). I believe Devinatz-Hopkins have thought about this problem, but do not know the answer. $\endgroup$ Commented Jul 24, 2014 at 0:08
  • $\begingroup$ I do not know if they (or anyone else) has worked on the problem of the existence of $A_\infty$-structures on generalized Moore spectra. As I understood, the obstruction for the Moore spectra was $E_1$-local. $\endgroup$ Commented Jul 24, 2014 at 0:09
10
$\begingroup$

Alright, I will be gutsy and try to provide the idea behind getting bounds for higher associativity of $M(p^i)$. I am taking the risk of prematurely displaying part of the work in my thesis in public domain. I will only give a sketch. As a warning I would say that this is just a rough idea, I haven't written down the details, so I can potentially write down something stupid.

Firstly, one can realize $M(p^i)$ as Thom spectra of a map

$$ 1+p^i: S^1 \to B \mathscr{G}_p$$ where $B\mathscr{G}_p$ is the classifying space of $p$-spheres. More details in modern language is explained in the answer to the Mathoverflow Question. If this map is $A(\infty)$, then this map is a loop map and lifts to $\Omega \mathbb{C}P^{\infty}$. Stably speaking, this is same as saying $$ p^i \in [\Sigma^{-2}\mathbb{C}P^{\infty}, b]$$ where $b$ is the suspension-spectrum of $\mathscr{G}_p$.

However using "Stasheff's version of bar construction" one should be 'able to argue' that if $1+p^i$ is $A(n)$ then it lifts to some truncation of bar construction, as a consequence (I think it should be true, but did not work out the details yet) $$ p^i \in [\Sigma^{-2}\mathbb{C}P^{n}, b].$$ To compute the value of $i$, we run Atiyah-Hirzebruch SS $$ H^*(\mathbb{C}P^n; \pi_{-*}(b)) \Rightarrow [\mathbb{C}P^n,b].$$ Things to note here is $$\pi_k(b) = \pi_k(S^0_p) $$ for $k>0$ and where $S^0_p$ is the $p$-adic sphere. Also $\pi_0(b)$ are strict units of $p$-adic integers. Notice that $E^2_{0,0}$ of the spectral sequence is $\mathbb{Z}$ and everytime there is a differential from $(0,0)$ spot the target is always $\mathbb{Z}/p^r$ (depending on the homotopy groups of sphere). So every differential from the $(0,0)$ kills of some power of $p$. I do not have any information about the differentials, except that there are finitely many. First elementary estimate can be obtained by assuming all possible differentials exists (the worst case scenario). It turns out that $M(p^i)$ is $A(n)$ if $i>o(n)$ where $$o(n) = \underset{k\leq 2n-3 \text{ and odd}}{\Sigma} \text{maximum $p$-torsion in } \pi_{k}(S^0_p)$$.

This is the first elementary approximation. One can get much better answer than this, however its not feasible for me to write it down here at this stage.

Again I may have made mistakes several places above, but this is more or less the idea.

$\endgroup$
9
$\begingroup$

I think the answer to your question is essentially unknown. As far as I'm aware the best known results are:

  • $M(p)$ admits an $A_{p-1}$ structure but never an $A_p$-structure. I learnt the following argument from Tyler Lawson (I'm not sure if this is the "folklore" proof). If $M(p)$ was $A_p$ then there would be an $A_p$-map $M(p) \to H\mathbb{Z}/p$. The induced map on homology is meant to hit $\xi_1 \in \mathcal{A}_*$, but there is a $p$-fold Massey product $\langle \tau_0,\ldots,\tau_0 \rangle = \xi_1$, which is a contradiction.
  • For $M(p^i)$ a few cases have been studied by Oka ("Multiplications on the Moore spectrum"). Namely, at the prime 2, $M(4)$ does not admit an $A_3$-structure, whilst $M(2^i)$ does admit an $A_3$ structure when $i>2$.

Prasit Bhattacharya is thinking about this problem and is worth speaking to.

Update: The argument alluded to in the first dot point above can be found in Angeltveit's "Topological Hochschild homology and cohomology of $A_\infty$ ring spectra" - http://msp.org/gt/2008/12-02/gt-2008-12-022s.pdf

$\endgroup$
5
  • $\begingroup$ I think $M(4)$ is homotopy associative. I think Theorem 2(d) in Oka's paper says so explicitly. $\endgroup$
    – Prasit
    Commented Jul 13, 2014 at 17:10
  • 1
    $\begingroup$ I would also like to suggest that the key to the above argument is that if $X$ is $A(n)$, then $n$-fold Massey product is defined. This is a work of Robert Hank and I joined that project to do a "space-level" version of Massey product. First one should note that the map $M(p) \to H\mathbb{Z}/p$ is as associative as $M(p)$ due to lack of homotopy groups in the target. Moreover if this map is $A(n)$ then it preserves $n$-fold Massey product. Above argument says the $p$-fold Massey product is not preserved. $\endgroup$
    – Prasit
    Commented Jul 13, 2014 at 17:15
  • 2
    $\begingroup$ Thanks for recognising my work. My Thesis will be all about finding associativity of $M(p^i)$. And so far I have a result that there is a function (which I may mess up if I try to state here), say $o(n)$ such that $M(p^i)$ is at least $A(n)$ if $i > o(n)$. I do not think this function $o(n)$ is sharp. Nor I have an answer to explicit obstruction, however work is in progress. $\endgroup$
    – Prasit
    Commented Jul 13, 2014 at 17:25
  • 1
    $\begingroup$ Hi @Prasit - that's certainly a very fascinating answer! You should definitely put it up as more than a comment and (if you want to) elaborate on it slightly. In particular, we know that we can calculate Massey products to show that it is NOT A(n) - but what ideas/tools did you use to show the positive case? $\endgroup$ Commented Jul 13, 2014 at 20:21
  • $\begingroup$ I was about to suggest talking to @Prasit as well! $\endgroup$ Commented Jul 16, 2014 at 16:42
6
$\begingroup$

The classical reference is as follows:

@article {MR760188,
    AUTHOR = {Oka, Shichir{\^o}},
     TITLE = {Multiplications on the {M}oore spectrum},
   JOURNAL = {Mem. Fac. Sci. Kyushu Univ. Ser. A},
  FJOURNAL = {Memoirs of the Faculty of Science. Kyushu University. Series
              A. Mathematics},
    VOLUME = {38},
      YEAR = {1984},
    NUMBER = {2},
     PAGES = {257--276},
      ISSN = {0373-6385},
     CODEN = {MFKAAF},
   MRCLASS = {55P45},
  MRNUMBER = {760188 (85j:55019)},
MRREVIEWER = {Donald M. Davis},
       DOI = {10.2206/kyushumfs.38.257},
       URL = {http://dx.doi.org/10.2206/kyushumfs.38.257},
}

Here is the review by Don Davis:

For any positive integer $q$, let $M_q$ denote the Moore spectrum whose only nontrivial homology group is $\mathbb{Z}/q$ in dimension 0. The main result is the following theorem: (a) The number of homotopy classes of multiplications on $M_q$ is $4$ if $q≡0 (4)$, 1 if $q$ is odd, and $0$ if $q≡2 (4)$. (b) These multiplications are commutative if and only if $q≡0$ (8) or $q$ is odd, and are associative if and only if $q\not\equiv 2 (4)$ and $q\not\equiv\pm 3 (9)$. A number of more technical results are proved, many involving premultiplications and regularity.

$\endgroup$
4
$\begingroup$

I randomly ran into this old question - the state of the art is now the following paper of Robert Burklund where he proves that many Moore spectra do, in fact, have $A_\infty$ -structures. For example, at odd primes, he proves that $\mathbb S/p^{n+1}$ admits an $\mathbb E_n$-structure.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .