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As discussed for example in this paper (and this MO thread), there are obstructions for the existence of $A_n$-structures on Moore spectra $M(p^r):=\mathbb{S}/p^r$, which in general don't vanish. In particular, for odd primes, $M(p)$ does not admit an $A_p$-strucutre, and if I understand correctly, it is expected that for no value of $r$ the spectrum $M(p^r)$ admits an $A_\infty$-structure.

I am interested in the analogous question for the $d$-truncated Moore spectra $M(p^r,d) := \tau_{\le d}(\mathbb{S}/p^r)$:

What is known about the existence of $A_n$ (or perhaps even $E_n$) structures on $M(p^r,d)$?

For $d=0$ one gets $M(p^r,0) \simeq H\mathbb{Z}/p^r$, which is an $E_\infty$-ring. However, I am not sure what is the situation even for $d=1$. Specifically, I am interested in the case where $p$ and $d$ are fixed and $r \gg 0$. It would be best if one could get an $E_\infty$-strucutre (the goal being to approximate $\tau_{\le d}(\mathbb{S}_p)$ by $\pi$-finite $E_\infty$-rings).

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    $\begingroup$ For $p$ odd, $r\geq 1$, and $d=1$ the truncation is the same as for $d=0$ hence you get an $E_\infty$-structure. The only non-trivial case would then be $p=2$. $\endgroup$ Feb 22, 2022 at 14:25
  • $\begingroup$ The reason why you don't get anything for $p=2$, $r=1$, and any $d\geq 1$ is the same as in mathoverflow.net/questions/87919/… $\endgroup$ Feb 22, 2022 at 15:17
  • $\begingroup$ @FernandoMuro. This much is clear to me, but thanks for making it explicit. I am mainly interested in results that are uniform in $p$ and $d$ and allow arbitrarily large $r$. So for $d=1$, the question is what can be said about $M(2^r,1)$. $\endgroup$
    – KotelKanim
    Feb 22, 2022 at 20:12
  • $\begingroup$ Would you also be interested in higher associativity / commutativity of truncated Moore spaces? Not that I have any insight either way, but if you're ultimately interested in something about $\pi$-finite spaces, then maybe truncated Moore spaces are relevant too? $\endgroup$
    – Tim Campion
    Feb 22, 2022 at 20:56
  • $\begingroup$ Also, Prasit's Main Theorem 1.3 in the paper you linked seems to do the sort of thing you want, doesn't it? It gives a lower bound for the associativity of $M(p^r)$ when $r$ is large. If I'm reading it right, it says that as $r \to \infty$, the associativity level $n$ of $M(p^r)$ (with $p$ fixed) increases to $\infty$ as well. So a fortiori, the same is true after truncation. Are you really hoping that $n$ actually reaches $\infty$ at some finite $r$ or something? $\endgroup$
    – Tim Campion
    Feb 22, 2022 at 21:03

2 Answers 2

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EDIT:

In case you missed it, there's been a huge breakthrough in this area by Robert Burklund.


I believe that Prasit Bhattacharya's methods, as linked to in the question, can be used to show that for fixed $p,d$ and $r$ sufficiently large, the truncated Moore spectrum $M(p^r,d)$ is $A_\infty$, and probably even $E_\infty$.

Prasit shows that $M(p^r)$ is the Thom spectrum of any map $f_{p,r,u} : S^1 \to BGL_1(\mathbb S^\wedge_p)$ which picks out $1 + p^r u \in \mathbb Z_p \subset \mathbb Z/(p-1) \times \mathbb Z_p = \pi_1(BGL_1(\mathbb S^\wedge_p)$ for any unit $u \in \mathbb Z^\times_p$. His bounds on $A_n$-ness come from finding a choice of unit $u$ for which the map $f_{p,r,u}$ is $A_n$.

Now, if we're interested in $M(p^r,d)$, then if I'm not mistaken, from facts like $\tau_{\leq d} \Sigma^\infty_+ X = \tau_{\leq d} \Sigma^\infty_+(\tau_{\leq d} X)$ we can deduce that $M(p^r, d)$ is the $\tau_{\leq d}$-truncation of the the Thom spectrum of the map $S^1 \xrightarrow{f_{p,r,u}} BGL_1(\mathbb S^\wedge_p) \to\tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p)$. [1] Since passage to Thom spectra takes $O$-maps to $O$-algebras for any operad $O$, it will suffice to show that $u$ can be chosen so that $S^1 \xrightarrow{f_{p,r,u}} BGL_1(\mathbb S^\wedge_p) \to \tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p)$ is an $A_n$ map.

As Prasit shows, this means we have to lift the composite map

$$S^2 \to \Sigma \tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p) \to B\tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p)$$

through the map $S^2 \to \mathbb C\mathbb P^n$ (the significance of $\mathbb C \mathbb P^n$ is that it's the $n$-truncated bar construction on $S^1$). Prasit shows that this is possible before taking $\tau_{\leq d+1}$ for some $n = n(r)$ if $r$ is sufficiently large. But then if we're truncating, because the maps $\mathbb C \mathbb P^n \to \mathbb C \mathbb P^{n+1}$ are increasing in connectivity, we can just automatically extend, as long as $2n(r) \geq d$. Since $n(r) \to \infty$ as $r \to \infty$, this is possible.

I believe the obstruction theory to get an $E_\infty$ map works similarly in that the obstructions lie in higher and higher homotopy groups, so as long as you get some lift up to level $d+1$, you can extend to an $E_\infty$ structure when you're $d$-truncating. But perhaps somebody who is actually familiar with the relevant $E_\infty$ obstruction theory could say something more definitive.

[1] The argument I have in mind constructs the Thom spectrum as a bar construction $M(f) = |\Sigma^\infty_+ F \wedge \Sigma^\infty_+ BGL_1(\mathbb S^\wedge_p)^\bullet|$ where $F$ is the fiber of $f$. So $\tau_{\leq d} M(f) = \tau_{\leq d} |\tau_{\leq d} \Sigma^\infty_+ \tau_{\leq d} F \wedge \Sigma^\infty_+ \tau_{\leq e} BGL_1(\mathbb S^\wedge_p)^\bullet|$ so long as $e \geq d$. When we take $e = d+1$, we observe that because the homotopy groups of $S^1$ (the domain of $f$) are easy, we have a fiber sequence $\tau_{\leq d} F \to S^1 \to \tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p)$. So the bar construction we're taking is now, up to $\tau_{\leq d}$-truncation, exactly the same bar construction as for the Thom spectrum of the map $S^1 \to \tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p)$. There's probably a nicer way to say this, though.

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  • $\begingroup$ That looks good! I'll have to think about it a little. $\endgroup$
    – KotelKanim
    Feb 22, 2022 at 22:48
  • $\begingroup$ I had a little moment of panic where I thought that the suspension messed things up, but luckily the map continues on to $B$, which preserves truncatedness! $\endgroup$
    – Tim Campion
    Feb 22, 2022 at 22:54
  • $\begingroup$ Actually, there's at least one funny thing -- I think the argument as it stands is implicitly assuming that $\tau_{\leq d+1} BGL_1(\mathbb S^\wedge_p) = B GL_1 R$ for some ring spectrum $R$, and maybe because of that it doesn't actually make sense... I think it's okay because $\tau_{\leq d} GL_1(\mathbb S^\wedge_p)$ acts on $\tau_{\leq d} S^\wedge_p$, though. $\endgroup$
    – Tim Campion
    Feb 22, 2022 at 22:58
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    $\begingroup$ Any obstruction to extending a unital $A_{n-1}$-structure to a unital $A_n$-structure on $M(p^r)$, will lie in $\pi_{2n-3}(M(p^r))$. So if one truncates to a degree before you see any real obstruction, then you extend the multiplicative structure on the truncation of $M_{p^r}$ an $A_{\infty}$-structure. This is because the obstruction classes will be forced to be zero in the truncated version. This is quite similar in spirit to the answer above. $\endgroup$
    – Prasit
    Feb 22, 2022 at 23:28
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    $\begingroup$ Yes, I think so. I believe that the proof above maybe modified to study $E_{\infty}$-structure using the fact that $S^1$ admits an $E_{\infty}$-structure ($S^1 \simeq \Omega^{\infty-1}H\mathbb{Z}$). One might be able to show that for $r$ large enough $M(p^r, d)$ admits an $E_{\infty}$-structure. $\endgroup$
    – Prasit
    Feb 23, 2022 at 15:43
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The $1$-type of $M\mathbb{Z}/(2^r)$ does have an $E_\infty$-ring structure for $r> 1$. I'm going to show it by using the algebraic models for $1$-truncated connective commutative ring spectra from:

MR2405894 Reviewed Baues, Hans-Joachim; Jibladze, Mamuka; Pirashvili, Teimuraz Third Mac Lane cohomology. Math. Proc. Cambridge Philos. Soc. 144 (2008), no. 2, 337–367.

MR2793446 Reviewed Baues, Hans-Joachim; Muro, Fernando The algebra of secondary homotopy operations in ring spectra. Proc. Lond. Math. Soc. (3) 102 (2011), no. 4, 637–696.

These algebraic models, called $E_\infty$-quadratic pair algebras, give rise to bipermutative categories (see Remarks 5.9, 6.10, and 9.14 in the second paper) which, in turn, can be used to construct 1-truncated connective commutative ring spectra as per:

A. D. Elmendorf and M. A. Mandell, Rings, modules, and algebras in infinite loop space theory, Adv. Math. 205 (2006) 163–228.

I'm sure you can find other proofs, e.g. via an explicit topological construction or lifting the first $k$-invariant to topological André-Quillen cohomology.

Consider the following $E_\infty$-quadratic pair algebra

$$\begin{array}{ccc} \mathbb{Z}/2\times \mathbb{Z}&\stackrel{\partial}\longrightarrow& \mathbb{Z}\\ \nwarrow^P&&\swarrow_H \\ &\mathbb{Z}& \end{array}$$

Here, $$ \begin{array}{rcl} \partial([a],n)&=&2^rn,\\ P(a)&=&([a],0),\\ H(n)&=&\frac{n(n-1)}{2}. \end{array} $$

Moreover, $\mathbb{Z}$ is endowed with the usual product (in a quadratic pair algebra it could have a product which is only right distributive), but it doesn't in this case) and $\mathbb{Z}/2\times \mathbb{Z}$ is endowed with the usual left and right product by elements of $\mathbb{Z}$ (again, the product from the left need not be distributive in general). There could be an additional $\smile_1$ operation that in this case is trivial because all previous products commute.

The $k$-invariant of such a structure is the homomorphism $$\operatorname{coker}\partial\otimes\mathbb{Z}/2\longrightarrow \ker\partial$$ defined by $$[a]\otimes[1]\mapsto P(H(2a)-2H(a)).$$

In the previous example, this morphism is $$\mathbb{Z}/2^r\otimes\mathbb{Z}/2\cong \mathbb{Z}/2$$ which coincides with the map $$\pi_0M\mathbb{Z}/2^r\otimes\pi_1S\longrightarrow \pi_1M\mathbb{Z}/2^r\colon [f]\otimes[g]\mapsto [fg].$$ This is the first $k$-invariant of $M\mathbb{Z}/2^r$.

You may wonder what fails for $r=1$. For the second equation of the second set in Definition 6.1 (in the second of the aforementioned papers) we need $PH\partial=0$. This holds iff $r>1$ since $$PH\partial([a],n)=\left(\left[\frac{2^rn(2^rn-1)}{2}\right],0\right)=([2^{r-1}n],0).$$

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    $\begingroup$ whoa... this is super-surprising to me! $\endgroup$
    – Tim Campion
    Feb 22, 2022 at 22:52
  • $\begingroup$ @TimCampion thanks for your appreciation, your excellent answer is actually more complete! Actually, I didn’t know that paper. $\endgroup$ Feb 23, 2022 at 6:09

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