Multiplicative structures on truncated Moore spectra

Question

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).

Answer 1

EDIT:

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

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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.

Answer 2

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 $$PH\partial([a],n)=\left(\left[\frac{2^rn(2^rn-1)}{2}\right],0\right)=([2^{r-1}n],0).$$