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Introduction

This question is about Picard spectra for the symmetric monoidal $\infty$-category of spectra. We say that a spectrum $X$ is invertible if there is another spectrum $Y$ such that $X\wedge Y\simeq S^0$. It is well-known that any such $X$ is equivalent to $S^n$ for some $n\in\mathbb{Z}$, and that the space of endomorphisms of $S^n$ is $QS^0=\lim_{\to k}\Omega^kS^k$. This space has $\pi_0(QS^0)=\mathbb{Z}$, and we write $Q_{\pm 1}S^0$ for the union of the two components corresponding to $1$ and $-1$, which is the space of self-homotopy equivalences of $S^n$. We write $\text{pic}(S)$ for the $K$-theory spectrum of the symmetric monoidal category of invertible spectra, so $\text{pic}(S)$ is $(-1)$-connected with $\pi_0(\text{pic}(S))=\mathbb{Z}$ and $\Omega^{\infty + 1}(\text{pic}(S))=Q_{\pm 1}S$.

Given an invertible spectrum $S^{2n}$, we have a naively homotopical commutative ring spectrum $R(n)=\bigvee_{k\in\mathbb{Z}}S^{2nk}$ with ${\pi_{\ast}}(R)={\pi_{\ast}}(S)[x,x^{-1}]$ where $|x|=2n$. One might like to build a strictly commutative (or $E_\infty$) version of $R(n)$, but this is not obviously possible.

Various people have studied the strict Picard spectrum $\text{spic}(S)$, which is the $(-1)$-connected cover of $F(H\mathbb{Z},\text{pic}(S))$; in particular there is the paper On the Strict Picard Spectrum of Commutative Ring Spectra by Carmeli. There it is shown (amongst many other things) that $\pi_0(\text{spic}(S))$ maps trivially to $\pi_0(\text{pic}(S))=\mathbb{Z}$, and that the strict Picard spectrum is related to the realisation problem mentioned above, so that $R(n)$ cannot be made $E_\infty$ unless $n=0$. However, we only get to this conclusion after developing an extensive theory.

The question

Is there a simple direct proof that $R(n)$ cannot be made $E_\infty$ for $n\neq 0$?

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    $\begingroup$ I believe this is a repeat of this question $\endgroup$ Commented Feb 5, 2023 at 19:36

2 Answers 2

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An $E_{\infty}$ structure extending the $E_1$ structure on $R(n)$ in particular yields maps $(\mathbb{S}^{2n})^{\otimes p}_{hC_p} \to \mathbb{S}^{2pn}$ splitting the inclusion of the bottom cell. The left hand spectrum can be viewed as homotopy orbits of the $C_p$ action on the representation sphere $\mathbb{S}^{2n \rho}$, which can be interpreted as Thom spectrum on $BC_p$. The existence of a map as above is precluded by the action of the Steenrod algebra on $\mathbb{F}_p$ cohomology, which can be computed through the Thom isomorphism.

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The following solution to the question, which is very close to the one given by Achim, is basically what leads to the entire proof that $spic(\mathbb{S}) \simeq \widehat{\mathbb{Z}}$ (which is proven in the referred paper using more stuff in order to generalize to all spherical Witt vectors).

So we have a $C_p$-equivariant multiplication map $R(n)^{\otimes p} \to R(n)$, which restricts to a map $(\mathbb{S}^{2n})^{\otimes p} \to \mathbb{S}^{2np}$. In fact, forgeting the $C_p$-action this map is the identity of $\mathbb{S}^{2np}$ so it is an isomorphism of spectra with $C_p$-action. To see that this can not be the case, apply the Tate-construction: the source become $\mathbb{S}_p^{2n}$ via the Tate-diagonal, while the target become $\mathbb{S}_p^{2np}$ via the canonical map. Since these are different spectra there is no such an isomorphism and there is no $E_\infty$-structure.

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