**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](https://arxiv.org/abs/2208.03073)* 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 $n\neq 0$?



Why are ordinary spheres not strictly invertible?
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.


