**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 $n\neq 0$?