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


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.


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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