In classical algebra, there is a notion of "monoid rings" such that the functor taking monoids to the monoid rings is the left adjoint to the forgetful functor from the category of commutative rings to that of commutative monoids forgetting the addition. Moreover, there is an equivalence between the category of representations of a monoid and the category of the monoid ring of the monoid.

Question 1: Let a functor $F$ from the $\infty$-category of $E_\infty$-rings of spectra to the $\infty$-category of $E_\infty$-spaces be the functor post-compositing infinitely looping functor $\Omega^\infty$(we see algebras as $\infty$-operad maps and $\Omega^\infty:Sp \rightarrow S$ is a lax monoidal functor). Then, is there a right adjoint functor to $F$? ($Sp$ is the $\infty$-category of spectra and $S$ is the $\infty$-category of spaces.)

Question 2: If the Question 1 is true, then we will denote the functor by $\mathbb{S}[-]$. Let $M$ be an $E_\infty$-space. Is there an equivalence between the $\infty$-category $Fun(BM, Sp)$ of repesentations of $M$ and the $\infty$-category $Mod_{\mathbb{S}[M]}(Sp)$ of modules of the monoid ring $\mathbb{S}[M]$? ($BM$ is an $\infty$-category which has a single object $*$ and $Map(*, *)$ = $M$ as $A_\infty$-spaces.)


1 Answer 1


I think you made a sign mistake, and asked for a left adjoint to $\Omega^\infty$ since the monoid ring is a left along to the forgetful functor.

If so then the answer is yes. $\Sigma^\infty_+:\mathrm{Space}→\mathrm{Spectra}$ is a symmetric monoidal left adjoint, so it can be extended to an adjunction $$\Sigma^\infty_+:\mathrm{CAlg}(\mathrm{Space})→\mathrm{CAlg}(\mathrm{Spectra})\dashv \Omega^\infty:\mathrm{CAlg}(\mathrm{Spectra})→\mathrm{CAlg}(\mathrm{Space})$$ (see Higher Algebra, The functor $\Sigma^\infty_+$ is what you want to call $\mathbb{S}[-]$.

The answer to your question 2 is also true, since $\mathrm{Fun}(BM,\mathrm{Spectra})$ has a compact generator ($\mathbb{S}[M]$ with the obvious $M$-action, which represents simply the evaluation at the basepoint of $BM$) and then you can just apply Schwede-Shipley Morita theory.

  • $\begingroup$ Thank you very much for your answering! As you said, I made a typo. I will learn the Schwede-Shipley Morita theory. Is the theory in the paper "Stable model categories are categories of modules"? $\endgroup$ Sep 3, 2019 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.