Suppose I am given a morphism $f:BG\to BGL_1(R)$ for $R$ some at least $E_1$-ring spectrum and $G$ a loop space. Then This corresponds, I believe, to an action of $G$ on $R$, coming from a morphism $G\to GL_1(R)$. The first part of my question is: does this imply a map of spectra $R[G]\wedge R\to R$ or something like this? The second part of my question is regards the Thom spectrum $Mf$ associated to $f$. There are several standard constructions of $Mf$ from the given data, but I'm particularly interested in the interpretation of $Mf$ as $R/G$, the "quotient" of $R$ by the $G$ action (the first place I saw this discussed was in the preprint of Ando, Blumberg, Gepner, Hopkins and Rezk on units of ring spectra). Specifically, can $R/G$ be constructed by some specific bar construction, or something along these lines, in the $\infty$-category of spectra? ABGHR seem to indicate that the construction of the Thom spectrum as the colimit of the $BG$-shaped diagram inside of $R$-modules makes it obvious that we should call it $R/G$, but it's not as clear to me.

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    $\begingroup$ Re: the last sentence, take $G$ discrete and work in an ordinary category. An action of $G$ on an object in a category $C$ is precisely a diagram $BG \to C$ and the categorical quotient of that object by that action is precisely the colimit of this diagram in $C$. The ABGHR definition is a natural generalization of this. $\endgroup$ Jan 28, 2015 at 4:11

1 Answer 1


For your first question, the answer is yes. A map $BG\to BGL_1(R)$ gives you a map $G\to GL_1(R)$. Since $GL_1(R)$ is a set of component of $\Omega^{\infty}(R)$, this gives you a map $G\to \Omega^{\infty}(R)$ or equivalently a map $\mathbb{S}[G]=\Sigma_+^\infty G\to R$. The multiplication map $R\wedge R\to R$ gives you by adjunction a map $R\to F(R,R)$. Postcomposing with that map gives you a map $\mathbb{S}[G]\to F(R,R)$. The target has an $R$-module structure given by the map:

$$F(R,R)\wedge R\cong F(R,R)\wedge F(\mathbb{S},R)\to F(R,R\wedge R)\to F(R,R)$$ where the second map is just smashing the two functions spectra. We have an adjunction $-\wedge R:Spec\leftrightarrows Mod_R:forget$. Thus the map $\mathbb{S}[ G]\to F(R,R)$ is the same data as a map of $R$-module spectra $R[G]:=R\wedge\Sigma_+^\infty G\to F(R,R)$. Again by adjunction, you get from this a map $R[G]\wedge R\to R$.

Regarding your second question. A map $BG$ to an $\infty$-category $C$ is exactly the data of an object $X$ of $C$ together with a map $G\to Map_{C}(X,X)$. Now by definition the colimit of this diagram in $C$ is an object $Y$ of $C$ with a map $X\to Y$ which is $G$-equivariant when you give $Y$ the trivial $G$ action and which is initial with that property. In other word, the data of an other object $Z$ with a $G$-equivariant map $X\to Z$ (where $Z$ is equipped with the trivial $G$-action) should be the same as the data of a map $Y\to Z$. I think it is clear that $Y$ has to be $X/G$.

One explicit point set level method for constructing $R/G$ is the following. Pick a model for $G$ that is a strictly associative simplicial group and a map $G\to Map(R,R)$ (say you work in symmetric spectra). Then you can do a Bar construction $B(*,G,R)$. This is a simplicial object in the category of symmetric spectra whose $n$-simplices are the spectrum $\Sigma_+^{\infty}G^n\wedge R$. Then if $R$ is a cofibrant spectrum, the resulting simplicial object is Reedy cofibrant and computes $R/G$.

  • $\begingroup$ So, just to be clear, one cannot equivalently construct $R/G$ by something like $B(G,R,S)$? $\endgroup$ Jan 28, 2015 at 11:56
  • $\begingroup$ sorry dunno if the above pings you or not. Anyway, I'm basically thinking of the construction of group quotients by the bar construction and wondering if something similar can be done here. $\endgroup$ Jan 28, 2015 at 12:02
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    $\begingroup$ @JonBeardsley. I have edited my answer and added something that should answer your question $\endgroup$ Jan 28, 2015 at 12:30
  • $\begingroup$ In spaces, $B(X,G,Y)$ is a model (derived functor) for $X\times_G Y$ where $X$ and $Y$ are both $G$-spaces. So the two sided bar construction you wrote down doesn't model what you are asking about. Also, how would $G$ obtain an $R$-action? (you don't need to reply as I gather Geoffroy's nice answer helped clear things up.) $\endgroup$ Feb 2, 2015 at 16:00

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