Bar Construction Model of Ring Spectrum Quotient

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.

• 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. – Qiaochu Yuan Jan 28 '15 at 4:11

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 $R/G$.
• So, just to be clear, one cannot equivalently construct $R/G$ by something like $B(G,R,S)$? – Jonathan Beardsley Jan 28 '15 at 11:56
• 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.) – Sean Tilson Feb 2 '15 at 16:00