# Is there an ∞-categorical interpretation of the Quillen S⁻¹S construction?

The Quillen S⁻¹S construction (not to be confused with the Quillen Q-construction or the Quillen plus-construction), as defined by Grayson in Higher algebraic K-theory: II (page 219), takes as an input a symmetric monoidal groupoid S and produces a symmetric monoidal category S⁻¹S, whose objects are pairs (s,s') of objects in S and morphisms (s,s')→(t,t') are isomorphism classes of triples (A,A⊕s→t,A⊕s'→t').

Has this construction been investigated ∞-categorically, e.g., in the language of model categories?

I cannot even find a reference in the literature that shows that S⁻¹S represents (in the Thomason model structure) the homotopy group completion of S, so any references on this matter will be appreciated.

Once we do know that S↦S⁻¹S is the homotopy group completion functor, there is also the question of interpreting it as a left (derived) Quillen functor for some choice of a model structure on symmetric monoidal groupoids and (presumably) the Thomason model structure on categories.

A paper by Thomason (Beware the phony multiplication on Quillen's S⁻¹S) shows in particular that one cannot have an inverse functor x↦−x on S⁻¹S so that x⊕−x is naturally (strictly) isomorphic to 0. However, this does not preclude other ∞-categorical interpretations of the S⁻¹S construction.

• do you want to see $S\mapsto S^{-1}S$ as left adjoint or as homotopy left adjoint ? Would you be satisfied if you translate every thing in terms of $E_{\infty}$-spaces (since you have mentioned Thomason model structure). – Ilias Amrani Nov 30 '15 at 10:37
• @IliasAmrani: I might be wrong, but I think the implicit folklore understanding is that S⁻¹S computes the homotopy left adjoint. I'll add “derived” to the post. I will also be happy with an interpretation using group-like E_∞-spaces of any kind. – Dmitri Pavlov Nov 30 '15 at 15:15
• @IliasAmrani: I guess I misunderstood the intended meaning of “translate … E_∞-spaces”: showing that some other well-known formula for the homotopy group completion (e.g., ΩB) implements the homotopy group completion functor is not a part of the question. I am interested exclusively in S⁻¹S, not some other construction. – Dmitri Pavlov Nov 30 '15 at 20:36
• No problem, that happens :). You wrote "I cannot $\mathbf{even}$ find a reference in the literature that shows that S⁻¹S represents (in the Thomason model structure) the homotopy group completion of S, so any references on this matter will be appreciated." That is why i concluded that you were asking also if this functor is homotopy left adjoint under hypothesis that you know that it represent complition... – Ilias Amrani Nov 30 '15 at 20:44
• @IliasAmrani: I now see where the misunderstanding comes from. My intended meaning of “even” was that even if one does know that S⁻¹S is the homotopy group completion of S, there is still the question of which model structures to choose on both sides so that S⁻¹S preserves (acyclic) cofibrations (and hence is a left Quillen functor). – Dmitri Pavlov Nov 30 '15 at 20:50

Sketch of proof: We will use some facts and definitions (everything is written in the derived sense):

1. The model category $Cat$ with Thomason model structure is equivalent to the category of simplicial sets $sSet$ with the standard model structure via the nerve functor.
2. for any two symmetric monoidal categories $C$ and $D$, we define the simplicial set (or topological spaces) $Map(C,D)$ as $Map_{E_{\infty}}(|N_{\bullet}C|,|N_{\bullet}D|)$, where $|N_{\bullet}C|$ is the realization of the nerve of the category $C$, this is an $E_{\infty}$-space.
3. Now, the model category of $E_{\infty}$-space is pointed, thus we can define the suspension functor (in this case it is the bar construction $\mathrm{B}$) and the loop functor (derived right adjoint) which I will denote by $\Omega$.
4. The completion functor of an $E_{\infty}$-space is given by the $E_{\infty}$-map $$X\mapsto \Omega\mathrm{B}X$$.

since $\Omega\mathrm{B}X$ is $E_{\infty}$-space group like, we have that $$Map_{E_{\infty}}(\Omega\mathrm{B}X,Y)\simeq Map_{E_{\infty}}(\Omega\mathrm{B}X,Y^{int})$$ where $Y^{int}$ is defined as a pullback in the category of $E_{\infty}$-spaces as follows
$$Y^{int}=lim[\pi_{0}(Y)^{\ast}\rightarrow \pi_{0}Y\leftarrow Y]$$ here, $\pi_{0}(Y)^{\ast}$ is the group of ivertible elements in $\pi_{0}(Y)$. Therefore $Y^{int}$ is an $E_{\infty}$ space grouplike. Playing with the homotopy adjunctions, we obtain:

$$Map_{E_{\infty}}(\Omega\mathrm{B}X,Y)\sim Map_{E_{\infty}}(\Omega\mathrm{B}X,Y^{int}) \sim Map_{E_{\infty}}(\Omega\mathrm{B}X,\Omega\mathrm{B}Y^{int})\sim Map_{E_{\infty}}(\mathrm{B}\Omega\mathrm{B}X,\mathrm{B}Y^{int})\sim Map_{E_{\infty}}(\mathrm{B}X,\mathrm{B}Y^{int})\sim Map_{E_{\infty}}(X,\Omega\mathrm{B}Y^{int})\sim Map_{E_{\infty}}(X,Y^{int})$$ Thus we have shown that the (homotopy) right adjoint to the completion functor $\Omega\mathrm{B}$ is given by $(-)^{int}$.

Let $X$ be a connected topological CW complex, and $E_{\infty}(X)$ the free $E_{\infty}$ space generated by $X$, then there is nice consequence: the homotopy cofiber in the category of $E_{\infty}$-spaces of the natural $E_{\infty}$-map $$E_{\infty}(X)\rightarrow \Omega\mathrm{B} E_{\infty}(X)$$ is contractible.

The equivalence $$\Omega\mathrm{B}|N_{\bullet }S|\sim |N_{\bullet }S^{-1}S|$$ is true for sufficiently nice symmetric monoidal groupoid $(S,\oplus)$ I think you need to assume that the translation functors $s\oplus-: S\rightarrow S$ induce 'nice' functors for any object $s\in S$, but I don't remember the precise hypothesis and I don't have a precise reference... I hope it is helpful.

• The last displayed formula is an equivalent reformulation of my question, so how do you prove it? – Dmitri Pavlov Nov 30 '15 at 19:09
• @DmitriPavlov $\Omega\mathrm{B}|N_{\bullet }S|\sim |N_{\bullet }S^{-1}S|$ ? I don't have an exact reference for now... I thought your question was about a formulation of a left adjoint – Ilias Amrani Nov 30 '15 at 19:14
• No, I know many equivalent formulas for the homotopy group completion functor (including ΩB), but the purpose of this question is to find out whether S⁻¹S is one of them (or not). – Dmitri Pavlov Nov 30 '15 at 20:29