# $K$-theory backwards

Let $R$ be a ring. The $K$-theory of $(Mod(R)^{f.g.proj},\oplus)$ is obtained by first throwing out non-isomorphisms and then group completing. What happens if these steps are reversed?

That is, consider the symmetric monoidal $\infty$-category $C = (Mod(R)^{f.g.proj},\oplus)$ of finitely-generated projective $R$-modules under direct sum (of course, this is in fact an ordinary category). To compute $K(R)$, we first pass to the core $\iota(C)$ to get an $\infty$-groupoid (which in this case is a 1-groupoid) and then we group complete to get $K(R) = \iota(C)[C^{-1}]$.

I'm thinking of group completion as an $\infty$-categorical process: for a symmetric monoidal $\infty$-category $C$, the group completion $C[C^{-1}]$ is the free symmetric monoidal $\infty$-category on $C$ where every object has a monoidal inverse; this agrees with the usual notion for symmetric monoidal $\infty$-groupoids. So it makes sense to reverse these steps and consider $\iota(C[C^{-1}])$. So my question is:

Question:

1. If $C$ is a symmetric monoidal $\infty$-category, then do we have $\iota(C)[C^{-1}] = \iota(C[C^{-1}])$?

2. If not, does this at least hold in the case $C = Mod(R)^{f.g.proj}$, yielding an alternate description of $K(R)$?

Motivation: My motivation here is basically to explore how robust the definition of $K$-theory is to perturbation. For example, an alternate description of $K(R)$ that definitely does work is the following: $K(R) = \iota_\oplus(\iota(Mod(R))[R^{-1}])$: that is, take the groupoid of all $R$-modules and rather than group completing, monoidally invert the single object $R$ itself viewed as an $R$-module; finally, throw away all objects which are not monoidally invertible (note in particular that since we're not monoidally inverting everything, we've sidestepped the Eilenberg swindle). I like this description because you don't have to put the "finitely-generated projective" condition in by hand -- it falls out naturally just from knowing that that the symmetric monoidal category $(Mod(R),\oplus)$ has a distinguished object $R$. If the answer to my question is "yes", then one could simplify this further to $K(R) \overset{?}{=} \iota_\oplus(\iota(Mod(R)[R^{-1}]))$; then even the choice of the $\oplus$ monoidal structure becomes more "canonical" in the sense that it has a universal property in $Mod(R)$, unlike in $\iota(Mod(R))$. Other variants start to suggest themselves, too.

• ... "yroeht-K?" – Greg Martin Dec 14 '17 at 17:59

I should probably have thought a bit more before posting this question: the answer to (2) (and a fortiori to (1)) is no. In fact, $C[C^{-1}]$ is the terminal category in the case of interest and so $\iota(C[C^{-1}])$ is contractible.

Recall that a commutative rig category $(C,\otimes,I,\oplus,0)$ has symmetric monoidal structures $(C,\oplus,0)$ and $(C,\otimes,I)$, with distributivity $A\otimes 0 = 0$ and $A\otimes(B\oplus C) = (A\otimes B) \oplus (A \otimes C)$ coherently. Alternatively, it is an $E_\infty$-object in symmetric monoidal categories. Let us write $A^\vee$ for a $\otimes$-dual to $A$ and $-A$ for a $\oplus$-dual to $A$.

Lemma: Let $(C,\otimes,I,\oplus,0)$ be a commutative rig category. If $0^\vee$ exists, then $0$ is a zero object (i.e. it is both initial and terminal).

Proof: By adjunction, we have $C(X,0^\vee) = C(X,0^\vee \otimes I) = C(0 \otimes X, I) = C(0,I)$, naturally in $X$. Taking $X = 0^\vee$ and $1 \in C(0^\vee,0^\vee)$, we obtain a map $X \to 0^\vee$, natural in $X$. Dually, we have a map $0^\vee \to Y$ natural in $Y$, and by composing, we have a map $0_{X,Y}:X \to Y$ natural in $X$ and $Y$. So there is a zero object $\emptyset$ in (the idempotent completion of) $C$. Because $0$ is dualizable, tensoring with it preserves limits and colimits, so we have $\emptyset = \emptyset \otimes 0 = 0$ by distributivity.

Lemma: Let $(C,\otimes,I,\oplus,0)$ be a commutative rig category. If $0$ is a zero object and $-A$ exists, then $A=0$.

Proof: Because $0$ is a zero object, both the unit and counit maps $A \oplus (-A)^\to_\leftarrow 0$ are zero maps. Then a triangle identity shows that $0_{A,A} = 1_A$, so that $A=0$.

Corollary: If $C$ is a commutative rig category with additive duals $-A$ for all objects $A$, and the multiplicative dual $0^\vee$, then $C$ is the terminal category.

Corollary: If $C = (Mod(R)^{f.g.proj},\oplus)$, then $C[C^{-1}]$ is the terminal category. Hence $\iota (C[C^{-1}])$ is contractible.

Proof: $(C,\otimes,\oplus)$ is a commutative rig category (now $\otimes$ is the usual tensor product of modules). Group completion is a strong monoidal functor on symmetric monoidal categories, so $C[C^{-1}]$ is likewise a commutative rig $\infty$-category, and $C \to C[C^{-1}]$ is a morphism thereof. Therefore, because $0$ is $\otimes$-self-dual in $C$, it is likewise self-dual in $C[C^{-1}]$. Moreover, $C[C^{-1}]$ has additive duals (even better, it has additive inverses). So by the previous Corollary, $C[C^{-1}]$ is the terminal category.

• An alternative argument for the second lemma: if $-A$ exists, then $A \oplus(-)$ preserves limits and colimits, so $A = A \oplus 0 = 0$. – Tim Campion Dec 22 '17 at 4:26

The only way I know to speak of group-completion' of a symmetric monoidal $\infty$-category involves not only adjoining inverses of all objects but also inverses of all morphisms. In that sense, group-completion corresponds to first taking the classifying space (rather than the core) and then group-completing. For example, if $\mathcal{C}$ has an initial or terminal object, its classifying space is trivial, so its group completion is trivial.

Of course there are plenty of decent symmetric monoidal categories where every object has an inverse but which are not groupoids (for example, the full subcategory of $\text{Mod}_R$ on invertible modules). But maybe this does not deserve to be called group-complete' because the inverse is not functorial (at least, not covariantly).

If there is a functorial inverse $-\text{id}:\mathcal{C}\rightarrow\mathcal{C}$ such that $-\text{id}\otimes\text{id}$ is the zero functor, then the category is a groupoid. Specifically, each morphism $f:X\rightarrow Y$ has an inverse (up to equivalence) $X\otimes Y\otimes(-f):Y\rightarrow X$.

EDIT: Why would we want the inverse to be covariantly functorial? Let's say for example that $\mathcal{R}$ is a semiring $\infty$-category which has additive inverses. Then multiplication by any element (in particular -1) is functorial $\mathcal{R}\rightarrow\mathcal{R}$, so $\mathcal{R}$ is an $\infty$-groupoid as above.

Thus group-complete symmetric monoidal $\infty$-categories are exactly grouplike symmetric monoidal $\infty$-groupoids, also known as grouplike $\mathbb{E}_\infty$-spaces or connective spectra. The universal example of one of these is the sphere spectrum $\mathbb{S}$. Modules over $\mathbb{S}$ (in the sense of modules over a semiring $\infty$-category) are precisely the connective spectra, and the group-completion operation you want is $$-\otimes\mathbb{S}:\text{SymMon}\rightarrow\text{Mod}_\mathbb{S}\cong\text{Sp}_{\geq 0},$$ analogous to $$-\otimes\mathbb{Z}:\text{ComMon}\rightarrow\text{Mod}_\mathbb{Z}\cong\text{Ab}.$$ I discuss all of this in section 4 of my paper https://arxiv.org/abs/1606.05606

• I'm confused -- let $Pic_R$ denote the full subcategory of $Mod_R$ on the $\otimes$-invertible objects. Then the $\otimes$-inverses in $Pic_R$ are functorial, since $L^{-1} = Hom(L,R)$, and $Hom(-,R): (Pic_R)^{op} \to Pic_R$ is a functor. But $Pic_R$ has non-invertible morphisms. – Tim Campion Dec 14 '17 at 16:38
• In fact, more generally, if $(C,\otimes)$ has duals for objects, taking the dual is a functor $C^{op} \to C$. In particular, if $(C,\otimes)$ has inverses for objects, then taking the inverse is a functor $C^{op} \to C$. Are you requiring there to be a functor $C \to C$ rather than $C^{op} \to C$? – Tim Campion Dec 14 '17 at 16:53
• I was assuming the inverse is a covariant functor. How would you write down axioms for a grouplike symmetric monoidal $\infty$-category with a contravariant inverse? We would like to have an axiom like $-X\otimes X\cong 1$, but the functoriality is wrong. – John Berman Dec 14 '17 at 17:00
• Because of this issue with mixed functoriality, you might have trouble making sense of the free "group-completion" construction. A similar question was asked a couple years ago mathoverflow.net/questions/214767/… – John Berman Dec 14 '17 at 17:07
• Maybe you're looking for the notion of a dinatural transformation. In any event, let's say that $(C,\otimes)$ is objectwise group complete if every object has an $\otimes$-inverse. The objectwise group complete symmetric monoidal $\infty$-categories are closed among all symmetric monoidal $\infty$-categories under products and pullbacks and filtered colimits, and so by the adjoint functor theorem there is a left adjoint to the inclusion functor. This left adjoint is what I'm referring to above. – Tim Campion Dec 14 '17 at 17:23