# Group completion of $\mathbb{E}_{\infty}$-monoids via tensor products

$$\newcommand{\K}{\mathrm{K}}$$The abelian group completion functor $$\K_0\colon\mathsf{CMon}\to\mathsf{Ab}$$ satisfies $$\K_0(A) \cong \mathbb{Z}\otimes_{\mathbb{N}}A,$$ naturally in $$A\in\mathrm{Obj}(\mathsf{CMon})$$, where

• $$\mathbb{Z}$$ is the additive monoid of integers (i.e. $$\K_0(\mathbb{N})$$, the group completion of $$\mathbb{N}$$);
• $$\otimes_\mathbb{N}$$ is the tensor product of commutative monoids.

Question. Does the $$\mathbb{E}_{\infty}$$-group completion functor $$\K_0\colon\mathsf{Mon}_{\mathbb{E}_\infty}(\mathcal{S})\to\mathsf{Grp}_{\mathbb{E}_\infty}(\mathcal{S})$$ similarly satisfies $$\K_0(X)\cong QS^{0}\otimes_\mathbb{F}X,$$ where now

• $$QS^0$$, the stabilization of $$S^0$$, is the $$\mathbb{E}_{\infty}$$-group completion of $$\mathbb{F}=\coprod_{n\in\mathbb{N}}\mathbf{B}\Sigma_{n}$$, the groupoid of finite sets and permutations;
• $$\otimes_{\mathbb{F}}$$ is the tensor product of $$\mathbb{E}_{\infty}$$-spaces?

Yes, for the same reason. Let me sketch a proof.

1- $$QS^0\otimes X$$ is group-complete. Indeed, its $$\pi_0$$ is $$\mathbb Z\otimes \pi_0(X)$$, and that's a group for the usual reasons. Another way to prove it is to prove that the shear map for $$X\otimes Y$$ is (the shear map of $$X)\otimes Y$$, which can be seen by noting that $$\otimes$$ commutes with coproducts and hence finite products in each variable.

2- There is a natural transformation $$X\to QS^0\otimes X$$ given by tensoring $$\mathbb F\to QS^0$$ by $$X$$, and this induces a natural transformation $$X^{gp}\to QS^0\otimes X$$.

3- Both sides commute with colimits (a colimit of grouplike $$E_\infty$$-spaces is grouplike so I don't have to worry about whether I'm talking about colimits in monoids or grouplike monoids), therefore to check that this map is an equivalence, it suffices to do so for $$X= \mathbb F$$, and for that one it is a tautology.

Another way to phrase this is to use the following sequence of natural equivalences (and using point 1- for the last one):

$$X^{gp} = QS^0\otimes_{QS^0} X^{gp} = (QS^0\otimes_\mathbb F X)^{gp}= QS^0\otimes X$$

The second natural equivalence comes from the fact that group completion is symmetric monoidal, and $$(QS^0)^{gp}\simeq QS^0$$.

• Thanks, Maxime! This is really nice!
– Théo
Aug 8 '21 at 20:09