9
$\begingroup$

$\newcommand{\K}{\mathrm{K}}$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?
$\endgroup$
9
$\begingroup$

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 $(QS^0)^{gp}\simeq QS^0$.

$\endgroup$
1
  • $\begingroup$ Thanks, Maxime! This is really nice! $\endgroup$
    – Théo
    Aug 8 '21 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.