# Is the classifying space of a symmetric monoidal category an infinite loop space?

Wikipedia states:

The classifying space (geometric realization of the nerve) of a symmetric monoidal category is an infinite loop space.

If my mind is correct, Segals delooping machine gives a functor $Sp$ from topological symmetric monoidal categories $C$ to prespectra (In this question, a presepectrum is a sequence of based spaces $X_0,X_1$,... with structure maps $\Sigma X_i\rightarrow X_{i+1}$) with $Sp(C)_0=BC$ the classifying space of the category.

If I remember correctly, one of Segals results is that the adjoints of the structure maps $Sp(C)_i\rightarrow \Omega Sp(C)_{i+1}$ are weak homotopy equivalences for $i\ge1$ and for $i=1$ this is the case iff $\pi_0(Sp(C)_0)=\pi_0(BC)$ is an abelian group with the induced multiplication which comes from the fact that $BC$ is a commutative monoid up to homotopy.

So the classifying space of a topological symmetric monoidal category $C$ can in this sense infinitely delooped only if $\pi_0(BC)$ is a group.

I ask myself:

1.) Did I say anything stupid or wrong?

2.) Is Wikipedia wrong?

• What you say is correct, so Wikipedia was wrong. I have added a minimal correction to the statement in Wikipedia, but it would be nice if someone could expand on it. – Neil Strickland Dec 3 '14 at 17:44
• Wouldn't this be a more appropriate discussion for the Wikipedia talk page? – Ryan Budney Dec 3 '14 at 17:44
• in general $BC$ is an $E_{\infty}$-space and in the case where $\pi_{0}BC$ is a group (abelian in this case) then $BC$ is an infinite loop space. – Max Dec 3 '14 at 18:24
• @RyanBudney People are a lot more likely to notice a question here on MO than on the Wikipedia page, I think. There have been a number of similar questions about issues with Wikipedia articles, and I think they are reasonable questions. Usually they get answered quickly and have the positive effect of improving Wikipedia. – Dan Ramras Dec 3 '14 at 18:39
• Hi Dan, it looks to me like this question was asked and answered on the talk page a few months ago. – Ryan Budney Dec 3 '14 at 18:43

## 1 Answer

You need group Completion, indeed.