By the Barratt-Priddy-Quillen theorem, the space $B \Sigma_\infty^+$ is the infinite loop space $\Omega^\infty \Sigma^\infty S^0$. I'm curious about a "high-concept" reason that $B \Sigma_\infty^+$ (and "more generally" $BGL_\infty^+(R)$ for a ring $R$) should be infinite loop spaces (well, almost that). For instance, I read about a theorem as follows:

Define an $\mathcal{I}$-functor $F$ as the following data: $F$ is a functor from the category of finite-dimensional inner product spaces over $\mathbb{R}$ and isometric linear imbeddings to the category of pointed spaces and closed imbeddings, and there are functorial maps $F(V) \wedge F(W) \to F(V \oplus W)$ for each $V, W$ satisfying associativity and commutativity relations. Then it is a theorem that the connected component $F(\mathbb{R}^\infty) = \varinjlim F(\mathbb{R}^n)$ is an infinite loop space. This is because the functorial maps given above lead to an action on $F(\mathbb{R}^\infty)$ of the linear isometries operad, which is an $E_\infty$-operad, and one can invoke May's recognition principle. This provides additional explanation why the infinite orthogonal group is an infinite loop space.

The data above is sort of analogous to the theory of orthogonal spectra. Namely, it states that $F$ is a commutative algebra object in a certain monoidal functor category from inner product spaces to pointed topological spaces; the functor $F(V) :=S ^V$ is another one and orthogonal spectra are precisely modules over this functor.

I'm curious whether there is an analog for symmetric spectra. Namely, let's say we have a functor $G$ from finite sets to pointed spaces, together with maps $G(A) \wedge G(B) \to G(A \sqcup B)$ satisfying associativity and commutativity conditions. I am curious whether the connected component of $G(\mathbb{N}) = \bigcup G([n])$ is an infinite loop space. By similar reasoning, $G(\mathbb{N})$ admits an action of the "injection operad," which is however only a set-valued operad. So if this is true, perhaps Segal's delooping machinery might be relevant (from his paper "Categories and cohomology theories"). Can it be proved that $G(\mathbb{N})$ is a $\Gamma$-space, in his terminology?

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    $\begingroup$ Dear Akhil: You need to take the plus construction before you can get an infinite loop space. For example, we have $\pi_1(B\Sigma_\infty)=\Sigma_\infty$, while $\pi_1(QS^0)=0$. $\endgroup$
    – Tom Church
    Commented Jan 1, 2012 at 3:54
  • $\begingroup$ Dear Tom: Whoops! Thanks for the correction. $\endgroup$ Commented Jan 1, 2012 at 4:34

1 Answer 1


You are looking at the telescope of maps $BG_n\longrightarrow BG_{n+1}$ where the coproduct of the $G_n$ (thought of as categories) has a structure of permutative category. The group completion property of infinite loop space machines defined on permutative categories shows easily that there is a canonical map from $BG_{\infty}$ to the component of the unit (for the permutative structure) in the zeroth space of the associated spectrum that induces an isomorphism on homology (with any ring of coefficients). Passing to the plus construction, it becomes an equivalence.

In the kind of examples you refer to with the linear isometries operad, the input looks quite a lot different. The fact that you get equivalent spectra when both constructions apply is non-trivial. See my paper

The spectra associated to $\mathcal{I}$-monoids.

Math. Proc. Camb. Phil. Soc. 84(1978), 313--322.


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