5
$\begingroup$

Let $E$ be an operad in topological spaces. $E$ is usually called an $E_{\infty}$-operad, if all the spaces $E_n$ are contracticle. If $E$ acts on a space $X$, then by the recognition principle, $X$ turns out to be an infinite loop space.

How much of the theory is preserved, if I replace contractible by weakly contractible? Do I still have deloopings, if $E$ acts on a space?

Improve this question
$\endgroup$
2
  • $\begingroup$ It's unlikely since if $E_n$ is weakly contratible but not contractible then $E_n$ cannot be cofibrant (i.e. a retract of a CW-complex), so transfer results for algebras over operads begin to fail... $\endgroup$
    – Fernando Muro
    Commented Dec 14, 2011 at 8:12
  • $\begingroup$ Can't you take the realization of the singular simplicial sets of all spaces in question (this produces a contractible space out of a weakly contractible one) and get an action of an $E_{\infty}$-operad on $|Sing_{\bullet}X|$ which is weakly equivalent to $X$ and hence an infinite loop space? $\endgroup$
    – Johannes Ebert
    Commented Dec 14, 2011 at 9:18

2 Answers 2

Reset to default
6
$\begingroup$

To answer your question and complete Justin's answer, one can pick a cofibrant replacement of $E$ to get a genuine $E_\infty$-operad $F$ acting on the space $X$, and conclude that $X$ has an infinite delooping up to group completion issues.

But the other way round fails: not all infinite loop spaces are acted on by $E$. Example: any group like commutative monoid is an infinite loop space, but not all infinite loop spaces are commutative monoids (otherwise no non-trivial Dyer Lashof operation would exist).

Remark: If $E$ is weakly-contractible, then the constant maps $c: E_n\rightarrow pt$ define an acyclic fibration, in the category of topological operads, from the operad $E$ towards the operad of commutative monoids $C$. If you have a preferred (cofibrant) $E_\infty$-operad $F$, then you can use the LLP to get an operad weak-equivalence $f: F\xrightarrow{\sim} E$, making $F$ a cofibrant replacement of $E$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ @BF: I appreciate your elaboration. I do want to point out that one does not require that the operad to be cofibrant in May's theory, only $\Sigma$-cofibrant. Although we can always replace an operad by a cofibrant replacement, it is often useful to compare algebras over an operad that naturally acts on them (such as the little n-cubes operad or the Barratt-Eccles operad). $\endgroup$
    – Justin Noel
    Commented Dec 14, 2011 at 13:41
  • 1
    $\begingroup$ Sure: The $\Sigma$-cofibrant condition is sufficient to make homotopy categories equivalent. Every space $X$ acted on by a $\Sigma$-cofibrant $E_\infty$-operad is weakly to a space $Y$ acted on by the Barratt-Eccles operad (for instance). This does not mean however that $X$ itself is acted by the Barratt-Eccles operad, and cofibrant operad come at this point, in order to get homotopy invariant structures. $\endgroup$
    – user19952
    Commented Dec 14, 2011 at 14:08
  • $\begingroup$ One could use cofibrant operads, this is a perfectly good way to go, but it is not required, nor does it appear in the reference I quoted. All $E_\infty$ operads produce weakly equivalent deloopings. The comparison merely requires taking the product of operads and using the bar construction. $\endgroup$
    – Justin Noel
    Commented Dec 14, 2011 at 16:02
4
$\begingroup$

Some clarifications: 1) You need that $X$ is grouplike (so the induced multiplication makes $\pi_0 X$ a group). This condition is always satisfied for a loop space, but not satisfied by the discrete $E_\infty$ space $\mathbb{N}$.

2) In order for $\mathcal{L}$ to be an $E_\infty$ operad we require more than $\mathcal{L}(n)\simeq *$, otherwise we could have just used the commutative operad and all grouplike $E_\infty$ spaces would be topological abelian groups. We also require that $\mathcal{L}(n)$ is a cofibrant $\Sigma_n$-space, so that it has a free $\Sigma_n$ action and consequently homotopy equivalent to $E\Sigma_n$.

In May's 'Geometry of iterated loop spaces' especially Ch. 3 http://www.math.uchicago.edu/~may/BOOKS/geom_iter.pdf it is shown that the delooping machine does not depend on the choice of $E_\infty$ operad.

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ Everything Justin says is correct. I like model categories too, nowadays, but, as a practical matter, I know of nothing of real use in iterated loop space theory that actually requires use of a model structure on the category of topological operads. Admittedly, the use of products of operads feels like a cheap trick, but it is nice to be able to prove some things cheaply, without the slightest use of categorical homotopy theory. Younger mathematicians should understand that a huge amount of important algebraic topology that is still current predates the widespread use of model categories. $\endgroup$
    – Peter May
    Commented Dec 14, 2011 at 21:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .