# Snaith splitting for operads in spectra?

Let $$O$$ be an augmented operad. Then there is a functor $$J^O: Top_\ast \to Top_\ast$$ which takes $$X$$ to the free $$O$$-algebra on $$X$$ subject to the condition that the nullary operation coincides with the basepoint. The Snaith splitting tells us that $$\Sigma^\infty J^O(X) = \Sigma^\infty \vee_{n \geq 1} O(n)_+ \wedge_{\Sigma_n} X^{\wedge n}$$. Is there a version of this where $$X$$ is a pointed spectrum rather than a pointed space?

That is, let $$\mathbb S \to X$$ be a spectrum equipped with a map from the sphere spectrum. If $$O$$ is an augmented operad, there should be an $$O$$-algebra $$J^O(X)$$ such that the data of a map of $$O$$-algebra spectra $$J^O(X) \to A$$ is equivalent to a map of spectra $$X \to A$$ commuting with the maps from $$\mathbb S$$.

Question: Is there an easy formula for $$J^O(X)$$ when $$X$$ is a spectrum equipped with a map $$\mathbb S \to X$$? Can one at least say when $$J^O(X)$$ is nonzero?

Apparently such a formula will have to be a little more complicated than when $$X$$ is a space -- in particular, the dependence on the choice of map $$\mathbb S \to X$$ will be more subtle. For instance, if $$O$$ is an $$E_n$$ operad and $$\mathbb S \to X$$ is the zero map (or more generally smash-nilpotent), then $$J^O(X) = 0$$, rather than some large sum of wedges of $$X$$.

I think there's at least a colimit formula analogous to the James construction (which is, after all, the case $$O = E_1$$). For instance, when $$O = E_\infty$$, it should be the case that $$J^O(X)$$ is a certain colimit indexed over the category of finite sets and injections sending $$n \mapsto X^{\wedge n}$$. However, what I'd really like is a formula which (like the formula of the Snaith splitting) allows one to easily see that $$J^O(X) \neq 0$$ in reasonable cases, and so far this colimit formula has been a little too complicated for me to say this.

• You mean for O to be unital, not augmented, so that you’re asking about the left adjoint to the forgetful functor from O-algebras to E_0 algebras. I don’t think there is a splitting for this functor or any formula easier than the defining pushout of O-algebras (which can also be computed via a bar-like construction). It does come with a filtration whose associated graded is free on X/S, if that helps. (This gives you a sseq and you can try to see if 1 gets hit by a diffl). – Dylan Wilson May 30 '19 at 19:57

## 1 Answer

The best proofs of the `Snaith splitting' that you write down at the beginning, uses the $$J^{\mathcal O}$$ construction on spectra with units. Here is a complete proof:

Apply $$J^{\mathcal O}$$ to the zig-zag of equivalences of spectra (under and over $$S$$): $$\Sigma^{\infty} X_+ \xrightarrow{\sim} X \times S \xleftarrow{\sim} X \vee S.$$ On the left one gets $$\Sigma^{\infty}J^{\mathcal O}(X)_+$$ and on the right one gets precisely the wedge you wrote down (with an extra $$S$$ wedged on). QED

The origins of this proof go back to a short paper by Ralph Cohen around 1980. In the form I just presented it - making it very clear that everything is natural in the data - it appears in one of my papers (maybe my paper on stable splittings and the diagonal?).

• But this is precisely the case where you apply J^{O} to something equivalent to a free E_0-algebra (because of that disjoint basepoint) which is the opposite of the ‘bad’ situations Tim is asking about. (Those can’t arise from suspension spectra of pointed spaces because the basepoint splits off stably... hence the Snaith splitting). – Dylan Wilson May 31 '19 at 2:09
• @NicholasKuhn Are you saying that the Snaith splitting is natural even in maps between suspension spectra which are not the stabilization of an unstable map? – Tim Campion Jun 1 '19 at 22:27
• @TimCampion No. – Nicholas Kuhn Jun 2 '19 at 20:51