In 1997, Elmendorf, Kriz, Mandell, and May wrote a book Rings, Modules, and Algebras in Stable Homotopy Theory in which they introduced the category of $S$-modules as a model for the stable homotopy category. The category of $S$-modules is a closed symmetric monoidal model category whose monoidal product descends to the usual product on the stable homotopy category. Its unit is the sphere spectrum. All objects are fibrant but the unit is not cofibrant. Every operad $O$ is admissible, meaning the category of $O$-algebras has a transferred model structure where a morphism $f$ of $O$-algebras is a weak equivalence (resp. fibration) if and only if $U(f)$ is in $S$-modules. A reference is Proposition 1.5 in "Moduli Spaces of Commutative Ring Spectra" by Goerss and Hopkins. The proof uses that $S$-modules have a structured interval object.

For the operad $O = Ass$, Theorem VII.6.2 in EKMM proves that if $A$ is a cofibrant $S$-algebra then the unit $S \to A$ is a cofibration of $S$-modules. Just after, they write "In the commutative case, the argument fails because we must pass to orbits over actions of symmetric groups."

The property that I want, that a cofibrant $O$-algebra should forget to a cofibrant spectrum, has sometimes been called "convenient" but that word is overloaded. A related notion, studied by Pavlov and Scholbach, is to say that an operad $O$ is strongly admissible, meaning that, if $a: A\to A'$ is a cofibration of $O$-algebras with $A$ cofibrant, then $U(a)$ is a cofibration and $U(A)$ is cofibrant. So this implies what I want, and even more.

Unfortunately, in the category of $S$-modules with the EKMM model structure, the commutative monoid operad does not have the property I want. $S$ is a cofibrant commutative monoid that is not cofibrant as an $S$-module. We have the same problem with $O = Ass$. Interestingly, because $S$ is not cofibrant in $S$-mod, the associative operad is not $\Sigma$-cofibrant, as $Ass(n)$ is a coproduct of copies of $S$, one for each $\sigma$ in the symmetric group $\Sigma_n$.

There might still be a hope that $\Sigma$-cofibrant operads are strongly admissible, but normally proving this requires the unit to be cofibrant. In particular, I don't know whether $S$-modules satisfy the Pavlov-Scholbach condition of "symmetric h-monoidality" or the related conditions in Section 6 of Bousfield Localization and Algebras over Colored Operads. If they do, then we know that cofibrant $O$-algebras forget to cofibrant $S$-modules for operads $O$ whose spaces $O(n)$ are cofibrant.

There are several other good models of spectra, including symmetric and orthogonal spectra, and various tweaks to their stable model structures. In the positive model structure on symmetric or orthogonal spectra, all operads are admissible. For symmetric spectra, this is proven in Theorems 8.3.1 and 6.1.1 of Bousfield Localization and Algebras over Colored Operads, among other places. For orthogonal spectra this is proven in Corollary 5.15 of Right Bousfield Localization and Eilenberg-Moore Categories, among other places. However, in general, a cofibrant commutative monoid need not forget to a cofibrant object in these categories, by Proposition 4.2 in Shipley's paper A convenient model category for commutative ring spectra. For this reason, Shipley introduced what is now called the positive flat stable model structure on symmetric spectra (and the same works for orthogonal spectra, see Remark 5.14 in the Right Bousfield paper above) where, indeed, cofibrant algebras over entrywise cofibrant colored operads (and commutative monoids) forget to cofibrant objects in these positive flat model structures, e.g., by Section 6 of the Left Bousfield paper above.

(1) Can we tweak the EKMM model structure on $S$-modules in some way so that cofibrant commutative monoids forget to cofibrant $S$-modules?

(2) Can we do the same for some class of operads? Maybe in the case of $\Sigma$-cofibrant operads, we don't have to tweak the EKMM model structure at all. If so, I'd love to know.

  • $\begingroup$ Given that strong admissibility is defined as preservation of cofibrations with cofibrant source by the forgetful functor, which immediately implies preservation of cofibrant objects (since the identity map on any cofibrant object is always a cofibration), I would very much like to know where in our work we "must assume the unit of the base model category M is cofibrant". Would you mind giving a reference? $\endgroup$ Jun 26 at 16:05
  • $\begingroup$ You can tweak the EKMM model structure to make the unit cofibrant. Would this suffice? $\endgroup$ Jun 26 at 16:09
  • $\begingroup$ @DmitriPavlov I'm looking at Lemma 6.2 in your paper. In (2), it tells me that $U$ sends cofibrations with cofibrant source to cofibrations. In (3) it says "if the unit $1\in C$ is cofibrant, $U$ also preserves cofibrant objects." $\endgroup$ Jun 26 at 17:31
  • $\begingroup$ @DmitriPavlov Ah, I understand now what I had wrong. Being out of practice with this stuff, I interpreted "U preserves cofibrations with cofibrant source" to mean "if A -> B is a cofibration in O-alg, where A is a cofibrant O-algebra, then U(A)->U(B) is a cofibration." But I see now in your Defn 5.1 you also require that in the conclusion U(A) is cofibrant in M. I edited the question to avoid misrepresenting your definition. $\endgroup$ Jun 26 at 17:44
  • $\begingroup$ I think Corollary 5.5 in Berger-Moerdijk "Axiomatic homotopy theory for operads" tells us that, if $O$ is a $\Sigma$-cofibrant operad in $S$-modules then a cofibrant algebra forgets to a cofibrant $S$-module. So this answers my question (2). $\endgroup$ Jun 26 at 18:06

1 Answer 1


Here is something that I think is reasonably difficult to get around. As you observe, the unit is the initial object of commutative monoids, and so your request includes that the unit is cofibrant.

Suppose $(C,\otimes,1)$ is a symmetric monoidal category with colimits where the symmetric monoidal structure preserves colimits in each variable. Then the free commutative monoid on $X$ is $$ \coprod_{n \geq 0} X^{\otimes n}/\Sigma_n. $$ In the case when $X$ is the monoidal unit $1$, however, the unit axioms imply that the action of $\Sigma_n$ on $1^{\otimes n}$ is trivial, and so the free commutative monoid on $1$ is $$ \coprod_{n \geq 0} 1. $$ However, if there are function spaces, the free $E_\infty$ algebra on $1$ should instead have the homotopy type of $$ \coprod_{n \geq 0} 1_{h\Sigma_n}. $$

For us, this means that we need some kind of compromise, especially if we are hoping for a Quillen adjunction between objects of C and commutative monoids.

  • Maybe we need to accept that there's not going to be a Quillen adjunction between commutative monoids and objects of C (non-positive model structures on symmetric spectra).
  • Maybe we need to accept that the unit isn't cofibrant (positive model structures, S-modules).
  • Maybe we need to accept that commutative monoids simply aren't equivalent to algebras over an $E_\infty$ operad, and so the homotopy theory of commutative monoids is the homotopy theory of some completely new type of object ($\Gamma$-spaces, topological spaces, equivariant spectra).
  • Maybe we get lucky and this distinction between orbits and homotopy orbits is not actually a problem (characteristic zero).
  • Maybe we go beyond the pale and find a model where the symmetric monoidal structure interacts poorly with colimits (?).
  • 1
    $\begingroup$ In the application I have in mind, I probably want to force the unit to be cofibrant, and give up that commutative monoids are equivalent to $E_\infty$ things. $\endgroup$ Jun 26 at 17:46
  • $\begingroup$ @DavidWhite that you can do under mild assumptions, in particular for $S$-modules sciencedirect.com/science/article/pii/… I don’t know which of Tyler’s points would fail, though. I suspect that the first one. $\endgroup$ Jun 26 at 21:50
  • $\begingroup$ Yes, Theorem 3 there does seem to apply (and you already did it, in your Example 6 there). Next is to check that all operads in the new model structure are admissible. We still have the good interval object, but now not all objects are fibrant. It's an exercise. The hard part (to know cof $O$-alg forget to cof obj) is checking good behavior of the functor $(O,A) \to O_A$. The only way I know to do that is to check that, for $X$ with a $\Sigma_n$ action, $X$ cofibrant in $M$, $X \otimes_{\Sigma_n} f^{\Box n}$ is some kind of cofibration, where $f$ is a cofibration. That's really hard. $\endgroup$ Jun 27 at 5:14
  • $\begingroup$ @DavidWhite That's certainly a possibility. Hopefully then you have some intrinsic interest in the types of O-algebras that can be modeled in your new category! $\endgroup$ Jun 27 at 15:46
  • $\begingroup$ Yes, indeed, there are several general theorems that would be true for this new model that are not known in general. Like lifting quillen equivalences between models of spectra to categories of commutative ring spectra or commutative ideals. There are lots of results where you want a commutative O algebra to forget to a commutative underlying object. But this is too far afield for what i need right now $\endgroup$ Jun 27 at 19:44

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.