# What is a simplicial commutative ring from the point of view of homotopy theory?

Let $k$ be a field. There are two natural categories to consider:

• The category of simplicial commutative $k$-algebras.
• The category of connective $E_\infty$ $k$-algebras (i.e., chain complexes of $k$-vector spaces in nonnegative dimensions with a coherently associative and commutative multiplication law).

These categories are not the same if $k$ does not have characteristic zero. Simplicial commutative $k$-algebras are rather special, and (for example) not every commutative dga over $k$ (which determines an $E_\infty$-algebra over $k$) comes from a simplicial commutative $k$-algebra. (The homotopy groups of a simplicial commutative ring have divided powers, by an explicit construction that I don't really understand.) The category of $E_\infty$-algebras over $k$ has a nice interpretation via homotopy theory: it is the category of commutative algebra objects (in an appropriate sense) in the category of connective $k$-module spectra. (In particular, it is monadic over connective $k$-module spectra, in the $\infty$-categorical sense.) I don't know how to think of simplicial commutative rings in this way; all I know for motivation is that they form a nice homotopy theory (e.g. presented by a fairly concrete model category) that allows you to extend the category of ordinary commutative rings (e.g., to resolve non-smooth objects by smooth ones).

Is there an analog of the above discussion for $E_\infty$-algebras that works for simplicial commutative $k$-algebras? In particular, can they be described as algebras over a nice monad for $k$-module spectra?

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Such a construction can't be too general, because it relies on the existence of "symmetric power" functors on simplicial $k$-modules that refine the homotopical symmetric powers (and those don't exist in general). The monad you'd want to write down is basically built out of these symmetric power functors, and I've heard it said that a refinement to a simplicial commutative ring is equivalent to being equipped with these symmetric powers. I don't know the argument. –  Tyler Lawson Jan 10 '13 at 5:51
Simplicial commutative algebras are commutative algebra objects in the category of simplicial modules. This is nice! –  Fernando Muro Jan 10 '13 at 7:10
@Fernando: I think that Akhil wants an answer that only depends on the $\infty$-category of simplicial modules, and not on any particular 1-category model of it. –  André Henriques Jan 10 '13 at 12:15
@Tyler: I guess I don't see why it's impossible to form the symmetric power functors. If $X \to Y$ is a morphism of simplicial $k$-modules which is a homotopy equivalence, then the same is true for $(X^{\otimes n})_{\Sigma_n} \to (Y^{\otimes n})_{\Sigma_n}$, no? –  Akhil Mathew Jan 10 '13 at 14:04
One clue to think about: if k has finite characteristic, then simplicial commutative rings come with a natural Frobenius endomorphism. E-infinity rings don't generally have anything like that. –  Charles Rezk Jan 10 '13 at 15:15

I don't know a really satisfying answer to this question, but here are a few observations.

1) The $\infty$-category of simplicial commutative $k$-algebras is monadic over the $\infty$-category of connective $k$-module spectra. The relevant monad is the nonabelian left derived functor of the "total symmetric power" on ordinary $k$-modules, which is different from the construction $M \mapsto \bigoplus_{n} (M^{\otimes n})_{h \Sigma_n}$ unless $k$ has characteristic zero.

2) The $\infty$-category of simplicial commmutative rings is freely generated under sifted colimits by the ordinary category of finitely generated polynomial algebras over $k$. In other words, it can be realized as the $\infty$-category of product-preserving functors from the ordinary category of $k$-schemes which are affine spaces to the $\infty$-category of spaces.

3) Let $X$ be the affine line over $k$ (in the sense of classical algebraic geometry). Then $X$ represents the forgetful functor {commutative $k$-algebras} -> {sets}. Consequently, $X$ has the structure of a commutative $k$-algebra in the category of schemes. Also, $X$ is flat over $k$.

Now, any ordinary scheme can be regarded as a spectral scheme over $k$: that is, it also represents a functor {connective E-infty algebras over k} -> {spaces}. In general, products in the category of ordinary $k$-schemes need not coincide with products in the $\infty$-category of spectral $k$-schemes. However, they do agree for flat $k$-schemes. Consequently, $X$ can be regarded as a commutative $k$-algebra in the $\infty$-category of derived $k$-schemes. In particular, $X$ represents a functor {connective E-infty algebras over k} -> {connective E_infty algebras over k}. This functor has the structure of a comonad whose comodules are the simplicial commutative $k$-algebras.

You can summarize the situation more informally by saying: derived algebraic geometry (based on simplicial commutative $k$-algebras) is what you get when you take spectral algebraic geometry (based on E-infty-algebras over $k$) by forcing the two different versions of the affine line to coincide.

4) The forgetful functor {simplicial commutative $k$-algebras} -> {E-infty algebras over $k$} is both monadic and comonadic. In particular, you can think of a simplicial commutative $k$-algebra $R$ as an E-infty algebra over $k$ with some additional structure. As Tyler mentioned, one way of thinking about that additional structure is that it gives you the ability to form symmetric powers of connective modules. Of course, if $M$ is any $R$-module spectrum, you can always form the construction $(M^{\otimes n})_{h \Sigma_n}$. However, this doesn't behave the way you might expect based on experience in ordinary commutative algebra: for example, if $M$ is free (i.e. a sum of copies of $R$) then $(M^{\otimes n})_{h \Sigma_n}$ need not be free (unless $R$ is of characteristic zero). However, when $R$ is a simplicial commutative ring, there is a related construction on connective $R$-module spectra, given by nonabelian left derived functors of the usual symmetric power. This will carry free $R$-module spectra to free $R$-module spectra (of the expected rank).

It is possible to describe the $\infty$-category of simplicial commutative $k$-algebras along the following lines: a simplicial commutative $k$-algebra is a connective E-infty algebra over $R$ together with a collection of symmetric power functors Sym^{n} from connective $R$-modules to itself, plus a bunch of axioms and coherence data. I don't remember the exact statement (my recollection is that spelling this out turned out to be more trouble than it was worth).

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Thanks -- this is very helpful. Characterization 2 of simplicial commutative rings is a nice parallel to the description of connective $HR$-modules as freely generated under sifted colimits by the finitely generated free ones. –  Akhil Mathew Jan 10 '13 at 19:09

Given a symmetric monoidal $\infty$-category $\mathcal C$, you can consider $E_\infty$-objects in it, and that's it -- nothing more. For example, for a spectrum, there's nothing more than $E_\infty$ that you can ask it to be.

Now, if the symmetric monoidal structure is the cartesian product, then you can also consider what Jacob Lurie calls very commutative'' objects in $\mathcal C$. These are more commutative than $E_\infty$.

A very commutative object in $\mathcal C$ is an object along with a lift of the corresponding representable presheaf $\mathcal C^{op}\to \mathcal Spaces$ to a functor $\mathcal C^{op}\to$ {connective $H\mathbb Z$-modules} = {topological abelian groups}. By contrast, an $E_\infty$-object in $\mathcal C$ (again I'm assuming that the monoidal structure is cartesian) is a lift to a functor $\mathcal C^{op}\to$ {connective $\mathbb S$-modules} = {infinite loop spaces}.

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Interesting. So presumably the infinite symmetric product on a space the free "very commutative" monoid on it, which would explain the Dold-Thom theorem. –  Akhil Mathew Jan 10 '13 at 14:07
This answer also suggests that there is no simple relationship between simplicial $k$-algebras and $k$-module spectra. Do you know of a characterization of the $\infty$-category of simplicial $k$-algebras though? –  Akhil Mathew Jan 10 '13 at 14:08
What do you mean by "characterization of the $\infty$-category"? –  André Henriques Jan 10 '13 at 14:37
I'm looking for something analogous to "spectra are the free stable $\infty$-category (with colimits) on an object" and "$E_\infty$-algebras are commutative algebra objects in spectra." –  Akhil Mathew Jan 10 '13 at 14:45

This is based on conversations with Tom Goodwillie about Tyler and Jacob's answer.

André points out an analogous situation: topological abelian groups are to spectra as simplicial commutative rings are $E_\infty$-algebras. A third example is the LMS theory of $E_\infty$-$G$-rings, which refine $E_\infty$-rings in the category of $G$-spectra, but in a different direction. One motivation for the two ring examples is the units; and this is a way to tell that they really are different theories: The units of an $E_\infty$-$\mathbb Z$-algebra form a spectrum, while the units of a simplicial commutative ring are lifted to chain complexes; similarly, the units of a commutative $G$-ring are lifted to $G$-spectra, while the units of $E_\infty$-algebra in $G$-spectra is something weaker, maybe a diagram of fixed spectra.

Jacob gave two ways in which symmetric powers enter. In (1), like Tyler, he looked at the monad of simplicial commutative rings over $\mathbb Z$-modules, while in (4), he looked at the monad of simplicial commutative $A$-algebras over $A$-modules as recording the simplicial commutative structure on $A$. In general, not just for $\mathbb Z$, this monad is built of symmetric power operations. Just as an $E_\infty$-$A$-algebra structure on $B$ involves lifting the multiplication $B^{\otimes n}\to B$ to $(B^{\otimes n})_{h\Sigma_n}$, the tensors being over $A$, the simplicial commutative ring structure lifts it further to some kind of symmetric power functor. Since an $E_\infty$-algebra $A$ can lift in multiple ways to a simplicial commutative ring, this shows that the symmetric power operations on $A$-modules are not determined by the symmetric monoidal structure on $A$-modules. Thus, as Tyler says, we need extra structure on $\mathbb Z$-modules. In fact, there is a unique such structure, since there is a unique simplicial commutative ring structure on $\mathbb Z$, but to understand what the structure is, we should probably understand it on other categories of modules.

Recall that putting an $E_\infty$ structure on an associative algebra $A$ is equivalent to putting a symmetric monoidal structure on its category of modules. Is there a refinement of this structure on the modules that is equivalent to a simplicial commutative ring structure on the algebra? I have seen no proof, but I imagine that phrased this way it is formal. The question is whether this structure can be understood, perhaps in terms of generators and relations. The point of my answer is to suggest a better set of generators.

Symmetric power operations are necessary, since they are constituents of the monad, but I think they are secondary. On the category of modules of an $E_\infty$-algebra there are symmetric power operations $M\mapsto (M^{\otimes n})_{h\Sigma_n}$, but they are a consequence of the symmetric group action on $M^{\otimes n}$. I interpret André's example as lifting the space with $\Sigma_n$-action to a $\Sigma_n$-space, that is, a diagram of fixed spaces (eg, $(Y^{\Sigma_2} \to Y) = (\Delta\colon X\to X^2)$). From the $\Sigma_n$-spaces, one can extract the quotient objects to build the monad, but I expect the coherence is nicer than in the case of quotients, although I don't expect this to be a good answer, but more of a hint in the right direction.

That is, I think that there should be a refinement of a symmetric monoidal category in which the $n$th power functors coherently lift from objects with $\Sigma_n$-action to $\Sigma_n$-objects. Then one can extract symmetric power operations and thus define commutative monoids respect to this structure. There should be a trivial refinement in which the action is free and the commutative monoids recover the $E_\infty$-objects. On a category whose tensor product is the categorical product, there should be two structures, one trivial giving $E_\infty$-objects and a canonical non-trivial one built from diagonals giving very commutative objects. Unfortunately, when one passes from spaces to spectra, things get more complicated; if I recall correctly, and one may need to replace diagrams of fixed point objects with an analogue of full-fledged $\Sigma_n$-spectra, certainly for $G$-rings and maybe for simplicial commutative rings. Regardless of whether we can understand coherence, we should understand how much structure we can put on the $n$th power funtors.

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@Ben: thanks for these comments. I have been wondering for a little while what the role of symmetric powers is in homotopy theory, especially after reading the discussion at mathoverflow.net/questions/37647/…. Although I do not know equivariant homotopy theory, I believe that the outcome of that discussion was essentially that for a space $X$, the power $X^n$ can be promoted to an "honest" $\Sigma_n$-space. (Elmendorf's theorem states that to do this, you need to specify the fixed points of subgroups, but these are iterated diagonals.) –  Akhil Mathew Jan 11 '13 at 0:04
In particular, the construction of the infinite symmetric product in the Dold-Thom theorem is something that makes sense in any $\infty$-category, though using the cartesian product. Presumably this is related to André's answer above about "very commutative" objects. –  Akhil Mathew Jan 11 '13 at 0:07
<<Recall that putting an E∞ structure on an associative algebra A is equivalent to putting a symmetric monoidal structure on its category of modules.>> I'm confused: I know how to put an E∞ structure on the category of modules over the ring of 2x2-matrices (over a field), but that ring is not commutative, and in particular not E∞. –  André Henriques Jan 11 '13 at 10:49
An $E_\infty$-structure with the property that the ring itself is the unit. –  Ben Wieland Jan 11 '13 at 16:21
I think I agree with your approach, but I suspect that asking for the target category to be "$\Sigma_n$-objects" may be underemphasizing the difference between nonequivariant and equivariant homotopy theory. It seems to me that the target category itself should probably be part of the data. (Side question that I've asked several people: what's the most refined structure that you can put on a smash power over $R$, where $R$ is some commutative object?) –  Tyler Lawson Jan 12 '13 at 6:54

I have nothing non-trivial and non-digressive to say, but it might help to point out in an elementary way some things that may be relevant. One way to think about things is that there are distinctions in algebra that lack equivalents in spectra. This relates to answers from the $\infty$ category point of view of the original question, but is perhaps more explicit and concrete.

The statement of the question might be a little confusing, since there are perhaps six rather than two natural categories to consider, so let me pedantically make a fuller list, with everything over some commutative ring $k$. As a joke start, notice that simplicial commutative algebras are the same as commutative simplicial algebras. However, simplicial $E_{\infty}$ algebras make no sense (since we are not thinking about a homotopy theory on plain algebras), whereas $E_{\infty}$ simplicial algebras might make sense: as I understand it, that is where the problem about symmetric powers enters.

(1) commutative DG $k$-algebras

(2) $E_{\infty}$ DG $k$-algebras

(3) commutative simplicial $k$-algebras = simplicial commutative $k$-algebras

(4) $E_{\infty}$ simplicial $k$-algebras?

(5) commutative $Hk$-algebras in any good category of spectra.

(6) $E_{\infty}$ $Hk$-algebras in any good category of spectra

The term $k$-module spectra in the question and some answers means $Hk$-module, I presume, where $Hk$ is the Eilenberg-MacLane spectrum for $k$. I find it helpful to maintain a notational distinction. Mandell proved that $E_{\infty}$ $k$-algebras (algebra) are equivalent to $E_{\infty}$ $Hk$-algebras (topology). That is, (2) and (6) are equivalent: the algebraic and topological notions of $E_{\infty}$ are equivalent, of course restricting the latter to Eilenberg-MacLane algebras.

In algebra, the evident forgetful functor from commutative DG $k$-algebras to $E_{\infty}$ $k$-algebras is an equivalence for fields of characteristic 0 but not in general otherwise. That is, (1) and (2) are not equivalent.
Analogously, if sense can be made of (4), then there is a functor from commutative simplicial $k$-algebras to $E_{\infty}$ simplicial $k$-algebras which is not an equivalence.

In topology, the world of spectra, the smash product builds in higher homotopies and there is no distinction between commutative $Hk$-algebras and $E_{\infty}$ $Hk$-algebras: (5) and (6) are equivalent. Andre's answer is an $\infty$-categorical version of this: "in general, there is nothing more that $E_{\infty}$ to ask for". But the real topological reason in the case of spectra is a miracle of the modern theory of spectra. For good spectra in any good category of spectra, the natural map from the homotopy symmetric power to the symmetric power

$(E\Sigma_n)_{+}\wedge_{\Sigma_n} X^n \longrightarrow X^n/\Sigma_n$

is an equivalence, where $X^n$ is the $n$-fold smash power. (I think I've advertised this in answer to some other question, but it is worth repeating.) One cannot expect anything like this in algebra, except when working over a field of characteristic $0$.

In the model categorical sense, there does not seem to be a good homotopy theory of commutative DG $k$-algebras (ignoring the trivial rational case), and by analogy one does not expect such a good homotopy theory of simplicial commutative algebras. There is a good model theoretic homotopy theory of $E_{\infty}$ DG-algebras.

In the non-commutative case, the question has been answered, mainly by Brooke Shipley.
With the word commutative deleted and $E_{\infty}$ replaced by $A_{\infty}$, I'm pretty sure that all 6 homotopy categories make sense and are equivalent.

Several people have mentioned equivariant homotopy theory. Akhil, the cartesian power (or smash power in the based context) $X^n$ of a space $X$ is clearly a $\Sigma_n$-space; nothing to that. Maybe you are thinking of genuine spectra. Either way, Elmendorf's theorem is not particularly relevant. The earlier question you referred to asked for a homotopy colimit version of the infinite symmetric power of spaces. The equivalence above can be jacked up to say that in modern categories of spectra, the naive homotopy colimit and categorical colimit construction of infinite symmetric powers of spectra give equivalent answers, which is bizarre from a pre-1990s view of homotopy theory. There are interesting old-fashioned symmetric powers of spectra, and they are different. I don't understand them in the modern world.

In the world of $G$-spectra, things are much more interesting. There are many types of operadically defined commutative ring $G$-spectra. This is new work starting with Hill and Hopkins and continued by Blumberg and Hill, as yet not written down (at least not yet for public consumption). The comparison between algebra and topology here is fascinating and will be relevant to representation theory as well as to algebraic topology, or so I think. The old work by Lewis and myself is that part of the story that most directly mimics the nonequivariant case. I think I see how to understand equivariant units in the new context, and there is a relevant paper by Santhanam in the "classical" context, but there is more to be said even there, part of a thorough treatment of equivariant infinite loop space theory now in progress at Chicago.

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Thanks for this answer! I did not know that newer categories of spectra were designed to build higher homotopies into the smash product. –  Akhil Mathew Jan 12 '13 at 2:44