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It is known that the rational homotopy theory of spaces (e.g. simplicial sets) is equivalent in some sense to the homotopy theory of cdgas over $\mathbb{Q}$. This has been expressed in various forms in the literature. For instance, Felix-Halperin-Thomas show that homotopy classes of maps between simply connected rational spaces with finite-dimensional homology in each dimension can be computed via homotopy classes of cdga maps between Sullivan models for each of them. However, there is a stronger statement that I would like to be true, have heard asserted (without proof) that it is true, but haven't been able to figure out: I would like for that statement to be true without finite-dimensional hypothesis.

Consider the following two model categories:

First, we can take simplicial sets with the usual (Kan) model structure, and then left Bousfield localize at the class of "rational homology equivalences." In other words, the model structure is such that the cofibrations are the injections, weak equivalences are maps inducing isomorphisms on $H_*(\cdot, \mathbb{Q})$, and everything else is determined. That this model structure exists follows from a combination of the small object argument and the Bousfield-Smith cardinality argument (or one can probably appeal to general facts on existence of Bousfield localizations).

Second, we can take commutative dgas (nonnegatively graded) over $\mathbb{Q}$. The model structure is obtained by transfer from a slight variant of the model structure on nonnegatively graded chain complexes. In other words, a fibration of cdgas is a surjection, and a weak equivalence is a quasi-isomorphism. The cofibrations are thus determined; there is a standard generating set (basically, what one gets by applying the free functor to generating sets for chain complexes).

Edit: As Tyler Lawson observes below, this is not actually a model structure. I am not sure at the moment what the right one is. Perhaps we should relax "surjections" to "surjections in degrees $\geq 1$." Alternatively, we could consider the model category of all cdgas (in which case the functor below is obviously not anywhere near a Quillen equivalence).

Now, we have a Quillen adjunction between the first and the opposite of the second, which sends a simplicial set $X_\bullet$ to the "polynomial de Rham algebra" $A^{PL}(X_\bullet)$, a cdga which is quasi-isomorphic to the rational cochain algebra (and which, in particular, solves the commutative cochain problem).

Q1: Is this a Quillen equivalence? (As Tyler Lawson points out, there is a simple reason why this is not the case: namely, that elements in $H^0$ could be nilpotent. Is this, however, the "only" reason why it fails? For instance, what if one restricts to cdgas that are "connected," i.e. have only $\mathbb{Q}$ in degree zero?)

I have not seen this statement in what I've read (not that much). In fact, most authors seem to prefer to work only with simply connected spaces; the advantage is that there you can localize at the rational homotopy equivalences (which are the same as the rational homology equivalences by the mod $\mathcal{C}$ theory). Granted, simply connected spaces do not form a model category. Quillen dealt with it by working with 2-reduced simplicial sets (those with only one vertex and edge). Other authors (e.g. Felix-Halperin-Thomas) don't use the framework of model categories at all, but state something essentially equivalent to this in the case when one works with simply connected spaces with finite-dimensional homology groups.

Here's what I understand of the argument, and why I don't know how to extend it. The basic claim, as before, is that if $X_\bullet $ and $Y_\bullet$ are simplicial sets, with $Y_\bullet$ an abelian space and a rational Kan complex (if I'm not mistaken, these are fibrant in the model structure thus constructed), and $A, B$ are cofibrant (e.g. Sullivan) models in cdgas for $X_\bullet, Y_\bullet$, then homotopy classes of maps $[X_\bullet, Y_\bullet]$ are the same as homotopy classes of maps $ B \to A$. By using a Postnikov tower to express $Y_\bullet$ as a homotopy inverse limit under a whole bunch of fibrations with Eilenberg-MacLane spaces as fibers, we can assume that $Y_\bullet $ is a $K(V, n)$ for $V$ a vector space over $\mathbb{Q}$. (If I understand correctly, fibrations and homotopy inverse limits of spaces obtain good Sullivan models.) If $V$ is finite-dimensional, then we have an explicit Sullivan model (just a free (graded)-commutative algebra), and so we can compute homotopy classes of maps $B \to A$; they'll be the same as maps from $V$ into the cohomology of $X_\bullet$. But maps from $X_\bullet$ into an Eilenberg-MacLane space classify cohomology classes, so we're done.

But, this doesn't seem to work when $V$ is infinite-dimensional. I don't know what a good Sullivan model for a $K(V, n)$ is anymore. The cohomology in dimension $n$ is $V^\vee$, but, say, if $n$ is even, then the cohomology in dimension $2n$ should be $(V \otimes V)^{\vee}$, not what would be nice: $V^\vee \otimes V^\vee$. So, is this statement even true? At the very least, can we get some kind of equivalence of $\infty$-categories?

Quillen himself stated the result using a Quillen equivalence, but it's actually a somewhat complex series of them. He starts with reduced 2-simplicial sets, and then goes to reduced simplicial groups via the loop group construction, then takes the completed group ring dimensionwise to get a simplicial complete Hopf algebra, and then takes primitive elements to get a simplicial Lie algebra, and then applies the lax symmetric monoidal Dold-Kan functor to get dg-Lie algebras. So maybe dg-Lie algbras (or dg-coalgebras) are better than cdgas for describing rational homotopy theory.

Still, one thing that I would like to be true, but is not proved in Quillen's paper, is a direct Quillen equivalence starting with (localized) simplicial sets and ending in some algebraic category. The problem is, the Quillen equivalences I've described go in the opposite direction. The loop group is a left adjoint, but taking primitive elements is a right adjoint. So, while there is an honest Quillen equivalence between reduced simplicial sets and simplicial complete Hopf algebras, there is not proved (in this paper, as far as I can tell) the existence of a single Quillen equivalence (not a zig-zag) which ends in a category with no reference to topology.

Q2: Is there a direct Quillen equivalence of the rational homotopy category with a nice algebraic category?

For instance, it would be interesting if there was some coalgebra version of the de Rham complex.

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  • $\begingroup$ I find it very doubtful that you will find an identification Spaces=Algebras^op, but as I am not an expert I will leave my concerns as a comment. The point is that Spaces=Kan Simplicial Complexes is a (\infty,1) presentable category, whereas Algebras^op is not (Algebras is). My impression is that this is one of the main reasons to impose finiteness conditions. My impression moreover is that you can drop some of those finiteness conditions by using Coalgebras (which is presentable) rather than Algebras^op as your algebraic model. $\endgroup$ Commented Oct 27, 2011 at 20:22
  • $\begingroup$ (At least, it's clear that "cofree" takes simplicial sets to simplicial cocommutative coalgebras; recall that cofree is left-adjoint to forget. I'm less confident that I can write down a dg coalgebra. The alternating sum of face maps makes the cofree thing into a dg vector space, but it is not a derivation of the coalgebra structure, I think.) $\endgroup$ Commented Oct 27, 2011 at 20:34
  • $\begingroup$ Something like the following should work. I begin by trying to assign a dg coalgebra to each simplex, as follows. First, I assign a dg vector space, and the space I choose will be the one that computes (I think it's called) normalized homology: in homological degree $k>0$ I place the vector space spanned by the $k$-dimensional faces, and in homological degree $0$ I place the linear combinations of vertices with sum(coefficients)=0. This is a (co?)simplicial vector space, and hence a chain complex. Note that by construction it is an exact complex. Then I construct its cofree cocommutative (...) $\endgroup$ Commented Oct 27, 2011 at 21:01
  • $\begingroup$ (...) coalgebra. Recall that in characteristic $0$, the cofree cocommutative coalgebra of a vector space (or chain complex, or...) is the "symmetric algebra" but thought of as a coalgebra, not an algebra, and this works also in the category of chain complexes ("symmetric" and "cocommutative" and so on in the super sense). Then its homology is just Cofree(zero), which is the one-dimensional coalgebra. Ok, so I have assigned a cocommutative dg coalgebra to each standard simplex with the desired homology, and these package together to a cosimplicial cocommutative dg coalgebra. Then for any (...) $\endgroup$ Commented Oct 27, 2011 at 21:04
  • $\begingroup$ (...) space, thought of as a simplicial set, I can take the tensor product ("end" or "coend" or something) of my standard cosimplicial cocommutative dg coalgebra with your simplicial set. Since DG Coalgebras is tensored over Set in the natural way, this tensor product gives a dg coalgebra for each simplicial set. I think, but I haven't checked, that its homology is precisely what it should be, and that it plays a role analogous to PL Forms for a coalgebraic version. $\endgroup$ Commented Oct 27, 2011 at 21:07

2 Answers 2

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The answer to question 1 is no. For one thing, the functor to the category of commutative DGAs is not essentially surjective. For any space $X$, $H^0(X)$ is a product of copies of $\mathbb Q$, whereas it is relatively easy to rig up CDGAs that don't have this property.

More seriously, nonnegatively (cohomologically!) graded CDGAs don't form a model category using the structure you've listed. Try to construct a cofibrant replacement for $\mathbb Q[x]/x^2$ concentrated in degree zero, and you'll find that you want to adjoin classes in degree $-1$; being cofibrant means, by consideration of maps to DGAs concentrated in degree zero, that the algebra in degree zero has a strong lifting property that can't be altered.

One thing to note about question 2 is that the functor $\pi_0$ is still well-defined on the rational homotopy category, because rational equivalences are always isomorphisms on $\pi_0$. Any algebraic model that you construct, then, has to account for this. (This is not even beginning to worry about $\pi_1$, which always messes things up.)

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  • $\begingroup$ Thanks a lot for this. Embarrassingly, I was going to claim the model structure on cdgas as above in a talk in a couple of hours... $\endgroup$ Commented Oct 27, 2011 at 19:16
  • $\begingroup$ What's interesting is that Bousfield-Guggenheim seem to claim exactly that this is a model structure (see Def. 4.2 of "On PL de Rham theory and rational homotopy type"). Maybe I've misunderstood and one has to work in all degrees. $\endgroup$ Commented Oct 27, 2011 at 19:27
  • $\begingroup$ @Akhil: It is true without the boundedness assumption, which I don't see being made by Bousfield-Gugenheim. $\endgroup$ Commented Oct 27, 2011 at 21:08
  • $\begingroup$ (years later...) @TylerLawson, they do state the boundedness condition in the second paragraph of "§1The simplicial algebra nabla" (on page 1, total page 10). The claim of this model structure on dgca-s in non-negative degrees is reviewed for instance on pages 5-6 of Hess's arxiv.org/abs/math/0604626 . In your example, isn't $\mathbb{Q}[x]/x^2$ seen to already be cofibrant by having left lifting against quasi-isomorphic surjections? $\endgroup$ Commented Feb 22, 2017 at 10:02
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This is more like a comment, but the usual space allowed for comments is too small!

It seems (after Toën's work) that we cannot describe rational homotopy types in terms of algebra, but rather in terms of (derived) algebraic geometry; the fact that commutative dg algebras (i.e. affine derived schemes) are sufficient for nice simply connected spaces is really due to the lack of monodromy. This is one of the reasons why Toën developped his theory of affine homotopy types (partly following ideas which were already sketched in Grothendieck's Pursuing stacks): for any ring $k$ and any space $X$, there is the pro-unipotent completion of the $\infty$-groupoid corresponding to $X$, denoted by $(X\otimes k)^{uni}$, which is the higher analog of the pro-unipotent completion of a group. If $k$ is a field of characteristic zero, and if $X$ is $1$-connected, nilpotent and of finite type, then the stack $(X\otimes k)^{uni}$ is the spectrum (in the sense of derived algebraic geometry) of the commutative dg algebra of de Rham cohomology of $X$, so that this theory includes classical rational homotopy theory. The essential information given by the stack $(X\otimes k)^{uni}$ is essentially the $\infty$-category of $k$-linear local systems over $X$.

All this is explained in the paper

B. Toën, Champs affines, Selecta Math. (N.S.) 12 (2006), no. 1, 39-135.

(see in particular Cor. 2.4.11, Cor. 2.5.3 and Cor. 2.5.4, to see that this extends nicely classical rational homotopy theory to non-simply connected homotopy types).

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  • $\begingroup$ This is very interesting; thanks for the answer! I've been meaning to learn a little about derived algebraic geometry, and this is more motivation for me. $\endgroup$ Commented Oct 28, 2011 at 14:13

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