Most of the literature considers the standard model category structure on (graded) commutative differential algebras. But this generalizes to all (not-necessarily commutative) dg-algebras.

Details and references are listed here: nLab: model structure on dg-algebras.

What is known about how commutative dg-algebras sit inside all dg-algebras intrinsically ?

Do we have, for instance, a Quillen inclusion of the CommutativeDGAlgs into DGAlgs, such that its adjoint is a localization?

  • $\begingroup$ I think this is a very interesting question, but I'm not sure about the term localisation in this context, could you elaborate? I also wonder what happens when you forget char 0 and instead use simplicial algebras. It's not true in the commutative case, but is it true in the associative case that simplicial algebras are Quillen equivalent to dg-algebras? If so, how do simplicial commutative algebras sit in dg-(associative)-algebras? I think there must be subtleties that I'm missing because wouldn't that be computationally nice? $\endgroup$ – James Griffin Nov 11 '09 at 14:35
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    $\begingroup$ By localization I was thinking of <a href="ncatlab.org/nlab/show/Bousfield+localization">Bousfield localization</a> of model categories. But I was intentionally being a bit vague, because I really just want to know what one can say about the inclusion of CommDGAlg into all DGAlg and I didn't want to presuppose too much of a possible answer. But, as pointed out below, it's not a Bousfield localization. $\endgroup$ – Urs Schreiber Nov 12 '09 at 13:53
  • $\begingroup$ Sorry, I forgot that I can't use hrefs here: the missing bit should have said "I was thinking of Bousfield localization" and should have provided this link to further details: ncatlab.org/nlab/show/Bousfield+localization $\endgroup$ – Urs Schreiber Nov 12 '09 at 13:55

OK, first of all we had better be working over a field of characteristic 0 if we expect commutative DGAs to be a model category where the weak equivalences are homology isomorphisms and everything is fibrant. Otherwise it is impossible.

In this case, it is true that the forgetful functor from commutative DGAs to DGAs preserves fibrations and weak equivalences, so is a right Quillen functor. Its left adjoint of abelianization is therefore a left Quillen functor.

However, it is definitely not true that commutative DGAs are a localization of DGAs by this adjoint pair; the map from a DGA to its abelianization is not going to be a weak equivalence, as it won't be a homology isomorphism.

As I recall, an abelian object in the category of DGAs is not a commutative DGA; it is instead a square 0 extension. So if you do Quillen homology (derived functors of abelianization) in DGAs, you are supposed to recover Hochschild cohomology of DGAs, I believe.

On the other hand, it does seem to me that you could try to declare the map from a DGA to its abelianization to be a weak equivalence. I don't know what would happen then. Mark Hovey

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    $\begingroup$ You refer to the left adjoint as "abelianization", but it does not take a DGA to a commutative group object in DGAs (and unless you assume objects are augmented, the only such is the zero object), but rather the "commutativization" where you take the quotient of the DGA by all commutators. Your statement seems to be that this should preserve weak equivalences between cofibrant objects. If R is commutative and T -> R is a fibrant replacement, this factors through T_{comm} -> R; is this generally a weak equivalence? If not, what are the derived functors? $\endgroup$ – Tyler Lawson Nov 11 '09 at 14:17
  • $\begingroup$ I've heard it said a number of times that the derived functor of abelianisation is Hochschild homology, but in that case shouldn't it be functorial on algebras, which it classically isn't? $\endgroup$ – James Griffin Nov 11 '09 at 14:18
  • $\begingroup$ Sorry Tyler I didn't see your comment before posting mine. To address yours: Call the inclusion functor from comm-dg-algebras to dg-algebras I. It's left adjoint is the abelianisation functor (factoring out by commutators), call this A. Then A.I is the identity functor. Since this "derives to the identity functor" and since I does preserve weak equivalences then am I right in saying that A must preserve weak equivalences? If so then that's a "yes" to your first question. $\endgroup$ – James Griffin Nov 11 '09 at 14:30
  • $\begingroup$ Thaks for all the replies! I wish I could reply more globally, but I can't here on this forum, therefore this "comment" here. meanwhile I found out that every E-oo object in cochain complexes is weakly equivbalent as an E-oo object to a commutative dg-algebra. This suggests a different angle on the question: how does the forgetful Quillen functor from commutative dg-algebras to all dg-algebras relate to the forgetful (oo,1)-functor form E-oo objects to A-oo objects in cochain complexes? $\endgroup$ – Urs Schreiber Nov 11 '09 at 21:03

I don't know how to make a follow up as opposed to a new answer, but this is a follow up to my preceding answer and the comments on it.

I believe that the commutativization functor C, defined by C(A) = A/[A,A], does NOT preserve weak equivalences, and I believe that Tyler is right and there are derived functors, but I don't know what they are.

Here is a simple example. Take the CDGA A=Q[x]/(x^2), where x is in degree 2. If I take a cofibrant replacement B for this in the category of DGAs, I will need a generator x in degree 2 and a generator y in degree 5 in order to kill x^2.

But then x^3 will be killed by both xy and yx, which are different elements in the tensor algebra on x and y that I have so far.

So I think I need a new tensor generator z in degree 6 to kill xy - yx.

When I apply commutativization C, xy-yx will go to 0, but z won't. So z will be a cycle that is not a boundary in C(B), so the map

from C(B) to A will not be a weak equivalence.

The answer to James's question is that C has to preserve weak equivalences between cofibrant objects, but not necessarily all weak equivalences, as apparently it does not.

So I have lots of questions myself about this now. Is it even true that the homotopy category of CDGAs is a full subcategory of the homotopy category of all DGAs? You think it must be, because if you start with a CDGA you can think of it as a DGA. But we have just determined that if you start with A, think of it as a DGA and take a cofibrant replacement B, then C(B) is not weakly equivalent to A. So you can't use this argument.

  • $\begingroup$ Secondly, I would be skeptical about CDGAs being a full subcategory of the homotopy category of DGAs, as (from characteristic p considerations) I would have expected there to be multiple inequivalent commutative DGAs whose cofibrant replacements (as associative DGAs) became equivalent. But in characteristic zero there are no Steenrod operations and so I'm not sure how one might find some invariant not detected on the associative level. $\endgroup$ – Tyler Lawson Nov 11 '09 at 20:44
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    $\begingroup$ You mean A/A[A,A]A, rather than A/[A,A], right? (If we mod out the linear span of commutators, this does not really make thinks commutative [and derived functors compute cyclic homology, I believe].) The derived functors of this are briefly discussed (for some examples) in arxiv.org/abs/math/0610410. $\endgroup$ – Vladimir Dotsenko Nov 11 '09 at 23:32
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    $\begingroup$ I would definitely mean A/A[A,A]A, yes, since we have to get a CDGA out of our DGA A, so we need to mod out by the differential ideal generated by the commutators. The reference you gave refers to the derived functors of this as "higher abelianization", but as you say the discussion is very brief, and there are no references to anywhere it is discussed in depth. Tantalizing! Thanks for the reference. $\endgroup$ – Mark Hovey Nov 12 '09 at 14:29
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    $\begingroup$ The homotopy category definitely doesn't embed fully faithfully. Maybe the easiest example (working over a field C) is to take the usual polynomial ring R = C[x,y]. In commutative dgas this is already cofibrant, so [R,A] = H_0(A) x H_0(A) for any cdga A. In associative dgas a cofibrant replacement for R is the free associative algebra on x,y and z in hom. degree 1 with dz=xy-yx. So if A happens to be a cdga you get [R,A] = H_0(A) x H_0(A) x H_1(A). $\endgroup$ – Jacob Lurie Dec 28 '10 at 12:29

By the adjoint functor theorem the inclusion has a left adjoint iff it preserves limits, which it does, if I am not completely wrong. Weak equivalences and fibrations are exactly the same (defined via the underlying complexes) in both categories, so the inclusion functor does preserve fibrations and trivial fibrations, so you have a Quillen inclusion.

The adjoint is a localization in the sense that you could equivalently localize the category of dg-algs, by declaring the maps from a dg-algebra to its abelianization to be an equivalence and inverting those. By the universal property of the localised category you have a functor to the category of commutative dg-algebras and this functor should be an equivalence. But here you have an actual localization not a Bousfield localization.

My impression is that you can take the model structure on dg-Alg and Bousfield localize it with the above additional weak equivalences. Conveniently everything is fibrant, so the local objects would be exactly the commuative dg-algebras. Probably this was your actual question, but I am not sure about this - better wait for a hotshot to confirm my guess or tear it into pieces...


I can second Mark's example, however I would point out that the cocycle z lies in degree 8 and not 6. There's another one, w in degree 11 whose purpose is to kill y^2, which is 0 in the abelianisation. Not to mention xz in degree 10. The resultant commutative algebra seems to be of the form k[x]⊗k[V], where V has some kind of algebraic structure, I'm guessing a (co)lie-module after a suspension, but I can't work out what it should be.

I also have an interpretation of all of this. In working out an associative quasi-free presentation (TW, δ) we're actually working out the bar homology W of the algebra (actually including the A-infinity coalgebra structure as well). But there's a decomposition of the bar homology of a commutative algebra known as the λ-decomposition (see for instance Loday's Cyclic Homology book). If we were to take the "middle" piece W' of this decomposition we would get the commutative bar homology (and it's associated L-infinity coalgebra structure). And this is just what we need to get the resolution of the original algebra as a quasi-free commutative algebra (SW', δ'). All the extra bits (the V of the example) come from other pieces of the λ-decomposition. I think that this should impose strong structural conditions on (SW, δ'').

Going back the example we can work out that z is in the second part of the decomposition, although I'm not sure about w.

So what this means for the various homotopy functors between the homotopy categories in question I'm not sure, the following is speculative: It seems to me that it indicates that they're not full. There are homotopy morphisms from k[x]/x^2 to another commutative algebra that are not homotopy commutative algebra morphisms. They should fit into the λ-decomposition and their position there should indicate just how "not commutative" they really are.

Final note on "not commutativity": this notion can be made more rigorous using operads, but in a hand-wavey way it just means to what level the higher homotopies of the commutativity condition hold.

  • $\begingroup$ A morphism in the homotopy category of DGAs from a CDGA A to a CDGA B is a zig-zag C --> A and C --> B, where C --> A is a homology isomorphism and C is any DGA. So it makes sense that you should be able to put in noncommutative C's and get more morphisms from A to B then you would in the homotopy category of DGAs. But I would not be surprised if the functor were not faithful either. $\endgroup$ – Mark Hovey Nov 12 '09 at 16:53

in the rational case, the forgetful functor U: dgCAlg-------> dgAlg induces a faithful functor in homotopy category under the following conditions. R is connected (connective) differential commutative graded Q-algebra and S= C(X,Q) (the cochaine complex of a space with coefficients in Q, then

Ho(dgCAlg) (R,S)-----> Ho(dgAlg)(UR,US) is injective.


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