Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Ok, this question is much less ambitious than it might sound, but still:

Two commutative differential graded algebras (cdga's) are quasi-isomorphic if they can be connected by a chain of cdga quasi-isomorphisms. There is a similar definition for not necessarily commutative differential graded algebras (dga's).

  1. If two $\mathbf{Q}$-cdga's, $A$ and $B$, are quasi-isomorphic as dga's, are they necessarily quasi-isomorphic as cdga's? I suspect that the answer is no, but don't know any counter-examples, nor can prove that such counter-examples exist.

  2. Same question as 1 when $A$ and $B$ are the Sullivan $\mathbf{Q}$-polynomial cochain algebras of simply connected compact polyhedra. In other words, is the "rational noncommutative homotopy type" of compact simply-connected polyhedra the same as the usual rational homotopy type?

share|improve this question
1  
Do you want a cdga to be strictly commutative or differential-commutative? I'm thinking that Tyler answered the question with the former meaning, but I'm not sure. –  Greg Kuperberg Dec 17 '09 at 14:27
    
(I removed said answer because it had a serious error.) –  Tyler Lawson Dec 17 '09 at 14:41
    
Greg -- cdga's I'm interested in are graded commutative: $ab=(-1)^{deg a\deg b}ba$. –  algori Dec 17 '09 at 15:26
add comment

6 Answers 6

I think that the answer for the first question is "yes". I disagree with the Joey's argument because he uses a wrong definition of $Comm_\infty$ morphism. Let me recall the right definition first. A $Comm_\infty$ algebra $C$ is a coderivation $Q$ such that $Q^2=0$ on cofree LIE coalgebra cogenerated by $C[1]$. So, it should be a free Lie coalgebra, not COMMUTATIVE one. Now a $Comm_\infty$ morphism is just a map of from the two cofree Lie coalgebras which commutes with $Q$. It is a quis if it is, moreover, a quis of complexes.

Well, now suppose that $C$ and $D$ are two cdga placed in any degrees. If they are $A_\infty$ quis, it means that there is a map of cofree dg coalgebra cogenerated by $C[1]$ to the same guy cogenerated by $D[1]$. This is a map of coalgebras, then it defines a map between their primitive elements.

(Recall $x$ is primitive iff $\Delta x=1\otimes x+x\otimes 1$). Apriori this gives a map of free Lie coalgebras, but NOT dg Lie coalgebras. Moreover if $C$ and $D$ would be non-commutative, the primitive elements would be not closed under the differential. But if $C$ and $D$ are commutative, the maps of the Lie coalgebras of primitive elements is a map of dg Lie coalgebras.

It is only remains to note that this map of dg Lie coalgebras is quis; then it is by definition a $Comm_\infty$ quis map. But this easily follows from the PBW theorem. The tensor coalgebra as a vector space is the symmetyric (co)algebra of the free Lie coalgebra. If the map of Lie coalgebras would be not a quis, the corresponding map of their symmetric powers also would be not a quis. This would contradict that the initial map was an $A_\infty$ quis.

We are done.

share|improve this answer
    
Thanks, Boris! This seems to work. –  algori Dec 21 '09 at 0:07
    
I agree with this answer, but have two questions. 1) In order to define the primitive elements we need a unit, the bar complex does not usually have a unit, so where does this come from? 2) Concerning the final paragraph, we only need the map to be quis on the cogenerators, right? The bar complex is acyclic for a unital algebra, but we don't consider every unital algebra to be quis to 0. –  James Griffin Dec 22 '09 at 20:37
add comment

I can say that if A and B are commutative DGAs over Q which are connective (their homology groups are zero in negative degrees, using homological grading), then A and B are equivalent. The only proof I can come up with is a Postnikov argument based on the Hochschild-Kostant-Rosenberg theorem.

Namely, let PnA and PnB be Postnikov stages, which have the same homology as A and B respectively in degrees less than or equal to n and zero above; you can construct these by adding new polynomial generators to A and B with differentials that erase higher homology classes. Assume you've already constructed an equivalence from PnA to PnB.

Then Pn+1A and Pn+1B are classified by their next k-invariants. In the associative case this is an element in the Hochschild cohomology of PnA with coefficients in a shift of the module M = Hn+1A = Hn+1B, and similarly for B. In the commutative case this is an element in the Andre-Quillen cohomology instead. These are classified by maps from the Hochschild homology (strictly speaking, its augmentation ideal) and Andre-Quillen homology (a shift of the derived Kahler differentials) to this shift of M.

Since we're in characteristic zero, the HKR theorem tells us that the Hochschild homology is the (derived) free graded-commutative algebra over PnA on the Andre-Quillen homology. In particular, there is a retraction down. This implies that the collection of possible commutative k-invariants constructing an extension of PnA by M are a subset of the possible associative k-invariants.

As a result, if Pn+1A and Pn+1B have a fixed equivalence as associative algebras, that implies that their associative k-invariants are equal. Hence that their commutative k-invariants are equal, so this allows us to lift to an equivalence of commutative algebras.

This is a sneaky problem. The fact that I needed to assume connectivity for this argument is pretty annoying, and I am not sure whether it is material or simply a failing on my part to get something general. Perhaps someone else can do better.

ADDED LATER: I believe that the same basic argument works in the coconnective case under added hypotheses (e.g. the algebra being simply connected, having only Q in degree 0 and 0 in degree 1). Unfortunately you cannot construct Postnikov stages quite as easily for general cochain algebras.

For example, let A be the commutative DGA Q[x,y,s,t] ⊗ Λ[w], a free algebra on generators x,y,s,t in degree zero and z in (cohomological) degree 1, satisfying dx = dy = dw = 0, ds = xw, dt = wy. Then there is a Massey product <x,w,y> = sy - xt in cohomology degree 0 which is nonzero, and any map from A to a DGA which has zero cohomology in positive degrees must necessarily send w to zero, and hence also this Massey product.

I have this vague feeling that in this simply-connected case (and also in the connective case) there might be a much shorter argument involving minimal models.

share|improve this answer
    
Thanks, Tyler. Do you suppose one could make this work in the coconnective case instead? This corresponds to cochain algebras, which is what I'm mainly interested in. –  algori Dec 18 '09 at 3:55
    
I added some comments about the coconnective case above - roughly, I think it probably works under extra hypotheses. –  Tyler Lawson Dec 18 '09 at 13:45
    
ok good I have enough points to directly comment here ... As I mentioned below this is an open problem, so it would be nice to have a solution somewhere to refer to. –  Don Stanley Jan 28 '10 at 16:41
add comment

I think the answer to question 1 is "Yes." Here's why:

Fix a field k of characteristic 0 (you asked about k = Q).

Let C and D be (graded-)commutative DG k-algebras. Then there exists a noncommutative quasi-isomorphism from C to D iff there exists an $A_{\infty}$ Quasi-iso $F:C \to D$. (In fact more is true---there is a one-to-one correspondence between equivalence classes of noncommutative quasi-iso's and equivalence classes of $A_\infty$ morphisms.)

Since there exists a commutative quasi-isomorphism from C to D iff there exists a $C_{\infty}$ quasi-iso $G: C \to D$, your question can be rephrased as follows: "Suppose C,D are commutative DG k-algebras, and $F: C \to D$ is an $A_{\infty}$ map. Does this imply the existence of a $C_{\infty}$ map?"

We can describe $A_{\infty}, C_{\infty}$ maps as follows: for a dga A, we can consider the cofree (conilpotent) coalgebra $T^c(A)$, and we observe that the dga structure maps $d: A \to A, \cdot: A \otimes A \to A$ define a map $T^c(A) \to A$ by projecting onto $A \oplus (A \otimes A)$ and then doing $d$ on the first factor and $\cdot$ on the second. This map can be extended as a coderivation $D_A:T^c(A) \to T^c(A)$, and the claim is that $D_A^2 = 0.$ An $A_{\infty}$ map $F:A \to B$ is by a definition a dg-coalgebra map $F:(T^c(A), D_A) \to (T^c(B), D_B)$. Similarly, a commutative dga gives rise to a cofree cocommutative dg coalgebra, using $S^c$, and $C_{\infty}$ morphisms are dg-coalgebra maps between these. An $C_{\infty}$ map is actually an $A_{\infty}$ map with some symmetry properties (since $S^c \hookrightarrow T^c$).

Here's a first step in the argument "up to signs" that you can symmetrize the $A_{\infty}$ map F.

To get a $C_{\infty}$ map from F, write $F = f + f_2 + f_3 + \cdots$, where $f_n : C^{\otimes n} \to D$. Then the condition that $F$ is an $A_{\infty}$ morphism gives $(df_2)(x,y) = f(x) \cdot_{D} f(y) \pm f(x \cdot_{C} y)$ where $df_2$ should be understood as the differential in the Hom complex $Hom(C^{\otimes 2}, D)$. Since the products in C and D are commutative, we actually get: $(df_2)(x,y) = f(x) \cdot_{D} f(y) \pm f(x \cdot_{C} y) = \pm f(y) \cdot_D f(x) \pm f(y \cdot_C x) = (df_2)(y,x)$ so we see that at least $(df_2)$ is graded symmetric. Since k has characteristic 0, we can define

$g_2 := \frac{1}{2} \left( f_2 + f_2^{op} \right)$

and since

$(dg_2) = \frac{1}{2} d \left( f_2 + f_2^{op} \right) = \frac{1}{2} \left( df_2 + d(f_2^{op}) = \frac{1}{2}( df_2 \pm (df_2)^{op} ) = (df_2) \right)$

we see that $dg_2$ still provides the homotopy between $f( \cdot_C )$ and $f( ) \cdot f( )$.

I think an argument along these lines will also provide suitable $g_n$ for $n \geq 3$.

I wonder if someone can provide an "abstract nonsense" answer along the following lines: the functor i: Commutative DGAs -----> nonCommutative DGAs has a left adjoint which preserves quasi-iso's (does abelianization preserve maps inducing isomorphisms in Homology?), and so we can verify that the image of Commutative DGAs in (nonCommutative DGAs)$[W^{-1}]$ has the universal property that a localization (Commutative DGAs)$[W^{-1}]$ should have... (Where $W$ above should be taken to be the collection of maps inducing iso's in homology in each category).

share|improve this answer
    
Thanks, Joey! Indeed, if abelianization preserved weak equivalences, we would be done, but unfortunately it doesn't, see Mark Hovey's comments here mathoverflow.net/questions/5031/… –  algori Dec 18 '09 at 3:41
    
Right, that makes sense. I think my first argument will work. –  Joey Hirsh Dec 18 '09 at 6:56
    
This answer's fine and roughly equivalent to Tyler's but without needing the full power of HKR. I would say that it gets to the guts of the argument. The only thing lacking is the higher invariants. I doubt the existence of a better abstract nonsense argument because this really relies on a certain spliting that just happens to occur because we have a char 0 field. –  James Griffin Dec 18 '09 at 12:07
    
Can't seem to comment on Boris' answer, so I'll just say here that I agree with his corrections. Thank you. –  Joey Hirsh Dec 21 '09 at 7:29
add comment

As far as I know for positively graded connected and simply connected complexes with differential of degree $+1$ even in characteristic $0$ (corresponding to the rational homotopy theory case), this is an open problem.

So it would be nice to have a published paper that settles it.

share|improve this answer
    
Thanks, Don! Maybe one of the approaches suggested in this thread can be turned into a proof... –  algori Jan 28 '10 at 11:32
    
Is there a reference somewhere in the literature to this problem? –  Tyler Lawson Jan 28 '10 at 18:09
    
not that I know of –  Don Stanley Jan 28 '10 at 21:25
add comment

This is an attempt to rephrase some of Joey's answer and also to add in the higher operations required.

Suppose we have two cdgas A and C and that they're connected by an $A_\infty$ morphism. This is encoded as a map from $BA$, the associative bar construction on A to C. A $C_\infty$ morphism is encoded by a morphism from the commutative bar construction, $B_cA$, this takes the form of a quasi-free colie algebra on A. There is a natural transformation from the free coassociative algebra functor to the free colie algebra functor and this coincides with a natural transformation from the associative to commutative bar functor. This is ofcourse the wrong direction, the natural transformation says "every $C_\infty$ morphism is an $A_\infty$ morphism".

To go the other way we observe that the maps $BA\rightarrow B_cA$ split. I wont offer a proof here but the proof goes by constructing certain idempotents, it's known as the Hodge decomposition of the bar complex of a commutative algebra. We now have the composition

$B_cA\rightarrow BA \rightarrow C$

which is our $C_\infty$ morphism. Warning The $A_\infty$ morphism we get from this $C_\infty$ morphism is not necessarily the one we started off with, we have just stripped away all the "stuff" coming from other parts of a Hodge filtration.

Finally I'd like to add that your intuition was broadly right, there will be examples of non char 0 cases where the answer to your question will be "no".

share|improve this answer
add comment

I think I was wrong in my answer. The matter is that I have a map of dg associative coalgebras, it indeed induces a map on primitive elements, but these primitive elements are the cogenerators, that is elements of $A[1]$, not of $Lie(A[1])$. So I can not deduce that I have an induced map of free dg Lie coalgebras. I do not know is it the case or not, but my speculation certaily fails.

Dear James, concerning your question 2). Here is not any mistake. The difference with bar complex case is that it is not tensor power of generators as COMPLEX. In commutative case, the free Lie (co)algebra is closed under the differential, and the entire complex is the symmetric power (by PBW) of this Lie (co)algebra AS A COMPLEX. That is the only component of the differential comes from this free dg Lie (co)algebra, and we have the symmetric power of a complex. So the claim reads like: if two complexes are quis, their symmetric powers also are.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.