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

Question: Given two quasi-isomorphic dg commutative algebras (over a field of characteristic zero, if you like), to what extent do their various homological geometric data agree?

Example: Given a dg commutative algebra $A$, there is a dg Lie algebra $\operatorname{Der}(A)$ defined by understanding the notion of "derivation" internal to the category of dg vector spaces. If $A$ and $B$ are quasi-isomorphic dg commutative algebras, are $\operatorname{Der}(A)$ and $\operatorname{Der}(B)$ quasi-isomorphic dg Lie algebras? Note that any homomorphism $f: A \to B$ defines a (dg) vector space of "derivations relative to $f$", which is a bimodule for the Lie algebras $\operatorname{Der}(A)$ and $\operatorname{Der}(B)$ and receives maps as one-sided modules from each of these; one would expect these maps to be quasi-isomorphisms if $f$ is.

Remark: I intend my question to be somewhat open ended. As such, I would accept an answer that points me to the appropriate literature.

share|improve this question
You can speak of derivations $A\to X$, where $X$ is an $A$-module. The chain complex $Der(A,X)$ can be identified with $Hom(\Omega_A,X)$ where $\Omega_A$ is the module of differentials. This might be a useful way of thinking about these things. –  Tom Goodwillie Nov 28 '11 at 14:51
The derived statement should involve the first Hochschild cohomology group rather than just the derivations, probably, no? –  Mariano Suárez-Alvarez Dec 1 '11 at 23:43
@Mariano: Would that it were. My understanding is that Hochschild cohomology is the better-behaved object. But in my application I specifically care about things that are more like the chain complex of derivations for a given dga --- I really would like all this strictness, for example. –  Theo Johnson-Freyd Dec 3 '11 at 8:00
add comment

2 Answers

I don't know the full answer but it's not true that if $A$ and $B$ are quasi-isomorphic then so are $Der(A)$ and $Der(B)$. The following is over $\mathbb Q$. The simplest counterexample is the minimal model of $S^2$. $S^2$ is formal so its minimal model is quasi-isomorphic to the one for its cohomology. Note that $H^*(S^2)$ has no odd derivations being evenly graded. Yet, its minimal model is $M= (\Lambda (x,y), d)$ with $ deg (x) =2, deg (y)= 3$ and $dx=0, dy=x^2$. You can directly see that $H_{odd}(Der(M))$ is not zero. Specifically, there is a closed non cohomologous to zero derivation $\theta$ of degree $-3$ with $\theta(x)=0, \theta(y)=1$. In general there is a natural map $H(Der(M))\to Der(H(M))$ which is onto for formal spaces but it need not be injective. This map looks like it should be the edge homomorphism in some spectral sequence as it commutes two functors (which is common for some natural spectral sequences like the Eilenberg-Moore spectral sequence for diff Tor) but I don't know what that spectral sequence should be. It is however true that for positively elliptic spaces such as $S^2$ (they are all formal) one has that $H_i(Der(M))\to Der_i(H(M))$ is an isomorphism for negative even $i$. Here a space is called (rationally) elliptic if it has finite dimensional total rational cohomology and homotopy. Basic examples are homogeneous spaces and fiber bundles built out of them. A space is called positively elliptic if it's elliptic and has positive Euler characteristic (e.g. $S^2$). This is equivalent to saying that its homotopy Euler characteristic is zero, i.e. the total rank of all even homotopy groups is the same as the total rank of odd homotopy groups. See the book by [Félix, Halperin and Thomas on the basics of elliptic spaces.][1] [1]: http://www.ams.org/mathscinet/search/publdoc.html?r=1&pg1=CNO&s1=1802847&loc=fromrevtext

share|improve this answer
On your last paragraph: (1) What is a "positively elliptic space"? Googling doesn't help (and my knowledge of algebraic topology is spotty). (2) Certainly you must have switched signs somewhere, since in your first paragraph I would have thought that your derivation $D$ was a non-zero class in $\mathrm{H}_{-3}(\operatorname{Der}(M))$? –  Theo Johnson-Freyd Nov 28 '11 at 18:55
To fill in the details of your first paragraph (esp. for other readers), if $D = [\mathrm{d},E]$ for some derivation $E$, then $E$ must be degree $-4$; but every derivation is determined by what it does to $x$ and $y$ and so there are no degree $-4$ derivations. –  Theo Johnson-Freyd Nov 28 '11 at 18:57
I'm not sure what you mean about switching signs. I'm talking about cohomology of derivations in degree $-3$ and as you say the particular derivation I'm considering lives in $H_{-3}(Der(M))$. The signs look correct to me as they are?? and yes, the differential on derivations is just the Lie bracket with $d$. –  Vitali Kapovitch Nov 28 '11 at 20:58
@Vitali: Sorry. I must have missed the word "even" in "is an isomorphism for negative even i." –  Theo Johnson-Freyd Nov 29 '11 at 6:53
Thank you for all of this information, by the way. My intuition is not very good: do you expect a similar story if, say, "derivations" is replaced by "nth order differential operators", or would you expect more to go wrong? I will also look through the book by Félix, Halperin and Thomas, which I have now checked out from the library. –  Theo Johnson-Freyd Nov 29 '11 at 22:54
show 1 more comment

If you have two cdgas which are cofibrant (so built out of free cdgas and their cones via an iterated sequence of pushouts) and quasi-isomorphic then their homological invariants agree (one of the properties of cofibrant models is that any quasi-isomorphism between them admits a homotopy inverse). Vitali's example illustrates the problem in general: the minimal model for $H^* (S^2)$ is cofibrant, but $H^* (S^2)$ with trivial differential is not.

To clarify things a bit further, as Tom pointed out in the comments above, when we talk about derivations we usually have a map $A\rightarrow B$ of simplicial commutative rings (in characteristic zero we can use commutative dgas) and a $B$-module $M$. Since you didn't mention this data I assumed we were taking $A$ to be the unnamed field of characteristic zero and $B$ to be augmented over $A$ so that we could take the module $M$ to be $A$ (I think this is a common situation). Now $Der_A (B;A)$ is contravariantly functorial in $B$ as an augmented $A$ algebra and takes homotopic maps to the same map. So if I have maps of augmented $A$ algebras $B\rightarrow C\rightarrow B\rightarrow C$ such that the composite of each two maps is homotopic to the identity (such as when we have a quasi-isomorphism between two cofibrant $A$-algebras) I can apply $Der_A (-,A)$ to the sequence and obtain an isomorphism between $Der_A(B,A)$ and $Der_A(C,A)$.

You might find it helpful to read Quillen's 1970 paper: On the (co-)homology of commutative rings. These ideas are explained and generalized there.

share|improve this answer
I'm probably just being stupid --- is it clear that even if you have maps in both directions, that you can move the homology of Derivations (say) from one to the other? –  Theo Johnson-Freyd Dec 3 '11 at 7:58
Hi Justin, Your clarification is helpful. My impression is that it is quite common in the homotopy literature to work with augmented algebras, but my impression has always been that this is unsatisfying. In particular, when I say "derivation" of a commutative ring $B$, I mean a linear (over whatever ground ring) map $d: B \to B$ such that for $a,b\in B$, we have $d(ab) = da\,b + a\,db$. This is correct, of course, for degree-$0$ maps, and the usual sign rules are standard (and best handled with internal category theory a la Deligne–Morgan, rather than explicit sign rules). For example, a –  Theo Johnson-Freyd Dec 20 '11 at 5:35
cdga is nothing more nor less than a commutative graded algebra with a distinguished degree-$(\pm 1)$ (depending on conventions) self-commuting derivation. So I think this is subtly different than your setting. That said, I entirely agree that for any given module, it makes sense to take derivations into that module, and there will be some functoriality. I will think about what types of homotopies are necessary to use this functoriality to do what I'd like. Thank you for the Quillen reference. –  Theo Johnson-Freyd Dec 20 '11 at 5:39
@Theo: Thanks for the positive response. Might I suggest looking at $Der_A(B; B)$. This is the set of derivations of $B$ in your sense. Since this functor also respects homotopy equivalences (covariantly) in the $B$-module you can see that a homotopy equivalence of $A$-algebras $B\simeq C$ induces an isomorphism [ Der_A(B;B)\rightarrow Der_A(B;C)→Der_A(C,C).] –  Justin Noel Dec 20 '11 at 8:28
add comment

Your Answer


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.