Suppose we are given a map $f:A \rightarrow B$ between two dg-algebras which are formal. Is the map $f$ also "formal" in some sense?

More precisely can we find isomorphisms $\phi_A:A\rightarrow H^\bullet(A)$ and $\phi_B:B\rightarrow H^\bullet(B)$ in the derived category of dg-algebras such that

$$H^\bullet(f) \circ \phi_A= \phi_B \circ f $$ holds in the derived category of dg-algebras?

I fear that this may be wrong in general. What would be a counter example? Are there criteria which ensure that $f$ is "formal" in this sense?


Let me give you an example which has a topological flavour. Let us consider, the De-Rham complex of differential forms on a sphere $S^n$, we denote it $A^*(S^n)$, it is a formal commutative differential graded algebra. Let us look at the morphisms of commutative differential graded algebras between $A(S^2)$ and $A(S^3)$ into the derived category.

A good way to compute them is to take a cofibrant replacement of $A(S^2)$, such a cofibrant resolution is given by the algebra $(\mathbb{R}[x_2]\otimes\Lambda(x_3),D)$ where the differential satisfies $D(x_2)=0$ and $D(x_3)=x_2^2$.

And as $A(S^3)$ is formal you can replace it by its cohomology algebra which is the exterior algebra $\Lambda(y_3)$. It is easy to see that a morphism of algebra $\phi$ between these two algebras is completely determined by the image of $x_3$ thus by a real number $\lambda$ such that $\phi(x_3)=\lambda.x_3$. We have proved: $$[A(S^2),A(S^3)]_{CDGA}\cong\mathbb{R},$$
while $H(\phi)=0$.

This example has a topological flavour because it is a way (not the best of course) to show that $\pi_3(S^2)\otimes \mathbb{R}\cong \mathbb{R}$.

People in rational homotopy theory have studied the concept of formality for morphisms:

  • Vigué-Poirrier, Micheline Formalité d'une application continue. C. R. Acad. Sci. Paris Sér. A-B 289 (1979), no. 16

  • Félix, Yves; Tanré, Daniel Sur la formalité des applications. Publ. U.E.R. Math. Pures Appl. IRMA 3 (1981), no. 2, exp. no. 1, 45 pp.

  • Oprea, John F. DGA homology decompositions and a condition for formality. Illinois J. Math. 30 (1986), no. 1, 122–137.

Edit: if you want to play with dgas, then the only thing that you have to change in the example above is the cofibration resolution of the algebra $A(S^2)$. You start with $(T(x_2,x_3),D)$ the tensor algeba on $2$ generators with differential $D(x_2)=0$ and $D(x_3)=x_2\otimes x_2$ it has the same cohomology as $A(S^2)$ until the differential degree $5$ where you have created a new cycle :$[x_3,x_2]=x_3\otimes x_2-x_2\otimes x_3$ which is not a boundary. Then you have to add a generator $x_4$ to kill this cycle i.e. you put $D(x_4)=[x_3,x_2]$. And so on and so forth, it is easy to check that all the generators you add to $(T(x_2,x_3),D)$ in order to build a cofibrant resolution of $A(S^2)$ will have degree at least $4$.

With this resolution in hand you check that a homotopy class of morphism of dgas is also completely determined by the image of $x_3$.

  • $\begingroup$ Thanks for this nice example! Its only slight disadvantage is that it lives in the category of commutative dgas. Do you know a simple counterexample in the category of dgas? $\endgroup$ – Jan Weidner Nov 28 '12 at 15:40
  • $\begingroup$ I have edited my answer to take into account your remark. $\endgroup$ – David C Nov 28 '12 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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