MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a finite group and $\mathcal{A}$ be a $dg$-algebra. Assume $G$ acts on $\mathcal{A}$, i.e. there exists a homomorphism $G\to {\rm Aut}_{dg}(\mathcal{A})$.

Assume further there exists a $dg$-algebra $\mathcal{B}$ and an isomorphism $f:\mathcal{A}\to \mathcal{B}$, then you can find a $G$-action on $\mathcal{B}$ such that $f$ is equivariant.

What is if $f$ not an isomorphism but only a quasi-isomorphism. Is it possible to find a $G$-structure on $\mathcal{B}$ such that $\mathcal{A}$ and $\mathcal{B}$ are equivariantly quasi-isomorphic.

I guess no. But I have hope that following statement is true:

If a $dg$-algebra $\mathcal{A}$ with a $G$-action is formal, then it is also equivariantly formal?

Are any results in this direction known?

share|cite|improve this question
What is a quasi-isomorphism? – Turion Mar 25 '14 at 15:45
A homomorphism $f:A\to B$ is called a quasi-isomorphism, if it induces an isomorphism on cohomology. – Oliver Straser Mar 25 '14 at 15:48
up vote 3 down vote accepted

I believe your second statement is true when $\mathcal{A}$ is the minimal Sullivan model of a $G$-space $X$. Namely, if the $G$-space $X$ is rationally formal, then it is rationally $G$-formal, in the sense that it has a rational $G$-minimal model which can be constructed from $H^\ast(X;\mathbb{Q})$ with it's $G$-action.

This is discussed in Section 3.3 of the book Algebraic Models in Geometry by Félix, Oprea and Tanré, see in particular Remark 3.30(2). They refer to a paper of John Oprea: Lifting homotopy actions in rational homotopy theory, J. Pure Appl. Algebra 32 (1984), no. 2, 177–190.

share|cite|improve this answer
This is excellent. The existence of a minimal model requires $H^0(\mathcal{A})=\mathbb{Q}$. Is it possible to weaken this condition, say $ H^0(\mathcal{A})=\mathbb{Q}^n$? Best, Oliver – Oliver Straser Mar 26 '14 at 12:43
@Oliver: I don't know, sorry. Such an $\mathcal{A}$ would correspond to a space with $n$ path components. You could take a model of each component separately, but then it may be that the group action mixes them up somehow. – Mark Grant Mar 26 '14 at 14:32

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.