# Equivariant Formality

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?

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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

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 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