# Formality of commutative differential graded algebras

I want to understand the definition of a commutative differential graded algebra (CDGA) to be formal.
Actually I encountered two definitions, but I have trouble with both.

From Wikipedia:

A commutative differential graded algebra $$A$$, again with $$A^0 = \mathbb Q$$, is called formal if $$A$$ has a model with vanishing differential. This is equivalent to requiring that the cohomology algebra of $$A$$ (viewed as a differential algebra with trivial differential) is a model for $$A$$ (though it does not have to be the minimal model).

From Griffiths, Morgan:

A space is said to be formal if the homotopy type of the DGA of forms on the space is the same as the homotopy type of the cohomology ring of the space. Thus, if $$X$$ is formal and $$\mathcal m_X$$ is a minimal model for the forms on $$X$$, then there is a DGA map $$\mathcal m_X\to (H^*(X),d=0))$$ which induces the identity on cohomology.

As far as I know, a CDGA is a model for another CDGA, if they are quasi-isomorphic (i.e. isomorphic in cohomology). A CDGA is a minimal model if it is a model and it is minimal (free and decomposable).

So here are my issues with the definitions:

Wikipedia: To me it seems that the complex $$(H^*(A),d=0))$$ always is a model of $$A$$.

Griffiths, Morgan: We already have the existence of an isomorphism $$\mathcal (H^*(m_X,d_\mathcal m))\to (H^*(X),d=0))$$. But this is not strong enough, we need this to be the identity. But in order to speak about the identity, the spaces must be same, which they are only up to isomorphy.

It seems like I missed a crucial point of the definitions. But what is it?

• Two complexes can have isomorphic cohomology but not be quasi-isomorphic. – JHF Dec 5 '18 at 19:54
• @JHF: So what is the real definition of quasi-isomorphy? – klirk Dec 5 '18 at 21:18
• @klirk Being quasi-isomorphic means there is -map- (in the category) inducing an isomorphism on cohomology, not just that the cohomology algebras as abstractly isomorphic. Compare this, to some extent, to the context of Whitehead’s theorem for CW complexes; two spaces can have the same homotopy groups and not be homotopy equivalent (e.g. $RP^3\times S^2$ and $RP^2 \times S^3$), but if there is a map inducing the isomorphism, then the spaces are in fact homotopy equivalent. – Aleksandar Milivojevic Dec 6 '18 at 13:21
• Ah, I see. thx. – klirk Dec 6 '18 at 20:23

You need to distinguish between maps of complexes and maps of cdgas. I'll work over a field $$k$$. If $$(C,d)$$ is a complex of $$k$$-vector spaces, then it's true that $$C$$ is quasi-isomorphic to $$(H(C),0)$$, via taking sections (which we can do because every vector space is a projective module; over more general rings it is not true that every complex is quasi-isomorphic to its cohomology). If $$(C,d)$$ is in addition a cdga, then $$(H(C),0)$$ is a cdga as well. These two cdgas are quasi-isomorphic as complexes of vector spaces, but there may be no quasi-isomorphism between them that preserves the multiplicative structure. You can think of this as a derived version of the fact that a given vector space can admit many different algebra structures.
My go-to concrete example of a non-formal dga $$A$$ appears on page 13 of Hess's Rational Homotopy Theory: A Brief Introduction [https://arxiv.org/pdf/math/0604626.pdf ] as the exterior algebra on three elements $$u,v,w$$ of degrees $$3,3,5$$ respectively, and with $$d(w)=uv$$. You can check that there is no cdga quasi-isomorphism $$A \to H(A)$$.
• Yes, you need a quasi-isomorphism of cdgas. In other words, $A$ is a formal cdga if and only if there is a zigzag of cdga morphisms $A \leftarrow \tilde{A} \to H(A)$ with both maps quasi-isomorphisms. If you are working with nonnegatively graded cdgas you can take $\tilde A$ to be a minimal model of $A$. This is where Griffiths and Morgan's map comes from, although I am not sure what they mean by 'identity' (as opposed to just `isomorphism'). – Matt Booth Dec 5 '18 at 22:09