1
$\begingroup$

A cgda $(A,d)$ is formal if it is weakly equivalent to $(H(A),0)$. There are several equivalent conditions for this. Similarly, a space $X$ is formal if the cgda $(A_{PL}(X),d)$ of polynomial differential forms is formal.

So far, so good. There is however a sentence in the book "Rational Homotopy theory" by Felix, Halperin, Thomas, that confuses me: "Thus $(A,d)$ and $X$ are formal if and only if their minimal sullivan model can be computed directly from their cohomology algebras". The reason why this confuses me, aside from the fact that it is not immediately clear to me why this is true, is that it seems from this sentence that the property of being formal can be read directly from the cohomology algebra $(H(A),0)$. Is this true? More precisely:

Question: [Edited after HenrikRüping comment] suppose that $X$ and $Y$ are spaces with isomorphic cohomology. Assume that $X$ is formal. is it true that $Y$ is formal as well?

Either a motivation for why this is true, or an example of a non-formal space with th same cohomology of a formal one, would be greatly appreciated!

I am mostly interested in spaces with the cohomology of spheres, at most one of which is even dimensional, but any more information on the subject would help me understand this a little better.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ You could always pick $A,d$ as $H^*(B),0$. Thus you are asking whether all cgda's are formal. There are non-formal cdgas. $\endgroup$
    – HenrikRüping
    Commented Feb 17 at 15:03
  • $\begingroup$ You are right. I guess my question was sort of ill-posed. What I am really interested is more topological: if $X$ and $Y$ have the same cohomology and $X$ is formal, is $Y$ formal as well? I will update my question $\endgroup$
    – CuriousUser
    Commented Feb 17 at 15:23
  • 3
    $\begingroup$ Just take a non-formal space $Y$ whose cohomology ring has trivial cup products, and let $X$ be a wedge of spheres of the appropriate dimensions to make its rational cohomology the same as $Y$. $\endgroup$
    – Andy Putman
    Commented Feb 17 at 16:27
  • 1
    $\begingroup$ I don't see how the sentence you quote from Felix, Halperin and Thomas is claiming that the property of formality can be read from the cohomology algebra. $\endgroup$
    – Jeremy Rickard
    Commented Feb 18 at 8:22
  • 1
    $\begingroup$ @JeremyRickard That was my original interpretation, and the source of my confusion. The examples from Andy Putman and HenrikRüping have helped me better understand what was claimed by Felix Halperin Thomas $\endgroup$
    – CuriousUser
    Commented Feb 19 at 4:26

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .