Question: Given a finite dimensional positively graded algebra $A$ over some ring $R$ that satisfies Poincaré duality in some dimension $n$, is there necessarily a topological space $X$ such that $H^*(X;R) \cong A$?

I recognise this is some sort of realization question but I don't know much algebraic topology.

The case I am most interested in is when $R$ is a field. As vague motivation, I'm interested in whether, given such an $A$ over $\mathbb{Q}$, there is an elliptic Sullivan algebra $(\Lambda V, d)$ such that $H(\Lambda V, d) \cong A$. The converse appears in the textbook Rational Homotopy Theory by Felix et. al.:

Theorem: If $(\Lambda V,d)$ is an elliptic Sullivan algebra (i.e. $V$ and $H(\Lambda V, d)$ are finite dimensional vector spaces) over a field of characteristic 0, then $H(\Lambda V, d)$ satisfies Poincaré duality.

There is at least some Sullivan algebras $(\Lambda V, d)$ quasi-isomorphic to $A$ (since $A^0 \cong R$) but whether any of them are elliptic is the question. I may make this another post later.

  • $\begingroup$ It seems there might be something related to your question in Rational Homotopy Theory by Felix, Halperin, and Thomas. In particular, check Chapter 38 - Poincare Duality. $\endgroup$ – M Turgeon Jul 12 '12 at 18:19
  • $\begingroup$ Although I agree this won't answer your (very) general question. $\endgroup$ – M Turgeon Jul 12 '12 at 18:20
  • 1
    $\begingroup$ Ah, that is the motivation for the question actually (I just made an edit while you were commenting). $\endgroup$ – AnonymousCoward Jul 12 '12 at 18:25

The answer is, in general, no.

For example, the following is corollary 4L.10 of Hatcher's Algebraic Topology book (freely available):

If $H^\ast(X;\mathbb{Z})$ is a polynomial algebra $ \mathbb{Z}[\alpha]$, possibly truncated by $\alpha^m = 0$ for $m > 3$, then $|\alpha| = 2$ or $4$.

Here, $|\alpha|$ denotes the degree of $\alpha$, meaning $\alpha \in H^{|\alpha|}(X;\mathbb{Z})$.

  • $\begingroup$ I had not yet seen your edited question when I posted this. I'll think about the edited version... $\endgroup$ – Jason DeVito Jul 12 '12 at 18:34
  • $\begingroup$ Thanks, this answers the question of the post though, although I am looking at that chapter to see if $R$ not being a field has any core part in the example. $\endgroup$ – AnonymousCoward Jul 12 '12 at 18:36

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.