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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.

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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. –  M Turgeon Jul 12 '12 at 18:19
    
Although I agree this won't answer your (very) general question. –  M Turgeon Jul 12 '12 at 18:20
1  
Ah, that is the motivation for the question actually (I just made an edit while you were commenting). –  AnonymousCoward Jul 12 '12 at 18:25

1 Answer 1

up vote 3 down vote accepted

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 $\alpha \in H^{|\alpha|}(X;\mathbb{Z})$.

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I had not yet seen your edited question when I posted this. I'll think about the edited version... –  Jason DeVito Jul 12 '12 at 18:34
    
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. –  AnonymousCoward Jul 12 '12 at 18:36

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