Let $X$ be an elliptic simply connected space. Is it rationally homotopy equivalent to the suspension of some connected space $Y$? If not, is it rationally homotopy equivalent to a loop space?
Is a simply connected elliptic space rationally homotopy equivalent to a loop space or a suspension?
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1$\begingroup$ $\mathbb{CP}^2$ is elliptic, it's definitely not a rational suspension, and I think it's not a rational loop space. $\endgroup$– Qiaochu YuanCommented Oct 31, 2017 at 0:11
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$\begingroup$ Cup products on a suspension vanish. Given that elliptic simply connected spaces satisfy Poincaré duality over $\mathbb Q$, that severely limits your options for being rationally equivalent to a suspension. $\endgroup$– Arun DebrayCommented Oct 31, 2017 at 0:12
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$\begingroup$ that's right but it can exist $\endgroup$– tarikCommented Oct 31, 2017 at 0:19
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$\begingroup$ my question about loop space is equivalent to when the rational cohomology algebra of X is a free graded commutative algebra. $\endgroup$– tarikCommented Oct 31, 2017 at 0:31
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1 Answer
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As Qiaochu points out, $\mathbb{C}\mathbb{P}^2$ is a simply-connected elliptic space (its only non-zero rational homotopy groups being $\pi_2(\mathbb{C}\mathbb{P}^2)\otimes \mathbb{Q} = \mathbb{Q}$ and $\pi_5(\mathbb{C}\mathbb{P}^2) \otimes \mathbb{Q} = \mathbb{Q}$) that is not homotopy equivalent to a suspension, since it has non-trivial cup product. It is also not rationally homotopy equivalent to a loop space. Indeed, a loop space has the rational homotopy type of a product of Eilenberg-MacLane spaces $K(\mathbb{Q}, k)$. Since $\pi_2(\mathbb{C}\mathbb{P}^2)\otimes \mathbb{Q} = \mathbb{Q}$, there would be a factor of $K(\mathbb{Q}, 2)$ in the rational homotopy type of $\mathbb{C}\mathbb{P}^2$, but $K(\mathbb{Q}, 2)$ has infinite cohomology.
answered Oct 31, 2017 at 0:30