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Does there exist a path-connected non-contractible CW-complex $X$ such that suspension $\sigma\colon \pi_k(X)\to \pi_{k+1}(\Sigma X)$ is an isomorphism for all $k$?

If so is there also a simply-connected example? Or does there exist a finite CW-complex $X$ with this property? If not then does there exist an example with finitely many cells in every dimension?

Furthermore another interesting question is whether this also holds for the suspension $\Sigma X$.

ThorbenK
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  • Doesn't Eilenberg-Maclane space $K(G,2)$ for any nontrivial $G$ satisfy all of this? – freakish Jan 31 '19 at 20:56
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    No. This would imply that for example $K(\mathbb{Z}/2,3)$ is homotopy equivalent to $\Sigma K(\mathbb{Z}/2,2)$, but this would imply that all cup-products vanish in $K(\mathbb{Z}/2,3)$. Nevertheless the square of the non-zero class of $H^3(\mathbb{R}P^\infty;\mathbb{Z}/2)$ is non-zero, which contradicts the above by Brown representability. I'm pretty confident that you can build something similar for other rings as well and I don't see any reason why it would be true for rings but not for groups. – ThorbenK Jan 31 '19 at 21:12
  • About your second question, isn't an example automatically simply connected, because of $k=1$ and the fact that the suspension of a path connected space is simply connected ? – Maxime Ramzi Jan 31 '19 at 21:42
  • No. The suspension map is a map $\pi_k(X)\to \pi_{k+1}(\Sigma X)$ so you can't conclude that. – ThorbenK Jan 31 '19 at 22:02
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    Fascinating question. My gut says no such space exists, but it might also be a difficult question. – Cheerful Parsnip Jan 31 '19 at 22:23
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    @Cheerful Parsnip My gut says the same. Furthermore even though I don't really know why, I fell like this question might be related. – ThorbenK Jan 31 '19 at 22:31
  • @ThorbenK Please be aware that (as far as I remember) that answer is incorrect, but I still believe the claim. I have always intended to go back and fix it, but have not yet found the time. Maybe sometime in the next few months. –  Feb 01 '19 at 18:21

1 Answers1

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You are asking for $X$ such that the canonical map $X \to \Omega(\Sigma X), x \mapsto (t \mapsto (x,t))$ is a weak homotopy equivalence (where $\Sigma$ is reduced suspension and $\Omega$ is loop space).

By a theorem of Bott–Samelson, the homology of $\Omega \Sigma X$ with field coefficients is isomorphic to the free associative algebra on the vector space $\tilde{H}_*(X)$, the tensor algebra $T(\tilde{H}_*(X))$. Moreover the canonical map $X \to \Omega \Sigma X$ induces the canonical inclusion $H_*(X) \to T(\tilde{H}_*(X))$ on homology. (See e.g. Theorem 7.3.1 in Selick's Introduction to Homotopy Theory.) This inclusion is not an isomorphism unless $\tilde{H}_*(X) = 0$. It follows that under your assumptions, $X$ is acyclic.

(Credit due to Mike Miller for the last steps.) Then $\pi_1(X) = \pi_2(\Sigma X)$ is abelian, so $\pi_1(X) = H_1(X) = 0$ by Hurewicz. A simply connected acyclic CW-complex is contractible.

Najib Idrissi
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    This is enough to conclude: $\pi_1$ is abelian by assumption, and so you have demonstrated that $\pi_1 = H_1 = 0$. Simply connected acyclic spaces are contractible. –  Feb 01 '19 at 15:18
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    @MikeMiller Perfect! Thanks. – Najib Idrissi Feb 01 '19 at 16:01
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    It turns out the James construction isn't needed. I should brush up on my homotopy theory! – Najib Idrissi Feb 01 '19 at 16:42
  • How does the proof without the James construction work? – ThorbenK Feb 01 '19 at 16:49
  • @ThorbenK See my latest edit. You just need the Bott–Samelson theorem. It doesn't say anything about $JX$. (Well it does, but you don't need $JX$ for the statement or the proof.) – Najib Idrissi Feb 01 '19 at 16:51
  • This is the original Bott-Samelson paper. It seems they use a suitable filtration to reduce this to a computation at the $E^2$ page, and then argue that a certain property of the spectral sequence implies the result about being free on $\tilde H_*(X)$. It looks like it is mostly elementary (to today's students) and shouldn't be too hard to read, but I don't have time to really go through it now. I'll admit I didn't try to look at Selick. –  Feb 01 '19 at 18:05