Spaces such that $\Omega^2 X \cong X$

Question

We know from Bott Periodicityt that there is a space X such that $\Omega^2 X \cong X$ (homotopy equivalence) , but these spaces are rather complicated and I am curious, is there any easy example of a non-contractible, path-connected space X such that $\Omega^2 X \cong X$?

Answer 1

Observe that finding such a space is equivalent to finding a $2$-periodic cohomology theory. Indeed, if you have such a space $X$, then

$E^{2n}(Y) = [Y, X]$, $E^{2n+1}(Y) = [Y, \Omega X]$

is a reduced cohomology theory. Conversely, if you have a $2$-periodic cohomology theory $E^{*}$, then by Brown representability there will be spaces $X_{n}, n \in \mathbb{Z}$ such that

$E^{n}(Y) \simeq [Y, X_{n}]$

and the suspension axiom, together with periodicity, will give you $X_{0} \simeq \Omega ^{2} X_{2} \simeq \Omega ^{2} X_{0}$. One way to construct such cohomology theories en masse is taking $E$ to be any cohomology theory and defining

$\hat{E}^{n}(Y) = \prod _{k \in \mathbb{Z}} E^{n+2k}(Y)$.

The exactness and suspension axioms are clear, the additivity axiom follows from distributivity of products. Note that on the level of representing spaces this amounts to replacing $X_{n}$ by $\hat{X}_{n} = \prod _{k \in \mathbb{Z}} X_{n+2k}$.

For concreteness, let's work out what happens in the case of singular cohomology theory with coefficients in $\mathbb{Z}$. Here we have

$X_{n} = K(\mathbb{Z}, n)$, for $n \geq 0$

and

$X_{k} = \Omega ^{-k} K(\mathbb{Z}, 0) = \{ pt \}$ for $k < 0$. Hence, the "$\Omega^{2}$-periodic" space associated to this cohomology theory is $\prod _{n \geq 0} K(\mathbb{Z}, 2n)$.