Is $S^n\langle n\rangle$ a suspension?

Question

Let $S^n\langle n \rangle$ denotes a $n$-connected cover of $n$-sphere $S^n$, i.e. homotopy fiber of $S^n\to K(\mathbb Z, n)$. Is it true that there exists space $X$, such that $S^n\langle n \rangle\simeq \Sigma X$ ? This is true for $n=2$, since $S^2\langle 2 \rangle = S^3$ but I doubt that result holds for general $n$. To disprove it will be sufficient to find non-trivial cup-product in $H^* S^n\langle n \rangle$, but I haven't tried to construct such yet.

Answer 1

Rather than saying it is true for $n=2$ is because $S^2\langle 2 \rangle = S^3$, I would say it is because $K(\mathbb{Z}, 1) = S^1$. Consider the Serre spectral sequence for the homotopy fibration $K(\mathbb{Z}, n-1) \to S^n\langle n \rangle \to S^n$. When $n=2$, this is very simple. When $n>2$, though, $K(\mathbb{Z}, n-1)$ has more interesting cohomology, and in particular has lots of nontrivial cup products. If we let $\iota$ be the nonzero class in $H^{n-1}(K(\mathbb{Z}, n-1), \mathbb{F}_2)$, then I think that in general, $\mathop{Sq}^2 \iota$ and $\mathop{Sq}^3 \iota$ are permanent cycles and have a nontrivial cup product which is necessarily also a permanent cycle.

A bit more detail: when $m < \deg \iota$, $\mathop{Sq}^m \iota$ is a permanent cycle because there is nothing available for it to hit. You can handle the cases $\deg \iota=2$ and $\deg \iota=3$ by hand: when $\deg \iota=2$, for example, $\mathop{Sq}^2 \iota = \iota^2$, and since the differential satisfies the Leibniz formula and since we're working in characteristic 2, $d_3 (\mathop{Sq}^2 \iota) = 0$.