2
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It is known that the Hopf invariant for maps from $\mathbb{S}^{4n - 1} \to \mathbb{S}^{2n}$ is nontrivial (and captures the rational homotopy of the spheres). For $n = 1$, the Hopf fibration provides a nicely structured example of map having Hopf invariant one. The construction extends to $n = 2$ or $n=4$, and it known by a theorem of Adams that there is no map of Hopf invariant one for other values of $n$.

My question is whether there is a nice explicit construction of maps of Hopf invariant $2$ for any $n \ge 1$.

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    $\begingroup$ Yes, the Whitehead product of the identity with itself: Attaching a $4n$-cell (in your question you are off by one) to $S^{2n}$ with this map gives you $S^{2n}\times S^{2n}$ with $S^{2n}\times pt$ and $pt\times S^{2n}$ identified, and pulling back the generator of $H^{2n}$ and its square to $S^{2n}\times S^{2n}$ you get that the Hopf invariant is the coefficient of $xy$ in $(x+y)^2$, i.e. $2$. $\endgroup$ – Bertram Arnold Mar 1 at 11:25
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    $\begingroup$ I guess you would prefer to have a definition by an explicit formula. Is that correct? $\endgroup$ – Piotr Hajlasz Mar 1 at 12:41

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