Are there “nice” explicit representations of infinite order elements in $\pi_{4n-1}(S^{2n})$

Question

Motivation:

• The representation is very clean in the Hopf fibration case (2n=2,4,8).
• These maps are known since Serre's 1951 paper.
• There are clean generators for the only other infinite case $\pi_n(S^n)$.

Answer 1

You can write an explicit formula as follows. Consider a pair of finite-dimensional inner product spaces $U$ and $V$. We can define a map $$ f\colon S(U\oplus V) \to S(U\oplus\mathbb{R})\vee S(V\oplus\mathbb{R}) $$ by $$ f(u,v) = \begin{cases} \left(2\sqrt{\|v\|^2-\|u\|^2}\;u,\;2\|u\|^2-\|v\|^2\right)/\|v\|^2 \in S(U\oplus\mathbb{R}) & \text{ if } \|u\|\leq\|v\| \\ \left(2\sqrt{\|u\|^2-\|v\|^2}\;v,\;2\|v\|^2-\|u\|^2\right)/\|u\|^2 \in S(V\oplus\mathbb{R}) & \text{ if } \|u\|\geq\|v\| \end{cases} $$ The two clauses are consistent because they both give the basepoint $(0,1)$ when $\|u\|=\|v\|$. It is an algebraic exercise to check that both clauses give a unit vector in the indicated space. Further formulae that I will not give here identify the cofibre of $f$ with $S(U\oplus\mathbb{R})\times S(V\oplus\mathbb{R})$, which shows that $f$ is the attaching map for the top cell of this product. It is also interesting to identify the fibres of $f$. For any point $(x,t)\in S(U\oplus\mathbb{R})$ that is not equal to $(0,1)$, we can define $g_{(x,t)}\colon S(V)\to U\oplus V$ by $$ g_{(x,t)}(v) = \left(\frac{x}{\sqrt{(1-t)(3+t)}},\sqrt{\frac{2}{3+t}}v \right). $$ One can check that this lands in $S(U\oplus V)$ and gives a homeomorphism $S(V)\to f^{-1}\{(x,t)\}$. Similarly, the fibre of $f$ over any non-basepoint in $S(V\oplus\mathbb{R})$ is homeomorphic to $S(U)$.

Now take $V=U=\mathbb{R}^{2n}$ and compose $f$ with the fold map $S(U\oplus\mathbb{R})\vee S(U\oplus\mathbb{R})\to S(U\oplus\mathbb{R})$ to get a map $$ h\colon S^{4n-1} = S(U\oplus U) \to S(U\oplus\mathbb{R}) = S^{2n}. $$ This is a representative of the usual Whitehead homotopy class.