# Whitehead product with identity on homotopy groups of spheres

For $n\geq 2$ let $(S^n,p)$ be the $n$-sphere with a base point $p$. Let $1:S^n\to S^n$ denote the identity map. Let us define the map

$Wh_1: \pi_i(S^n,p)\to \pi_{i+n-1}(S^n,p), \alpha\mapsto [\alpha,1],$

where $[\cdot,\cdot]$ is the Whitehead bracket.

Question 1 : What is known about $Wh_1$ ?

Question 2 : If we let $Wh_f$ denote the corresponding map on homotopy groups for $f:(S^m,p)\to (S^n,p)$ then what is known about $Wh_f$ ?

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As you probably know, the Whitehead product is a degree $-1$ Lie bracket on homotopy groups, i.e. it is graded anticommutative and satisfies the graded Jacobi identity, but not $[x,x]=0$. In particular your maps are homomorphisms. As for the first one, it need not be trivial, if that's what you think. Since suspensions of Whitehead products vanish, your maps are trivial when the target is stable, e.g. for $n\geq i+1$ in the first case. Also, in the first case, it is the kernel of the suspension map in the critical dimension, i.e. $$\pi_n(S^n)\stackrel{Wh_1}\longrightarrow \pi_{2n-1}(S^n)\stackrel{\Sigma}\longrightarrow \pi_{2n}(S^{n+1})$$ is an exact sequence, e.g. for $n=2$ this looks like as follows $$\mathbb{Z}\stackrel{2}\longrightarrow\mathbb{Z}\twoheadrightarrow \mathbb{Z}/2.$$ This uses the Blakers-Massey theorem and the fact that $1\in\pi_n(S^n)$ is a generator. Hence, the analog exact sequence works in the second case only in the special case that $f\in \pi_m(S^n)$ is a generator. Probably more things can be deduced from the elementary properties of primary homotopy operation, but this is what comes to my mind right now.
Thanks for your answer! That's what (that $[x,x]=0$) I was thinking naively but I dug up some literature and realized that it's not the case. It's non-zero when $n$ is odd and generates a $\mathbb{Z}_2$. When $n$ is even it generates an infinite cyclic subgroup which splits on occasion. Judging from old papers, it seems that it would be too stupid (on my part) to expect a complete answer. Among other things, it would grossly underestimate the intricacies involved in knowing the homotopy groups of spheres! – Somnath Basu Sep 19 '12 at 12:13