Let $\iota_2 : \pi_2\mathbb{S}^2$ be a generator and let $\eta : \pi_3\mathbb{S}^2$ be the Hopf map.
The Whitehead product $[\iota_2, [\iota_2, \iota_2]] : \pi_4\mathbb{S}^2$ must be trivial, because $$[\iota_2, [\iota_2, \iota_2]] = [\iota_2, 2\eta] = 2[\iota_2, \eta]$$
But what is $[\iota_2, \eta]$?
According to proposition 10.48 in Lecture Notes in Algebraic Topology by Davis and Kirk, any Whitehead product has to be in the kernel of the suspension $\pi_4\mathbb{S}^2 \to \pi_5\mathbb{S}^3$.
If I am not mistaken, this means every Whitehead product in $\pi_4\mathbb{S}^2$ must be trivial, because the suspension $\pi_4\mathbb{S}^2 \to \pi_5\mathbb{S}^3$ is an isomorphism.
However, my present interest in triviality of $[\iota_2, \eta]$ is precisely that I am trying to prove that the suspension $\pi_4\mathbb{S}^2 \to \pi_5\mathbb{S}^3$ is an isomorphism... so this would be begging the question.
Are there other ways to understand why $[\iota_2, \eta]$ is trivial?
Let me attempt to elaborate on my current failed attempt to answer this question. I am also interested in other kinds of answers.
Under Pontryagin, an element of $\pi_4\mathbb{S}^2$ is supposed to correspond to a framed 2-manifold in $\mathbb{R}^4$. I believe it should be a framed torus (coming from the $\eta$) "linked with" a framed (2-)sphere (coming from the $\iota_2$.) The framing on the torus is such that -- at least if the torus were by itself -- we could do framed surgery along an equator (say the inner equator) but not along a meridian. But I don't see how we can do this surgery on the torus while it is linked around the sphere, or how to unlink them.
