Let $\mathbb{HP}^2$ denote the quaternionic projective plane. According to


[A note on $\mathcal{E}(\mathbb{HP}^n)$ for $n\leq 4$](http://escholarship.lib.okayama-u.ac.jp/mjou/vol33/iss1/17/), N. Iwase, K-I. Maruyama, S. Oka, Math. J. Okayama Univ. 33 (1991) , 163-176.


any homotopy self equivalence of $\mathbb{HP}^2$ is homotopic to the identity on the 4-cell. That seems to mean that the composition


$$S^7 \overset{\nu}\longrightarrow S^4 \overset{-1}\longrightarrow S^4 \longrightarrow \mathbb{HP}^2$$


is not nullhomotopic, where $\nu$ is the Hopf map, and $-1$ is a map of degree $-1$. This in turn seems to show that $(-1)\circ \nu$ is not in $\langle \nu \rangle$, and in particular is not $-\nu$ (negatives are taken by composing with a map of degree $-1$ in the source sphere).


In the stable homotopy groups of spheres, the composition product is graded-commutative, so $(-1) \circ \nu = \nu \circ (-1) = -\nu$. Have I made a mistake, or is it really true that this fails unstably?


If it is true, how can I detect the nontrivial class $\nu + (-1)\circ \nu \in \pi\_7(S^4)$?



Signs in the unstable homotopy groups of spheres
OK, after having made so many stupid comments, I felt obligated to remember what I knew about unstable homotopy theory in order to try to say something meaningful.


Recall that $\pi\_7(S^4\vee S^4)\cong\pi\_7S^4\oplus\pi\_7 S\_4\oplus \mathbb{Z}$, where the third summand is the kernel of the homomorphism
$$\pi\_7(S^4\vee S^4)\longrightarrow \pi\_7(S^4\times S^4)$$
induced by the inclusion of the coproduct in the product, generated by the map representing the Whitehead product operation.


Moreover, for any $\alpha\in\pi\_7S^4$, the failure in the commutativity of the following diagram
$$\begin{array}{ccc}
S^7&\stackrel{\alpha}\rightarrow&S^4\\\
\downarrow&&\downarrow\\\
S^7\vee S^7&\stackrel{\alpha\vee\alpha}\rightarrow&S^4\vee S^4
\end{array}$$
where the vertical arrows are the coproducts, is $(0,0,H(\alpha))\in\pi\_7(S^4\vee S^4)$, where $H$ denotes the Hopf invariant (see G. Whitehead's book). 


The map $\alpha+(-1)\alpha$ is the composite
$$S^7\stackrel{\rm coprod.}\longrightarrow S^7\vee S^7\stackrel{\alpha\vee\alpha}\longrightarrow S^4\vee S^4\stackrel{(1,-1)}\longrightarrow S^4.$$
Moreover, the composite 
$$S^4\stackrel{\rm coprod.}\longrightarrow S^4\vee S^4\stackrel{(1,-1)}\longrightarrow S^4$$
vanishes. Therefore $\alpha+(-1)\alpha$ it coincides with the image of $(0,0,H(\alpha))\in\pi\_7(S^4\vee S^4)$ by the homomorphism
$$\pi\_7(S^4\vee S^4)\longrightarrow \pi\_7S^4$$
induced by $(1,-1)$,
which is $H(\alpha)[1\_{S^4},1\_{S^4}]\in \pi\_7S^4$, where the bracket denotes the Whitehead product operation. In particular $$\nu+(-1)\nu=[1\_{S^4},1\_{S^4}]\in \pi\_7S^4$$
which generates the kernel of the suspension homomorphism (by the work of Blakers and Massey on homotopy groups of triads)
$$\pi\_7S^4\longrightarrow\pi\_8S^5$$
described in Lennart Meier's comment above.


