Signs in the unstable homotopy groups of spheres

Question

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

A note on $\mathcal{E}(\mathbb{HP}^n)$ for $n\leq 4$, 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 $\nu + (-1)\circ \nu \in \pi_7(S^4)$?

Answer 1

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.

Answer 2

I like this discussion: I just wanted to add that of course a similar thing happens for $\mathbb{O} P^2$ as well. The homotopy theory is very nicely discussed by Baues in section 9 of On the group of homotopy equivalences of a manifold .