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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 $\nu + (-1)\circ \nu \in \pi_7(S^4)$?

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I think, in general, precomposition and postcomposition in the homotopy groups of spheres agree only after one suspension. This can be already seen in $\pi_3S^2$ since complex conjugation on $S^3\subset \mathbb{C}^2$ has degree $1$, while complex conjugation on $\mathbb{CP}^1$ has degree $-1$. Therefore, $\eta$ postcomposed with $-1$ is $\eta$ again, but $\eta \neq -\eta$. On the other hand, by this argument we have $\Sigma \eta = - \Sigma \eta$. –  Lennart Meier Nov 22 '11 at 21:48
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Furthermore, $\pi_7 S^4 \cong \mathbb{Z}\oplus \mathbb{Z}/12 \to \pi_3^{st}\cong \mathbb{Z}/24$ is given by observing that the generator of $\mathbb{Z}$ is mapped to the generator of $\pi_3^{st}$ ($\nu$) and the generator $a$ of $\mathbb{Z}/12$ is mapped to twice $\nu$ (since $\eta^3\neq 0\in\pi_3^{st}$). The class $\nu+(-1)\circ\nu\in\pi_7 S^4$ is in the kernel of the stabilization map. Furthermore, the "non-torsion part" of $(-1)\circ \nu$ must be $\pm 1$. Therefore, $(-1)\circ \nu = \nu - a$, I think and $\nu + (-1)\circ \nu$ is detected by the Hopf invariant. –  Lennart Meier Nov 22 '11 at 22:12
    
The Hopf invariant satisfies $H(n\iota\circ f)=n^2 H(f)$, so this should tell you that the non-torsion part of $H((-1)\circ \nu))$ is exactly, $\nu$. –  Charles Rezk Nov 22 '11 at 22:17
    
Never mind, I guess you already have that. –  Charles Rezk Nov 22 '11 at 22:19
    
I claim $\nu+(-1)\nu=24\nu$! –  Fernando Muro Nov 22 '11 at 23:20
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2 Answers 2

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.

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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 .

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