Suspension of the third Hopf map

Question

There are Hopf maps in three dimensions, I denote their homotopy classes by $h_1 \in \pi_3(S^2)$, $h_2 \in \pi_7(S^4)$ and $h_3 \in \pi_{15}(S^8)$ respectively. For $h_1$ and $h_2$ it is true that their supensions generate the corresponding stable homotopy groups of spheres: $Sh_1$ generates $\pi^s(1) \approx Z_2$, and $Sh_2$ generates $\pi^s(3) \approx Z_{24}$. Question: Is this true for $h_3$ too?. I.e. does $Sh_3$ generate $\pi^s(7) \approx Z_{240}$?

Answer 1

I think this follows from elementary considerations, as in Chris Schommer-Pries's question How to see the quaternionic hopf map generates the stable 3-stem?

The long exact sequence of the Hopf fibration $$ S^7\to S^{15}\stackrel{\sigma}{\to} S^8$$ leads to a short exact sequence $$0\to \pi_{15}(S^{15})\stackrel{\sigma_\ast}{\to} \pi_{15}(S^8)\to \pi_{14}(S^7)\to 0$$ in which the kernel is isorphic to $\mathbb{Z}$ and the quotient to $\mathbb{Z}_{120}$. In fact this is split, as can be seen by looking up $\pi_{15}(S^8)\cong \mathbb{Z}\oplus\mathbb{Z}_{120}$ in tables (there should be a better way to see this). It follows that $\sigma$ generates the $\mathbb{Z}$ summand.

Now the Freudenthal Suspension Theorem implies that $$E: \mathbb{Z}\oplus\mathbb{Z}_{120}\cong\pi_{15}(S^8)\to \pi_{16}(S^9)\cong \pi_7^{st}\cong\mathbb{Z}_{240}$$ is surjective, and it is now an exercise in elementary algebra to check that $E\sigma$ generates.

Update: As András points out, the above answer is incomplete: it does not follow from what's written above that $E\sigma$ generates $\pi_7^{st}=\mathbb{Z}_{240}$. Here is an attempt to rectify this. It seems to work, modulo a claim made below which hopefully is true and maybe even well known.

There is a short exact sequence $$ 0\to \pi_{15}(S^{15})\stackrel{P}{\to} \pi_{15}(S^8)\stackrel{E}{\to} \pi_{16}(S^9)\to 0 $$ which is the portion of the EHP sequence mentioned in András' comment. The map $P$ is composition with the Whitehead product $[\iota_8,\iota_8]:S^{15}\to S^8$. We can cross this with the short exact sequence coming from the Hopf fibration above to get a cross diagram with exact diagonals $$ \begin{array}{ccccc} \pi_{15}(S^{15}) & & & & \pi_{15}(S^{15}) \newline & \searrow & & \swarrow & \newline \downarrow & & \pi_{15}(S^8) & & \downarrow \newline & \swarrow & & \searrow & \newline \pi_{16}(S^9) & & & & \pi_{14}(S^7) \end{array} $$ and by the Butterfly Lemma (Mosher and Tangora, Exercise 7.1) we are reduced to showing that $\partial [\iota_8,\iota_8]$ generates $\pi_{14}(S^7)$, where $\partial:\pi_{15}(S^8)\to \pi_{14}(S^7)$ is the connecting homomophism of the Hopf fibration.

Claim: The map $\partial: \Omega S^8\to S^7$ which induces this connecting homomorphism is an H-map.

Accepting this claim for a moment, we see that $\partial$ maps the Whitehead product $[\iota_8,\iota_8]$ to the Samelson product $\langle \iota_7,\iota_7\rangle\in \pi_{14}(S^7)$ which is defined using the multiplication on $S^7$. But this Samelson product generates $\pi_{14}(S^7)\cong \mathbb{Z}_{120}$, by a result of James (page 176 of

I. M. James, On $H$-spaces and their homotopy groups, Quart. J. Math. Oxford Ser. (2) 11 (1960), 161--179.)