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 $\pi^s(7) \approx Z_{240}$?
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asked Nov 20, 2015 at 6:46
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$\begingroup$ This is claimed at mathoverflow.net/questions/218549/…. $\endgroup$– Qiaochu YuanNov 20, 2015 at 7:15
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$\begingroup$ 2-locally I think you can see this using the EHP sequence, as in mathoverflow.net/questions/50452/… $\endgroup$– Drew HeardNov 20, 2015 at 8:16
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1 Answer
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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.)
answered Nov 20, 2015 at 9:10
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5$\begingroup$ A better way to see that the sequence splits is to note that the inclusion of the fiber is null and therefore its homotopy fiber is $\Omega S^8 \cong S^7 \times \Omega S^{15}$. $\endgroup$ Nov 20, 2015 at 13:19
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$\begingroup$ The suspension map $\pi_{14}(S^7) \to \pi_{15N (S^8)$ provides the splitting. $\endgroup$ Nov 21, 2015 at 10:32
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$\begingroup$ The suspension map $\pi_{14}(S^7) \to \pi_{15N (S^8)$ provides the splitting. This follows from the fact that image of the double supsension $\pi_ {14}(S^7) \to \pi^{st}(7)$ has index $2$, and so injective. (The latter follows from the fact, that in dimension $7$ there is a Hopf invarinat one map. To see this use the butterfly lemmma to the "cross diagram" formed by the part of EHP sequence $\pi_{14}(S^7) \to \pi_{15}(S^8) \to Z$ and the Freudenthal exact sequence: $ 0 \to Z<[id_{S^8}, id_{S^8}]> \to \pi_{15}(S^8) \to \pi^{st}(7) \to 0$..) $\endgroup$ Nov 21, 2015 at 10:54
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$\begingroup$ But just the diagram does not seem to imply the answer, why the generator $\sigma$ of $Z$ goes onto the generator of $Z_{240}$ $\endgroup$ Nov 21, 2015 at 11:00
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1$\begingroup$ The only thing you are missing in your next to last comment is that the component of $[\iota_8,\iota_8]$ along $h_3$ in the splitting is $\pm 2$. This can be seen by computing the classical $\mathbb Z$-valued Hopf invariant via the cohomology rings of the cones on these maps. The cone on $[\iota_8,\iota_8]$ is the 16-skeleton of $\Omega S^9$ by the James construction. $\endgroup$ Nov 21, 2015 at 22:49