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I am currently thinking about a physics model related to framed bordism $\Omega_3^{fr}=\mathbb{Z}/24=\pi^s_3$, and the first stable example is $\pi_8(S^5)$, so I was curious about the generator, and happened to see $\pi_8(SO(6))=\mathbb{Z}/24$ as well.

So my question is: is there a homomorphism sending $\pi_8(SO(6))$ to $\pi_8(S^5)$? More generally, what's the relation between homotopy groups $\pi_n(S^m)$ and $\pi_n(SO(m+1))$ in general?

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    $\begingroup$ en.m.wikipedia.org/wiki/J-homomorphism $\endgroup$
    – Fernando Muro
    Jul 3, 2015 at 16:56
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    $\begingroup$ Thanks, sorry for vague question. I've read that page, however, that was about another group, say here, $\pi_3(O)$, look irrelevant. $\endgroup$
    – Yingfei Gu
    Jul 3, 2015 at 16:59

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There is a fibration $p : SO(m+1) \to S^m$ with fibre $SO(m)$ which induces a long exact sequence in homotopy

$$\dots \to \pi_n(SO(m)) \to \pi_n(SO(m+1)) \xrightarrow{p_*} \pi_n(S^m) \to \pi_{n-1}(SO(m)) \to \dots$$

For your particular question, we have the fibration $p : SO(6) \to S^5$ with fibre $SO(5)$ which gives

$$\dots \to \pi_8(SO(5)) \to \pi_8(SO(6)) \to \pi_8(S^5) \to \pi_7(SO(5)) \to \pi_7(SO(6)) \to \pi_7(S^5) \to \pi_6(SO(5)) \to \dots$$

I don't know how to compute these groups myself, but they have been computed by others. From the bottom of page $3$ of this we see that $\pi_8(SO(5)) = 0$, $\pi_7(SO(5)) = \mathbb{Z}$, $\pi_7(SO(6)) = \mathbb{Z}$, and $\pi_6(SO(5)) = 0$. From here we see that $\pi_7(S^5) = \mathbb{Z}_2$ so we have

$$\dots \to 0 \to \pi_8(SO(6)) \xrightarrow{p_*} \pi_8(S^5) \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_2 \to 0 \to \dots$$

As the map $\mathbb{Z} \to \mathbb{Z}_2$ is surjective, it has kernel $2\mathbb{Z}$. So the map $\mathbb{Z} \to \mathbb{Z}$ must be multiplication by $2$ (or $-2$) which is injective. Therefore the map $\pi_8(S^5) \to \mathbb{Z}$ is the zero map, so $p_*$ is an isomorphism.

The fibration always gives rise to a homomorphism $p_* : \pi_n(SO(m+1)) \to \pi_n(S^m)$ but it is not necessarily an isomorphism as $p_* : \pi_7(SO(6)) \to \pi_7(S^5)$ demonstrates.

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  • $\begingroup$ Sorry, why does that sequence imply $p_*$ is an isomorphism? It is obviously injective, however, not obviously surjective to me. $\endgroup$
    – Yingfei Gu
    Jul 3, 2015 at 18:04
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    $\begingroup$ @Yingfei: $\pi_8(S^5)$ is finite (e.g. by work of Serre) so the map to $\mathbb{Z}$ is necessarily zero. $\endgroup$
    – Qiaochu Yuan
    Jul 3, 2015 at 18:13
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    $\begingroup$ @YingfeiGu: You can use the fact that Qiaochu Yuan stated. Alternatively, it follows from the exact sequence (in particular, you don't need to know that $\pi_8(S^5)$ is finite). I have added some explanation. $\endgroup$
    – Michael Albanese
    Jul 3, 2015 at 18:39
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    $\begingroup$ The groups $\pi_k S^n$ are finite if $k\neq n$ and $k \neq 2n-1$ if $n$ is even. There is no short argument for this -- it is known as the Serre finiteness theorem. $\endgroup$
    – Lennart Meier
    Jul 4, 2015 at 8:49
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    $\begingroup$ There is also a proof via rational homotopy theory which I prefer, but as, Lennart says, the argument is hardly short as it requires the set up of rational homotopy theory. $\endgroup$
    – Sean Tilson
    Jul 7, 2015 at 8:24

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