9
$\begingroup$

There is a very natural way to define generators of $\pi_{4n-1}(SO(4n))\cong \mathbb{Z}\oplus \mathbb{Z}$ in terms of quaternions when $n=1$ and octonions when $n=2$ (see for example Tamura, On Pontrjagin classes and homotopy types of manifolds, 1957). Since there are no normed division algebras in higher dimensions, it is not possible to do the same for $n>2$.

I was wondering whether there still is a natural identification between $\pi_{4n-1}(SO(4n))$ and $\mathbb{Z}\oplus\mathbb{Z}$ using explicit generators when $n>2$?

Improve this question
$\endgroup$
2
  • $\begingroup$ What about using the fiber bundle $$SO(4n-1)\to SO(4n)\to S^{4n-1}$$ isn’t it helpful? $\endgroup$
    – user51223
    Mar 4, 2021 at 11:22
  • 3
    $\begingroup$ Matthias Wendt's answer here may be of interest. You can identify one copy of $\mathbb{Z}$ with $\pi_{4n-1}(SO)$ and the other with $\pi_{4n}(S^{4n})$. $\endgroup$
    – Michael Albanese
    Mar 4, 2021 at 13:41

1 Answer 1

Reset to default
14
$\begingroup$

One way to think about this would be to consider the (injective!) map $\pi_{4n-1} SO(4n) = \pi_{4n} BSO(4n) \to \mathbf Z^2$ that sends the classifying map of a $4n$-dimensional oriented real vector bundle $\xi\colon E \to S^{4n}$ to the pair $(e,p_n)$ consisting of its Euler and $n$th Pontryagin class evaluted against the fundamental class of the base sphere.

This map is injective and for $n \neq 1,2,4$ its image is given by $2\mathbf{ Z} \times a_n(4n-1)!\mathbf{Z}$ where $a_n = 2$ if $n$ is odd, and $1$ otherwise: the Euler number of any bundle is even by the Hopf invariant $1$ problem, and the possible values of $p_n$ of (stable) vector bundles over$ S^{4n}$ is known from homotopy theory, see Theorem 3.8. in these notes by Levine, for instance.

The generators you are looking for are then simply given by two bundles that are sent to $(2,0)$ and $(0,a_n(2n-1)!)$. The first one is clearly the tangent bundle of $S^{4n}$. I am not sure if the second one, which has vanishing Euler class and minimal $p_n$, has a nice geometric description.

Improve this answer
$\endgroup$
1
  • 5
    $\begingroup$ For the second generator, one may take any element which maps to the generator of the stable range $\pi_{4n}BSO$, which is given by a suitable power of the Bott element in $\pi_8 BSO$ and the generator of $\pi_4 BSO$. Both of these have explicit descriptions in terms of Clifford algebras, compare the paper by Atiyah, Bott and Shapiro. It should be possible to extract an explicit vector bundle from this description. $\endgroup$
    – Bertram Arnold
    Mar 4, 2021 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.