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Consider principal $SO(9)$ bundles over $S^8$.They are in 1-1 correspondence with $$[S^8,BSO(9)]\cong \pi_7(SO(9))\cong \mathbb{Z}$$

Now pick up one such bundle $\xi$,we have the long exact sequence of homotopy groups $$\cdots\to \pi_8(S^8)\xrightarrow{\partial} \pi_7(SO(9))\to\cdots$$

Both groups are isomorphic to $\mathbb{Z}$,and the homomorphism should be a "$\times$ $n$" map.How to determine that $n$.This $n$ should depend on the classifying map of the principal bundle $\xi$.

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    $\begingroup$ I'm taking a stab in the dark, but why wouldn't $\partial$ take the generator of $\pi_8(S^8)$ to the element of $\pi_7(SO(9))$ classifying the bundle? $\endgroup$ – David Roberts Feb 27 '15 at 22:46
  • $\begingroup$ I think it is more natural to reduce this question to another one, namely, how to describe the number (say, $c(\xi)$) that the isomorphism [bundles over $S^8$] $\cong\mathbb Z$ assigns to $\xi$. This given, evidently $\partial$ sends (the homotopy class of) a map $f:S^8\to S^8$ to $c(f^*(\xi))$. Moreover I believe it is easier to switch from principal $SO(9)$-bundles to (oriented) 9-dimensional vector bundles. $\endgroup$ – მამუკა ჯიბლაძე Feb 27 '15 at 23:18
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To add two cents to what was said in the comments and Alex Degtyarev's answer:

First cent: the long exact homotopy sequence associated to a fiber sequence $F\to E\to B$ comes from taking homotopy classes of maps from $S^0$ into the sequence $\cdots\to \Omega F\to \Omega E\to \Omega B\to F\to E\to B$. Now start with the classifying map $S^8\to BSO(9)$, and successively take fibers, you get $$ \cdots\to\Omega BSO(9)\cong SO(9)\to E\to S^8\to BSO(9). $$ From this, you see that the long exact homotopy sequences associated to $SO(9)\to E\to S^8$ and $E\to S^8\to BSO(9)$ are the same. Therefore, $\partial:\pi_8(S^8)\to\pi_7(SO(9))$ is a looping of the classifying map, hence $\partial$ maps the identity to the element of $\pi_7(SO(9))$ classifying the bundle.

Second cent: $\pi_7(SO(9))\cong\pi_7(SO)$ is in the stable range because of the fiber sequences $SO(n)\to SO(n+1)\to S^n$. To actually determine $n$ you can therefore use your favourite method of describing the isomorphism $\pi_8(KO)\cong\mathbb{Z}$. If I am not mistaken, the stable isomorphism class of the oriented rank $9$ vector bundle associated to $\xi$ should be determined by its characteristic classes, which in the particular case $S^8$ should just be the second Pontryagin class. Then the favourite way of actually computing the corresponding number would depend on how the bundle $\xi$ is given.

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You pull back the "universal" isomorphism $[S^8,BSO(9)]\to\pi_7(SO(9))$, which is also $\partial$, by a map $S^8\to BSO(9)$, which induces multiplication by some $n$ at the level of $\pi_8$. On the other hand, the fibers are mapped homeomorphically, so the induced map on $\pi_7(SO(9))$ is an isomorphism. From the commutative diagram, the map in question is the multiplication by same $n$.

More intuitively, both the classifying map and $\partial$ are obstructions to the existence of a section; so, they must coincide.

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