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I think this may all be classical bundle-theory. But I'm trying to read some old papers on classifications of bundles and the following came up as questions I couldn't immediately answer:

Consider the group homomorphism $\theta:\pi_{n-1}(SO(n)) \to H^n(S^n) $ given by sending an element $\alpha$ in $\pi_{n-1}(SO(n))$ to the Euler class of the $n$ - plane bundle classified by $\alpha$.

Is there a name for this map?

If we let $n=4$, we have $\theta :\pi_{3}(SO(4))=Z^2 \to H^4(S^4)=Z$. What is this map explicitely?

Now look at the bundle $SO(n) \to SO(n+1) \to S^n$. If we take $id_n \in \pi_{n}S^n$ to be the (class of the) identity map and $\partial(id_n)$ its image in $\pi_{n-1}SO(n)$ (from the homotopy LES of a fibration), then...

...what is $\theta (\partial(id_n))?$

Does it just map to a generator of $H^n(S^n)?$

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If I recall correctly, the map Z^2->Z is (1,-1). For the second question, the answer is no, eg for n=3 $ \pi_2(SO(3))=0$. In general it goes to the Euler class of the tangent bundle, which is 0 or 2. –  Paul Oct 14 '10 at 0:04

1 Answer 1

The bundle classified by this element of $\pi_{n-1}SO(n)$ is the tangent bundle of $S^n$, so the image under $\theta$ is twice a generator if $n$ is even, zero if $n$ is odd.

To say what $\theta$ is explicitly in the case $n=4$ in terms of the $\mathbb Z$-basis for $\pi_3SO(4)$, we would have to know what basis you have in mind, but for the one I usually think of it takes both basis elements to the same generator.

Edit: Note also that the map $\theta:\pi_{n-1}SO(n)\to H^n(S^n)$ is isomorphic to the map $\theta:\pi_{n-1}SO(n)\to \pi_{n-1}(S^{n-1})$ induced by the bundle projection $SO(n)\to S^{n-1}$.

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Oh, the comment in the edit is nice, thanks. –  Romeo Oct 14 '10 at 2:48

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