Talk:Fibre homotopy trivial bundles (Ex)

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homotopy equivalences $f_k:S(\xi_k)\to S^4\times S^4$.
homotopy equivalences $f_k:S(\xi_k)\to S^4\times S^4$.
From another [[Tangent_bundles_of_bundles_(Ex)|exercise]] we know that stably $TS(\xi_k)\cong \pi^*\xi_k$.
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From [[Tangent_bundles_of_bundles_(Ex)|another exercise]] we know that stably $TS(\xi_k)\cong \pi^*\xi_k$.
From a third [[Obstruction_classes_and_Pontrjagin_classes_(Ex)|exercise]] we know that the first Pontryagin class of $\xi_k$ is $48k$.
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From [[Obstruction_classes_and_Pontrjagin_classes_(Ex)|a third exercise]] we know that the first Pontryagin class of $\xi_k$ is $48k$.
It follows that the first Pontryagin class of $S(\xi_k)$
It follows that the first Pontryagin class of $S(\xi_k)$

Revision as of 19:30, 29 May 2012

We consider 5-dimensional real vector bundles over S^4. Isomorphism classes of these are given by their clutching function in \pi_3(O_5).

Given that J:\pi_3(O_5)\to \pi_3(G_5) is isomorphic to the surjection \mathbb Z\to \mathbb Z/24, we see that the vector bundle \xi_k corresponding to 24k times the generator has a sphere bundle \pi:S(\xi_k)\to S^4 which is fiber homotopically trivial, so in particular we have homotopy equivalences f_k:S(\xi_k)\to S^4\times S^4.

From another exercise we know that stably TS(\xi_k)\cong \pi^*\xi_k. From a third exercise we know that the first Pontryagin class of \xi_k is 48k.

It follows that the first Pontryagin class of S(\xi_k) is non-trivial, since under f_k the map \pi just corresponds to projection to one factor. Hence f_k is a homotopy equivalence which doesn't preserve the first Pontryagin class, as S^4\times S^4 has stably trivial tangent bundle, hence trivial p_1.

Similarly one can argue with (4n+1)-dimensional vector bundles over S^{4n}; the J-homomorphism has always a non-trivial kernel, and the top Pontryagin class of the corresponding bundles are non-zero. This produces homotopy equivalences S(\xi_k)\to S^{4n}\times S^{4n} which do not preserve p_n.

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