# Talk:Fibre homotopy trivial bundles (Ex)

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We consider 5-dimensional real vector bundles over $S^4$$; We consider 5-dimensional real vector bundles over S^4. Isomorphism classes of these are given by their clutching function in \pi_3(O_5). Given that J:\pi_3(O_5)\to \pi_3(G_5) is isomorphic to the surjection \mathbb Z\to \mathbb Z/24, we see that the vector bundle \xi_k corresponding to k times the generator has a sphere bundle \pi:S(\xi_k)\to S^4 which is fiber homotopically trivial, so in particular we have homotopy equivalences f_k:S(\xi_k)\to S^4\times S^4. From [[Tangent_bundles_of_bundles_(Ex)|another exercise]] we know that stably TS(\xi_k)\cong \pi^*\xi_k. From [[Obstruction_classes_and_Pontrjagin_classes_(Ex)|a third exercise]] we know that the first Pontryagin class of \xi_k is k times the generator of H^4(S^4). It follows that the first Pontryagin class of S(\xi_k) is non-trivial, since under f_k the map \pi just corresponds to projection to one factor. Hence f_k is a homotopy equivalence which doesn't preserve the first Pontryagin class, as S^4\times S^4 has stably trivial tangent bundle, hence trivial p_1. Similarly one can argue with (4n+1)-dimensional vector bundles over S^{4n}; the J-homomorphism has always a non-trivial kernel, and the top Pontryagin class of the corresponding bundles are non-zero. This produces homotopy equivalences S(\xi_k)\to S^{4n}\times S^{4n} which do not preserve p_n. S^4$. Isomorphism classes of these are given by their clutching function in $\pi_3(O_5)$$\pi_3(O_5)$.

Given that $J:\pi_3(O_5)\to \pi_3(G_5)$$J:\pi_3(O_5)\to \pi_3(G_5)$ is isomorphic to the surjection $\mathbb Z\to \mathbb Z/24$$\mathbb Z\to \mathbb Z/24$, we see that the vector bundle $\xi_k$$\xi_k$ corresponding to $24k$$24k$ times the generator has a sphere bundle $\pi:S(\xi_k)\to S^4$$\pi:S(\xi_k)\to S^4$ which is fiber homotopically trivial, so in particular we have homotopy equivalences $f_k:S(\xi_k)\to S^4\times S^4$$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$$TS(\xi_k)\cong \pi^*\xi_k$. From a third exercise we know that the first Pontryagin class of $\xi_k$$\xi_k$ is $48k$$48k$ times the generator of $H^4(S^4)$$H^4(S^4)$.

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

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