Talk:Fibre homotopy trivial bundles (Ex)

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<wikitex>;
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 $\pi_3(O_5)$
+
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 $\mathbb Z\to \mathbb Z/24$,
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
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
has a sphere bundle $\pi:S(\xi_k)\to S^4$ which is fiber homotopically trivial, so in particular we have
homotopy equivalences $f_k:S(\xi_k)\to S^4\times S^4$.
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 $S(\xi_k)$
+
From another exercise we know that stably $TS(\xi_k)\cong \pi^*\xi_k$.
is non-trivial, since under $f_k$ the map $pi$ just corresponds to projection to one factor.
+
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
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 $p_1$.
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
+
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
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 $p_n$.
homotopy equivalences $S(\xi_k)\to S^{4n}\times S^{4n}$ which do not preserve $p_n$.
</wikitex>
</wikitex>

Revision as of 19:17, 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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