Obstruction classes and Pontrjagin classes (Ex)

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== Question ==
== Question ==
<wikitex>;
<wikitex>;
Take the stable vector bundle $\xi$ over the $4i$-sphere corresponding to a generator of $\pi_{4i}(BO) = \mathbb{Z}$.
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Take the stable vector bundle $\xi$ over the $4i$-sphere corresponding to a generator of $\pi_{4i}(BO) = \mathbb{Z}$. By defintion the the primary obstruction to trivialising $\xi^{4i}$ is an obstruction class $x \in H^{4i}(S^{4i})$ which generates $H^{4i}(S^{4i}) \cong \mathbb{Z}$.
{{beginthm|Question}}
{{beginthm|Question}}
What is the $i$-th integral Pontryagin class of $\xi^{4i}$, $p_i(\xi) \in H^{4i}(S^{4i}) \cong \mathbb{Z}$ ?
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What is the $i$-th integral Pontryagin class of $\xi^{4i}$, $p_i(\xi) \in H^{4i}(S^{4i})$ ?
</wikitex>
</wikitex>
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== Answer ==
== Answer ==
<wikitex>;
<wikitex>;

Revision as of 18:15, 17 March 2010

Contents

1 Question

Take the stable vector bundle \xi over the 4i-sphere corresponding to a generator of \pi_{4i}(BO) = \mathbb{Z}. By defintion the the primary obstruction to trivialising \xi^{4i} is an obstruction class x \in H^{4i}(S^{4i}) which generates H^{4i}(S^{4i}) \cong \mathbb{Z}.

Question 1.1.

What is the i-th integral Pontryagin class of \xi^{4i}, p_i(\xi) \in H^{4i}(S^{4i}) ?


2 Answer

Let a_j = (3 - (-1)^j)/2, let k! be the integer k-factorial and let x \in H^{4i}(S^{4i}) be a generator.

Theorem 2.1 [Kervaire1959]. There is an identity

\displaystyle  p_i(\xi^{4i}) = \pm a_i \cdot (2i-1)! \cdot x \in H^{4i}(S^{4i}).

3 Further discussion

...

4 References

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