Obstruction classes and Pontrjagin classes (Ex)

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== Question ==
<wikitex>;
<wikitex>;
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 $H^{4i}(S^{4i}) \cong \mathbb{Z}$.
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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 a cohomology class $x \in H^{4i}(S^{4i}; \pi_{4k-1}(O)) = H^{4i}(S^{4i})$ which generates $H^{4i}(S^{4i}) \cong \mathbb{Z}$.
{{beginthm|Question}}
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{{beginthm|Exercise}}
What is the $i$-th integral Pontryagin class of $\xi^{4i}$, $p_i(\xi) \in H^{4i}(S^{4i})$ ?
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Determine the $i$-th integral Pontryagin class of $\xi^{4i}$, $p_i(\xi) \in H^{4i}(S^{4i})$, in terms of $x$.
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{{endthm}}
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== Answer ==
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[[Category:Exercises]]
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[[Category:Exercises with solution]]
Let $a_j = (3 - (-1)^j)/2$, let $k!$ be the integer k-factorial and recall that $x \in H^{4i}(S^{4i})$ is a generator.
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{{beginthm|Theorem|{{cite|Kervaire1959}}}}
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There is an identity
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$$ p_i(\xi^{4i}) = \pm a_i \cdot (2i-1)! \cdot x \in H^{4i}(S^{4i}).$$
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Similarly, if $\eta$ denotes the complex vector bundle over $S^{2i}$ corresponding to a generator of $\pi_{2i}(BU)\cong \mathbb{Z}$, then its Chern class is given by
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$$ c_i(\eta) = \pm (i-1)! \cdot y\in H^{2i}(S^{2i}),$$
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where $y\in H^{2i}(S^{2i})\cong \mathbb{Z}$ is a generator.
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{{endthm|Theorem}}
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A way to prove the Theorem is to use the Chern character
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$$\tilde K_0(X)\to \tilde H^{ev}(X;\mathbb{Q})$$
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from complex topological $K$-theory. It can be defined using the explicit formula
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$$ ch(\xi)= \sum_{k>0} s_k(c_1(\xi),\dots,c_k(\xi))/k! $$
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for a virtual complex vector bundle $\xi$, where $s_k$ are the Newton polynomials. In the case $X=S^{2n}$ two special things occur:
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#The Chern character is injective with image $H^{2n}(S^{2n};\mathbb{Z})$. (This follows inductively from the case $n=0$ using Bott periodicity.)
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#A calculation shows that the image of a (virtual) complex vector bundle $\xi$ over $S^{2n}$ is given by $\pm c_n(\xi)/(n-1)!$.
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Hence, $c_i(\eta)$ is given by $\pm (n-1)!$ times a generator. This establishes the second part of the Theorem.
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The first part follows using the definition $p_i(\xi)=c_{2i}(\xi\otimes_\mathbb{R} \mathbb{C})$ together with the fact that complexification induces a map
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$$ - \otimes_\mathbb{R} \mathbb{C}\colon \widetilde{KO}^0(S^{4i})\to \tilde K^0(S^{4i}) $$
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which is given by multiplication by $a_i$, i.e. is a isomorphism in degrees $8i$ and multiplication by 2 in degrees $8i+4$.
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== Further discussion ==
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...
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== References ==
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{{#RefList:}}
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[[Category:Questions]]
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[[Category:Study questions]]
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Latest revision as of 01:00, 3 June 2014

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 a cohomology class x \in H^{4i}(S^{4i}; \pi_{4k-1}(O)) = H^{4i}(S^{4i}) which generates H^{4i}(S^{4i}) \cong \mathbb{Z}.

Exercise 0.1. Determine the i-th integral Pontryagin class of \xi^{4i}, p_i(\xi) \in H^{4i}(S^{4i}), in terms of x.

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