Obstruction classes and Pontrjagin classes (Ex)

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There is an identity
There is an identity
$$ p_i(\xi^{4i}) = \pm a_i \cdot (2i-1)! \cdot x \in H^{4i}(S^{4i}).$$
$$ p_i(\xi^{4i}) = \pm a_i \cdot (2i-1)! \cdot x \in H^{4i}(S^{4i}).$$
{{endthm}}
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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 $8i+4$.
</wikitex>
</wikitex>

Revision as of 15:52, 22 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 recall that x \in H^{4i}(S^{4i}) is 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}).

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

\displaystyle  c_i(\eta) = \pm (i-1)! \cdot y\in H^{2i}(S^{2i}),

where y\in H^{2i}(S^{2i})\cong \mathbb{Z} is a generator.

A way to prove the Theorem is to use the Chern character

\displaystyle \tilde K_0(X)\to \tilde H^{ev}(X;\mathbb{Q})

from complex topological K-theory. It can be defined using the explicit formula

\displaystyle  ch(\xi)= \sum_{k>0} s_k(c_1(\xi),\dots,c_k(\xi))/k!

for a virtual complex vector bundle \xi, where s_k are the Newton polynomials. In the case X=S^{2n} two special things occur:

  1. The Chern character is injective with image H^{2n}(S^{2n};\mathbb{Z}). (This follows inductively from the case n=0 using Bott periodicity.)
  2. 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)!.

Hence, c_i(\eta) is given by \pm (n-1)! times a generator. This establishes the second part of the Theorem.

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

\displaystyle  - \otimes_\mathbb{R} \mathbb{C}\colon \widetilde{KO}^0(S^{4i})\to \tilde K^0(S^{4i})

which is given by multiplication by a_i, i.e. is a isomorphism in degrees 8i and multiplication by 2 in degrees 8i+4.

3 Further discussion

...

4 References

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