Obstruction classes and Pontrjagin classes (Ex)

1 Question

Take the stable vector bundle $\xi$$<!-- COMMENT: To achieve a unified layout, along with using the template below, please OBSERVE the following: besides, $...$ and $$...$$, you should use two environments: - For statements like Theorem, Lemma, Definition etc., use e.g. {{beginthm|Theorem 1|(Milnor)}} ... ... ... {{endthm}}. - For references, use e.g. {{cite|Milnor1958b}}. END OF COMMENT --> == Question == <wikitex>; Take the stable vector bundle $\xi$ over the i$-sphere corresponding to a generator of $\pi_{4i}(BO) = \mathbb{Z}$. By defintion the the primary obstruction to trivialising $\xi^{4i}$ is an obstruction class $x \in H^{4i}(S^{4i})$ which generates $H^{4i}(S^{4i}) \cong \mathbb{Z}$. {{beginthm|Question}} What is the $i$-th integral Pontryagin class of $\xi^{4i}$, $p_i(\xi) \in H^{4i}(S^{4i})$ ? </wikitex> == Answer == <wikitex>; Let $a_j = (3 - (-1)^j)/2$, let $k!$ be the integer k-factorial and recall that $x \in H^{4i}(S^{4i})$ is a generator. {{beginthm|Theorem|{{cite|Kervaire1959}}}} There is an identity $$ p_i(\xi^{4i}) = \pm a_i \cdot (2i-1)! \cdot x \in H^{4i}(S^{4i}).$$ Similarly, if $\eta$ denotes the complex vector bundle over $S^{2i}$ corresponding to a generator of $\pi_{2i}(BU)\cong \mathbb{Z}$, then its Chern class is given by $$ c_i(\eta) = \pm (i-1)! \cdot y\in H^{2i}(S^{2i}),$$ where $y\in H^{2i}(S^{2i})\cong \mathbb{Z}$ is a generator. {{endthm|Theorem}} A way to prove the Theorem is to use the Chern character $$\tilde K_0(X)\to \tilde H^{ev}(X;\mathbb{Q})$$ from complex topological $K$-theory. It can be defined using the explicit formula $$ ch(\xi)= \sum_{k>0} s_k(c_1(\xi),\dots,c_k(\xi))/k! $$ for a virtual complex vector bundle $\xi$, where $s_k$ are the Newton polynomials. In the case $X=S^{2n}$ two special things occur: #The Chern character is injective with image $H^{2n}(S^{2n};\mathbb{Z})$. (This follows inductively from the case $n=0$ using Bott periodicity.) #A calculation shows that the image of a (virtual) complex vector bundle $\xi$ over $S^{2n}$ is given by $\pm c_n(\xi)/(n-1)!$. Hence, $c_i(\eta)$ is given by $\pm (n-1)!$ times a generator. This establishes the second part of the Theorem. The first part follows using the definition $p_i(\xi)=c_{2i}(\xi\otimes_\mathbb{R} \mathbb{C})$ together with the fact that complexification induces a map $$ - \otimes_\mathbb{R} \mathbb{C}\colon \widetilde{KO}^0(S^{4i})\to \tilde K^0(S^{4i}) $$ which is given by multiplication by $a_i$, i.e. is a isomorphism in degrees i$ and multiplication by 2 in degrees i+4$. </wikitex> == Further discussion == <wikitex>; ... </wikitex> == References == {{#RefList:}} <!-- Please add further headings according to your needs. --> [[Category:Questions]] [[Category:Study questions]]\xi$ over the $4i$$4i$-sphere corresponding to a generator of $\pi_{4i}(BO) = \mathbb{Z}$$\pi_{4i}(BO) = \mathbb{Z}$. By defintion the the primary obstruction to trivialising $\xi^{4i}$$\xi^{4i}$ is an obstruction class $x \in H^{4i}(S^{4i})$$x \in H^{4i}(S^{4i})$ which generates $H^{4i}(S^{4i}) \cong \mathbb{Z}$$H^{4i}(S^{4i}) \cong \mathbb{Z}$.

Question 1.1.

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

2 Answer

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

Theorem 2.1 [Kervaire1959]. There is an identity

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

3 Further discussion

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4 References

• [Kervaire1959] M. A. Kervaire, A note on obstructions and characteristic classes, Amer. J. Math. 81 (1959), 773–784. MR0107863 (21 #6585) Zbl 0124.16302