Talk:Surgery obstruction map I (Ex)
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Now $(\xi\oplus\xi,\phi * \phi)$ is the sum of $(\xi,\phi)$ and $(\xi,\phi)$ in $\mathcal N (X)=[X,G/O]$ with respect to the Whitney sum. | Now $(\xi\oplus\xi,\phi * \phi)$ is the sum of $(\xi,\phi)$ and $(\xi,\phi)$ in $\mathcal N (X)=[X,G/O]$ with respect to the Whitney sum. | ||
We compute | We compute | ||
− | $$\ | + | $$\theta(-(\xi,-\phi))+\theta(-(\xi,\phi)) - \theta(-(\xi\oplus\xi,\phi * \phi)) |
= 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -1 | = 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -1 | ||
= c \langle p_1(\xi)^2 ,[X] \rangle \ne 0, $$ | = c \langle p_1(\xi)^2 ,[X] \rangle \ne 0, $$ | ||
and so the surgery obstruction is not a group homomorphism with respect to the Whitney sum. | and so the surgery obstruction is not a group homomorphism with respect to the Whitney sum. | ||
</wikitex> | </wikitex> |
Revision as of 21:50, 29 May 2012
We take : The normal map gives the base point of . An element of is given by a bundle together with a fiber homotopy trivialization . Under the isomorphism , the pair corresponds to a normal map covered by . The surgery obstruction of a normal map covered by equals
so it depends only on the bundle over . There are fiber homotopically trivial bundles on corresponding to classes in which restrict to any given class in , since the corresponding Atiyah-Hirzebruch spectral sequence collapses. From another exercise we know that on we have such vector bundles with first Pontryagin class times the generator of . This means that on we have a vector bundle with whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence . Now is the sum of and in with respect to the Whitney sum. We compute
and so the surgery obstruction is not a group homomorphism with respect to the Whitney sum.