Talk:Surgery obstruction map I (Ex)
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<wikitex>; | <wikitex>; | ||
We take $X=\mathbb H P^2$: | We take $X=\mathbb H P^2$: | ||
− | The normal map $id_X$ gives the base point of $\mathcal N (X) | + | The normal map $id_X$ gives the base point of $\mathcal N (X)$. |
An element of $[X,G/TOP]$ is given by a bundle $\xi$ together with a fiber homotopy trivialization $\phi$. | An element of $[X,G/TOP]$ is given by a bundle $\xi$ together with a fiber homotopy trivialization $\phi$. | ||
− | Under the isomorphism $\mathcal N (X) \cong [X,G/ | + | Under the isomorphism $\mathcal N (X) \cong [X,G/TOP]$, the pair $(\xi,\phi)$ corresponds to a normal map $M\to X$ covered by $\nu_M\to \nu_X\oplus \xi$. |
The surgery obstruction of a normal map $M\to X$ covered by $\nu_M\to \eta$ equals | The surgery obstruction of a normal map $M\to X$ covered by $\nu_M\to \eta$ equals | ||
$$ sign(M)-sign(X)=\langle L(-\eta),[X]\rangle -1,$$ | $$ sign(M)-sign(X)=\langle L(-\eta),[X]\rangle -1,$$ |
Revision as of 21:47, 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.