# Talk:Surgery obstruction map I (Ex)

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− | We take $X=\mathbb H P^2$ | + | We take $X=\mathbb H P^2$: |

The normal map $id_X$ gives the base point of $\mathcal N (X)\cong [X,G/TOP]$. | The normal map $id_X$ gives the base point of $\mathcal N (X)\cong [X,G/TOP]$. | ||

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 $\phi$. |

## Revision as of 21:46, 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.