# Talk:Surgery obstruction map I (Ex)

(Created page with "<wikitex>; We take $X=\mathbb H P^2$ and consider various bundle reductions of the normal bundle: The normal map $id_X$ gives the base point of $\mathcal N (X)$. The surgery o...") |
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We take $X=\mathbb H P^2$ and consider various bundle reductions of the normal bundle: | We take $X=\mathbb H P^2$ and consider various bundle reductions of the normal bundle: | ||

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

− | The surgery obstruction of a normal map $M\to X$ covered by $\nu_M\to \ | + | An element of $[X,G/TOP]$ is given by a bundle $\xi$ together with a fiber homotopy trivialization $\phi$. |

− | $$ sign(M)-sign(X)=\langle L(-\ | + | Under the isomorphism $\mathcal N (X) \cong [X,G/O]$, 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 | ||

+ | $$ sign(M)-sign(X)=\langle L(-\eta),[X]\rangle -1,$$ | ||

so it depends only on the bundle over $X$. | so it depends only on the bundle over $X$. | ||

There are fiber homotopically trivial bundles on $X$ corresponding to classes in $[X,G/TOP]$ | There are fiber homotopically trivial bundles on $X$ corresponding to classes in $[X,G/TOP]$ | ||

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collapses. | collapses. | ||

From [[Fibre_homotopy_trivial_bundles_(Ex)|another exercise]] we know that on $S^4$ we have such vector bundles with first Pontryagin class $48k$ times the generator of $H^4(S^4)$. | From [[Fibre_homotopy_trivial_bundles_(Ex)|another exercise]] we know that on $S^4$ we have such vector bundles with first Pontryagin class $48k$ times the generator of $H^4(S^4)$. | ||

− | This means that on $X$ we have vector | + | This means that on $X$ we have a vector bundle $\xi$ with $p_1(\xi)=48$ whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence $\phi$. |

− | $\mathcal N (X)=[X,G/O]$ 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 | |

− | $$\sigma(\ | + | $$\sigma(-(\xi,-\phi))+\sigma(-(\xi,\phi)) - \sigma(-(\xi\oplus\xi,\phi * \phi)) |

− | = \langle L(TX\oplus\ | + | = 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$, $$ | |

+ | and so the surgery obstruction is not a group homomorphism with respect to the Whitney sum. | ||

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## Revision as of 21:44, 29 May 2012

We take and consider various bundle reductions of the normal bundle: 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

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and so the surgery obstruction is not a group homomorphism with respect to the Whitney sum.