# Talk:Surgery obstruction map I (Ex)

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We take $X=\mathbb H P^2$$; We take X=\mathbb H P^2 and consider various bundle reductions of the normal bundle: The normal map id_X gives the base point of \mathcal N (X). The surgery obstruction of a normal map M\to X covered by \nu_M\to \xi equals sign(M)-sign(X)=\langle L(-\xi),[X]\rangle -1, so it depends only on the bundle over X. There are fiber homotopically trivial bundles on X corresponding to classes in [X,G/TOP] which restrict to any given class in [S^4,G/Top], since the corresponding Atiyah-Hirzebruch spectral sequence collapses. From [[Fibre_homotopy_trivial_bundles_(Ex)|another exercise]] we know that on S^4 we have such vector bundles with first Pontryagin class k times the generator of H^4(S^4). This means that on X we have vector bundles \xi_1,\xi_2 whose sphere bundles are fiber homotopically trivial, by fiber homotopy equivalences \phi_i. Then (\xi_1\oplus\xi_2,\phi_i * \phi_2) is the sum of (\xi_1,\phi_1) and (\xi_2,\phi_2) in \mathcal N (X)=[X,G/O] with respect to the Whitney sum. Now we compute \sigma(\xi_1,\phi_1)+\sigma(\xi_2,\phi_2) - \sigma(\xi_1\oplus\xi_2,\phi_i * \phi_2) = \langle L(TX\oplus\xi_1), [X] \rangle + \langle L(TX\oplus\xi_2), [X] \rangle - \langle L(TX\oplus\xi_1\oplus\xi_2), [X] \rangle -\langle L(TX), [X] \rangle. X=\mathbb H P^2$ and consider various bundle reductions of the normal bundle: The normal map $id_X$$id_X$ gives the base point of $\mathcal N (X)$$\mathcal N (X)$. The surgery obstruction of a normal map $M\to X$$M\to X$ covered by $\nu_M\to \xi$$\nu_M\to \xi$ equals $\displaystyle sign(M)-sign(X)=\langle L(-\xi),[X]\rangle -1,$

so it depends only on the bundle over $X$$X$. There are fiber homotopically trivial bundles on $X$$X$ corresponding to classes in $[X,G/TOP]$$[X,G/TOP]$ which restrict to any given class in $[S^4,G/Top]$$[S^4,G/Top]$, since the corresponding Atiyah-Hirzebruch spectral sequence collapses. From another exercise we know that on $S^4$$S^4$ we have such vector bundles with first Pontryagin class $48k$$48k$ times the generator of $H^4(S^4)$$H^4(S^4)$. This means that on $X$$X$ we have vector bundles $\xi_1,\xi_2$$\xi_1,\xi_2$ whose sphere bundles are fiber homotopically trivial, by fiber homotopy equivalences $\phi_i$$\phi_i$. Then $(\xi_1\oplus\xi_2,\phi_i * \phi_2)$$(\xi_1\oplus\xi_2,\phi_i * \phi_2)$ is the sum of $(\xi_1,\phi_1)$$(\xi_1,\phi_1)$ and $(\xi_2,\phi_2)$$(\xi_2,\phi_2)$ in $\mathcal N (X)=[X,G/O]$$\mathcal N (X)=[X,G/O]$ with respect to the Whitney sum. Now we compute $\displaystyle \sigma(\xi_1,\phi_1)+\sigma(\xi_2,\phi_2) - \sigma(\xi_1\oplus\xi_2,\phi_i * \phi_2) = \langle L(TX\oplus\xi_1), [X] \rangle + \langle L(TX\oplus\xi_2), [X] \rangle - \langle L(TX\oplus\xi_1\oplus\xi_2), [X] \rangle -\langle L(TX), [X] \rangle.$