Talk:Surgery obstruction map I (Ex)

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We take X=\mathbb H P^2: 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. 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

\displaystyle  sign(M)-sign(X)=\langle L(-\eta),[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], as follows from the Puppe sequence with \pi_7(G/TOP)=0. From 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 a vector bundle \xi with p_1(\xi)=48 whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence \phi. 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

\displaystyle \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 = c \langle p_1(\xi)^2 ,[X] \rangle  \ne 0,

where the constant c can be computed from the L-genus to be -1/9.

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

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