# Talk:Surgery obstruction map I (Ex)

(Difference between revisions)

We take $X=\mathbb H P^2$$; We take X=\mathbb H P^2: The normal map id_X gives the base point of \mathcal N (X). An element of [X,G/TOP] is given by a bundle \xi together with a fiber homotopy trivialization \phi. Under the isomorphism \mathcal N (X) \cong [X,G/TOP], the pair (\xi,\phi) corresponds to a normal map M\to X covered by \nu_M\to \nu_X\oplus \xi. The surgery obstruction of a normal map M\to X covered by \nu_M\to \eta equals sign(M)-sign(X)=\langle L(-\eta),[X]\rangle -1, so it depends only on the bundle over X. There are fiber homotopically trivial bundles on X corresponding to classes in [X,G/TOP] which restrict to any given class in [S^4,G/Top], as follows from the Puppe sequence with \pi_7(G/TOP)=0. From [[Fibre_homotopy_trivial_bundles_(Ex)|another exercise]] we know that on S^4 we have such vector bundles with first Pontryagin class k times the generator of H^4(S^4). This means that on X we have a vector bundle \xi with p_1(\xi)=48 whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence \phi. Now (\xi\oplus\xi,\phi * \phi) is the sum of (\xi,\phi) and (\xi,\phi) in \mathcal N (X)=[X,G/O] with respect to the Whitney sum. We compute \theta(-(\xi,-\phi))+\theta(-(\xi,\phi)) - \theta(-(\xi\oplus\xi,\phi * \phi)) = 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -1 = c \langle p_1(\xi)^2 ,[X] \rangle \ne 0, where the constant c can be computed from the L-genus to be -1/9. So the surgery obstruction is not a group homomorphism with respect to the Whitney sum. X=\mathbb H P^2$: The normal map $id_X$$id_X$ gives the base point of $\mathcal N (X)$$\mathcal N (X)$. An element of $[X,G/TOP]$$[X,G/TOP]$ is given by a bundle $\xi$$\xi$ together with a fiber homotopy trivialization $\phi$$\phi$. Under the isomorphism $\mathcal N (X) \cong [X,G/TOP]$$\mathcal N (X) \cong [X,G/TOP]$, the pair $(\xi,\phi)$$(\xi,\phi)$ corresponds to a normal map $M\to X$$M\to X$ covered by $\nu_M\to \nu_X\oplus \xi$$\nu_M\to \nu_X\oplus \xi$. The surgery obstruction of a normal map $M\to X$$M\to X$ covered by $\nu_M\to \eta$$\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$$X$. There are fiber homotopically trivial bundles on $X$$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]$, as follows from the Puppe sequence with $\pi_7(G/TOP)=0$$\pi_7(G/TOP)=0$. From another exercise we know that on $S^4$$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$$X$ we have a vector bundle $\xi$$\xi$ with $p_1(\xi)=48$$p_1(\xi)=48$ whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence $\phi$$\phi$. Now $(\xi\oplus\xi,\phi * \phi)$$(\xi\oplus\xi,\phi * \phi)$ is the sum of $(\xi,\phi)$$(\xi,\phi)$ and $(\xi,\phi)$$(\xi,\phi)$ in $\mathcal N (X)=[X,G/O]$$\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$$c$ can be computed from the L-genus to be $-1/9$$-1/9$.

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