Talk:Surgery obstruction map I (Ex)

Revision as of 21:44, 29 May 2012

We take $<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 $\mathcal N (X)\cong [X,G/TOP]$. An element of $[X,G/TOP]$ is given by a bundle $\xi$ together with a fiber homotopy trivialization $\phi$. Under the isomorphism $\mathcal N (X) \cong [X,G/O]$, 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 $X$. 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 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 $$\sigma(-(\xi,-\phi))+\sigma(-(\xi,\phi)) - \sigma(-(\xi\oplus\xi,\phi * \phi)) = 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -1 = c \langle p_1(\xi)^2 ,[X] \rangle \ne 0, $$ and so the surgery obstruction is not a group homomorphism with respect to the Whitney sum. </wikitex>X=\mathbb H P^2$ and consider various bundle reductions of the normal bundle: The normal map $id_X$ gives the base point of $\mathcal N (X)\cong [X,G/TOP]$. An element of $[X,G/TOP]$ is given by a bundle $\xi$ together with a fiber homotopy trivialization $\phi$. Under the isomorphism $\mathcal N (X) \cong [X,G/O]$, 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

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

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 $[S^4,G/Top]$, since the corresponding Atiyah-Hirzebruch spectral sequence collapses. 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

$$\displaystyle \sigma(-(\xi,-\phi))+\sigma(-(\xi,\phi)) - \sigma(-(\xi\oplus\xi,\phi * \phi)) = 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -1 = c \langle p_1(\xi)^2 ,[X] \rangle \ne 0,$$

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